Transport I(CHE 311 Course Notes)

T.K. Nguyen

Chemical and Materials EngineeringCal Poly Pomona

(Fall 2014)

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Contents

Chapter 1: Introduction to Momentum and Heat Transfer1.1 Introduction 1-11.2 Transport Laws 1-2

Fourier’s Law of Heat Conduction 1-3Example 1.2-1: Heat transfer through a slab 1-4Example 1.2-2: Heat loss through a window 1-5Newton’s Law of Viscosity 1-6Example 1.2-3: Flow in a slider bearing 1-6Example 1.2-4: Location of maximum shear stress 1-8Example 1.2-5: Flow and fluid properties 1-8

1.3 Compressibility of Fluids 1-9Example 1.3-1: Volume change due to pressure change 1-11

1.4 Surface Tension 1-12Example 1.4-1: The capillary rise method 1-14Example 1.4-2: Pressure gage inside a water jet 1-15Example 1.4-3: Diameter of steel rod floating on water 1-16Example 1.4-4: Dynamic of capillary action 1-17

Chapter 2: Fluid Statics2.1 Variation of pressure with elevation 2-1

Example 2.1-1: Manometer system 2-2Example 2.1-2: Manometer pressure calculation 2-5Example 2.1-3: Manometer pressure calculation 2-5Example 2.1-4: Manometer calculation 2-6

2.2 Buoyancy forces 2-7Example 2.2-1: Liquid level in a lake 2-7Example 2.2-2: Water level in a pool 2-8

2.3 Pressure in Response to External Forces 2-9Vertical acceleration 2-9Horizontally Accelerating Free Surface 2-10Surface of a Rotating Liquid 2-11Acceleration in Uniform Circular Motion 2-11Example 2.3-1: Water displacement for a car ferry 2-12Example 2.3-2: Rotating U-tube 2-13Example 2.3-3: Pressure drop in a helix tube 2-14

Chapter 3: Fluid Properties3.1 Introduction 3-13.2 Rheology 3-2

Example 3.2-1: Shear force in a pipe. 3-83.3 Fully Developed Laminar Flow in Tube 3-9

Example 3.3-1: Viscosity evaluation 3-123.4 The Hagan-Poiseuille Equation 3-16

Example 3.4-1: Viscosity for shear thinning liquid 3-16Example 3.4-2: Cannon-Fenske Viscometer 3-18

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3.5 Mechanical Concepts of Energy 3-19Example 3.5-1: Effective range of a spear 3-20Example 3.5-2: Power required by a car 3-21

3.6 Bernoulli Equation 3-22Example 3.6-1: Height of water in a tank 3-24Example 3.6-2: Velocity head in a pipe 3-24Example 3.6-3: Flow rate into a tank 3-25Example 3.6-4: Pressure variation normal to a streamline 3-26Example 3.6-5: Velocity through a nozzle 3-28

Chapter 4: Fluid Kinematics4.1 The Velocity Field 4-1

Example 4.1-1: Velocity field representation 4-14.2 The Acceleration Field 4-2

Example 4.2-1: Acceleration along a streamline 4-6Example 4.2-2: Streamline representation 4-8

Chapter 5: Conservation Laws: Control-Volume Approach5.1 Macroscopic Mass Balance (Integral Relation) 5-1

Example 5.1-1: A mixing tank 5-1Example 5.1-2: Flow through a circular conduit 5-5

5.2 Microscopic Mass Balance (Differential Relation) 5-6Example 5.2-1: Divergence of a vector 5-8

5.3 Macroscopic Energy Balance 5-9Example 5.3-1: A water heater 5-11

5.4a Macroscopic Momentum Balance 5-15Example 5.4-1: Gravity flow tank 5-15Example 5.4-2: A compressed-air rocket 5-20

5.4b One-Dimensional Flow in a Tube 5-255.4c Definition of the Loss Coefficient and the Friction Factor 5-27

Example 5.4-3: Flow through a sudden expansion 5-28Example 5.4-4: A 90o horizontal reducing elbow 5-30Example 5.4-5: Friction factor correlation 5-31Example 5.4-6: Force on a horizontal bend 5-33Example 5.4-7: Flow through a 45o bend 5-34

5.5 Conservation of Angular Momentum 5-35Example 5.5-1: An ideal garden sprinkler 5-37

5.6 Momentum Balance on Moving Systems 5-38Example 5.6-1: Jet impinging against a cone 5-38Example 5.6-2: Jet-propelled boat 5-39

5.7 Microscopic Momentum Balance 5-41Example 5.7-1: Flow through a rectangular duct 5-47Shell BalanceExample 5.7-2: Flow between concentric cylinders 5-51Example 5.7-3: Axial flow between a rod and a cylinder 5-53:54

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Chapter 6: Dimensional Analysis6.1 Units and Dimensions 6-16.2 Dimensional Analysis 6-2

Example 6.2-1: Dimensional analysis 6-6Example 6.2-2: Dimensional analysis 6-7Example 6.2-3: Scale Up 6-9:10

Chapter 7: Flow in Closed Conduits7.1 Flow regimes 7-17.2 Generalized Mechanical Energy Balance Equation 7-2

Example 7.2-1: Reservoir piping system 7-57.3 Pipe Flow Problems 7-77.3a Unknown Driving Force 7-7

Example 7.3-1: Two reservoirs systems 7-87.3b Unknown Flow Rate 7-9

Example 7.3-2: Steady oil transferred to storage tank 7-10Example 7.3-3: Flow from a reservoir through a turbine 7-12Example 7.3-4: Flow from a reservoir through a piping system 7-14

7.3c Unknown Diameter 7-15Example 7.3-5: Steady oil transferred to storage tank 7-16

7.4 Other Pipe Flow Problems 7-19Example 7.4-1: Design for pipeline corrosion 7-19Example 7.4-2: Time to empty an IV bag 7-20Example 7.4-3: Power delivered by the heart 7-22

Chapter 8: Pumps and Compressors8.1 Introduction 8-18.2 Centrifugal Pumps 8-3

Example 8.2-1: Centrifugal pump arrangement 8-6Pump Head 8-8Net Positive Suction Head 8-9

8.2 Compressors 8-13Isothermal Compression 8-14Isentropic Compression 8-14Multistage Compressors 8-15Compressor Performance 8-16Example 8.3-1: Nitrogen transport 8-17Example 8.3-2: Pressure-Enthalpy diagram 8-18Example 8.3-3: Hydrogen transport 8-20

Chapter 9: Flow Measurement and Control Valves9.1 Introduction 9-19.2 The Pitot Tube 9-1

Example 9.2-1: Velocity measurement by pitot tube 9-29.3 The Venturi Meter and the Flow Nozzle 9-3

Example 9.3-1: Venturi meter 9-59.4 The Orifice Meter 9-7

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Incompressible Flow 9-8Compressible Flow 9-8Example 9.4-1: Draining with an orifice meter 9-11Example 9.4-2: Orifice meter error 9-12Example 9.4-3: Orifice meter diameter 9-13

9.5 The Rotameter 9-15Example 9.5-1: Rotameter reading correction 9-17

9.6 Control Valves 9-18Incompressible Flow 9-19Example 9.6-1: An oil transferring system 9-20Compressible Flow 9-22Example 9.6-2: Flow rate through a 3-in Masoneilan valve 9-23Example 9.6-3: Sizing a control valve 9-25:26

Chapter 10: Flow in Chemical Engineering Application10.1 Drag Force on Solid Particles in Fluids 10-110.2 Separations by Free Settling 10-2

Example 10.2-1: Terminal velocity of a sphere 10-3Example 10.2-2: Time to reach terminal velocity 10-4

10.3 Flow Through Packed Beds 10-9Example 10.3-1: Void fraction in a packed bed 10-13Example 10.3-2: Pressure drop across a packed bed 10-13

10.4 Laminar Flow Through Porous Medium 10-15Numerical Solution of the Laplace equation using Comsol 10-16

10.5 Flow Through Fluidized Beds 10-23Example 10.5-1: Minimum fluidizing conditions 10-26Example 10.5-2: Maximum flow rate through a packed bed 10-28

Appendix

A. Previous Exams (2007)Quiz 1 A-1Quiz 2 A-4Quiz 3 A-7Quiz 4 A-11Quiz 5 A-14

B. Previous Exams (2008)Quiz 1 B-1Quiz 2 B-5Quiz 3 B-9Quiz 4 B-12Quiz 5 B-16

Answers to 2008 quizzes B-19

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C. Previous Exams (2009)Quiz 1 C-1Quiz 2 C-4Quiz 3 C-7Quiz 4 C-10Quiz 5 C-13

Answers to 2009 quizzes C-16

D. Previous Exams (2010)Quiz 1 D-1Quiz 2 D-5Quiz 3 D-8Quiz 4 D-11Quiz 5 D-14

Answers to 2010 quizzes D-17

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Chapter 1 Introduction to Momentum and Heat Transfer

1.1 Introduction to Fluid Mechanics

Fluids mechanics is a branch of physics that seeks to describe the physical phenomena that involve the flow of liquids and gases. The fundamental principles that apply to the analysis of fluid flows are the conservation laws and the transport laws. We will learn to understand the physical meaning and mathematical representation of such quantities as velocity, stress, pressure, momentum, and energy. We will approach the study of fluid mechanics in the engineering point of views. Although science and math are important tools for engineers we often need to use our experience and judgment to obtain a quick estimate to obtain a working solution. Approximation based on sound physical principles and good engineering judgments are invaluable to the engineer. This requires a thorough understanding of any studied system so that an engineer can organize and apply the obtained information to analyze and/or design similar systems on a different scale.

We first review some of the basic definitions necessary when applying the conservation laws to study a system.

'Rate''Rate' implies an element of speed, how fast an event happens, and time.

'System'A system is any designated region of a continuum of fixed mass. The

boundaries of a system may be deformable but they always enclose the same mass. In thermodynamics, the universe can be divided into two parts. One part is the system and the other part is the rest of the universe called the surroundings.

Surroundings

Boundary

System

Figure 1.1 Schematic diagram of the "universe", showing a system and the surroundings.

'Control volume'A 'control volume' is also any designated region of a continuum except that it

may permit matter to cross its boundaries. If the boundaries of a control volume are such that matter may not enter or leave the control volume, the control volume is identical to a system. In these respects, a 'system' is a subset of a 'control volume'.

1-2

'Equilibrium''Equilibrium' means that there are no spatial differences in the variables that

describe the condition of the system, also called the 'state' of a system, such as its pressure, temperature, volume, and mass (P, T, V, m), and that any changes which occur do so infinitesimally slowly.

'Stress'In fluid mechanics, stress is force per unit area. There are two types of

stresses: normal and shear stresses. As the names imply, normal stress results when a force acts normal to a surface and shear stress occurs when a force acts tangentially to the surface.

1.2 Transport Laws

Transport laws govern the rate at which conserved quantities (such as mass, energy, momentum, etc.,) are transported from one region to another in a continuous medium. These are called phenomenological laws because they are based upon observable phenomena and logic but they cannot be derived from more fundamental principles. The transport laws can be expressed in the general form as

Rate of transport = = Conductance Driving forceResistance

forceDriving

The rate of transport of any conserved quantity Q per unit area normal to the direction of transport is called the flux of Q. The driving force for the transport is the negative of the gradient (with respect to the direction of transport) of the concentration of Q denoted by Qc as shown in Figure 2.2-1.

Qc

xx1

dQ /dxc

Q (x)c

Figure 2.2-1 The concentration gradient at x1.

Figure 2.2-1 shows the concentration gradient of Q, , which is the slope of the curve of dx

dQc

Qc versus x. Since Q is transported from a region of higher concentration to a region of lower concentration, the slope is negative. The driving force is defined as the negative of the

concentration gradient, , so that the flux is in the positive direction. dx

dQc

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Flux of Q in the x direction = KT (1.2-1)dx

dQc

In this expression, KT is the transport coefficient for the quantity Q. For molecular transport, KT is a property only of the medium.

Fourier’s Law of Heat Conduction

Fourier's law (1822), developed from observed phenomena, states that the rate of heat

transfer in the 'n' direction is proportional to the temperature gradient nT

qn nT

where n is the direction of heat transfer and is the change of distance in the ndirection n.

n is the unit normal vector, and t is the unit tangential vector with the following properties,

n= 1; t= 1; nt = 0; nn = 1; tt = 1

qn = C , where C = AknnT

qn = knA nT

qn

qt

nT

qn

1-4

where kn = thermal conductivity in 'n' direction, [W/mK]A = area of surface perpendicular to n through which qn flows

The minus is a sign convention so that qn is positive in the direction it transfers. In this text, we will usually consider the isotropic materials where the thermal conductivity k is independent of direction. For one dimensional heat transfer in the x-direction only, the heat transfer rate is then

qx = kA (1.2-2)dxdT

or in terms of the heat flux "xq

= k (1.2-3)"xq

dxdT

For system with constant density and constant heat capacity Cp, equation (1.2-3) can be written as

= = (1.2-4)"xq

pCk

dxTCd p )(

dxTCd p )(

Equation (1.2-4) has the form of equation (1.2-1)

Flux of Q in the x direction = KT (1.2-1)dx

dQc

Q in this case is just the energy or heat so that = flux of Q in the x direction. KT = is "xq

called the thermal diffusivity in m2/s (SI). Qc = CpT = concentration of energy.

Example 1.2-1 ----------------------------------------------------------------------------------A large slab with thermal conductivity k and thickness L is maintained at temperatures T1 and T2 at the two surfaces. Determine the heat flux through this material at steady-state condition.

Solution ------------------------------------------------------------------------------------------

1-5

The x-coordinate is assigned in the direction normal to the slab with x = 0 at the left surface where the temperature is T1. Since the temperature varies across the slab or the x-direction, a differential control volume with the same cross-sectional area A as that of the slab and a thickness dx will be considered. An energy balance (first law) is then applied to this differential control volume

= qin – qout + qgendtdE

For steady state with no heat generation

qin = qout A = Axxq"

dxxxq

"

= 0dx

qqxxdxxx

""

In the limit when dx approaches zero

= = 0 = - k = constantdx

qqxxdxxx

""

dxdqx

""xq

dxdT

If the thermal conductivity k is a constant,

= = constantdxdT

012

LTT

Therefore the heat flux through the slab is simply"xq

= k = constant"xq

LTT 21

Example 1.2-2 ----------------------------------------------------------------------------------The inner and outer surface temperatures of a glass window 5 mm thick are 15 and 5oC. What is the heat loss through a window that is 1 m by 3 m on a side? The thermal conductivity of glass is 1.4 W/mK. (1.51)Solution ------------------------------------------------------------------------------------------

= k q = A = kA = kA"xq

dxdT "

xqdxdT

LTT 21

q = (1.4)(13) = 8,400 W3

(15 5)5 10

1 Fundamentals of Heat Transfer by Incropera and DeWitt.

1-6

Newton’s Law of Viscosity

y

xv (y)x

Vo

Figure 2.2-2 Transport of momentum from lower plate to upper plate.

Figure 2.2-2 illustrates two horizontal parallel plates with a fluid between them. If the top plate is fixed while the bottom plate is moving with velocity Vo, there will be a transport of momentum from the lower plate to the upper plate. Newton’s equation for momentum transfer for constant density can be written as follows in a manner similar to equation (1.2-1):

(yx)mf = – = – (1.2-5)

dyvd x )(

dyvd x )(

In this equation (yx)mf is the flux of x-momentum in the y direction in and = is 2

kg m/sm s

the kinetic viscosity in m2/s (SI). If Fx is the force acting in the x-direction on the lower plate to make it move, it must also be the driving force for the transport of x-momentum, which flows from the faster to the slower fluid. The force Fx is then given by

Fx = yxA (1.2-6)

In this equation yx is the shear stress acting on the y-surface in the x-direction and A is the surface area of the plate. The shear stress is the negative of the momentum flux in the y direction

yx = – (yx)mf = (1.2-7)dydvx

Example 1.2-3 ----------------------------------------------------------------------------------A slider bearing consists of a sleeve surrounding a cylindrical shaft that is free to move axially within the sleeve. Grease is in the gap between the sleeve and the shaft to isolate the metal surfaces and support the stress resulting from the shaft motion. The diameter of the shaft is 2.0 cm and the sleeve have an inside diameter of 2.04 cm and a length of 20 cm.

Shaft

Sleeve

Sleeve

Vo

L

Ds

1-7

1) If you want to limit the total force on the sleeve to less than 2.0105 dyne, when the shaft is moving at a velocity of 10 m/s, what should the viscosity of the grease be?2) If the grease viscosity is 200 cP, what is the force exerted on the sleeve when the shaft is moving at 10 m/s?3) The sleeve is cooled to a temperature of 35oC, and it is desired to keep the shaft temperature below 90oC. What is the cooling rate in W? Thermal conductivity of grease is 0.40 W/moK.Solution ------------------------------------------------------------------------------------------1) Determine the viscosity of the grease

Since the gap is small compared to the diameter of the shaft, we can approximate the flow within the gap by a flow between two parallel plates.

Let h = the gap, then h = (2.04 2.00)/2 = 0.02 cm. The magnitude of the shear stress is given as

|yx | = | | = dydvx

hVo

= = LD

Fs h

Vo

os LVDhF

= = 0.0312 g/cms = 3.12 cP)1000)(20)(04.2(

)102)(02.0( 5

2) If the grease viscosity is 200 cP, what is the force exerted on the sleeve when the shaft is moving at 10 m/s?

F = = 2 = 1.28108 dyneh

LDV so02.0

)20)(04.2)(1000(

3) The sleeve is cooled to a temperature of 35oC, and it is desired to keep the shaft temperature below 90oC. What is the cooling rate in W? Thermal conductivity of grease is 0.40 W/moK.

The heat flux is given as

q” = k = 0.40 = 110,000 W/m2s h

TT sleeveshaft 100/02.03590

The cooling rate is thenQ

= q”(DsL) = (110,000)()(0.0204)(0.2) = 1.41108 WQ

------------------------------------------------------------------------------------------------------

1-8

Example 1.2-4 ----------------------------------------------------------------------------------The velocity distribution for laminar flow between the two parallel plates is given by

vx = cy2

where c = 4000 s-1m-1. The distance between the plates is 5 mm, and the origin is placed at the lower plate. Consider the flow of water with a viscosity of 0.0012 Ns/m2. Determine the maximum magnitude of the shear stress and its location.Solution ------------------------------------------------------------------------------------------

Plotting out the velocity profile we obtain the following graph, which indicates that the highest change in velocity occurs at the top plate. This is the location of the maximum shear stress.

y vx

We can confirm this result by using equation (1.2-7) to determine the shear stress

yx = = 2cy (1.2-7)dydvx

The only variable in this equation is y. The shear stress will have the maximum value when y is at its highest allowable value, which is 5 mm. Therefore

yx = (2)(0.0012)(4000)(0.005) = 0.048 Pa

Example 1.2-5 ----------------------------------------------------------------------------------

Which of the quantities listed below are flow properties and which are fluid properties?Pressure Temperature Velocity DensityStress Speed of sound Specific heat Pressure gradient

Solution ------------------------------------------------------------------------------------------

Flow properties (exist only if the fluid is moving): Velocity, Stress, Pressure gradient.

Fluid properties (exist independent of fluid motion): Pressure, Temperature, Density, Speed of sound, Specific heat.

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Chapter 1

1.3 Compressibility of Fluids

The bulk modulus (or bulk modulus of elasticity) is a property used to determine the change in volume of fluid when there is a change in pressure. The bulk modulus is defined as

Ev = /

dPdV V

In this expression dP is the differential change in pressure required to create a differential change in volume d of a volume . For a given mass m of fluid where the volume and V Vdensity can change, we have

m = 0 = d + dV V V

Therefore

d / = d/V V

The bulk modulus can then be written as

Ev = /

dPd

From the definition, the bulk modulus has the dimension of pressure. For an ideal gas we have

P = TgRMw

In this equation, Mw is the gas molecular weight and Rg is the universal gas constant. You should notice that some texts use the gas constant R in the ideal gas law where

P = RT

In this equation, the gas constant R = is equal to the universal gas constant divided by gRMw

the gas molecular weight. In this text we always use the universal gas constant even though it might not have the subscript g. For an isothermal process with ideal gas

= T = constantP

gRMw

The bulk modulus for an isothermal process is then

1-10

Ev = = P/

dPd

For isentropic compression, the system is adiabatic and reversible. The change in internal energy is equal to the work supplied to the system

dU = dW CvdT = PdV

In this equation V = = volume. We now use ideal gas law to obtain an expression for PdV Vin terms of , d, T, and dT

PV = RgT PdV = RgTdV/V = RgT d/

Therefore

Cv = Rg Cvln = Rg ln = dTT

d

f

i

TT

f

i

f

i

TT

/g vR Cf

i

For an ideal gas Rg = Cp Cv = 1 = k 1g

v

RC

p

v

CC

In this equation, k is the ratio of the specific heat: k = . In terms of k, we havep

v

CC

= f

i

TT

1kf

i

From ideal gas law

= = = f

i

TT

f

i

PP

i

f

1kf

i

f

i

PP

kf

i

The above expression can be rearranged as

= = = constant = (constant)kk-1fkf

P

iki

P k

P

dPd

The bulk modulus for an isentropic with ideal gas is then

Ev = = (constant)kk-1 = (constant)kk = kP/

dPd

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Example 1.3-1 ----------------------------------------------------------------------------------In water the bulk modulus is nearly constant and has a value of 300,000 psi. Determine the percentage volume change in water due to a pressure change of 3000 psi.

Solution ------------------------------------------------------------------------------------------

Ev = = /

dPd v

dPE

d

Since m = V 0 = dV + Vd

Therefore

dV/V = d/ = = v

dPE

dVV v

PE V

V

= = 0.01VV

3,000300,000

The percentage volume change is 1%.

------------------------------------------------------------------------------------------

Due to the compressibility of the fluid, disturbances generated at some point in the fluid will propagate at a finite velocity. The velocity at which these small disturbances propagate is called the acoustic velocity or the speed of sound. It can be shown from mass and momentum balances that the speed of sound c is given by the following equation:

c = dPd

In terms of the bulk modulus Ev = , the speed of sound is written as/

dPd

c = vE

For ideal gases undergoing an isentropic process, Ev = kP, we have

c = = kP

gkR TMw

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1.4 Surface Tension

Surface tension is caused by a net attractive force in the interior of the liquid. It can be defined with reference to Figure 1.4-1. A molecule I, located in the interior of the liquid, is attracted equally in all directions by its neighbors. However, a molecule S, located in the surface, is drawn toward the bulk of the liquid because there are no liquid molecules in the other direction to pull them outward. Therefore energy is required to bring an interior molecule to the surface to overcome the net attractive force into the bulk of the liquid.

Free surface

Liquid

Molecule S

Molecule I W

L

Newly created surface

Figure 1.4-1 (a) Surface tension is caused by the attractive force between molecules;(b) newly created surface caused by moving the tension through a distance L.

Consider the surface tension per unit distance W of a line drawn in the surface as shown in Figure 1.4-1 (b). An amount of work or energy WL is required to create an area WL of fresh surface due to the movement of the surface tension over a distance L. Therefore surface tension has units of energy per unit area or force per unit length.

We now want to find the pressure difference pi po between the pressure pi inside a liquid droplet of radius r, shown in Figure 1.4-2, and the pressure po of the surrounding vapor.

Liquid droplet

Vapor

rpi

po

po

pi

Figure 1.4-2 Pressure difference across a curved surface.

The force due to the pressure difference between the pressure inside the droplet and the vapor outside is given by

(pi po)r2

The force due to surface tension, which acts on the circumference of length 2r, is equal to

1-13

2r

At equilibrium, these two forces are equal, giving

(pi po)r2 = 2r pi po = 2r

In general, the increase in pressure is given by

pi po = 1 2

1 1r r

In this equation, r1 and r2 are the principle radii of curvature. By convention, the radius of curvature is positive on the concave side and negative on the convex side. The surface tension can be measured with the device shown in Figure 1.4-3.

Film of liquid

Weight

Sliding partof frame

Figure 1.4-1 Surface tension from liquid film1.

A wire frame with one movable side is dipped into a liquid and carefully removed so that a film of liquid is formed in the space of the frame. The film tends to draw the movable part of the frame inward. The force required to resist this motion is measured by a weight. It has been observed that the ratio of the force to the length of the sliding part of the wire is always the same for a given liquid at a given temperature independent of the size of the frame. The liquid film in the frame has two surfaces (front and back), therefore the surface tension is given by

= 2mg

L

In this equation, m is the mass of the weight and L the length of the sliding part of the frame. The principle of the device shown in Figure 1.4-1 is easy to understand. However it is difficult to find the right weight to balance the liquid frame. One of the simplest method to measure surface tension is the measurement of capillary rise as shown in Figure 1.4-4. A narrow capillary tube of inside diameter Dc is dipped into a liquid that wets the tube. If the contact angle (the angle between the free surface and the wall) is , the meniscus will be

1 Noel de Nevers, Fluid Mechanics for Chemical Engineering, McGraw Hill, 3rd Edition, 2005, pg. 15.

1-14

approximated by part of the surface of a sphere with radius Dc/(2cos ). The surface tension is then given by

= 4cos

cghD

Figure 1.4-4 Rise of a liquid in a capillary.

Example 1.4-1 ----------------------------------------------------------------------------------

We observe that the liquid rises in the tube, above the level of the free surface with a contact angle = 60o. We want to derive a mathematical model that relates the height of capillary rise h to the surface tension and the inner capillary diameter Dc. This model (Fig. 1.4-4) will suggest another method for measuring the surface tension2. Determine the height of capillary rise if the liquid is water with surface tension = 0.072 N/m and capillary diameter Dc is 10-3 m.

Solution ------------------------------------------------------------------------------------------

If we approximate the meniscus as a part of a sphere surface with diameter Dc, then the two principal radii of curvature r1 and r2 are equal to each other, and are given by

r1 = r2 = Dc1

2cos( )

The pressure difference across the meniscus is then

patm – ps = p = 4 cos( )

cD

Where ps is the pressure at the meniscus in the liquid phase. Since the system is in static equilibrium, the pressure p1 must equal patm, for these two pressures occur at points that lie in the same horizontal plane in the same fluid.

ps + gh = p1 = patm

Therefore

2 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998, pg. 43.

1-15

gh = patm – ps = p = 4 cos( )

cD

or

h = 4 cos( )

cgD

h = = 0.0145 m = 1.45 cm m10s/m8.9m/kg10

)60cos(m/N072.043233

Example 1.4-2 ----------------------------------------------------------------------------------

A 10-mm diameter water jet discharges into the atmosphere. Determine the difference in pressure inside and outside the jet.

Solution ------------------------------------------------------------------------------------------

Assuming a long jet with L >> 2R, we have

(Pi Po)(2RL) = 2L

Therefore

Pi Po = /R

From TK ‘s program (Prop4), the surface tension of water at 25oC is 72.2 dyne/cm or 0.0722 N/m. Therefore

Pi Po = 0.0722/.005 = 14.4 Pa

Capillary rise h = 1.45 cm

R

L

1-16

Example 1.4-3 ----------------------------------------------------------------------------------

A small steel rod with diameter D and length L is placed on a surface of water. What is the maximum diamter that the rod can have before it will sink? The specific weight of steel is 490 lbf/ft3.

Solution ------------------------------------------------------------------------------------------

W

LLL = rod length

In order for the rod to float, the surface tension force must be greater or equal to the weight of the rod:

2L D2Lsteel4

In this equation, the surface tension force along the ends of the cylinder is neglected (2D << 2L). Thus the maximum diameter the rod could have for it to float is

Dmax = 1/ 2

8

steel

For water at 25oC, = 0.0722 N/m = 4.9510-3 lbf/ft (T.K. ‘s Prop4 program)

Dmax = = 0.0051 ft = 0.061 in1/ 238 4.95 10

490

A standard paper clip has a diameter of 0.036 in, which is less than 0.061 in, which is less than 0.061 in. It should float. Partially unfold a paper clip and see if you can get it to float on water.

1-17

Example 1.4-4 ----------------------------------------------------------------------------------A diagnostic device makes use of a thin rectangular channel to draw in a sample of blood. The length of the channel is L and its width is W. The separation of the channel is 2H. The volumetric flowrate through the channel is given by

Q = 3

02 ( )3

LWH p pL

In this equation, p0 is the inlet and pL the outlet pressure. The blood sample has a viscosity of 3 cP and the plates forming the channel are separated by a distance of 1 mm, estimate the time for the sample of blood to travel a distance of 15 mm in the channel. Assume the blood has a surface tension of 0.06 N/m and that the contact angle is 70o.7

Solution ------------------------------------------------------------------------------------------

Blood

p0

pL

patm

2H

For a rectangle channel with 2H separation distance, the two principal radii of curvature r1 and r2 are given by

r1 = and r2 = cos( )

H

The pressure difference across the meniscus is then

patm – pL = p = cos( )H

Where pL is the pressure at the meniscus in the liquid phase. The pressure p0 is equal patm, for these two pressures occur at points that lie in the same horizontal plane in the same fluid. Therefore

p0 – pL = p = cos( )H

Substituting the above expression for p1 – pi into the equation for the volumetric flowrate we have

7 Fournier R., Basic Transport Phenomena in Biomedical Engineering, Taylor & Francis, 2007, p. 157.

1-18

Q = = 2WH22 cos( )3

WHL

dLdt

Separating the variables and integrating of the above equation with the initial condition L(t = 0) = 0 we obtain

t = = 3cos( )H

0

LLdL

232 cos( )

LH

t = = 0.0987 s2(3)(0.003)(0.015)

2(0.0005)(0.06)cos(70)

2-1

Chapter 2

Fluid Statics

2.1 Variation of pressure with elevation

Fluid at rest cannot support shearing stress. It can only support normal stress or pressure that can result from gravity or various other forces acting on the fluid. Pressure is an isotropic stress since the force acts uniformly in all directions normal to any local surface at a given point in the fluid. An isotropic stress is then a scalar since it has magnitude only and no direction. By convention, pressure is considered a negative stress because it is compressive, whereas tensile stresses are positive. The direction of pressure force is always pointing inward the control volume. We now investigate how the pressure in a stationary fluid varies with elevation z as shown in Figure 2.1-1.

P|z

P|z+ z

z

z

A

Figure 2.1-1 Forces acting on control volume Az

Applying a momentum or force balance in the z-direction on the control volume Az we obtain

Fz = maz = 0

AP|z AP|z+z g Az = 0

Dividing the equation by the control volume Az and letting z approach zero we obtain

= = g (2.1-1)0

limz z

PP zzz

||

dzdP

Equation (2.1-1) is the basic equation of fluid statics. It can be integrated if the density and the acceleration of gravity are known functions of elevation. We will assume g a constant since the change in elevation is usually not significant enough for g to vary. The integration will depend on the variation of density. If the density is not a constant, a relation between and z or P must be obtained. For constant density fluids, equation (2.1-1) can be easily integrated

2-2

= g P2 P1 = g(z2 z1)2

1

P

PdP

2

1

z

zdz

This equation can also be written as

P2 + gz2 = P1 + gz1 = = constant

The sum of the local pressure P and static head gz is called the potential that is constant at all points within a given incompressible fluid.

Example 2.1-1. 1 ----------------------------------------------------------------------------------The manometer system shown in Figure 2.1-2 contains oil and water, between which there is a long trapped air bubble. For the indicated heights of the liquids, find the specific gravity of the oil. The two sides of the U-tube are open to the atmosphere.

h = 2.5 ft1

h = 0.5 ft2

h = 1.0 ft3

h = 3.0 ft4

Oil

Air

Water

P1P2

Figure 2.1-2 A manometer system with oil, air, and water

Solution ------------------------------------------------------------------------------------------

The pressure P1 at the water air interface on the left side of the U-tube has the same value as the pressure P2 in the water at the same elevation on the right side of the U-tube.

P1 = Patm + ogh1 + agh2

P2 = Patm + wg (h4 h3)

If the pressure due to the weight of air is neglected compared to that of oil and water, we can determine the specific weight so of oil as follow

ogh1 = wg (h4 h3) so = = = = 0.80w

o

1

34

hhh

5.20.10.3

1 Wilkes, J., Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 28

2-3

If the fluid can be described by the ideal gas law then = and equation (2.1-1) RTPM

becomes

= g (2.1-2)dzdP

RTPM

If the temperature is constant for all z, equation (2.1-2) can be integrated

= 2

1

P

P PdP

RTMg

2

1

z

zdz

ln = (z2 z1) = z1

2

PP

RTMg

RTMg

Hence P2 = P1exp (2.1-3)

z

RTMg

For an isentropic process (adiabatic and reversible), the temperature and pressure are not constants. Therefore an expression for temperature as a function of pressure is required to integrate equation (2.1-2). We can accomplish this by applying an energy balance (First Law of Thermodynamics) to the system and then use ideal gas law to substitute volume V in terms of pressure P and temperature T. For an adiabatic system, the change in internal energy is equal to the work supplied to the system

dU = dW CvdT = PdV

We now use ideal gas law to obtain an expression for PdV in terms of P, dP, T, and dT

PV = RT PdV + VdP = RdT PdV = VdP RdT

Therefore CvdT = VdP RdT = dP RdTP

RT

We can separate the variables to obtain

(Cv + R)dT = dP Cp = (Cp Cv)PRT

TdT

PdP

Since = = , where k = p

vp

CCC

vp

vp

CCCC/

1/ k

k 1

v

p

CC

= TdT

kk 1

PdP

2-4

Integrating the equation we obtain

= (2.1-4)1T

T kk

PP

1

1

From equation (2.1-2)

= g (2.1-2)dzdP

RTPM

= = = dzdP

1RTMgP kk

PP /)1(

1

1RTMgP kkP /)1( kkP /)1(

1

1RTMg kP /1 kkP /)1(

1

Separating the variables yields

dP = 2

1

/1P

P

kP1RT

Mg kkP /)1(1

2

1

z

zdz

Integrating over the limits we obtain

= (z2 z1)1kk kkkk PP /)1(

1/)1(

2

1RTMg kkP /)1(

1

= zkkkk PP /)1(1

/)1(2

k

k 1

1RTMg kkP /)1(

1

= [1 z] (2.1-5a)kkP /)1(2

kkP /)1(1

kk 1

1RTMg

P2 = P1 (2.1-5b))1/(

1

11

kk

RTzMg

kk

From equation (2.1-4)

= 1

2

TT k

k

PP

1

1

2

The pressures in equation (2.1-5) can be replaced with the temperatures to give

T2 = T1

1

11RT

zMgk

k

2-5

A

B

C

Example 2.1-2. ----------------------------------------------------------------------------------

Consider the manometer system shown on the right. S is the specific gravity.

a) If the pressure of water at the left sphere (15 cm above A) is 4 kPa, determine the pressure at A.

b) If the pressure at A is 20 kPa, determine the pressure at C.

Solution ------------------------------------------------------------------------------------------

a) The pressure at A is

PA = 4 + 9.81×0.15 = 5.471 kPa

b) The pressure at C is

PC = 20 − (13.6)(9.81)(0.1) − (0.68)(9.81)(0.2) = 5.324 kPa

Example 2.1-3. ----------------------------------------------------------------------------------

If the gage pressure at C (interface between Hg and liquid with S = 0.8) is 3 kPa, determine the gage pressure in water at A.

A

B

C

(S = 13.6)

Solution ------------------------------------------------------------------------------------------

The gage pressure in water at A is

PA = 3000 + (1000)(9.81)(0.10.8 0.071.59 0.05) = 2202.4 Pa

2-6

Example 2.1-4. ----------------------------------------------------------------------------------

The U-tube shown below has legs of unequal internal diameters d1 = 10 mm and d2 = 5 mm, which are partly filled with immiscible liquids of density 1,800 kg/m3 and 1,200 kg/m3, respectively, and are open to the atmosphere at the top.

a. If an additional 1.0 cm3 of the second liquid is added to the right-hand leg, determine the change in hA. b. If the level hB falls by 0.5 cm, determine the rise in level hC.c. If hC = 3 cm and hA = 2 cm, determine the gauge pressure at the bottom of the manometer.d. If hC = 3 cm and hA = 2 cm, determine hB.

Solution ------------------------------------------------------------------------------------------

a. hA will change by an amount given by

= = 5.093 cm2

1( )(0.25 )

b. Level hC will rise by a distance .

= 0.5 = 0.125 cm25

10

c. The gauge pressure Pbot at the bottom of the manometer is

Pbot = (1.8)(0.03)(9.81) = 0.53 kPa = 530 Pa

d. If hC = 3 cm and hA = 2 cm, we have

530 = (1800)( hB)(9.81) + (1200)(.02)(9.81)

hB = = 0.0167 m = 1.67 cm294.56(1800)(9.81)

hB

hAhC

2-7

Chapter 2

2.2 Buoyancy forces. Archimedes’s law states that the buoyant force exerted on a submerged body is equal to the weight of the displaced fluid and acts in direction opposite to the gravity force. Consider the buoyant force on a submerged circular cylinder of height h and cross-sectional area A, shown in Figure 2.2-1.

P + gh

P

h

z

A

Solid Liquid

Figure 2.2-1 Buoyant force on a submerged cylinder.

Making a force balance in the vertical (z) direction, the net upward force due to the difference between the opposing pressures on the bottom and top surfaces is

Fb = (P + gh P)A = ghA

This net upward force is the buoyant force (Fb) that is the weight of the displaced liquid. Thus the effective net weight of a submerged body is its actual weight less the weight of an equal volume of the fluid. This result also applies to a body submerged in a fluid that is subject to any acceleration. For example, a solid particle of volume submerged in a fluid sV~

within a centrifuge at a point r where the angular velocity is is subjected to a net radial force equal to (s f)2r .sV~

Example 2.2-1. ----------------------------------------------------------------------------------Consider the situation shown in Figure 2.2-1, in which a person and a rock rest on a boat that floats in a lake. The rock is then pushed off the boat. If the rock sinks to the bottom, determine whether the water level in the lake will rise, fall, or remain constant, relative to the initial level in (a).

Lake Lake

Vi Vb

(a) Initial (b) Final Vr

Figure 2.2-1 A person with (a) and without (b) a rock in a boat floats in a lake.

2-8

Solution ------------------------------------------------------------------------------------------

Let the mass of the boat and the person be Mb and the mass of the rock be Mr. Let Vi be the initial volume of water displaced then

(Mb + Mr)g = Vig (E-1)

When the boat does not contain the rock, the volume of water displaced by the boat is given by

Mbg = Vbg (E-2)

Since the rock sinks, the weight of the water displaced by the rock no longer suffices to support the weight of the rock. Therefore

Mrg > Vrg (E-3)

Adding (E-2) and (E-3) we have

(Mb + Mr)g > (Vb + Vr)g (E-4)

Comparison (E-1) with (E-4) shows that

Vi > Vb + Vr

Since the volume of water in the lake is constant, and the total displaced volume is reduced, the level of the surface falls.

Example 2.2-2. ----------------------------------------------------------------------------------A rowboat is in a circular swimming pool with diameter 10 ft. The person in the rowboat throws overboard a 100 lbm block of wood, SG = 0.8. Determine The change in the water level in the pool. Water density is 62.3 lb/ft3.

Solution -----------------------------------------------------------------------------------------------------

In the boat, the part of the boat's displacement due to the block is

V m

water

100lbm

62.3 lbm / ft3 1.605 ft 3

In the water, the displacement due to the block is

V m

water

100lbm

62.3 lbm / ft3 1.605 ft 3

So the water level stays the same.

2-9

2.3 Pressure in Response to External Forces

The only stress that can exist in a fluid at rest is pressure. This also applies to fluids in motion provided there is no relative motion within the fluid since the shear stresses are determined by the velocity gradients or relative motion within the fluid. However the fluid will experience additional normal force if the motion involves acceleration.

Vertical Acceleration

Consider the cylinder of fluid illustrated in Figure 2.3-1. The fluid is accelerating upward with an acceleration of az. Applying a momentum or force balance in the z-direction on the control volume Az we obtain

Fz = maz

AP|z AP|z+z gAz = azAz

Dividing the equation by the control volume Az and letting z approach zero we obtain

= = (g + az) (2.3-1)0

limz z

PP zzz

||

dzdP

P|z

P|z+ z

z

z

A

az

Figure 2.3-1 Vertical accelerating fluid.

The effect of a superimposed vertical acceleration (az) is equivalent to an increase in the gravitational acceleration by az. In general, an acceleration (ai) in the i direction will result in a pressure gradient within the fluid in the i direction, of magnitude ai

= ai (2.3-2)ix

P

2-10

Horizontally Accelerating Free Surface

x

yax

Figure 2.3-2 Horizontally accelerating car.

Suppose we have a liquid in the bed of a pickup truck that accelerates in the x-direction with constant acceleration ax. The pressure within the fluid is now a function of both x and y since it experiences both the acceleration of gravity in the y-direction and the car acceleration in the x-direction.

P = P(x, y) dP = dx + dyxP

yP

dP = axdx gdy

The slope of the water surface can be determined since dP|s = 0 along the surface.

axdx|s gdy|s = 0

Therefore

= = tan sdx

dygax

The slope of the surface is independent of fluid properties.

2-11

Surface of a Rotating Liquid

Consider a liquid in a cylindrical container, as shown in Figure 2.3-3. The system is at steady state with an angular velocity (radian/s), and the axis is vertical. The pressure within the fluid is a function of z and r, therefore

P = P(r, z) dP = dr + dzrP

zP

z

r

zo

Figure 2.3-3 A container of liquid rotates about its axis.

The pressure gradient in the r-direction is due to the centripetal acceleration (r2) experienced by an element of liquid at radial position r. The centripetal acceleration is negative since it points toward the center while the unit vector for the r-direction is positive pointing outwards. Therefore

dP = r2dr gdz

This equation can be integrated along the free surface where dP|s = 0

2 g = 0 z = zo + rrdr

0 z

zo

dzg

r2

22

Acceleration in Uniform Circular Motion

We now want to derive the formula for the centripetal acceleration used in the previous section (a = r2). The acceleration of an object in uniform circular motion is not zero since the velocity vector changes direction. In Figure 2.3-4, a particle is located at the point determined by the angle at time t. At a later time t + t, the particle’s angular position is given by + . The direction of the velocity vector is always tangential to the circle and therefore changes continuously with time. The velocity vectors v(t) and v(t + t) have the same magnitude. The change in the velocity vector over the time period t is given by

v = v(t + t) v(t)

2-12

o o

v(t)

v(t+ t) r(t)r

r(t+ t)

vaav

v(t)

v(t+ t)

x

y

x

y

Figure 2.3-4 A container of liquid rotates about its axis.

This vector difference is drawn to occur at the midway point (time t + t/2). We see that v, and hence aav = v/t, points toward the center of the circle. In the limit in which t goes to zero, the ratio v/t gives us the instantaneous acceleration that points precisely to the center of the circle. The magnitude of the acceleration is

a = 0

limt t

v

The angle between v(t) and v(t + t) is the same as the angle between r(t) and r(t + t), therefore

= v = rvv

rr

rv

The acceleration is then

a = = v = rv

0limt t

r

rv

rv2

In terms of the angular velocity v = r, hence a = r2.

Example 2.3-1. ----------------------------------------------------------------------------------A car ferry is essentially rectangular with dimensions 8 m wide and 100 m long. If 70 cars, with an average mass per car of 1600 kg, are loaded on the ferry, how much farther will it sink into the water?

Solution ------------------------------------------------------------------------------------------

Let h be the distance the feery will sink farther into the water, we have

(8)(100)(h)(1000) = (1600)(70) => h = = 0.14 m(1.6)(70)800

2-13

Example 2.3-22. ----------------------------------------------------------------------------------A vertical U-tube manometer is open to the atmosphere and contains a liquid that has an SG of 0.87 and a vapor pressure of 450 mmHg at the operating temperature. The vertical tubes are 4 in. apart, and the level of the liquid in the tubes is 6 in. above the bottom of the manometer. The manometer is then rotated about a vertical axis through its centerline. Determine the required rotation rate (in rpm) for the liquid to start to boil.

z

r

(1)

(2)

Figure E2.3-1 A rotating U-tube manometer.

Solution ------------------------------------------------------------------------------------------

The pressure P(r, z) within the manometer fluid is a function of both r and z. Therefore

dP = dr + dz = r2dr gdzrP

zP

For a rotating fluid, the pressure P1 at r = 0 is the minimum. Therefore P1 = vapor pressure of liquid for it to boil.

= 2 dr g2

1

P

PdP

Rr

0 2

1

z

zdz

P2 P1 = 2 g(z2 z1)2

2R

Solving for the rotation speed , we have

2 = [ + g(z2 z1)]22

R )( 12 PP

2 = = 3.397104 rad2/s22)54.22(

2

54.26980

87.076010013.1)450760( 6

= 195 rad/s = 1861 rpm

2 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 103

2-14

Example 2.3-3. ----------------------------------------------------------------------------------The pressure gradient required for water flowing through a straight horizontal ¼ in. ID tube at a rate of 5 gpm is 1.4 psi/ft. Consider this same tube coiled in an expanding helix with a vertical axis. Water enters the bottoms of the coil and flows upward at a rate of 5 gpm. A mercury manometer is connected between two pressure taps on the coil, one near the bottom where the coil radius is 8 in., and the other near the top where the coil radius is 14 in. The taps are 1.6 ft apart in the vertical direction, and there is a total of 6 ft of tubing between the two taps. (1 ft3 = 7.48 gal). Note: pressure must be in unit of lbf/ft2, g = 32.2 ft/s2, lbf = 32.2 lbmft/s2.

a) Determine the pressure difference between the two taps due to gravity force only.b) Determine the pressure difference between the two taps due to centrifugal force only.c) Determine the pressure difference between the two taps due to frictional loss only.

Solution ------------------------------------------------------------------------------------------

a) The pressure difference between the two taps due to gravity force only is

Pg = (62.4)(32.2)(1.6)/32.2 = 99.84 lbf/ft2

b) The pressure difference between the two taps due to centrifugal force only is

Pc = = = v2ln(r2/r1) 2

1

2r

rr dr

2

1

2r

r

v drr

v = = 32.68 ft/s2

(5)(4)(7.48)(60)( )(0.25/12)

Pc = (62.4)(32.68)2ln(14/8)/32.2 = 1,158.3 lbf/ft2

c) The pressure difference between the two taps due to frictional loss only is

Pf = (1.4)(6)(144) = 1,209.6 lbf/ft2

3-1

Chapter 3

Fluid Properties

3.1 Introduction

A fluid is defined as a substance that deforms continuously when a shear stress is applied to it. Figure 3.1-1 shows the fluid motion when a force F is applied to a plate on top of the fluid causing the plate to move. As long as there is movement of the plate, the fluid continues to flow or deform since the fluid next to the plate is under the action of a shear stress equal to the force F divided by the surface area of the plate. When a fluid is at rest (no relative motion between the fluid elements), there can be no shear stress.

Fluid

FPlate

Figure 3.1-1 Fluid moves with the plate.

Both liquids and gases are fluids even though they are quite different at the molecular level. In liquids the molecules are held close together by significant attraction forces; in gases molecules are relatively far apart and have very weak attraction forces. Figure 3.1-2 shows a PV diagram for water where the isotherms are plotted with the isotherm of highest temperature on the top. An isotherm is a curve that relates pressure to volume at a constant temperature. As temperature and pressure increase, the differences between liquid and gas become less and less, until the liquid and gas become identical at the critical point. For water the critical point occurs at 374.14oC and 22.09 MPa. Because of their closer molecular spacing, liquids normally have higher densities, viscosities, and other physical properties than gases.

Figure 3.1-2 PV diagram for water.

3-2

3.2 Rheology

Rheology is the study of the deformation and flow behavior of fluids. For a Newtonian fluid, we have a linear relationship between shear stress () and the shear rate ( ) or rate of shear strain.

= (3.2-1)

In this equation, the proportional constant is called the viscosity of the fluid. The viscosity is the property of a fluid to resist the rate at which deformation takes place when the fluid is acted upon by a shear forces. As a property of the fluid, the viscosity depends upon the temperature, pressure, and composition of the fluid, but is independent of the shear rate. Most simple homogeneous liquids and gases are Newtonian fluid.

y

x

(v | - v | ) tx y+ y x y

Element at time t

Element attime t+ t

y

xFigure 3.2-1a Deformation of a fluid element.

The rate of deformation of a fluid element for a simple one-dimensional flow is illustrated in Figure 3.2-1a. The flow parallel to the x-axis will deform the element if the velocity at the top of the element is different than the velocity at the bottom. The shear rate at a point is defined as

= dtd

= dtd

0,,lim

tyx tttt

= dtd

0,,lim

tyx t

ytvvyxyyx

2/]/)arctan[(2/

= dtd

0,,lim

tyx t

ytvvyxyyx

]/)arctan[(

3-3

For small angle , arctan() = , therefore

= = dtd

0,,lim

tyx t

ytvvyxyyx

/)(

0,,lim

tyx y

vvyxyyx

)(

= = dtd

dydvx

The shear stress for this simple flow is the negative of the molecular momentum flux in the y-direction and is given as

yx = = (yx)mf (3.2-2a)dydvx

The subscript yx on yx denotes the viscous flux of x momentum in the y direction. In this equation, the shear stress is defined to have the opposite sign of the momentum flux (yx)mf. We have defined yx in terms of the force exerted on a plane of constant y by fluid at greater y, positive value of yx correspond to transfer of x momentum in the y direction as shown in Figure 3.2-1b

x

y

x

y

Momentum flux in -yForce in positive x direction

Momentum flux in yForce in negative x direction

F F

Figure 3.2-1b Stress in terms of the force per unit area on the top surface.

We can also use the opposite sign convention to defined yx in terms of the force exerted on a plane of constant y by fluid at lesser y, positive value of yx correspond to transfer of x momentum in the y direction as shown in Figure 3.2-1c

x

y

x

y

Momentum flux in -yForce in negative x direction

Momentum flux in yForce in positive x direction

F F

Figure 3.2-1c Stress in terms of the force per unit area on the bottom surface.

3-4

For this sign convention

yx = = (yx)mf (3.2-2b)dydvx

The shear stress in this case is just the viscous flux of x momentum in the y direction and we have analogy between transport in simple flows and the transport of heat or chemical species.

We may use equation (3.2-2) to obtain an expression for shear stress as a function of the fluid velocity and the system dimension. Consider the situation shown in Figure 3.2-2 where a fluid is contained between two large parallel plates both of area A. The plates are separated by a distance h. The system is initially at rest then a force F is suddenly applied to the lower plate to set the plate into motion in the x direction at a constant velocity V. Momentum is transferred from a region of higher velocity to a region of lower velocity. As time proceeds, momentum is transferred in the y direction to successive layers of fluid from the plate that is in motion in the x direction.

x

y

t < 0rest

t = 0lower plate moves

t > 0velocitydevelops

t >> 0steady velocityprofile

h

V V V F

Figure 3.2-2 Velocity profile development for a flow between two parallel plates.

The velocity profile of the fluid between the parallel plates may be obtained by applying the momentum balance, which states that

Time rate of changeof linear momentum =

within the CV

rate of linearmomentum enters

the CV

rate of linearmomentum exits +

the CV

sum of externalforces acting on

the CV

Since the velocity in the x direction vx is dependent on the y direction, we choose the control volume CV to be Ay as shown in Figure 3.3-3.

3-5

y

y+ yyx mf y+ y) |

yx mf y) |

Figure 3.3-3 x-Momentum entering and leaving the CV = Ay

Applying the x-momentum balance on the CV yields

( Ay vx) = A At

ymfyx )(

yymfyx )(

Dividing the equation by Ay and letting y 0, we obtain for constant physical properties

= = (3.2-3)t

vx

0limy y

ymfyxyymfyx

)()(

ymfyx

)(

Substituting (yx)mf = into equation (3.2-3) yields a second order partial differential yvx

equation (PDE)

= = (3.2-4)t

vx

2

2

yvx

tvx

2

2

yvx

where = / is the kinematic viscosity of the fluid. Equation (3.2-4) can be solved with the following initial and boundary conditions:

Initial condition: t = 0, vx = 0 (3.2-4a)

Boundary conditions: y = 0, vx = V and y = h, vx = 0 (3.2-4b)

Equation (3.2-4) with the auxiliary conditions (3.4-2a,b) can be solved by the separation of variables method with the following result

vx = V(1 ) exp sinhy

V2

1

1n n

2

22

htn

hyn

The solution can also be expressed in dimensionless form with vx* = , y* = , t* =

Vvx

hy

2ht

3-6

vx* = 1 y* exp sin(n y*) (3.2-5)

2

1

1n n

*22 tn

Table 3.2-1 lists the Matlab program to plot dimensionless velocity profiles at various dimensionless times. The results are shown in Figure 3.2-4.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

vx/V

y/h

t*=.01

t*=.05

t*=.1

t*=1

t*=.2

Figure 3.2-4 Dimensionless velocity profiles for flow between two parallel plates.

____ Table 3.2-1 Matlab program to plot vx* = 1 y* exp sin(n y*)___

2

1

1n n

*22 tn

yoh=0.05:.05:.95; np=length(yoh);u=yoh;thetav=[.01 .05 .1 .2 1]; nt=length(thetav);n=1:20; ns=n.*n;hold onfor k=1:nt theta=thetav(k);for i=1:np y=yoh(i); sum1=(1.0./n).*exp(-ns*pi*pi*theta)*sin(n*pi*y)'; u(i)=(1-y)-2*sum1/pi;endyp=[0 yoh 1];up=[1 u 0]; plot(up,yp)endxlabel('v_x/V');ylabel('y/h'); grid on

3-7

As time approaches infinity, the system reaches steady state and the summation terms in equation (3.2-5) become zero. The steady state velocity profile is then

vx* = 1 y* (3.2-6)

The steady state solution can also be obtained directly from equation (3.2-4) by setting the temporal derivative equal to zeo.

= 0 (3.2-7)2

2

dyvd x

Integrating equation (3.2-7) twice, we obtain

vx = Ay + B

The two constants of integration are evaluated from the boundary conditions:

y = 0, vx = V and y = h, vx = 0

Therefore B = V and A = V/h

Hence vx = V(1 y/h) vx* = 1 y*

The shear rate at any position y in the fluid is given as = = dydvx

hV

The force to pull the lower plate at velocity V can be evaluated: F = A = A0yyx

hV

Fluids are classified as Newtonian or non-Newtonian, depending upon the relation between shear stress and shear rate. In Newtonian fluids the relation is linear while in non-Newtonian fluids, the shear stress is not a linear function of shear rate as shown in Figure 3.2-5.

Yieldstress

Shear rate

Ideal plastic

Real plastic

Pseudo plastic

Newtonian fluid

Dilatant

Figure 3.2-5 Behaviors of Newtonian and non-Newtonian fluids.

3-8

The slope of the Newtonian fluid line is the viscosity. For the non-Newtonian fluids, the slope is not constant therefore its value at a given shear rate is called the apparent viscosity. The apparent viscosity of a dilatant fluid increases with shear rate while the apparent viscosity of a pseudo platic decreases with shear rate. The ideal or Bingham plastic has a linear shear stress-shear rate relation for stresses greater than the yield stress. Real plastic or Carson fluid also flows with stresses greater than the yield stress. The apparent viscosity however decreases with shear rate and at some point the Carson fluid behaves as a Newtonian fluid. Heterogeneous fluids that contain a particulate phase that forms aggregates at low rates of shear require a yield stress.

Blood is a heterogeneous fluid with the particulates consisting primarily of the red blood cells. Therefore blood follows the curve shown for real plastic. At low shear rates, red blood cells clump together to form aggregates. This behavior results in high value of apparent viscosity. However, at shear rate higher than 100/s, red blood cells do not clump together, therefore blood behaves as a Newtonian fluid with an apparent viscosity of about 3 cP. The properties of blood change rapidly if removed from the system and so it is extremely difficult to perform experiments on it under laboratory conditions.

Example 3.2-1. ----------------------------------------------------------------------------------A fluid of viscosity 0.4 kg/ms flows steadily through a 0.10 m diameter pipe with a velocity profile given by vz = 2.0[1 – (r/R)2] (m/s). The pipe is 25 m long.

a) Determine the shear stress at the inside surface of the pipe.b) If the shear stress at the inside surface of the pipe is 6 N/m2, determine the shear force

on the pipe.

Solution ----------------------------------------------------------------------------------------------------- a) The shear stress at the inside surface of the pipe is

rz = = 4r/R2 w = 4 /R = 4(0.4)/0.05 = 32 Paszdvdr

b) If the shear stress at the inside surface of the pipe is 6 N/m2, the shear force on the pipe is

F = (6)()(0.10)(25) = 47.12 N

3-9

3.3 Fully Developed Laminar Flow in Tube

We want to develop a relationship for shear stress-shear rate given volume flow rate Q and pressure drop P across the horizontal tube as shown in Figure 3.3-1. We use cylindrical coordinates with the following assumptions: the length of the tube (L) is much larger than the tube radius (R) (i.e. L/R > 100) to eliminate entrance effect; steady incompressible and isothermal flow; one-dimensional flow in the z direction only, therefore vz = vz(r); and no-slip boundary condition at the wall.

z

z

Rr

Pz Pz+ z

rz

L

PoPL

Figure 3.3-1 Forces acting on a cylindrical fluid element within a tubey.

Consider the control volume r2z shown in Figure 3.3-1. For steady flow, the summation of the viscous and pressure forces acting on the control volume must be equal to zero.

P|zr2 + rz2rz P|z+zr2 = 0

Dividing the equation by the control volume yields

= rzzPP

zzz

r2

In the limit when z 0, we obtain the differential equation for the shear stress distribution

= rz (3.3-1)dzdP

r2

Since rz = , the right hand side of equation (3.3-1) is a function of r only and the left drdvz

hand side of equation (3.3-1) is a function of z only. They both must be equal to a constant

= rz = dzdP

r2

LPP Lo

The equation is rearranged to

rz = = (3.3-2)2r

dzdP

LrPP Lo

2

The shear stress vanishes at the centerline of the tube and achieves its highest value, w, at the wall.

3-10

w = rz|r=R = = (3.3-3)2R

dzdP

LRPP Lo

2

Equations (3.3-2) and (3.3-3) are valid for both Newtonian and non-Newtonian fluids since we has not specified a relationship between the shear stress and shear rate. Solving equations

(3.3-2) and (3.3-3) for yieldsdzdP

= rz = wdzdP

r2

R2

Therefore rz = w = (3.3-4)Rr

LrPP Lo

2

If = = constant, the fluid is a Newtonian fluid and is called the viscosity. If =

rz

rz

constant, the fluid is a non-Newtonian fluid and is called the apparent viscosity. We will follow different procedures to determine a relationship between shear stress and shear rate depending on whether or not and rz are known directly.

A) and rz are not known directly

We want to find a general relationship between the shear rate and some function of the shear stress in terms of the measurable quantities Qmeas., Po PL, L, and R. That is:

= = (rz) (3.3-5)drdvz

We can follow the following procedures to obtain a relationship between the shear rate and shear stress rz.

1) We calculate the volumetric flow rate from the axial velocity profile as follows.

Qcal = 2 rdrR

z rv0

)(

2) We express Qcal in terms of shear rate using integration by part.

d(uv) = vdu + udv du = dv v )(uvd u

Let v = vz(r) dv = dvz(r)

du = rdr u = r221

3-11

Therefore Qcal = 2 r2 2 dr = dr21 R

zv0

Rr

0

2

21

drdvz

Rr

0

2

drdvz

3) Next, we change the integration variable from r to rz using equation (3.3-4)

rz = w r = rz dr = drzRr

w

R w

R

Qcal = drz , (Note: = )rz R

0

22

w

rz

w

R

drdvz

Qcal = drz (3.3-6)3

3

w

R

rz

0

2rz

4) We then assume a relationship between and rz (for example = ).

2/1rz

Equation (3.3-6) is then integrated to obtain Qcal. We will accept the assumed expression between and rz if Qcal Qmeas. Otherwise step (4) is repeated.

B) and rz are known directly

The shear stress and shear rate can be determined using a cup-and-bob or Couette viscometer. As the name implies, the Couette viscometer consists of two concentric cylinders as shown in Figure 3.3-2. The fluid is in the annular gap between the outer cylinder (cup) and the inner cylinder (bob).

T

L

RiRo

Figure 3.3-2 Couette viscometer.

The outer cylinder is rotated at a fixed angular velocity (). The shearing force is transmitted to the fluid, causing it to deform or flow. The inner cylinder is kept stationary by a torque (T)

3-12

that can be measured by a torsion spring. The shear stress at any position r within the gap (Ri r Ro) is determined by a balance of moments on a cylindrical surface 2rL

T = r(2rL)r

Solving for the shear stress, we have

r = (3.3-7)Lr22

T

Setting r = Ri gives the stress on the bob surface (i), and setting r = Ro gives the stress on the cub surface (o). If the gap is small [i.e., (Ro Ri)/Ro 0.02], the flow in the annular gap can be approximated by the flow between two parallel plates. In this case, an average shear stress should be used

r = where = (Ri + Ro)/2 (3.3-8)2

oi LR 22

T R

The average shear rate is given by

= = = (3.3-9)drdv

io

io

RRVV

io

o

RRR

oi RR /1

Equations (3.3-8, 9) provide the experimental values for the shear stress and the shear rate that can be fitted by a non-Newtonian fluid model.

Example 3.3-1.1 ----------------------------------------------------------------------------------The viscosity of a fluid sample is measured in a cup-and-bob viscometer. The bob is 15 cm long with a diameter of 9.8, and the cup has a diameter of 10 cm. The cup rotates, and the torque is measured on the bob. The following data were obtained:

(rpm) 2 4 10 20 40T (dyncm) 3.6105 3.8105 4.4105 5.4105 7.4105

(a) Determine the viscosity of the sample.(b) Fit the data with the following model equations

= o + (Bingham Plastic Model)

and = m n (Power Law Model)

(c) Determine the viscosity of this sample at a cup speed of 100 rpm in the viscometer using the above models.

1 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 74

3-13

Solution ------------------------------------------------------------------------------------------

Since (Ro Ri)/Ro = (10 9.8)/10 = 0.02, we can use the equation (3.3-8, 9) to determine the shear stress and the shear rate

r = , and = LR 22

T oi RR /1

The apparent viscosity is determined from

=

r

Table 3.3-1 lists the results from the calculation. Table 3.3-2 lists the Matlab program to fit data with the Bingham plastic and power law model.

Table 3.3-1 Fluid apparent viscosity at different shear rates(rpm) T (dyncm) (1/s) r(dyne/cm2) (Poise = g/cms)24102040100

360000380000440000540000740000

10.520.952.5105209524

156165191234320

14.897.863.642.231.53

For the Bingham plastic model, we obtain

(dyne/cm2) = o + = 147 + 0.827 (1/s)

For the power law model, we obtain

(dyne/cm2) = m n = 83.2 0.234

At 100 rpm or = 524 s-1, for the Bingham plastic model

= 147 + 0.827524 = 580 dyne/cm2

= 580/524 = 1.11 g/cms

For the power law model

= = 83.2 0.2341 = 0.69 g/cms

r

______ Table 3.3-2 Matlab program to fit shear stress and shear rate data ______

3-14

% Example 3.3-1%rpm=[2 4 10 20 40];Torque=[36 38 44 54 74]*1e4;ndata=length(rpm);Ri=9.8/2;Ro=10/2;L=15;omega=rpm*2*pi/60;shear_rate=omega/(1-Ri/Ro)Rave=(Ri+Ro)/2;stress=Torque/(2*pi*Rave^2*L)vis=stress./shear_rateco=polyfit(shear_rate,stress,1); tao=co(2)vis_inf=co(1)x=log(shear_rate);y=log(stress);co=polyfit(x,y,1); n=co(1)m=exp(co(2))drate=(shear_rate(ndata)-shear_rate(1))/25;s_rate=shear_rate(1):drate:shear_rate(ndata);tao1=tao+vis_inf*s_rate;vis1=tao1./s_rate;vis2=m*s_rate.^(n-1);loglog(shear_rate,vis,'d',s_rate,vis1,s_rate,vis2,':')xlabel('Shear rate (1/s)');ylabel('Viscosity (Poise)')legend('Data','Bingham plastic','Power law')grid on% Evaluate the correlation coefficientvis_ave=mean(vis);St=(vis-vis_ave)*(vis-vis_ave)';tao1=tao+vis_inf*shear_rate;vis1=tao1./shear_rate;vis2=m*shear_rate.^(n-1);S1=(vis-vis1)*(vis-vis1)';r1=sqrt(1-S1/St);S2=(vis-vis2)*(vis-vis2)';r2=sqrt(1-S2/St);fprintf('Correlation coefficient for Bingham plastic = %8.4f\n',r1)fprintf('Correlation coefficient for Power law = %8.4f\n',r2)

>> e3d1shear_rate = 10.4720 20.9440 52.3599 104.7198 209.4395stress = 155.8910 164.5516 190.5334 233.8365 320.4426vis = 14.8865 7.8568 3.6389 2.2330 1.5300tao = 147.2304vis_inf = 0.8270n = 0.2337m = 83.1932Correlation coefficient for Bingham plastic = 1.0000Correlation coefficient for Power law = 0.9938

3-15

A crude measure of the how well the data is fitted by an expression is given by the correlation coefficient r, which is defined as

r = tS

S1

In this expression St = is the spread of the data around the mean of the

N

ii YY

1

2)( Y

dependent variable and S = is the sum of the square of the difference between

N

iii yY

1

2)(

the data (Yi) and the calculated value (yi).

Figure 3.3-3 shows a plot of viscosity versus flow rate for the Bingham plastic and the Power law models. The Bingham plastic model fits the data better as evident by its higher correlation coefficient (1.0) in comparison with that (0.9938) of the Power law model.

101

102

103

100

101

102

Shear rate (1/s)

Vis

cosi

ty (P

oise

)

DataBingham plasticPower law

Figure 3.3-3 Behavior of non-Newtonian fluid.

---------------------------------------------------------------------------------------------------

3-16

3.4 The Hagan-Poiseuille Equation

We now consider the case of a Newtonian fluid flowing through a capillary. The shear rate-

shear stress relation (rz) = is substituted into equation (3.3-6) to obtain

rz

Q = drz = = 3

3

w

R

rz

rz

0

33

3

w

R w

rz

04 3

3

w

R

4

4w

Q = w = =

4

3R

4

3R L

RPP Lo

2

LPPR Lo

8

4

The velocity profile inside the capillary can also be obtained by integrating equation (3.3-2)

rz = = (3.3-2)drdvz

LrPP Lo

2

= zv

zdv0

LPP Lo

2

r

Rrdr

vz = L

RPP Lo

4

2

2

1Rr

Example 3.4-1. 2 ----------------------------------------------------------------------------------You are asked to measure the viscosity of an emulsion, so you use a tube flow viscometer similar to that shown below, with the container open to the atmosphere.

h

L

Q

Po

PL

The length of the tube is 10 cm, its diameter is 2 mm, and the diameter of the container is 3 in. When the level of the sample is 10 cm above the bottom of the container the emulsion drains through the tube at a rate of 12 cm3/min, and when the level is 20 cm the flow rate is

2 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 80

3-17

30 cm3/min. The emulsion density is 1.3 g/cm3. What can you tell from the data about the viscous properties of the emulsion?

Solution ------------------------------------------------------------------------------------------

Equation (3.3-3) provides a relation between the wall shear stress and the pressure drop across the tube

w = rz|r=R = = (3.3-3)2R

dzdP

LRPP Lo

2

This equation is valid for any Newtonian or non-Newtonian fluid. Po is essentially the pressure at the bottom of the container and PL is the ambient pressure Patm. Therefore

Po PL = gh

The magnitude of the wall shear stress is then given by

w = L

ghd4

where d is the inside diameter of the tube.

When h = 10 cm, w1 = = 63.7 dyne/cm2)10)(4(

)2.0)(10)(980)(3.1(

When h = 20 cm, w2 = = 127.4 dyne/cm2)10)(4(

)2.0)(20)(980)(3.1(

If the fluid is a Newtonian fluid then

Q = w4

3R

so that = = 21

2

QQ

1

2

W

W

From experimental data

= = 2.5exp1

2

QQ

1230

Therefore the viscosity at high shear rate is smaller than that at lower shear rate. The emulsion is a shear thinning liquid.

3-18

Example 3.4-2. ----------------------------------------------------------------------------------A Cannon-Fenske Viscometer is used to measure the viscosity of an unknown liquid. This liquid is drawn into the tube to a level above the top etched line. The time is then measured for the liquid to drain to the bottom etched line. The kinematic viscosity, , is then obtained form the equation = KR4t where K is a constant, R is the radius of the capillary tube and t is the drain time. When glycerine at 20oC ( = 0.119 cm2/s) is used to calibrate a particular viscometer, the drain time is 1,430 s. When a liquid having a density of 0.97 g/cm3 is tested in the same viscometer, the drain time is 900 s. What is the dynamic viscosity of this liquid?

Etched lines

Capillary

Solution ------------------------------------------------------------------------------------------

Using the relation = KR4t for glycerine we have

0.119 cm2/s = KR4(1,430 s)

KR4 = 8.3210-5 cm2/s2

For unknown liquid with t = 900 s

= (8.3210-5 cm2/s2)(900s) = 0.0749 cm2/s

The dynamic viscosity of the unknown liquid is then

= = (0.0749 cm2/s)(0.97 g/cm3) = 0.0727 g/cms

= 7.27 cP (centipoise)

3-19

Chapter 3

3.5 Mechanical Concepts of Energy

3.5-1 Kinetic Energy

Consider a moving body with velocity V shown in Figure 3.5-1. The body is acted on by a resultant force F, which could be a function of s, a distance along the path from point 0. Fs is a component of F along the path.

dsV

Fs

Fn

F

s

Path

0

y

x

Body

Figure 3.5-1 Force acting on a moving body.

The magnitude of the component Fs is related to the change in the magnitude of V by Newton’s second law of motion:

Fs = ma = m (3.5-1)dVdt

Using the chain rule, the above expression can be written as

Fs = m = m = mV (3.5-2)dVdt

dVds

dsdt

dVds

We have used the definition V = . Separating the variables and integrating from s1 to s2 dsdt

gives

= (3.5-3)2

1

V

VmVdV

2

1

s

ssF ds

m = (3.5-4)12 2 2

2 1V V2

1

s

sd AF s

3-20

The quantity mV2 is the kinetic energy, KE, of the body. The integral is the work 12

2

1

s

ssF ds

of the force Fs acting on the body from s1 to s2. Fsds is the scalar product of the force vector F and the displacement vector ds. Both kinetic energy and work are scalar quantity. The work done on the body can be considered a transfer of energy to the body, where it is stored as kinetic energy.

Example 3.5-1 ----------------------------------------------------------------------------------A spear of mass, m = 0.3 kg, is propelled horizontally from a spear gun by a scuba diver. The initial speed of the spear is Vo = 30 m/s. The force that resists its motion through the water is given by FD = kV2, where k = 0.033 N.sec2/m2. The spear is effective against sharks when its speed is above V = 10 m/s. Estimate the effective range of the spear.Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.System: spear.

SpearmaFD

Step #2: Find equation that contains s, the distance the spear travels.

Conservation of momentum: F = ma = mdtdV

In this equation s is related to V, speed of the spear by V = dtds

Step #3: Apply the momentum balance on the spear.

m = m = FDdtdV

dsdV

dtds

Step #4: Specify the boundary conditions for the differential equation.

The effective range of the spear is the distance from s = 0 where the initial speed of the spear is 30 m/s to the distance s where the speed of the spear is 10 m/s.

Step #5: Solve the resulting equation and verify the solution.

m = FDdsdV

dtds

mV = kV2

dsdV

= km

V

Vo VdV

sds

0

3-21

s = ln = lnkm

oVV

km

VVo

s = ln = 10.0 m033.03.0

1030

----------------------------------------------------------------------------

3.5-2 Potential Energy

Gravitational work can be defined as the work due to gravitational force field. The gravitational work, Wg, required to raise a body of mass m vertically from z1 to z2 is

Wg = = = mg = mg(z2 z1)2

1

z

gzF dz

2

1

z

zmgdz

2

1

z

zdz

The quantity mgz is the gravitational energy, PE. The change in gravitational potential energy, PE, is

PE = PE2 PE1 = mg(z2 z1)

Potential energy is regarded as an extensive property of the body. For this course, it is assumed that elevation differences are small enough that the gravitational force can be considered constant.

Example 3.5-2 ----------------------------------------------------------------------------------Consider a 3000-lb car cruising steadiy on a level road at 65 mile/h. Now the car starts climbing a hill that is sloped 6% from the horizontal. Determine the additional power delivered by the engine in order for the car to maintain its speed at 65 mile/h.Solution ------------------------------------------------------------------------------------------

The additional power required is simply the gravitional work that must be done per unit time to reais the elevation of the car, which is equal to the change in the potential energy of the car per unit time:

= mgz/t = mgVverticalgW

Vvertical = 65 mile/h = (65)(5280)(sin 6o)/3600 = 9.965 ft/s

= mgVvertical = = 29,895 lbfft/sgW2

2m

f

(3000 lb)(32.2 ft/s )(9.965 ft/s)lb ft/s32.2

lb

Since 1 hp = 550 lbfft/s

= 29,895/550 = 54.4 hpgW

3-22

3.6 Bernoulli Equation

Consider a moving body with velocity V shown in Figure 3.6-1. The velocity vector V is defined as the time rate of change of the position of the particle.

dsV

s

0

z

x

Body

n

R = R(s)

Streamlines

Figure 3.6-1 Streamline and normal coordinates.

The lines that are tangent to the velocity vectors throughout the flow field are called streamlines. For many applications it will be simpler to describe the flow in terms of the streamline coordinates based on the streamlines shown in Figure 3.6-1. The particle motion is described in terms of its distance, s = s(t), along the streamline from some arbitrary origin and the local radius of curvature of the streamline, R = R(s). The distance along the streamline is related to the particle’s speed by V = ds/dt. The radius of curvature is related to the shape of the streamline. For two-dimensional flow in the x-z plane, the acceleration has two components: one along the streamline, as, the streamwise acceleration, and one normal to the streamline, an, the normal direction. The component of acceleration in the s direction is given by

as = = = V (3.6-1)dVdt

Vs

dsdt

Vs

The component of acceleration in the n direction is given by

an = (3.6-2)2V

R

In this equation, R is the local radius of curvature of the streamline. There is acceleration normal to the streamline because the particle does not flow in a straight line. Newton’s second law can be applied to a fluid particle along the streamline to obtain the following equation of motion:

dp + d(V2) + gdz = 0 (3.6-3)12

3-23

For constant g, steady, inviscid, and incompressible flow, equation (3.6-3) can be integrated to obtain the famous Bernoulli equation

p + V2 + gz = constant along streamline (3.6-4)12

The equation of motion along the normal direction is given by

+ + g = 0 (3.6-5a)pn

2VR

dzdn

Eq. (3.6-5a) can be written as

g = (3.6-5b)pn

dzdn

2VR

In this form, the physical meaning of Eq. (3.6-5b) is that the change in the direction of a fluid particle is due to the actions of the pressure gradient and particle weight normal to the streamline. If gravity is neglected or if the flow is in a horizontal (dz/dn = 0) plane, Eq. (3.6-5) becomes

+ = 0 (3.6-6)pn

2VR

For flow between points 1 and 2 on the same streamline, or for any two points in a fluid under static equilibrium (in which case the velocities are zero, Eq. (3.6-4) can be written as

+ gz1 + = + gz2 + (3.6-7)2

1

2V 1p

22

2V 2p

Eq. (3.6-7) states that although the kinetic, potential, and pressure energies may vary individually, their sum remains constant. Eq. (3.6-7) can also be divided by the gravitational acceleration g to have unit of length:

+ z1 + = + z2 + = H (3.6-8)2

1

2V

g1pg

22

2V

g2pg

The terms , z1, , and H are called the velocity head, static head, pressure head, and 2

1

2V

g1pg

total head, respectively. The static head after a pressure rise across a pump will indicate the height that the liquid can be delivered by the pump.

3-24

Example 3.6-1. ----------------------------------------------------------------------------------Water is flowing into the top of a tank at a rate of 5.0 ft3/s. The tank is 18 in. in diameter and has a 4 in. diameter hole in the bottom, through which the water flows out. If the inflow rate is adjusted to match the outflow rate, what will the height of the water be in the tank if friction is negligible?

Solution -----------------------------------------------------------------------------------------------------

(1)

(2)

h

Applying Bernoulli equation between points 1 and 2 gives

gz1 + 0.5u12 = gz2 + 0.5u2

2 h = z1 z2 = ( u22 u1

2)12g

From the mass balance we have

u2 = u1 = 20.25u1 u1 = = 2.8294 ft/s2

1

2

DD

2

5(9 /12)

The height of the water in the tank is

h = 2.82942 = 50.85 ft220.25 1

(2)(32.2)

If we neglect u12 compared to u2

2, the height of the water in the tank is

h = 2.82942 = 50.97 ft220.25

(2)(32.2)

Example 3.6-2. ----------------------------------------------------------------------------------Water (density = 1000 kg/m3) is flowing inside a 0.12 m diameter pipe with a flow rate of 0.14 m3/s. Determine the velocity head in the pipe.

Solution -----------------------------------------------------------------------------------------------------

Velocity in the pipe is V = = 12.38 m/s2

0.140.06

The velocity head is then: = = 7.81 m2

2V

g

212.382 9.81

3-25

Example 3.6-3.3 ----------------------------------------------------------------------------------A stream of water of diameter d = 0.1 m flows steadily from a tank of diameter D = 1.0 m as shown below. Determine the flowrate, Q, needed from the inflow pipe if the water depth remains constant, h = 2.0 m.

Solution -----------------------------------------------------------------------------------------------------

Applying the Bernoulli equation between points 1 and 2 we have

+ gz1 + = + gz2 + (E-1)2

1

2V 1p

22

2V 2p

Since p1 = p2 = patm, z1 = h, and z2 = 0, Eq. (E-1) becomes

+ gh = (E-2)2

1

2V 2

2

2V

From the steady mass balance for incompressible fluid we have

A1V1 = A2V2 V1 = V2

2dD

Substituting V1 = V2 into Eq. (E-2) we have2d

D

V22 + gh = 1

2

4dD

22

2V

Solving for V2 gives

3 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 115

3-26

V2 = = = 6.26 m/s

1/ 2

42

1 /gh

d D

1/ 2

4(2)(9.81)(2)1 0.1/1

Q = A2V2 = (0.052)(6.26) = 0.0492 m3/s

For <0.4 we can neglect compared to since the error in velocity calculation is dD

21

2V 2

2

2V

less than 1%. The velocity V2 is then

V2 = 1/ 22gh

Example 3.6-4.4 ----------------------------------------------------------------------------------Shown in Fig. E3.6-4-1 are two flow fields with circular streamlines. The velocity distributions are

V(r) = C1r for case (a)

V(r) = for case (b)2Cr

In these two equations C1 and C2 are constants.

Figure E3.6-4-1 Rotational and irrotational flows.

Determine the pressure distributions, p = p(r), for each, given that p = p0 at r = r0.

Solution -----------------------------------------------------------------------------------------------------

Case (a) represents forced vortex or rotational flow as in solid rotation of a rigid body. Case (b) represents free vortex or irrotational flow such as a tornado or the swirl of water in a drain. Figure E3.6-4-2 illustrates the position of a fluid particle for cases (a) and (b). For rotational flow the fluid element will rotate in a circular motion while for irrotational flow there will be no rotation of the fluid element during circular motion.

4 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 103

3-27

Case (a)

Rotational flow

Case (b)

Irrotational flow

r

n

Figure E3.6-4-2 Orientation of a fluid element during rotational and irrotational flows.

Assuming steady, inviscid, and incompressible flow with streamlines in the horizontal plane (dz/dn = 0) the equation of motion in the normal direction is given as

+ + g = 0 (E-1)pn

2VR

dzdn

We have = 0, = , R = r, Eq. (E-1) becomesdzdn

pn

pr

+ = 0pr

2Vr

Case (a)

= = = C12rp

r

2Vr

21C r

r

Integrating the equation we obtain

p = C12(r2 r0

2) + p012

Case (b)

= = pr

2Vr

22

3

Cr

p = C22 + p0

12 2 2

0

1 1r r

3-28

Example 3.6-5. ----------------------------------------------------------------------------------The flanged joint shown in Figure E3.6-5 bolts a nozzle onto a pipe. The flowing fluid is water. The cross-sectional area perpendicular to the flow at point 1 is 12 in2 and at point 2 is 4 in2. At point 2 the flow is open to the atmosphere1. The pressure at point 1 is 36 psig.

Pipe Nozzle

Water

1

2

Figure E3.6-5. Nozzle, bolted to pipe.

Neglecting friction loss, determine the velocity of water leaving the nozzle

Solution -----------------------------------------------------------------------------------------------------

Applying Bernoulli equation between points 1 and 2 we have

+ gz1 + = + gz2 + (E-1)2

1

2V 1p

22

2V 2p

Since p2 = patm = 0 psig, z1 = z2 = 0, Eq. (E-1) becomes

+ = (E-2)1p

21

2V 2

2

2V

From the steady mass balance for incompressible fluid we have

A1V1 = A2V2 V1 = V2 = 2

1

AA

2

3V

Substituting V12 = into Eq. (E-2) and solving for V2 we have

22

9V

= 0.5(V2)2 (1 1/9) = (4/9)(V2)21p

V2 = = 77.58 ft/s0.536 144 32.2 9

62.4 4

1 Noel de Nevers, Fluid Mechanics for Chemical Engineering, McGraw Hill, 3rd Edition, 2005.

4-1

Chapter 4 Fluid Kinematics

In fluid kinematics we study fluid motions without being concerned with the actual force necessary to produce the motion. We will consider the velocity and acceleration of the fluid, and the description and visualization of its motion.

4.1 The Velocity Field

We use continuum model and consider fluids consisting of fluid particles that interact with each other and with their surroundings. We will study the motion of fluid particles rather than individual molecules. This motion can be described in terms of the velocity and acceleration of the fluid particles. The representation of velocity or acceleration as functions of the spatial coordinates is called a field representation of the flow. The velocity field is given by the expression

V = u(x, y, z, t) + v(x, y, z, t) + w(x, y, z, t)i j k

In this equation u, v, and w are the x, y, and z components of the velocity vector. We can obtain the field description of the velocity vector V = V(x, y, z, t) by plotting the velocity for all of the particles.

Example 4.1-11 ----------------------------------------------------------------------------------A velocity field is given by V = Vo(x y ) where Vo is a constant. At what location in the i jflow field is the speed equal to Vo. Make a sketch of the velocity field in the first quadrant (x 0, y 0) by drawing arrows representing the fluid velocity at various locations.

Solution ------------------------------------------------------------------------------------------

This is a two-dimensional flow with u = Vox and v = Voy. The fluid speed is given by

V = (u2 + v2)1/2 = Vo(x2 + y2)1/2

The speed is V = Vo at any location on the circle of radius 1 centered at the origin as shown in Figure E4.1-1. The direction of the fluid velocity relative to the x-axis is given in terms of = arctan(v/u). For this flow

tan = = = vu

o

o

V yV x y

x

To make a sketch of the velocity field, we need to draw velocity vectors at various locations. For example, at x = 2, y = 1, the velocity vector is plotted in Figure E4.1-2 for Vo =1. Since it is tedious to sketch the velocity field, we can use Matlab to plot the velocity field. Table E4.1-1 list the Matlab program to produce Figure E4.1-1.

1 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 152

4-2

Figure E4.1-1 Velocity field of V = Vo(x y )i j

0

0

1

2

3

1 2 3 4

2-1

y

xFigure E4.1-2 Velocity vector at x = 2, y = 1.

*** Table E4.1-1 Matlab program to plot Velocity field of V = Vo(x y ) ***i j

clear % Clear variablesclf % and figuresVo=1;[x,y]=meshgrid(0:.5:2,0:.5:2); % Define a grid% The grid defines the locations where the velocity vectors are plotted.%Vx=Vo*x; % Evaluate the x-component of the velocityVy=-Vo*y; % Evaluate the y-component of the velocityquiver(x,y,Vx,Vy); % Plot vectors

4-3

% The quiver command will plot the vectors with the correct components at % the locations defined by the grid [x,y]%title('Velocity field of V=Vo(xi-yj)')xlabel('x'); ylabel('y'); grid on% plot the circle centered at the origin with radius r = 1hold on % Continue to plot on the same graphxp=0:0.02:1; yp=(1-xp.*xp).^0.5;plot(xp,yp)

--------------------------------------------------------------------------

4.2 The Acceleration Field

4.2a Partial and Total Derivatives

Let consider the unsteady one dimensional heat transfer when temperature T(x, t) depends on x and t then

dT = dt + dx = + tT

xT

dtdT

tT

xT

dtdx

The partial derivative denotes the change in temperature with respect to time at a fixed tT

location x. The total derivative denotes the change in temperature with respect to time at dtdT

a location moving with velocity c = . Therefore when = 0 or dtdx

dtdT

+ c = 0 (4.2-1)tT

xT

Equation (4.2-1) indicates that temperature T is a constant at the location moving in the x direction with velocity c. Consider the equation

+ = 0 (4.2-2)tu

xu

where u(x,t) is the unknown function. Let u(x,t) = (x – t)2, then

= 2(x – t) and = – 2(x – t)xu

tu

Therefore u(x,t) = (x – t)2 is a solution of the PDE (4.2-2). Other functions: , 3sin(x – 2)( txe

t), 2cos(x – t), … are also solution of (4.2-2). In general the solution of u(x,t) has the form f(x – t) or

4-4

u(x,t) = f(x – t)

To obtain a unique solution to the PDE (4.2-2), a boundary or initial condition is required.

Example 4.2-1. ---------------------------------------------------------------------------------

Find solution of + = 0 that satisfies the initial condition u(x,t) = .tu

xu

2xe

Solution --------------------------------------------------------------------------------------------

Since u(x,t) = f(x – t) is the solution of the PDE, this solution must satisfy the initial condition u(x,0) = . Therefore x must be replaced by x – t for the solution to have the form f(x – t)

2xe

that satisfies the initial condition.

u(x,t) = 2)( txe

For any fixed value of t, the graph of u(x,t) = , as a function of x, represents a snapshot 2)( txe

of a waveform as shown in Figure E4.2-1

-3 -2 -1 0 1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

u

1 2 3 4 5 6 70

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

u

Figure E4.2-1a. at t = 0 Figure E4.2-1.b. at t = 42)( txe 2)( txe

4.2b Material Derivative

Consider a fluid particle moving along its path-line as shown in Figure 4.2-12. In general, the particle velocity VA is a function of both location and time and is written as

VA = VA[xA(t), yA(t), zA(t), t]

In this expression, VA is the velocity of particle A. xA(t), yA(t), and zA(t) define the location of the particle. The total derivative of VA with respect to time is the acceleration of the particle

aA(t) = = + + + (4.2-1)AddtV A

tV A

xV dx

dtA

yV dy

dtA

zV dz

dt

2 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 160

4-5

Since the particle velocity components are given by uA = , vA = , and wA = , Eq. dxdt

dydt

dzdt

(4.2-1) becomes

aA = + uA + vA + wAA

tV A

xV A

yV A

zV

In general for any particle we have

a = + u + v + w (4.2-2)t

V

xV

yV

zV

Figure 4.2-1 Velocity and position of particle A at time t.

The acceleration a is a vector with components ax, ay, and az in the x, y, and z directions. In the component form we have

ax = + u + v + wut

ux

uy

uz

ay = + u + v + wvt

vx

vy

vz

az = + u + v + wwt

wx

wy

wz

Eq. (4.2-2) can be written in vector notation as

a = DDtV

4-6

The operator = + u + v + w is called the material derivative or DDt

t

x

y

z

substantial derivative. The vector notation of the material derivative is given by

= + (V)( ) (4.2-3) DDt

t

The dot product of the velocity vector, V, and the gradient operator ( ) is the spatial derivative terms appearing in the Cartesian coordinate representation of the material derivative.

(V)( ) = u + v + w x

y

z

The material derivative of any variable is the rate at which that variable changes with time as seen by one moving along with the fluid.

Example 4.2-23. ---------------------------------------------------------------------------------An incompressible, inviscid fluid flows steadily past a sphere of radius R, as shown in the following figure.

The fluid velocity along streamline A–B is given by

V = u(x) = V0i3

31 Rx

i

where V0 is the upstream velocity far ahead of the sphere. Determine the acceleration experienced by fluid particles as they flow along this streamline.

Solution --------------------------------------------------------------------------------------------Along the streamline A-B there is only x component of velocity, therefore

a = + u = + ut

V

xV u

t

ux

i

3 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 162

4-7

or ax = + uut

ux

Since u(x) = V0 is a function of x only, = 0. The acceleration becomes3

31 Rx

ut

ax = u V0 V0[R3( 3x-4)] = 3(V02/R)u

x

3

31 Rx

3

4

1 //R x

x R

Let ax* = ax/(V02/R) and x* = x/R the acceleration is in dimensionless form

ax* = 3

3

4

1 1/ **x

x

The following Matlab program plots ax* as a function of x*

% Example 4.2-2xs=-4:0.02:-1;axs=-3*(1+1.0./xs.^3)./xs.^4;plot(xs, axs)grid on; xlabel('x/R'); ylabel('a_x/(V_0^2/R)')

4-8

Example 4.2-34 ----------------------------------------------------------------------------------A velocity field is given by V = Vo(x y ) where Vo is a constant. Make a sketch of the i jstreamline in the first quadrant (x 0, y 0) by first obtain the equation for the streamline.

Solution ------------------------------------------------------------------------------------------

This is a two-dimensional flow with u = Vox and v = Voy. The fluid speed is given by

V = (u2 + v2)1/2 = Vo(x2 + y2)1/2

Since a streamline is everywhere tangent to the velocity field, the streamlines are given by solution of the equation

= = = dydx

vu

yx

dyy

dxx

Integrating the equation we obtain

ln y = ln x + C’ or yx = C

We can plot various streamlines in the x-y plane using the following Matlab program:

% Example 4.2-3x=0.2:0.1:4;y1=1.0./x; y2=4.0./x; y3=9.0./x;plot(x,y1,x,y2,x,y3)xlabel('x');ylabel('y')

4 Munson, Young, and Okiishi, Fundamentals of Fluid Mechanics, Wiley, 2006, pg. 156

5-1

Chapter 5 Conservation Laws: Control-Volume Approach

5.1 Macroscopic Mass Balance (Integral Relation)

The general balance equation can be written as

Accumulation = Input + Generation - Output - Consumption

The terms (Generation - Consumption) are usually combined to call Generation with positive value for net generation and negative value for net consumption.

LetM = total mass (kg) of A within the system at any time.

= rate (kg/s) at which A enters the system by crossing the boundaries.inm= rate (kg/s) at which A leaves the system by crossing the boundaries.outm= rate (kg/s) of generation of A within the system by chemical reactions.genr= rate (kg/s) of consumption of A within the system by chemical reactions.consr

Then the mass balance on species A can be written as

= + (5.1-1)dt

dMinm genr outm consr

If there is no chemical reaction, the mass balance equation is simplified to

= (5.1-2)dt

dMinm outm

Example 5.1-1. ----------------------------------------------------------------------------------A tank contains 2 m3 of pure water initially as shown in Figure 5.1-1. A stream of brine containing 25 kg/m3 of salt is fed into the tank at a rate of 0.02 m3/s. Liquid flows from the tank at a rate of 0.01 m3/s. If the tank is well mixed, what is the salt concentration (kg/m3) in the tank when the tank contains 4 m3 of brine.

Fi , Ai

F , A

A

V(t = 0) = 2 cubic meter

Fi = 0.02 m3

Ai = 25 kg/m3

/s

Figure 5.1-1 A tank system with input and output.

5-2

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

System: salt and water in the tank at any time.

Step #2: Find equation that contains A, the salt concentration in the tank at any time.The salt balance will contain A.

Step #3: Apply the salt balance around the system.

= FiAi - FA = 0.5 – 0.01A (E-1)dtVd A )(

where V is the brine solution in the tank at any time. Need another equation to solve for V and A.

= Fi – F = 0.02 – 0.01 = 0.01 m3/s (E-2)dtdV

Step #4: Specify the boundary conditions for the differential equation.

At t = 0, V = 2 m2, A = 0, at the final time t, V = 4 m2

Step #5: Solve the resulting equations and verify the solution.Integrate Eq. (E-2)

= 0.01 , to obtain V = 2 + 0.01tV

dV2

tdt

0

When V = 4 m3, t = 200 sec

The LHS of Eq. (E-1) can be expanded to

V + A = 0.5 – 0.01Adtd A

dtdV

Hence

(0.01t + 2) + 0.01A = 0.5 - 0.01Adtd A

The above equation can be solved by separation of variables

5-3

= A

Ad

02.05.0 201.0 t

dt

- ln (0.5 – 0.02A) = ln(0.01t + 2) + C102.01

01.01

- ln (0.5 – 0.02A) = ln(0.01t + 2) + C (E-3)21

at t = 0, A = 0, hence

- ln(0.5) = ln(2) + C (E-4)21

Eq. (E-3) - Eq. (E-4)

- = 21

5.02.5.0ln A

2201.0ln t

1 - 0.04A = (1 + 0.005 t)-2

Finally

A = 25 - (E-5)2)005.01(25

t

at t = 200 sec

A = 25 - = 18.75 kg/m32)200005.01(

25

Verify the solution

At t = 0, from (E-5); A = 0, as t , A = 25 kg/m3

---------------------------------------------------------------------------------------------

5-4

We now consider the general open system or control volume fixed in space and located in a fluid flow field, as shown in Figure 5.1-2. The streamline of a fluid stream is the curve where the velocity at any point is tangent to it. For a differential element of area dA on the control surface, the rate of mass efflux from this element = (v)( dAcos), where ( dAcos) is the area dA projected in a direction normal to the velocity vector v, is the angle between the velocity vector v and the outward-directed unit normal vector n to dA, and is the density.

v

n

Control volume

Control surface

Normal to surface dA

dAStreamlines offluid stream

Figure 5.1-2 Flow through a differential area dA on a control surface.

(v)(dAcos) is the scalar or dot product of (vn)dA. Since the normal vector n is pointing outward, the mass (efflux) leaving the control volume is positive ( < 90o) and the mass (influx) entering the control volume is negative ( > 90o). If we now integrate this quantity over the entire control surface A, we have the net outflow of mass across the control surface or the net mass efflux from the entire control volume .V~

=

volumecontrolfromeffluxmassnet

volumecontrolfromoutputmassofrate

volumecontrolfrominputmassofrate

= = (vn)dA

volumecontrolfromeffluxmassnet

A

dAv cos A

Since the rate of mass accumulation in control volume is = , the mass balance dt

dM

V

dVt

with no chemical reaction is

= (vn)dA (5.1-3)

V

dVt

A

Equation (5.1-3) has the same physical meaning as equation (5.1-2) = - . dt

dMinm outm

However equation (5.1-3) is more general since it can account for the variation of density over the control volume and the variation of velocity v over the control surface A.V~

5-5

Example 5.1-2. ----------------------------------------------------------------------------------Water is flowing through a large circular conduit with inside radius R and a velocity profile given by the equation

v(fps) = 8

2

1Rr

Determine the mass flow rate through the pipe and the average water velocity in the 2.0 ft pipe.

6 ft 2 ft

r

Solution ------------------------------------------------------------------------------------------

Since v is a function of r, we first need to determine the mass flow rate through the differential area dA = 2rdr

d = (v)(dAcos) = (v)( 2rdr) since cos = cos(0) = 1m

The mass flow rate through the area R2 is then obtained by integrating over the area

= 2 rdr (E-1)m

R

Rr

0

2

18

Let z = dr = Rdz, equation (E-1) becomesRr

= 16R2 zdz = 16R2m 1

0

21 z m1

0

42

42

zz

= 4R2 = 462.432 = 7057 lb/sm

The average velocity in the 2-ft pipe is

vave = = = 36 ft/s2Rm

214.627057

5-6

5.2 Microscopic Mass Balance (Differential Relation)

x

z

y

z

x

y

v |x x+ xv |x x

v |y y

v |y y+ y

Figure 5.2-1 Illustration of a differential element in Cartesian coordinates.

Appling the conservation of mass to a 3-D control volume xyz in Cartesian coordinates we obtain

= xyz = (5.2-1)dt

dMt

inm outm

The mass flow into and out of each surface are given by

- = yz[(vx)|x (vx)|x+x] + xz[(vy)|y (vy)|y+y]inm outm + xy[(vz)|z (vz)|z+z]

Therefore

xyz = yz[(vx)|x (vx)|x+x] + xz[(vy)|y (vy)|y+y]t

+ xy[(vz)|z (vz)|z+z]

Divide the equation by xyz and take the limit as xyz 0

In the limiting process of making x, y, z 0

5-7

x dxy dyz dz

=

000

lim

zyx

t

xvv xxxxx

|)(|)(

yvv yyyyy

|)(|)(

zvv zzzzz

|)(|)(

This limit process produces partial derivatives

= 0

limitx x

vv xxxxx

|)(|)(

xvx

)(

= 0limit

y yvv yyyyy

|)(|)(

yv y

)(

= 0limit

z zvv zzzzz

|)(|)(

zvz

)(

We then obtain the differential mass balance or the continuity equation. This equation must be satisfied at all points within any flowing fluid.

= (5.2-2)t

xvx

)(

yv y

)(

zvz

)(

If the fluid is incompressible, is a constant and equation (5.2-2) becomes

+ + = 0 (5.2-3)xvx

yv y

zvz

Equation (5.2-2) can be written in vector notation as

= vt

The vector differential or del operator is defined in RCCS (Rectangular Cartesian Coordinate System) as

= ex + ey + ez x

y

z

In this expression, ex, ey, and ez are the units vector in RCCs with the properties

5-8

ei ej = 0 for i j and ei ei = 1 for i = j where i, j = x, y, or z

Therefore

v = (ex + ey + ez )( exvx + eyvy + ezvz)x

y

z

v = + + xvx

)(

yv y

)(

zvz

)(

Since = + vx , v = v + vxvx

)(

xvx

x

v is the divergence of vector v and is given in RCCS as

v = + + xvx

yv y

zvz

is the gradient of the scalar and is given in RCCS as

= ex + ey + ezx

y

z

The gradient of is a vector in the direction in which increases most rapidly with distance. The term v is the overall or net rate of mass loss. v is the rate of outflow of volume (per unit volume). This situation could occur if the fluid were expanding due to a decrease in pressure. This would result in an outflow of volume across the boundaries of a fixed unit volume. Therefore v is the rate of mass loss due to expansion. v is the net loss of mass due to flow if there is an increase in density in the gradient direction. There will be more mass flow out of the system than the flow in.

Example 5.2-1. ----------------------------------------------------------------------------------Evaluate the divergence of the velocity vector v = vxex + vyey, where

vx = ceky sin(kx t), vy = ceky cos(kx t); c, k, and are constants

Solution ------------------------------------------------------------------------------------------

v = + = [ ceky sin(kx t)] + [ceky cos(kx t)]xvx

yv y

x

y

v = ckeky cos(kx t) + ckeky cos(kx t) = 0

The flow is incompressible since v = + = 0.xvx

yv y

5-9

Chapter 5

5.3 Macroscopic Energy Balance

The general energy balance equation has the form

= +

Energyof

onAccumulati

Energyof

Input

Energyof

Output

SystemtoaddedHeat

SystembydoneWork

Let Esys be the total energy (internal + kinetic + potential) of a system, be the mass flow inmrate of the system input stream, and be the mass flow rates of the system output stream, outmthen

= + (5.3-1)dtd syspsysksys EEU ,, inm

in

inin gzVu

2

2

outm

out

outout gzVu

2

2

Q W

whereUsys = system internal energyEk,sys = system kinetic energyEp,sys = system potential energy

, = internal energies per unit mass of the system inlet and outlet streamsinu outu, = average velocity of the system inlet and outlet streamsinV outV = rate of heat added to the systemQ

= rate of work done by the system W

The net rate of work done by the system can be written as

= + W sW fW

where = rate of shaft work = rate of work done by the system through a mechanical sW

device (e.g., a pump motor) = rate of flow work = rate of work done by the system fluid at the outlet minus fW

rate of work done on the system fluid at the inlet

Pin Pout

Rate of work = Force = Force velocitydistancetime

Rate of flow work done on the system fluid = PinAinVin = Pin inF

5-10

Rate of flow work done by the system fluid = PoutAoutVout = Pout outF

Eq. (5.3-1) becomes

= dtd syspsysksys EEU ,, inm

in

inin gzVu

2

2

outm

out

outout gzVu

2

2

+ + Pin Pout (5.3-2)Q sW inF outF

The internal energy can be combined with the flow work to give the enthalpy

+ Pin = = ininF inu inF ininF

in

inin

Pu ininF inh

In terms of enthalpies and inh outh

= + (5.3-3)dtd syspsysksys EEU ,, inm

in

inin gzVh

2

2

outm

out

outout gzVh

2

2

Q sW

The internal energy and the enthalpy can be related to the heat capacities where

Cp = , and Cv = pT

h

vTu

For constant values of Cp and Cv

h = Cp(T - Tref) and u = Cv(T - Tref)

For solid and liquid Cp Cv

If the system is at steady state with one inlet and one exit stream = = , equation m inm outm(5.3-3) is simplified to

hout hin + g(zout zin) + = (5.3-4) 22

21

inout VV mQ

mWs

Let = (“out”) (“in”), and q = , w = be the heat added to the system and work done mQ

mWs

by the system, respectively, per unit mass flow rate. Equation (5.3-4) becomes

h + gz + V2 = q w (5.3-5)21

5-11

This equation also applies to a system comprising the fluid between any two points along a streamline within a flow field. If these two points are only infinitesimal distance apart, the differential form of the energy equation is obtained

dh + gdz + VdV = q w (5.3-6)

The d() notation represents a total or “exact” differential and applies to the change in state properties that are determined only by the initial and final states of the properties. The () notation represents an “inexact” differential and applies to the change in properties that depend upon the path taken from the initial to the final point of the properties. The forms of energy can be classified as either mechanical energy, associated with motion or position, or thermal energy, associated with temperature. Mechanical energy is considered to be an energy form of higher quality than thermal energy since it can be converted directly into useful work. Mechanical energy includes potential energy, kinetic energy, flow work, and shaft work. Internal energy and heat are thermal energy forms that cannot be converted directly into useful work. For systems that involve significant temperature changes, the mechanical energy terms are usually negligible compared with the thermal terms. In such cases the energy balance equation reduces to a “heat or enthalpy balance”, i.e. dh = q.

Example 5.3-1. ----------------------------------------------------------------------------------In a residential water heater, water at 60oF flows at a constant 5 GPM into the 100 gallons tank and leaves at 3 GPM. Initially the tank has 10 gallons of 75oF water in it. The tank gas heater heats the tank contents at a constant rate of 800 Btu/min. Assume perfect mixing, determine the temperature of the discharge water after 20 min. of operation.

Water: Cp = Cv = 1 Btu/(lb.oF), density = 62.4 lb/ft3. Unit conversion 1 ft3 = 7.481 gal.

QFo, To F, T

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

Step #2: Find an equation that contains the temperature of the discharge water.

The energy balance for the system gives the desired equation.

5-12

Step #3: Apply the energy balance on the system with the reference temperature Tref = 0oF. Neglect the changes in kinetic and potential energies compared with the changes in thermal energies.

(VCpT) = FoCpTo - FCpT + dtd Q

(VT) = FoTo - FT + dtd

pCQ

= 5 - 3 = 2 => V = 10 + 2tdtdV

Step #4: Specify the initial condition for the differential equation.

At t = 0, T = 75oF

Step #5: Solve the resulting equation and verify the solution.

V + T = FoTo - FT + dtdT

dtdV

pCQ

(10 + 2t) + 2T = (5)(60) - 3T + dtdT

)1)(4.62()481.7)(800(

(2t + 10) = 395.91 - 5TdtdT

= - ln = ln

T

TdT

75 591.395

t

tdt

0 102 51

75591.395591.395 T

21

10102t

395.91 - 5T = 20.91 => T = 79.182 - 4.182 5.2

10102

t 5.2

10102

t

at t = 20 min., T = 79.1oF

-----------------------------------------------------------------------------------------------------

Equation (5.3-5) and its differential form, equation (5.3-6), are not convenient for solving engineering problems.

h + gz + V2 = q w (5.3-5)21

dh + gdz + VdV = q w (5.3-6)

5-13

We can use thermodynamics relations to convert the enthalpy term into a form that involves temperature, pressure, and density changes across the system.

du = Tds Pd(1/)

dh = du + d(P/) = Tds Pd(1/) + d(P/)

dh = Tds Pd(1/) + Pd(1/) + = Tds + (5.3-7)

dP

dP

For an idealized reversible process in which no energy dissipation occurs, the entropy change arises from heat transfer across the system boundaries

Tds = q

In any real system, the process is irreversible and there is dissipation of energy, therefore

Tds = q + ef = du + Pd(1/) (5.3-8)

dh = q + ef +

dP

In this equation ef represents the thermal energy generated due to the irreversibility of the

system. Substituting dh = q + ef + into the differential energy balance, Eq. (5.3-6),

dP

gives

+ gdz + VdV + ef = w

dP

This equation can be integrated along a streamline from the inlet to the outlet of the system to give

+ g(zo zi) + (oVo2 iVi

2) + ef + w = 0 (5.3-9)o

i

P

P

dP 2

1

where ef = , w = , and = kinetic energy correction factor, = 2 for laminar fe w

flow, = 1 for turbulent flow.The kinetic correction factor is due to the fact that the velocity profile is not uniform over the cross-sectional area of flow. For uniform flow, the rate of kinetic energy entering a C.V. is given as

= VA kE

2

21 V

The kinetic energy per unit mass flow rate is then

5-14

= V2VAEk

21

For turbulent flow, the velocity profile is almost flat, therefore

V2 = 1 for turbulent flowturbulent

k

VAE

21

Laminarvelocity profile

Uniformvelocity profile

V

vz

Figure 5.3-1 Laminar velocity profile in a pipe.

The velocity for laminar flow in a pipe is given as

vz = 2V = 2V (1 2), where =

2

1Rr

Rr

The rate of kinetic energy entering a C.V. is

= 2rdrkE

R

zv0

2

21

Therefore = rdr = R2 dkE R

zv0

3 R

zv0

3

Rr

Rr

= R2 V3(1 2)3d = 8R2V3 dkE 1

08

1

0

32 )1(

= 8R2V3 = R2V3 = A V3kE

81

Therefore = V2, and = 2 for laminar flowarla

k

VAE

min

22

5-15

Chapter 5

5.4a Macroscopic Momentum Balance. The equation for the conservation of momentum with respect to a control volume (CV) can be written as follows:

rate of accumulationof momentum in CV

= rate of momentuminto CV

- rate of momentumout of CV

+ sum of forcesacting on CV

(m )cv = + (5.4-1)dtd V

in

iVm )(

out

oVm )(

CVon

F

The total force acting on the control volume consists both of surface forces and body forces. The surface force is due to interaction between the control volume and its surrounding through direct contact at the boundary. The body force is due to the location of the control volume in a force field. The gravitational field and its resultant force is the most common example of the body force that will act on the entire control volume and not just at the control surface.

Example 5.4-1. ----------------------------------------------------------------------------------

h

Qo

Q

Figure 5.4-1 Gravity-flow tank

The above figure shows a tank into which an in compressible liquid is pumped at a volumetric flow rate Qo (ft3/s). The height of liquid in the vertical tank is h (ft). The flow rate out of the tank is Q (ft3/s).The length of the exit line is L (ft) and its inside diameter is Dp (ft). The cylindrical tank has an inside diameter Dt (ft). The liquid level h is determined as a function of time given an initial height ho and initial velocity Vo of the liquid in the pipe.

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

We assume plug-flow conditions and incompressible liquid for the liquid flowing through the pipe. Therefore all the liquid is moving at the same velocity, more or less like a

5-16

solid rod. The velocity V (ft/s) is equal to the volumetric flow Q divided by the cross-sectional area Ap of the pipe.

0 LF0 FL

Figure 5.4-2 Exit line of the gravity-flow tank

Let control volume (CV) be the fluid inside the exit line then the mass m of fluid inside the pipe is ApL.

Step #2: Find equation that contains the liquid level h.

The liquid level can be obtained from the mass balance however the volumetric flow rate must be obtained from the momentum balance.

Step #3: Apply the momentum balance on system.

= ( ApLV)

CVinmomentumofonaccumulatiofrate

dtd

= QV

CVtoinput

momentumofrate

CVofout

momentumofrate

= Fo FL Ff , where

CVonacting

forcesofsum

Fo = Aphg (hydraulic force) + ApPatm (static force due to surrounding pressure)

FL = ApPatm (static force due to surrounding pressure)

Ff = KfLV2 is the frictional force due to the viscosity of the liquid pushing in the opposite direction from right to left and opposing the flow.

The momentum equation for the fluid inside the exit line becomes

( ApLV) = Fo FL Ff = Aphg KfLV2dtd

or

= h V2 (E-1)dtdV

Lg

p

f

AK

To describe the system completely a total continuity equation (mass balance) on the liquid in the tank is also needed.

5-17

At = Qo Q = Qo ApVdtdh

or

= V (E-2)dtdh

t

o

AQ

t

p

AA

Step #4: Specify the boundary conditions for the differential equations.

We need to solve two coupled ordinary differential equations (E-1) and (E-2). Physical dimensions, parameter values, and initial flow rate and liquid height are given in Table 5.4-1

Table 5.4-1 Gravity-flow tank data_______________________________________________________________________Pipe: ID = 3 ft Area = 7.06 ft2 Length = 3000 ftTank: ID = 12 ft Area = 113 ft2 Height = 7 ftInitial values h = 3.0 ft V = 2.0 ft/sParameters density = 62.4 lb/ft3 Kf = 0.0281 lbfs2/ft3

_______________________________________________________________________

Step #5: Solve the resulting equations and verify the solution.

Substituting the numerical values of parameters into Eqs. (E-1) and (E-2) give

= 0.0107h 0.00205V2 (E-3)dtdV

= 0.06248V (E-4)dtdh

113oQ

The above equations can be solved by Matlab with initial values of V and h at a given Qo. Table 5.4-2 lists the Matlab programs and results with Qo = 35.1 ft3/s. Figure 5.4-3 shows the results in graphical forms.

Table 5.4-2 Gravity-flow tank results

% Example 5.4-1, gravity flow tank% Main programxspan=0:10:500;[x,y]=ode45('ex541',xspan,[2 3]);plot(x,y(:,1),'-',x,y(:,2),':')grid on

5-18

xlabel('t(s)')ylabel('V,h')legend('V(ft/s)','h(ft)')

% Example 5.4-1, gravity flow tank function yy = ex541(x,y)% y(1)=V, y(2)=hyy(1,1)=0.0107*y(2) - 0.00205*y(1)^2;yy(2,1)=35.1/113- 0.06248*y(1);

t (sec) V (ft/s) h (ft) 0 2.0000 3.0000

10.0000 2.3225 4.7643 20.0000 2.7830 6.2814 30.0000 3.3314 7.4812 40.0000 3.9112 8.3254 50.0000 4.4704 8.8100 60.0000 4.9660 8.9622 70.0000 5.3712 8.8320 80.0000 5.6730 8.4842 90.0000 5.8703 7.9809 100.0000 5.9718 7.3821 110.0000 5.9925 6.7461 120.0000 5.9491 6.1199 130.0000 5.8571 5.5362 140.0000 5.7314 5.0198 150.0000 5.5878 4.5889 160.0000 5.4381 4.2515 170.0000 5.2896 4.0072 180.0000 5.1502 3.8529 190.0000 5.0272 3.7803 200.0000 4.9246 3.7785 210.0000 4.8437 3.8344 220.0000 4.7843 3.9347 230.0000 4.7467 4.0652 240.0000 4.7295 4.2115 250.0000 4.7300 4.3626 260.0000 4.7450 4.5096 270.0000 4.7711 4.6440 280.0000 4.8045 4.7591 290.0000 4.8418 4.8514 300.0000 4.8804 4.9201 310.0000 4.9177 4.9653 320.0000 4.9514 4.9883 330.0000 4.9799 4.9917 340.0000 5.0023 4.9790 350.0000 5.0190 4.9540 360.0000 5.0297 4.9203 370.0000 5.0349 4.8817

5-19

380.0000 5.0354 4.8420 390.0000 5.0319 4.8034 400.0000 5.0254 4.7677 410.0000 5.0168 4.7364 420.0000 5.0072 4.7109 430.0000 4.9973 4.6918 440.0000 4.9877 4.6789 450.0000 4.9789 4.6717 460.0000 4.9712 4.6697 470.0000 4.9649 4.6720 480.0000 4.9602 4.6776 490.0000 4.9572 4.6856 500.0000 4.9556 4.6951

*********************************************************************

0 50 100 150 200 250 300 350 400 450 5002

3

4

5

6

7

8

9

t(s)

V,h

V(ft/s)h(ft)

Figure 5.4-3 Velocity and liquid height in an unsteady-state gravity flow tank.

5-20

The following example presents a system where energy and momentum balance must be applied to obtain the results.

Example 5.4-21 ----------------------------------------------------------------------------------

Advertised is a small toy, aerocket, that will send up a signal flare and the operation “is so simple that it is amazing” (Fig. 5.4-4).

piston

trigger

Figure 5.4-4 Aerocket

Our examination of this device indicates that it is a sheet metal tube 7 ft long and 1 in.2 area. A plug shaped into the form of a piston fits into the tube and a mechanical trigger holds it in place 2 ft above the bottom. The mass of the piston is 3.46 lb. To operate the device, the volume below the piston is pumped up to a pressure of about 4 atm absolute with a small hand pump, and then the trigger is depressed allowing the piston to fly out the top. The pyrotechnic and parachute devices contained within the piston are actuated by the acceleration force during ejection.

When we operated this toy last summer, the ambient temperature is 90oF. Assuming no friction in the piston and no heat transfer or other irreversibility in the operation, how high would you expect the piston to go? What would be the time required from the start to attain this height?

Solution ------------------------------------------------------------------------------------------

Step #1: Define the system.

System: The piston.

Step #2: Find equation that contains h, the height the piston will attain.

The momentum balance will provide the velocity of the piston as it leaves the metal tube. An energy balance can then be applied to determine h.

Step #3: Apply the momentum balance on the system.

1 Tester, J. W. and Modell M., Thermodynamics and Its Applications, Prentice Hall, 1997, p.58

5-21

piston

trigger

zzi

v

+

compressed air

Let the positive direction be pointing upward, the momentum balance on the piston becomes

m = (P Patm) Ap mgdt

dVvel

wherem = mass of the pistonVvel = velocity of the pistonP = pressure of the compressed air within the metal tubePatm = surrounding atmospheric pressureAp = cross-sectional area of the pistong = acceleration of gravity

We need to obtain the velocity of the piston as a function of the vertical distance z. First, the relation between the pressure P and the volume V of the moles N of compressed air within the metal tube must be determined. For the adiabatic expansion of air

dU = dW => NCvdT = PdV

From ideal gas law: PV = NRT => d(PV) = NR dT

N dT = (PdV + VdP)R1

PdV + VdP = PdV => CvVdP = (R + Cv)P dV = CpPdVR

CV

RCV

= P

dPV

p

CC

VdV

5-22

Integrating the equation from the initial conditions (Pi, Vi) to (P, V) gives

ln (P/Pi) = ln(V/Vi) , where = V

p

CC

Therefore

P = Pi = Pi = Pi

iVV

pi

p

AzzA

izz

From the momentum equation

m = (P Patm) Ap mgdt

dVvel

m = Pi Ap Patm Ap mgdt

dVvel

izz

= z- dt

dVvel

mzAP ipi

g

mAP patm

Let a = , and b = m

zAP ipi

gm

AP patm

The momentum equation becomes

= az- b => = az- b => Vvel = az- bdt

dVvel

dzdVvel

dtdz

dzdVvel

Step #4: Specify the boundary conditions for the differential equation.

At z = zi = 2 ft, Vvel = 0At z = 7 ft, Vvel = ?

Step #5: Solve the resulting equation and verify the solution.

How high would you expect the piston to go?

Integrate the momentum equation to obtain

= z1- bz + C121 2

velV1

a

5-23

At z = zi = 2 ft, Vvel = 0 => C1 = bzi zi1-

1a

The numerical values of a, b, and C1 can be evaluated

a = = = 1.4441103 ft2.4/s2m

zAP ipi

46.3)2.32)(2)(1)(7.14)(4( 4.1

Note the conversion factor from lbf to lb

b = = + 32.2 = 169 ft/s2gm

AP patm 46.3

)2.32)(1)(7.14)(1(

C1 = bzi zi1- = (169)(2) + 2-0.4 = 3,074 ft2/s2

1a

4.0104441.1 3

Therefore

Vvel = 20.5(3,074 3,610z-0.4 169z)0.5

At z = 7 ft, Vvel = 21.6 ft/s

The height h attained by the rocket when it leaves the cylinder is evaluated from the conservation of energy. At the maximum height, Vvel = 0, all the kinetic energy becomes the potential energy.

Vvel2 = gh => h = = = 7.2447 ft21

21

gVvel

2

21

2.326.21 2

The total height is 14.24 ft from the ground.

What would be the time required from the start to attain this height?

The total time consists of the time the piston spent within the cylinder and the time spent outside the cylinder.

Time spent within the cylinder can be obtained by integrating

Vvel = = 20.5(3,074 3,610z-0.4 169z)0.5 dtdz

from z = 2 ft to z = 7 ft. This equation is rearranged and integrated numerically using Simpson’ formula with 5 points.

5-24

t = =

7

2 5.04.05.0 )169610,3074,3(2 zzdz

7

2)( dzzf

z(ft) f(z)2.003.254.505.757.00

f1 = 0.047909f2 = 0.026712f3 = 0.022637f4 = 0.020751f5 = 0.019641

t = (f1 + 4f2 + 2f3 + 4f4 + f5) = 0.126 sec12

27

Time spent outside the cylinder is obtained by applying Newton’s second law to the piston with the positive direction upward.

m = mg => = g => Vvel = = gt + Vvel,0dtdVvel

dtdVvel

dtdz

at t = 0 , Vvel = Vvel,0 = 21.6 ft/s

Integrate the equation again

z = 0.5gt2 + Vvel,0t + zo

at t = 0, z = zo = 0

z = 0.5gt2 + Vvel,0t = 16.1t2 + 21.6t

at z = 7.2447 ft = 16.1t2 + 21.6t

Solve this equation for t => t = 0.67 sec

Time to attain a height of 14.24 ft is: 0.126 + 0.67 = 0.796 sec

*********************************************************************

5-25

Chapter 5

5.4b One-Dimensional Flow in a Tube.

m

dx

x

V x

V +dVx

x

P

P+dP

A w

g

z

dx dz

Figure 5.4-5 Momentum balance on the CV of Adx.

We will consider the steady state plug flow in a tube with cross-sectional area A as illustrated in Figure 5.4-5. Applying the x-momentum balance on the control volume Adx gives

(m )cv = + = 0dtd

xV xVm )( xx dVVm CVon

xF

= 0 (5.4-2)CVon

xF xdVm

The forces acting on the control volume (fluid) result from pressure (dFP), gravity (dFg), wall drag (dFw), and possible external “shaft” work (Fext).

= dFP + dFg + dFw + Fext (5.4-3)CVon

xF

Each of the force in the above expression is given by

dFP = A[P (P + dP)] = AdP

dFg = Adxgcos = Adzg (since dxcos = dz)

dFw = wdAw = wWpdx (Wp = perimeter of the wall)

Fext = dxW

5-26

In these expressions, w is the stress exerted by the fluid on the wall and W is the shaft work performed by the fluid. Substituting the expressions for the forces from Eq. (5.4-3) into the momentum balance equation (5.4-3) gives

AdP Adzg wWpdx = 0dxW

xdVm

Dividing the equation by ( A), we obtain

+ gdz + dx + + dV = 0

dPA

Wpw

AdxW

AVA

In this equation V = Vx. Let w = = work done per unit mass of fluid, the equation AdxW

becomes

+ gdz + dx + w + VdV = 0

dPA

Wpw

Integrating this expression from the inlet to the outlet gives

+ g(zo zi) + dx + w + (oVo2 iVi

2) = 0 (5.4-4)o

i

P

P

dP

L pw

AW

0

21

where w = , and = kinetic energy correction factor, = 2 for laminar flow, = 1 for w

turbulent flow. Comparing equation (5.4-4) with the energy equation (5.3-9)

+ g(zo zi) + (oVo2 iVi

2) + ef + w = 0 (5.3-9)o

i

P

P

dP 2

1

shows that they are identical, provided

ef = dxL pw

AW

0

For steady flow in a uniform conduit

ef = = , where Dh = 4A

LWpw

w

hDL4

pWA

In this expression Dh is called the hydraulic diameter for a conduit of any cross-sectional shape. For a circular tube Dh is the same as the tube diameter D

5-27

Dh = 4 = 4 = DpW

AD

D

4

2

For this special case of one-dimensional fully developed flow in a straight uniform pipe, the energy and momentum equations provide the same results for the relationship between temperature, pressure, density, velocity, and dissipated energy (ef) across the system. From the energy balance, ef represents the lost energy associated with irreversible effects. From the momentum balance, ef represents the work required or friction loss to overcome the frictional force at the wall. In general, the momentum balance gives additional information relative to the forces exerted on and/or by the fluid in the system through the boundaries. This information cannot be obtained by the energy equation.

5.4c Definition of the Loss Coefficient and the Friction Factor

The energy equation can be made dimensionless by dividing it by any term within the equation.

+ g(zo zi) + (oVo2 iVi

2) + ef + w = 0 (5.3-9)o

i

P

P

dP 2

1

If the dividing term is the kinetic energy per unit mass of fluid, we obtain the dimensionless loss coefficient, Kf

Kf = 2/2V

e f

A loss coefficient can be defined for any element that causes resistance to flow such as a length of pipe, a valve, a pipe fitting, a contraction, or an expansion. The total friction loss is then calculated from the sum of the losses in each element.

ef = i

ifi VK 2

21

The pipe wall stress (w) is the flux of momentum from the fluid to the tube wall. Therefore it can be made dimensionless by dividing by the flux of momentum V2 carried by the fluid along the conduct. However a factor of ½ is chosen for the flux of momentum so that it represents the kinetic energy per unit volume (½V2). The Fanning friction factor f is defined as

f = 2/2V

w

Mechanical and civil engineers usually use the definition of the Darcy friction factor fD given by

5-28

fD = 4 = 4f2/2V

w

The advantage of this definition is that the lost energy will have a form that shows a direct dependence on the kinetic energy per unit mass. Since

w = fDV281

The lost energy in terms of the Darcy friction factor fD is given by

ef = = = fDV2A

LWpw

w

hDL4

21

hDL

Thus, it is important to know which definition is implied when data for friction factors are used. The Fanning friction factor can be related to the loss coefficient

Kf = = = = 2/2V

e f2

2V

w

hDL4

22

V 2

2Vf

hDL4

hDfL4

Example 5.4-32 ----------------------------------------------------------------------------------

Consider the flow in a sudden expansion from a small conduit to a larger one. The conditions upstream of the expansion (point 1) are known, as well as the areas A1 and A2. Find the velocity and pressure downstream of the expansion (V2 and P2) and the loss coefficient, Kf.

AV

1

1P1

AVP

2

2

2

P1a

x

Figure 5.4-6 Flow through a sudden expansion.

Solution ------------------------------------------------------------------------------------------Applying the mass balance to the fluid in the shaded area give

V2 = V12

1

AA

The downstream pressure (P2) can be obtained from the energy equation

2 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 124

5-29

+ (V22 V1

2) + ef = 0

12 PP 21

Since the above equation contains two unknowns, P2 and ef. We need another equation, the steady-state momentum balance

(V2 V1) = 0CVon

xF m

The forces acting on the control volume are substituted into the equation to obtain

P1A1 + P1a(A2 A1) P2A2 + Fwall = A1V1(V2 V1)

In this equation P1a is the pressure force from the solid surface on the left-hand boundary of the system, and Fwall is the drag force of the wall on the fluid at the horizontal boundary of the system. Since the fluid pressure cannot change suddenly, P1a P1. Neglect Fwall because the horizontal boundary of the system is relatively small, we have

(P1 P2)A2 = A1V12

1

2

1

AA

From the energy equation

ef = + (V12 V2

2) = V12 + V1

2

21 PP 21

2

1

AA

1

2

1

AA

21

2

2

11AA

ef = V12 = V1

221

2

2

1

2

1

2

2

1 122AA

AA

AA

21

2

2

11

AA

The loss coefficient for the flow through a sudden expansion is defined as Kf = , 2

2

11

AA

therefore

ef = KfV12

21

The experimental value of the loss coefficient for a sharp 90o contraction is Kf = 0.5(1 2)

where = . For most systems, the loss coefficient cannot be determined accurately from 2

1

DD

applying the macroscopic momentum and energy balances. They must be determined from experimental data.

5-30

Example 5.4-4 ----------------------------------------------------------------------------------

Figure 5.4-7 A 90o horizontal reducing elbow.

Water is flowing through a 90o horizontal reducing elbow as shown. Determine the retaining forces Fx and Fy required to keep the elbow in place. Neglecting frictional loss, D1 = 0.20 m, D2 = 0.15 m, P1 = 1.5 bar (gage), V1 = 6.0 m/s. 1 bar = 105 Pa. Water density is 1000 kg/m3.

Solution ------------------------------------------------------------------------------------------

From the mass balance

V2 = V1 = 6.0 = 10.67 m/s2

1

AA 2

15.20.

The downstream pressure (P2) can be obtained from the energy equation with ef = 0

+ (V22 V1

2) + ef = 0

12 PP 21

P2 = P1 + (V12 V2

2) = 1.5105 + 500(62 10.672) = 1.11105 Pa21

Applying the x-momentum balance on the reducing elbow

Fx + P1A1 + A1V1(V1x V2x) = 0

Fx = A1(P1 + V12) = (.102)( 1.5105 +100062) = 5843 N

The force Fx is actually in the negative x-direction. A similar calculation for the y-momentum balance gives Fy = 3974 N. [Fy P2A2 + A1V1(V1y V2y) = 0]

5-31

Chapter 5

Example 5.4-5 ----------------------------------------------------------------------------------

A pasta-producing machine extrudes raw pasta dough through an annular semicircular die of length L as shown above. Calculate the pressure, in units of psig, required to produce 10 kg/h of pasta under the following conditions: = 10 Pas, = 1200 kg/m3, L = 5 cm, R1 = 0.5 cm, R2 = 1 cm. The friction factor for the flow an annular semicircular area can be determined from Figure 5.4-83.

Figure 5.4-8 fRe curves for Newtonian fully developed laminar flowSolution ------------------------------------------------------------------------------------------

Making a force balance around the channel for shaping pasta yields

LWpw = AP

In this equation, L is the length, Wp is the perimeter, A is the cross-sectional area, and P is the required pressure drop to push the pasta out of the channel at velocity U. Solving for the shear stress at the wall gives

w = LP

pWA

3 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998, p. 234

5-32

From the definition of the friction factor and the hydraulic diameter we have

f = , and Dh = 42/2U

w

pWA

Therefore

= fU2LP

4hD

21

The friction factor f can be determined from Figure 5.4-8 for annular sector with = = 2

1

RR

0.5. We obtain fRe = 18.8.

= = 4

hD

pWA

)(2)()(5.0

2121

21

22

RRRRRR

= = 0.2062 cm4

hD)5.01(2)5.01(

)5.01(5.0 22

The Reynold number is defined as Re = . The properties are given as

UDh

= 10 Pas = 100 Poise; = 1200 kg/m3 = 1.2 g/cm3

U = = = 1.96 cm/sA

m

)5.01)(2/)(2.1(1000)3600/10(

22

Hence Re = = 0.0195 f = = 966.5100

)96.1)(2062.04)(2.1( 0195.0

8.18

The pressure drop required is now evaluated

P = fU2 = (0.5)(966.5)(1.2)(1.962)21

4/hDL

2062.05

P = 5.43104 dyne/cm2 = 5.43104 = 0.80 psig6107.14

5-33

Example 5.4-6 ----------------------------------------------------------------------------------

Find the horizontal force of the water on the horizontal bend shown. The inside diameter of the pipe is 8.0 cm and the inside diameter of the nozzle is 4.0 cm. The inlet velocity V1 is 10 m/s. You can assume that the friction loss of the bend is negligible.

V1

V2

+F

Figure 5.4-8 Force on a horizontal bend.

Solution ------------------------------------------------------------------------------------------

From the mass balance A1V1 = A2V2. Therefore, the nozzle exit velocity is

V2 = V1 = 10 = 40 m/s2

2

1

DD 2

48

Momentum balance in the x-direction gives

F + A1V1V1x A2V2V2x + P1A1 = 0

F = A1V1[V1 ( V2)] + P1A1

Since the friction loss is negligible, we can use Bernoulli’s equation to determine the inlet pressure P1.

+ = P1 = (V22 V1

2)2

21V

1P

2

22V

2

P1 = (500)(402 102) = 750,000 Pa

The force of the bend on the water is then

F = A1V1(V1 + V2) + P1A1 = A1[P1 + A1V1(V1 + V2)]

F = (0.042)[750,000 + 100010(10 + 40)] = 6283 N

The force of the water on the bend is F = 6283 N

5-34

Example 5.4-7 ----------------------------------------------------------------------------------

Water flows through a horizontal bend and discharges into the atmosphere as shown. The pressure gage reads 10 psi, and the flow rate Q is 5.0 ft3/s. Determine the retaining force FAx and FAy.

Determine the volumetric flow rate Q if the retaining force FAx is 1440 lbf.

Solution ---------------------------------------------------

The inlet velocity is V1 = = = 25 ft/s and the outlet velocity is V2 = 50 ft/s.1A

Q2.00.5

Momentum balance in the x-direction gives

FAx + A1V1V1x A2V2V2x + P1A1 = 0

Since V2x = V2cos(/4)

FAx + Q[V1 + V2cos(/4)] + P1A1 = 0

FAx = 5[25 + 50 cos(/4)] + 101440.2 = 873 lbf2.324.62

Momentum balance in the y-direction gives

FAy A2V2V2y = 0

FAy = QV2y = Q[ V2cos(/4)]

FAy = Q[V2cos(/4)] = 5[50 sin(/4)] = 343 lbf2.324.62

Momentum balance in the x-direction gives

FAx + A1V1V1x A1V1V2x + P1A1 = 0

A1V12 A1V1[ 2V1cos(/4)] = 1440 101440.2

A1V12[1 + 2cos(/4)] = 8144

V1 = = 35 ft/s2/1

)]4/cos(1[2.04.622.321448

The volumetric flow rate is then Q = 0.235 = 7.0 ft3/s

5-35

Chapter 5

5.5 Conservation of Angular Momentum

R

m

V

OR

Or

dr

Figure 5.5-1 A point mass m and a distributed mass M are rotating at a uniform angular velocity .

The linear momentum of a mass m moving in the x direction with a velocity Vx is mVx. The angular momentum (L) of a point mass m rotating with an angular velocity rad/s in an arc having a radius of curvature R is mVR = mR2. The angular momentum has dimension of “length times momentum,” and is therefore called the “moment of momentum.” If the mass is not a point but a rigid distributed mass (M) rotating at a uniform angular velocity, the angular momentum of the differential mass dM is given by

dL = r2dM

The total angular momentum of the mass M is obtained by integration over the entire mass

L = = = I (5.5-1)M

dMr 2 M

dMr2

where the moment of inertia I is defined as I = M

dMr2

As an example, we want to determine the moment of inertia for the flywheel of width W, radius R, and mass M, whose cross section is shown in Figure 5.5-1.

dM = 2rdrW

I = = 2rdrW = 2W = WR2 = MM

dMr2 M

r2 R

drr0

32

2R2

2R

For a fixed mass, the conservation of linear momentum is equivalent to Newton’s second law:

5-36

= m = m = F

adtVd

dtVmd )(

Let be the torque acting on the system, the conservation of angular momentum is written as

= = = I RF dtId )(

dtd

For a flow system, angular momentum may enter or leave the system by convection. As an example, consider the impeller of a centrifugal pump, whose cross section is shown in Figure 5.5-2.

V2

r2r1 V1

Figure 5.5-2 Cross section of pump impeller.

The impeller rotates with angular velocity , and its rotation causes the fluid to be thrown radially outwards between the vanes by centrifugal action. The fluid enters at a radial position r1 and leaves at a radial position r2; the corresponding tangential velocity V1 and V2 denote the inlet and exit liquid velocities relative to a stationary observer. Making an angular momentum balance on the impeller yields

= ( r1V1)in ( r2V2)out + dtId )( m m

For steady-state system, there is no accumulation of angular momentum, and the torque required to rotate the impeller is

= (r22 r1

2)m

The corresponding power required to drive the pump is the product of the angular velocity and the torque

Power =

5-37

Example 5.5-14 ----------------------------------------------------------------------------------Consider an ideal garden sprinkler shown in Figure 5.5-3. The central bearing is well lubricated, so the arms are free to rotate about the central pivot. Each of the two nozzles has a cross-sectional area of 5 mm2, and each arm is 20 cm long.

20 cm 20 cm

u

u

Figure 5.5-3 An ideal garden sprinkler.

If the water supply rate to the sprinkler is 0.0001 m3/s, determine:(a) The velocity u(m/s) of the water jets relative to the nozzles.(b) The angular velocity of rotation, , of the arms, in both rad/s and rps.(c) The applied torque required preventing the arms from rotating.

Solution ------------------------------------------------------------------------------------------

(a) The velocity u(m/s) of the water jets relative to the nozzles

u = = = 10 m/sA

Q2 26

34

1052/10

msm

(b) The angular velocity of rotation, , of the arms, in both rad/s and rps.

Making an angular momentum balance on the steady state sprinkler yields

0 = ( r1V1)in ( r2V2)out + m m

= [(r2V2)out (r1V1)in]m

Water enters at a radial position r1 and leaves at a radial position r2; the corresponding tangential velocity V1 and V2 denote the inlet and exit water velocities relative to a stationary observer. We assume that (r2V2)out >> (r1V1)in, therefore

(r2V2)outm

Since the arm is free to rotate, there is no applied torque or = 0 and V2 = 0. The water discharge velocity is zero as seen by a stationary observer. Let unoz be the velocity of the nozzle, then

V2 = u unoz = 0 unoz = u = 10 m/s

The angular velocity is then

4 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 103

5-38

= = = 50 rad/s2r

unoz

2.010

= = 7.96 rps2

50

(c) The applied torque required preventing the arms from rotating.

If the arm is not rotating then unoz = 0

V2 = u unoz = 10 0 = 10 m/s

(r2V2)out = (0.0001 m3/s)(1000 kg/m3)(0.2 m)(10 m/s) = 0.2 Nmm

5.6 Momentum Balance on Moving Systems

We will consider a system that is moving with constant velocity while streams carrying momentum and energy may flow into and out of the system. The absolute stream velocity in the x direction Vx is related to the system velocity Vsx and the stream velocity relative to the system Vrx by the relation

Vx = Vsx + Vrx

The analysis for a system moving with a constant velocity will be simplified if a momentum balance is applied to a control volume moving with the system velocity.

Example 5.6-1 ----------------------------------------------------------------------------------Figure 5.6-1 shows a plan of a jet of water impinging against a cone that is held stationary by a force F opposing the jet, which divides into several radially outwards streams, each leaving at an angle of 30 degree with respect to the horizontal. The velocity of the water jet is 18 m/s and its diameter is 8.0 cm.

Water F60o

V1

V2

V2

Figure 5.6-1 Jet impinging against a cone.

Determine the force need to:(a) Hold the cone stationary.(b) Move the cone away from the jet at 5 m/s.

Solution ------------------------------------------------------------------------------------------

5-39

(a) Force required holding the cone stationary

Applying the x-momentum balance on the control volume shown with the dash line

F + A1V1V1x A2V2V2x = 0

We will assume that the area of flow at the inlet is the same as that at the outlet: A1 = A2, therefore V1 = V2. For steady state system

= A1V1 = A2V2m

Therefore

F = A1V1(V1x V2x) = A1V1(V1 V1cos 30o) = A1V12(1 cos 30o)

F = (1000)(0.042)(182)(1 cos /6) = 218.2 N

(b) Force required moving the cone away from the jet at 5 m/s.

We now choose the control volume that moves with the cone so that the velocity of the water jet relative to the control volume is

V1r = V1 Vs = 18 5 = 13 m/s

The x-momentum balance is now written with the relative velocities

F + A1V1rV1xr A2V2rV2xr = 0

Since A1 = A2, therefore V1r = V2r and = A1V1r = A2V2rm

The force required for this case is then

F = A1V1r(V1xr V2xr) = A1V1r(V1r V1rcos 30o) = A1V1r2(1 cos /6)

F = (1000)(0.042)(132)(1 cos /6) = 113.8 N

Example 5.6-24 ----------------------------------------------------------------------------------

As shown in Figure 5.6-2, a boat of mass M = 1,000 lbm is propelled on a lake by a pump that

takes in water and ejects it, at a constant velocity of V2r = 30 ft/s relative to the boat, through a pipe of cross-sectional area A = 0.2 ft2. The resistance force F of the water is proportional to the square of the boat velocity Vs, which has a maximum value of 20 ft/s. What is the acceleration of the boat when its velocity is Vs = 10 ft/s?

4 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 98

5-40

MVs

V2r

F

M

V2r

FV1r

+

Figure 5.6-2 Jet-propelled boat.

Solution ------------------------------------------------------------------------------------------

Choose the control volume C.V. moving with the boat then the velocity of water entering the C.V. is

V1r = Vs

The mass flow rate of water entering or leaving the C.V. is

= A2V2r = (62.4)(0.2)(30) = 374.4 lbm/sm

The x-momentum balance is now written with the relative velocities

= M = Ma = V1r V2r + kVs2

dtMVd s )(

dtdVs m m

At maximum velocity Vs = 20 ft/s, there is no acceleration, therefore

kVs2 = (V2r V1r) k = (V2r V1r) m 2

sVm

k = = 220)2030(4.374

404.374

The acceleration at Vs = 10 ft/s is given by

a = (V1r V2r) + Vs2

Mm

Mk

a = (10 30) + (102) = 6.55 ft/s21000

4.374100040

4.374

5-41

Chapter 5

5.7 Microscopic Momentum Balance

We now consider the general open system or control volume fixed in space and located in a fluid flow field, as shown in Figure 5.7-1. The streamline of a fluid stream is the curve where the velocity at any point is tangent to it. For a differential element of area dA on the control surface, the rate of momentum efflux from this element = (v)(vdAcos), where (dAcos) is the area dA projected in a direction normal to the velocity vector v, is the angle between the velocity vector v and the outward-directed unit normal vector n to dA, and is the density.

v

n

Control volume

Control surface

Normal to surface dA

dAStreamlines offluid stream

Figure 5.7-1 Flow through a differential area dA on a control surface.

(v)(vdAcos) is the scalar or dot product of v(vn)dA. Since the normal vector n is pointing outward, the momentum (efflux) leaving the control volume is positive ( < 90o) and the momentum (influx) entering the control volume is negative ( > 90o). If we now integrate this quantity over the entire control surface A, we have the net outflow of momentum across the control surface or the net momentum efflux from the entire control volume .V~

=

volumecontrolfromeffluxmomentumnet

volumecontrolfromoutputmomentumofrate

volumecontrolfrominputmomentumofrate

= = v(vn)dA

volumecontrolfromeffluxmomentumnet

A

dAvv cos A

Since the rate of momentum accumulation within the control volume is = vdV, dt

dM

Vt

the momentum balance is written as

vdV = v(vn)dA + (5.7-1)

Vt

A

C.V.on

F

5-42

The above equation can also be written as

= + (5.7-2)0

lim zyx zyx

dVt

v

0

lim zyx zyx

dA

)( nvv

0

lim zyx zyx

C.V.on

F

Each of the above terms will be evaluated separately and substituted into equation (5.7-2).

The rate of momentum accumulation within the control volume. The rate of momentum accumulation within the control volume is given by

= = = + v (5.7-3)0

lim zyx zyx

dVt

v

zyxzyxt

v)/( v

t

tv

t

Net rate of momentum transported through the control volume. The net rate of momentum flux into the control volume illustrated in Figure 5.7-2 is

= + 0

lim zyx zyx

dA

)( nvv

zyxzyvv xxxxx

)||( vv

zyxzxvv yyyyy

)||( vv

+ zyx

yxvv zzzzz

)||( vv

x

z

y

z

x

y

vv |x x+ xvv |x x

vv |y y

vv |y y+ y

Figure 5.7-2 Momentum flux through a differential control volume

= + + (5.7-4)0

lim zyx zyx

dA

)( nvv

xvx

)( v

yvy

)( v

zvz

)( v

Applying the product rule to the terms on the right-hand side of the equation yields

5-43

= v + 0

lim zyx zyx

dA

)( nvv

zv

yv

xv zyx )()()(

zv

yv

xv zyx

vvv

The above equation can be simplified by the aid of the continuity equation (5.2-2)

= (5.2-2)t

xvx

)(

yv y

)(

zvz

)(

=

zv

yv

xv zyx )()()(

t

Therefore

= v + (5.7-5)0

lim zyx zyx

dA

)( nvv

t

zv

yv

xv zyx

vvv

Sum of external forces acting on control volume.

x

z

y

z

x

y

xx x+ x|

xy x+ x|

yy y+ y|

yx y+ y|

xx x|

xy x|

yy y|

yx y|

Figure 5.7-3 Forces acting on a differential control volume.

The forces acting on the control volume are those due to surfaces forces such as frictional forces and pressure force, and body forces such as gravitational force. Summing the forces in the x direction we obtain

= (xx|x+x xx|x)yz + (yx|y+y yx|y)xz + (zx|z+z zx|z)xyC.V.on

xF

+ gxxyz

In this equation gx is the component of the gravitational acceleration in the x direction. In the limit as xyz 0, we obtain

5-44

= + + + gx0

lim zyx zyx

C.V.on

xF

xxx

yyx

zzx

Similarly, we will obtain the following expressions for the summation of the forces in the y and z directions

= + + + gy0

lim zyx zyx

C.V.on

yF

xxy

yyy

zzy

= + + + gz0

lim zyx zyx

C.V.on

zF

xxz

yyz

zzz

Therefore

= ex + ey + ez0

lim zyx zyx

C.V.on

F

0

lim zyx zyx

C.V.on

xF

0

lim zyx zyx

C.V.on

yF

0

lim zyx zyx

C.V.on

zF

= + + ex + + + ey0

lim zyx zyx

C.V.on

F

xxx

yyx

zzx

x

xyy

yy

zzy

+ + + ez + (gxex + gyey + gz ez)

xxz

yyz

zzz

= + g = (P + ) + g (5.7-6)0

lim zyx zyx

C.V.on

F

The total stress at any point within a fluid is composed of both the isotropic pressure and the anisotropic stress components. By convention, pressure is considered a negative stress because it is compressive. The differential momentum equation becomes

= + 0

lim zyx zyx

dV

v

0

lim zyx zyx

dA

)( nvv

0

lim zyx zyx

C.V.on

F

+ v = v + + gt

vt

t

zv

yv

xv zyx

vvv

+ vv = P + t

v + g (5.7-7)

5-45

The three components of the momentum equation in rectangular coordinates system are

= + + + + gx (5.7-8a)

zvv

yvv

xvv

tv x

zx

yx

xx

xP

xxx

yyx

zzx

= + + + + gy (5.7-8b)

z

vv

yv

vxv

vt

v yz

yy

yx

y

yP

xxy

yyy

zzy

= + + + + gz (5.7-8c)

zvv

yvv

xvv

tv z

zz

yz

xz

zP

xxz

yyz

zzz

In the momentum equation, the terms on the left-hand side represent the time-rate of change of momentum, and the terms on the right-hand side represent the forces.

The term represents the time rate of change of vx at a fixed location, and is called the t

vx

local acceleration. The terms are called the convective

zvv

yvv

xvv x

zx

yx

x

acceleration that represents the change in velocity from location to location. The sum of local and convective acceleration is the total acceleration. The four terms

is the derivative following the motion of the fluid. This

zv

yv

xv

t zyx

derivative is called the substantial or material derivative and is denoted by .DtD

= DtD

zv

yv

xv

t zyx

When the substantial derivative notation is used, equations (5.7-8) become

= + + + + gx (5.7-9a)Dt

Dvx

xP

xxx

yyx

zzx

= + + + + gy (5.7-9b)Dt

Dv y

yP

xxy

yyy

zzy

= + + + + gz (5.7-9c)Dt

Dvz

zP

xxz

yyz

zzz

Equations (5.7-9) are valid for any type of fluid since we have not assumed any relationship between shear stress and the shear rate. For laminar flow and Newtonian fluid we have the following relations between shear stress components and the strain rates in rectangular coordinate form

5-46

xy = yx = (5.7-10a)

xv

yv yx

yz = zy = (5.7-10b)

yv

zv zy

zx = xz = (5.7-10c)

zv

xv xz

xx = P (5.7-10d)

v-

322

xvx

yy = P (5.7-10e)

v-322

yv y

zz = P (5.7-10f)

v-

322

zvz

These relations are called the constitutive equations that relate the local stress components to the flow or deformation of the fluid in laminar flow. If equations (5.7-10) are substituted into equations (5.7-9) and simplified, we will obtain the following relations for incompressible flow where v = 0

= + + gx (5.7-11a)Dt

Dvx

xP

2

2

2

2

2

2

zv

yv

xv xxx

= + + gy (5.7-11b)Dt

Dv y

yP

2

2

2

2

2

2

zv

yv

xv yyy

= + + gz (5.7-11c)Dt

Dvz

zP

2

2

2

2

2

2

zv

yv

xv zzz

Hence, the momentum equation for incompressible, laminar flow of a Newtonian fluid in vector notation is given by

= P + 2v + gDtDv

5-47

Chapter 5

Example 5.7-15 ----------------------------------------------------------------------------------

Consider the flow through a rectangular duct whose width is very large in the z direction when compared to the gap (2h) in the y direction. Such a situation could occur in a die when a polymer is being extruded at the exit into a sheet, which is subsequently cooled and solidified. Determine the relationship between the flow rate and the pressure drop between the inlet and exit, together with other relevant quantities.

Largewidth

2h

y

x

z

Inlet

Exit

LFigure 5.7-6 Flow through a rectangular duct.

Solution ------------------------------------------------------------------------------------------

We can analyze the problem by referring to a cross section of the duct, shown in Figure 5.7-7, taken at any fixed value of z.

P1 P2

vx

x

y h

h

L

Inlet Exit

Figure 5.7-7 A cross section of the duct with flowing fluid.

We make the following assumptions regarding the flow through the duct

1. The flow is steady with Newtonian fluid and constant physical properties.

2. There is only one nonzero velocity component in the x direction so that vy = vz = 0.

5 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 274

5-48

3. The entrance and exit effects are negligible and there is no variation of velocity in the z

direction because of the large width so that = 0.zvx

4. Gravity force is in the negative y direction; hence, gy = g and gx = gz = 0.

5. There is no slip at the boundary (the wall), so that vx = 0 at y = h.

The differential mass balance or continuity equation ( = v) is simplified to v = 0 t

for constant density

v = + + = 0xvx

yv y

zvz

Since vy = vz = 0, we have = 0. Therefore vx = vx(y) is a function of y only. We now xvx

consider the momentum equation in the x direction

= + + gx

zvv

yvv

xvv

tv x

zx

yx

xx

xP

2

2

2

2

2

2

zv

yv

xv xxx

This equation is simplified to

0 = + (E-1)xP

2

2

yvx

If the dependence of pressure on y is neglected, then P = P(x) is a function of x only. Equation (E-1) becomes

= = (E-2)2

2

dyvd x

dxdP

LPP 12

A first integration gives

= y + C1 (E-3)dydvx

1

dxdP

Since the velocity is a maximum at y = 0, we have

0 = (0) + C1 C1 = 01

dxdP

A second integration of equation (E-3) with C1 = 0 gives

5-49

vx = y2 + C221

dxdP

The constant of integration C2 can be obtained from the boundary condition that vx = 0 at y = h.

0 = h2 + C2 C2 = h221

dxdP

21

dxdP

The velocity profile is finally

vx = (y2 h2)21

dxdP

The volumetric flow rate Q for the system with a width W can be obtained by first consider the differential flow rate through an element Wdy

dQ = vxWdy

Hence Q = 2 = 2Wh

xWdyv0 2

1dxdP

h

dyhy0

22 )(

Q = = W

dxdP

h

yhy

0

23

3

32 3Wh

dxdP

The maximum velocity occurs at the centerline y = 0

vx,max = 2

2h

dxdP

The average velocity is just the volumetric flow rate Q divided by the area of flow

vx,ave = = WhQ

2 3

2h

dxdP

Therefore vx,ave = vx,max32

The shear stress at any location within the fluid is given by

yx = = = y

xv

yv yx

yvx

dxdP

5-50

Shell Balance

We can also apply the momentum balance directly to a differential fluid element to obtain the differential equation required for the calculation. Consider a fluid element with dimensions of x and y in the plane of diagram as shown in Figure 5.7-8. The element has a depth of W in the direction normal to the plane of the diagram.

P|x P|x+ x

yx y+ y|

yx y|

xy

x

y

Figure 5.7-8 Momentum balance on a differential fluid element xyW.

The convective momentum transfers through the left-hand and right-hand face are

vxvxyW|x vxvxyW|x+x = 0

We have assume that vx = vx(y) is a function of y only. Applying the momentum balance on the fluid element xyW yields

P|xyW P|x+xyW + yx|y+yxW yx|yxW = 0

Dividing the equation by the control volume xyW gives

+ = 0x

PP xxx

||

yyyxyyyx

||

In the limit as xyW 0, we obtain

= (E-4)dy

d yxdxdP

The shear stress at any location within the fluid is given by

yx = =

xv

yv yx

dydvx

Equation (E-4) becomes

5-51

= dyd

dydvx

dxdP

For constant physical properties, we have

= 2

2

dyvd x

dxdP

This equation is the same as the one derived from simplifying the following component of the Navier Stokes equations.

= + + gx

zvv

yvv

xvv

tv x

zx

yx

xx

xP

2

2

2

2

2

2

zv

yv

xv xxx

Example 5.7-2 ----------------------------------------------------------------------------------

The geometry of angular drag flow between two concentric cylinders is shown in Figure 5.7-9. The inner and outer cylinders have radii of R and R respectively. The inner cylinder is rotating with constant angular velocity while the outer one is fixed. Find the velocity distribution v(r) for the fluid between the cylinders and the torque required to rotate the inner cylinder.

z

r

L

Figure 5.7-9 Flow between concentric cylinders.

Solution ------------------------------------------------------------------------------------------

The r-component of the Navier Stokes equations for Newtonian isothermal incompressible flow in cylindrical coordinates is given as

= + + gr

zvv

rvv

rv

rvv

tv r

zr

rr

2

rP

2

2

22

2

221)(1

zvv

rv

rrv

rrrrr

r

5-52

For steady state flow with only one component of the velocity v(r), the equation is simplified to

= r

v2

rP

We have a radial pressure gradient due to centrifugal acceleration. The -component of the Navier Stokes equations is given as

= + + g

zvv

rvvv

rv

rvv

tv

zr

r

r1

P

2

2

22

2

221)(1

zvv

rv

rrv

rrrr

Discarding the zero terms in the above equation yields

0 =

)(1

rvdrd

rdrd

Integrating the equation gives

= a d(rv) = ardr rv = ar2 + b)(1rv

drd

r 21

v = ar + 21

rb

The two constants of integrations a and b can be evaluated using the boundary conditions.

At r = R v = R = aR + 21

Rb

At r = R v = 0 = aR + 21

Rb

Solving two equations for two unknowns a and b: b = and a =

/1

2R

/12

The velocity distribution is

v = + =

/12

2r

/1

2Rr1

/1R

Rr

rR

The shear stress r is given by

r = = r

rv

rrv

rr 1

rv

drd

5-53

= r

v

/1R

RrR 1

2

Taking the derivative of the above expression gives

=

rv

drd

/1R

32

rR

The shear stress is then

r = r = 2

rv

drd

/1 2

2

rR

At the surface of the inner cylinder r = R, therefore

r|R = 2 =

/1

121

2

The force applied on the inner surface of the cylinder is (2RL)r|R. The torque on the cylinder is the product of this force time the moment arm of radius R.

= (2RL)r|R(R) = (22R2L) = 212

2

22

14

RL

We have assumed laminar flow that is achievable for concentric cylinders as long as the rotational speed is below a value that satisfies

<

2R2/3)1(

40

Example 5.7-3 ----------------------------------------------------------------------------------

Find the velocity profile for the fluid between a rod and a cylinder shown in Figure 5.7-10. There is no axial pressure gradient. The rod is pulled through the cylinder with velocity V.

RR

V

v(r)Figure 5.7-10 Axial flow between a rod and a cylinder.

Solution ------------------------------------------------------------------------------------------

The z-component of the Navier Stokes equations for Newtonian isothermal incompressible flow in cylindrical coordinates is given as

5-54

= + + gz

zvvv

rv

rvv

tv z

zzz

rz

zP

2

2

2

2

211

zvv

rrvr

rrzzz

For steady state flow with only one component of the velocity vz(r), the equation is simplified to

0 = (E-1)

drdvr

drd

rz1

The two boundary conditions required to solve the differential equation are

r = R, vz = V and r = R, vz = 0

Integrating equation (E-1) once gives

= Cdrdvr z

Integrating the above equation gives

vz = Cln r + D

The two constants of integrations C and D can be obtained from the boundary conditions

At r = R vz = V = Cln R + D

At r = R vz = 0 = Cln R + D

Solving for C we have

C = = RR

Vlnln ln

V

D = ln

ln RV

The axial velocity profile is then

vz = ln r = Vln

Vln

ln RVln

)/ln( Rr

6-1

Chapter 6 Dimensional Analysis

6.1 Units and Dimensions

The dimensions of a quantity define the physical character of that quantity, e.g, force (F), mass (M), length (L), time (t), temperature (T), electric charge (e), etc. On the other hand, units identify the reference scale by which the magnitude of the respective quantity is measured. Many different units can be defined for a given dimension; for example, the dimension of length can be measured in units of miles, centimeter, meter, yards, etc.

Dimensions can be classified as either primary (fundamental) or derived. It happens that seven primary quantities are needed to completely describe all natural phenomena1. Primary dimensions cannot be expressed in term of other dimensions and include length (L), time (t), temperature (T), mass (M), and/or force (F) (depending upon the system of dimensions used.) Derived dimensions can be expressed in terms of primary dimensions, for example, area ([A]) = L2), energy ([E] = ML2/t2), etc. The decision as to which quantities are primary is arbitrary. The units of the primary quantities in the SI units (Système International d'Unitès, translated Internal System of Units) and their symbols are listed in Table 6.1-1 and defined arbitrarily as follows:

Meter: the length of the trajectory traveled by light in a vacuum per 1/299,792,458 s,

Kilogram: the mass of the platinum cylinder deposited at the International Office for Weights and Measures, Sèvres, France,

Second: 9,192,631,770 times the period of radiation in energy level transitions in the fine spectral structure of 133Cs,

Kelvin: 1/273.16 of the triple point temperature of water with naturally occurring amounts of H and O isotopes,

Amperes: the current which, on passing through two parallel infinite conducting wires of negligible cross section, separated by 1 m and in vacuum, induces a force (per unit length) of 2×10-7 N/m,

Mole: the amount of a matter containing the number of particles equal to the number of atoms in 0.012 kg of the pure isotope 12C,

Candela: the amount of perpendicular light (luminosity) of 1/60×10-6 m2 of the surface of an absolute black body at the melting temperature of platinum and a pressure of 101,325 Pa.

1 Thompson, E. V., A Unified Introduction to Chemical Engineering Thermodynamics, Stillwater Press, Orono, Maine, 2000.

6-2

Table 6.1-1 The seven primary quantities and their units in SI

Primary quantity UnitLength MassTimeTemperatureElectric current Amount of matterAmount of light

Meter (m)Kilogram (kg)Second (s)Kelvin (K)Ampere (A)Mole (mol)Candela (cd)

Several of the derived quantities with units are listed in Table 6.1-2. A derived unit is a quantity expressed in terms of a product and/or quotient of two or more primary units.

Table 6.1-2 The derived quantities and their units in SI

Derived quantity UnitCp, specific heat capacity E, energy F, forcek, thermal conductivityp, pressure q, heat transfer rateq", heat flux

, heat generation rate per unit volumeq , viscosity, density

J/kg·KJ = N·m/s, jouleN = kg·m/s2, newtonW/m·KPa = N/m2, pascalW = J/s = kg·m2/s3, wattW/m2 = J/s·m2

W/m3 N/m2=kg/s·mkg/m3

6.2 Dimensional Analysis

Dimensional analysis arises from the requirement that every term in an equation must have the same units. It is possible to learn a great deal about a complicated situation through dimensional analysis if you can identify the essential features of the problem. An example is the well-known story of how G. I. Taylor was able to deduce the yield of the first nuclear explosion from a series of photographs of the expanding fireball in Life magazine. The picture gave him the radius r(t) of the fireball, which is a strong shock, expanding into an undisturbed air at various time. He suggested that the important quantities in determining r(t) was the initial energy released, E, time, t, and the density of undisturbed air, . In other words, the radius depends on E, t, and

r(t) = f(E, t, )

He wanted to find an expression that can estimate the radius at various time based on E, t, and . Table 6.2-1 lists the units of r, E, t, and .

Table 6.2-1 r E t L ML2/t2 t M/L3

6-3

The unit of r is length, therefore the combination of the quantities E, t, and must also have unit of length. E and must come in as E/ to cancel out the mass. E/ has the dimensions L5/t2, so the only possible combination was

r(t) (6.2-1)5/12

Et

Taylor did a log-log plot of r versus t. The graph was a straight line with a slope of 2/5, which verified the theory. E/ could be obtained from the value of log r when log t = 0. Since the air density was known, the initial energy released E could be estimated to within a factor of order one. It is important to realize that the process of dimensional analysis in general cannot tell how the variables describing a system are related. The relationship must be determined either theoretically by application of basis scientific principles or empirically by measurements and data analysis. Equation (6.2-1) would never be confirmed without the data of r versus t.

Dimensional analysis can be applied to arrange the variables and/or parameters that are relevant in a given situation into a set of dimensionless groups. This has two important advantages. First, the information obtained from a dimensionless relation can be applied to geometrically and dynamically similar systems of any size or scale. This allows the scale up of equipment from laboratory models to plant equipment. Second, dimensional analysis will reduce the number of variables required for a system to a relatively smaller number of groups of variables.

We will consider the situation when there is no existing model or solution and we want to know the number of dimensionless groups that are important to the system. The general approach will be illustrated by a specific example of pipe flow as shown in Figure 6.2-1. We will determine an appropriate set of dimensionless groups to represent the wall shear stress w in pipe flow as a function of the distance x from the inlet, D, V, , , and the wall roughness . That is

w = f(x, D, V, , , )

D

w

V

xFigure 6.2-1 Wall shear stress for pipe flow.

The procedure is as follows.

Step 1: Identify the important variables in the system. In this example, all the relevant variables are given. In general, this step is the most challenging step in the process and can only be done if you have a good understanding of the system. You can use your experience, judgment, intuition or by examining the basic equations that describe the fundamental physical principles governing the system along with appropriate boundary conditions. You can only include those primary variables that are not derivable from others through basic

6-4

definitions. For example, the fluid velocity (V), the pipe diameter (D), and the volumetric flow rate (Q) are related by the definition Q = D2V/4. Therefore these three variables are not independent, it would be necessary to include only two of the three.

Step 2: Create a dimensional matrix. This dimensional matrix is simply a table with each column representing the exponents of the fundamental units of the primary variable.

w x D V M/Lt2 L L L/t M/L3 M/Lt L

MLt

112

010

010

01

1

130

111

010

The number of dimensionless groups that will be obtained is equal to the number of variables minus the rank of the dimensional matrix. The rank of a matrix is the highest order of a non zero determinant that can be obtained from the matrix. The number of dimensionless groups is usually equal to the number of variables less the minimum number of fundamental dimensions involved in these variables (7 3 = 4 groups in this example)

Step 3: Choose a set of reference variables. Choose the number of reference variables equal to the number of fundamental dimensions (3 in this case). Together they form what is known as the core group. The choice of variables is arbitrary, except that the following criteria must be satisfied:

(1) No two reference variables should have exactly the same dimensions. For example: x and D should not be both reference variables.

(2) All the fundamental dimensions that appear in the problem variables must also appear collectively in the dimensions of the reference variables.

In general, the algebra is simpler if the reference variables chosen have the least combination of dimensions, consistent with the preceding criteria. Another way to choose the core group is to exclude from it those variables whose effect one desirers to study. In the present problem it would be desirable to have the wall shear stress in only one dimensionless group, hence it will not be in the core. We choose D, V, and as the reference variables and form the core group by raising the variables to certain exponents.

Core group = DaVbc

Step 4: Form the dimensionless groups. We do this by dividing each of the remaining non-reference variables with the core group:

1 = , 2 = ,cbaw

VD

cbaVDx

3 = , 4 = cbaVD

cbaVD

6-5

Step 5: Solve for the exponents in each of the dimensionless groups. For the group 1

Dimensions of w = Dimension of DaVbc

ML-1t-2 = (L)ab

tL

c

3LM

Equating the exponent on M yields: 1 = c

Equating the exponent on L yields: 1 = a + b 3c

Equating the exponent on t yields: 2 = b

Solving for a, b, and c we obtain: b = 2, c = 1, and a = 0.

Therefore 1 = 2Vw

Applying similar procedure for 2, 3 , and 4 we obtain

2 = , 3 = , and 4 = Dx

DV D

The relationship between the dimensionless group is not known, therefore

f(1, 2, 3, 4) = 0

The above equation can also be written as

1 = = f(2, 3, 4) 2Vw

= f = f

= f 2Vw

, ,x

D V D D

DD

xVD ,,

DDx ,Re,

You should note that the results of the foregoing procedure are not unique because the reciprocal of each group is just as valid as the initial group. In fact, any combination of these groups will be dimensionless and will be just as valid as any combination as long as all of the original variables are represented among the groups. For a position far away from the

entrance, the flow is fully developed so that is not importantDx

= f developedfully

w

V 2

DRe,

6-6

For scale-up problems, the two systems must possess both geometrical similarity and dynamical similarity. Geometrical similarity requires that the two systems have the same shape, and dynamical similarity requires them to have equality of the appropriate dimensionless groups.

Example 6.2-1 ----------------------------------------------------------------------------------A classroom2 demonstration on the time to drain water from a set of initially filled inverted bottles yielded the following results;

Bottle volume, V (liters) 2 1.5 0.7 0.4 0.36 0.18Time to drain, t (s) 18.6 17.3 9.5 4.2 4.8 2.9

The bottles were not geometrically similar, but the same cap size fit each bottle.

Hence the neck diameter was the same in each case.In a study of bottle emptying, Whalley [Int. J. Multiphase Flow, 17, 145 (1991)]

presents a correlation of data of the form

= constanttDg

V2/52/1

4

where V is the filled bottle volume (L), D is the internal diameter of the bottle neck (m), g is the acceleration of gravity (9.8 m2/s), and t is the time to empty (s). The constant is different for each bottle shape.a. For our experiments, how should t depend on V, according to Whalley’s correlation? (Is the relationship linear, or does some other power relationship apply? If the latter, give the power n in t = KVn.)b. For geometrically similar bottles, how should t depend on V, according to Whalley’s correlation? (Is the relationship linear, or does some other power relationship apply? If the latter, give the power n in t = KVn.)c. Plot our data as a test of Whalley’s correlation. Do the data agree with the correlation?Solution ------------------------------------------------------------------------------------------a. For our experiments, how should t depend on V, according to Whalley’s correlation?

For our experiment D = constant, therefore

t = V = KV, where K = = constant2/52/1.4

DgConst 2/52/1.4

DgConst

t depends linearly on V.

b. For geometrically similar bottles, how should t depend on V?

For geometrically similar bottles we have V D3 or V = aD3 . Therefore

= Const. = Const.tDg

V2/52/1

4tDg

aD2/52/1

34

2 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998

6-7-2 -1.5 -1 -0.5 0 0.5 1

0.5

1

1.5

2

2.5

3

3.5

Log(V)

Log(

t)

DataFitted

t = = bV1/6, where b = constant2/1

2/1

.4

gConstaD

c. Plot our data as a test of Whalley’s correlation. Do the data agree with the correlation?

t = KVn

log t = log K + nlog V

Table 6.2-1 lists the Matlab program to plot and fit log V versus log t. The exponent n is the slope of the line, which is obtained from the program

n = 0.84 1

The slope is close to the value predicted by Whalley’s correlation.

Table 6.2-1 ---------------------------------------------------------------------------------------% Example 6.2-1V=[2 1.5 0.7 0.4 0.36 0.18];t=[18.6 17.3 9.5 4.2 4.8 2.9];logV=log(V);logt=log(t);coef=polyfit(logV,logt,1);n=coef(1);fprintf('n = slope = %g\n',n)x=log([2 0.18]);y=polyval(coef,x);plot(logV,logt,'o',x,y)xlabel('Log(V)');ylabel('Log(t)')grid onlegend('Data','Fitted')

>> e2d1n = slope = 0.837214

6-8

Example 6.2-2 ----------------------------------------------------------------------------------(5.203) Consider a gas bubble rising slowly through a viscous liquid. When the bubble reaches the surface it will burst and collapse, and under some conditions the collapse will create liquid droplets that are projected above the surface. A study of the phenomena by Takahashi et al. [Int. Chem. Eng., 21, 251 (1981)] led to a dimensionless correlation of the critical (minimum) bubble diameter that will yield drops on bubble collapse. Assume that the critical bubble diameter DB depends on the viscosity and density of the liquid, on the interfacial tension between the liquid and the gas bubble, and on the gravitational constant g. Define a dimensionless group proportional to DB, and state what other group(s) this dimensionless critical bubble diameter could depend on. These other groups should not include DB.Solution ------------------------------------------------------------------------------------------

The dimensional matrix for the problem is given as

DB gL M/t2 M/L3 M/Lt L/t2

MLt

010

10

2

130

111

01

2

The reference variables are , , and g. Therefore the core group is abgc and we have two dimensionless groups:

1 = , 2 = cbaB

gD cba g

Dimensions of DB = Dimension of abgc

M0L1t0 = a

3LM

b

2tM

c

2tL

Equating the exponent on M yields: 0 = a + b

Equating the exponent on L yields: 1 = 3a + c

Equating the exponent on t yields: 0 = 2b 2c

Solving for a, b, and c we obtain: b = 0.5, c = 0.5, and a = 0.5

Therefore 1 = 2/1

2/12/1

gDB

Applying similar procedure for 2 we have

3 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998

6-9

Dimensions of = Dimension of abgc

M1L-1t-1 = a

3LM

b

2tM

c

2tL

Equating the exponent on M yields: 1 = a + b

Equating the exponent on L yields: 1 = 3a + c

Equating the exponent on t yields: 1 = 2b 2c

Solving for a, b, and c we obtain: a = 0.25, b = 0.75, and c = 0.25

2 = 4/34/1

4/1

g

Hence 1 = f(2) or = f 2/1

2/12/1

gDB

4/34/1

4/1

g

Example 6.2-3 ----------------------------------------------------------------------------------(4.214) A one-third scale model pump (impeller diameter D1 = 0.5 ft) is to be tested when pumping Q1 = 100 gpm of water (1 = 62.4 lb/ft3) in order to predict the performance of a proposed full-size pump (D2 = 1.5 ft) that is intended to operate at rotational speed of N2 = 750 rpm with a flow rate of Q2 = 1,000 gpm when pumping an oil density 2 = 50 lb/ft3. The two important dimensional groups are:

1 = and 2 = where P is the power supplied by the pump53DNP

3NDQ

(a) Determine the rotational speed of the scale model.(b) If P1 = 1.20 hp, determine the power needed for the full size pump.Solution ------------------------------------------------------------------------------------------

Model Full-size pumpD1 = 0.5 ftQ1 = 100 gpm 1 = 62.4 lb/ft3

N1 = ?P1 = 1.20 hp

D2 = 1.5 ftQ2 = 1,000 gpm2 = 50 lb/ft3

N2 = 750 rpmP2 = ?

4 Wilkes, J., Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 219

6-10

(a) Determine the rotational speed of the scale model.

For dynamical similarity between the model and the full-size pump, all the pertinent dimensionless groups must have the same value.

1|model = 1|full and 2|model = 2|full

= 311

1

DNQ

322

2

DNQ

N1 = N2

2

1

QQ

3

1

2

DD

N1 = 750 = 2,025 rpm

1000100 3

5.05.1

The rotation speed of the scale model is 2,025 rpm

(b) If P1 = 1.20 hp, determine the power needed for the full size pump.

Equating the dimensionless group containing the power we have

= 51

311

1

DNP

52

322

2

DNP

P2 = P1

1

2

3

1

2

NN

5

1

2

DD

P2 = 1.2 = 11.87 hp

4.6250 3

2025750

5

5.05.1

11.87 hp is required for the full size pump.

7-1

Chapter 7 Flow in Closed Conduits

7.1 Flow regimes

Water

Dye

Valve

Water

Dye

Valve

(a) Re < 2000 (b) Re > 4000Figure 7.1-1 Reynolds’ experiment.

Laminar or well-ordered type of flow exists when adjacent fluid layers slide smoothly over another. Mixing between layers occurs only on a molecular level. Turbulent flow exists when packets of fluid particles are transferred between layers, giving the flow a fluctuating nature. Osborn Reynolds first described the existence of laminar and turbulent flow quantitatively through his classic experiment in 1883. As shown in Figure 7.1-1, water was allowed to flow through a transparent pipe at a rate controlled by a valve. Reynolds introduced a dye having the same specific gravity as water into the flow to observe what was happening. He found that at low flow rates the dye pattern was regular and formed a single line of color as show in Figure 7.1-1(a). The pressure drop was also found to directly proportional to the flow rate. As the flow rate was increased a point was reach where the dye trace was seen to be unstable and it broke up after a short distance. At still higher flow rates the dye almost immediately dispersed throughout the pipe cross section. The relationship between pressure drop and flow rate now became almost quadratic instead of linear.

The stable flow observed initially was called laminar flow. The unstable flow pattern, characterized by high degree of mixing between the fluid elements, was called turbulent flow. There is a transition region in between laminar and turbulent where the flow is unstable but not thoroughly mixed. Laminar flow in a tube persists up to a point where the value of the Reynolds number is about 2000. Reynolds number is defined as

NRe = =

DVDV

V/

2

The Reynolds number is a ratio of the inertial momentum flux (V2) in the flow direction to the viscous shear stress or viscous momentum flux in the transverse (V/D) direction. Turbulent flow occurs when Reynolds number is greater than about 4000. Viscous forces are a manifestation of intermolecular attractive forces that stabilize the flow. Therefore stable laminar flow should occur at low Reynolds numbers where viscous forces dominate.

7-2

7.2 Generalized Mechanical Energy Balance Equation

For laminar flow of a fluid in a cylindrical tube of radius R and length L, the Hagan-Poiseuille equation provides a relationship between volumetric flow rate and pressure drop across the tube as follows.

Q = w = =

4

3R

4

3R L

RPP Lo

2

LPPR Lo

8

4

z1 z2

PV

1

1

PV

2

2

Figure 7.2-1 A general piping system.

For a general piping system shown in Figure 7.2-1, we need the generalized relationship, equation (7.2-1). This equation can account for the effect of pressure drop on incompressible fluid flow, changes in elevation, tube cross section, changes in fluid velocity, sudden contractions or expansions, and friction loss through pipe and fittings such as valves and flow meters.

+ gz1 + + wp = + gz2 + + ef (7.2-1)

1P2

211V

2P

2

222V

Each term in this equation has units of energy per unit fluid mass flow rate or (length/time)2.

P = pressure = fluid densityg = acceleration of gravityz = elevation relative to a reference surfaceV = average fluid velocity = kinetic energy correction factor

= 2 for laminar flow = 1 for turbulent flow

wp = work done per unit mass flow rate = pump efficiency ( < 1)ef = friction loss due to piping and fitting

7-3

The friction loss is given by the following equation

ef = 4 + Kfitting,j (7.2-2)i

ifi

i

DL

2

2iV

j

jV2

2

where

fi = = friction factor in tube segment i with length Li and diameter Di.2

21 V

w

Vi = average velocity within tube segment i.

Kfitting = friction loss factor or loss coefficient for pipe fittings, some typical values are given in Table 7.2-1. The velocity Vj in the summation is for the fluid in the smaller pipe for sudden contraction and sudden expansion.

Table 7.2-1 Friction loss factor for various pipe fittings.Fitting Kfitting

Globe valve, wide openAngle valve, wide openGate valve, wide openGate valve, half openStandard 90o elbowStandard 45o elbowTee, through side outletTee, straight throughSudden contraction(turbulent flow)

Sudden expansion(turbulent flow)

7.53.80.154.40.70.351.50.4

0.42

1

21

AA

2

2

11

AA

A1 A2

A1 A2

Sudden contraction

Sudden expansion

The friction factor for laminar flow (NRe = < 2000) is given by

VD

f = (7.2-3)Re

16N

The friction factor for turbulent flow (Re > 4000) can be estimated by

f = { 1.737 ln[0.269 ln (0.269 + )]}-2 (7.2-4)D

Re

185.2N D

Re

14N

7-4

In this equation is the surface pipe roughness and D is the inside pipe diameter. Representative values for surface roughness are given in Table 7.2-2.

Table 7.2-2 Surface roughnessSurface (ft) (mm)ConcreteCast ironWrought ironGalvanized ironCommercial steelDrawn tubing

0.001-0.010.000850.000150.00050.000150.000005

0.3-3.00.250.0450.150.0460.0015

Equation (7.2-5) developed by Churchill1 adequately predicts the Fanning fiction factor over the entire range of Reynolds number including a reasonable estimate for the transition region between laminar and turbulent flow.

f = 2 (7.2-5)

12/1

2/3

12

Re )(18

BAN

In this equation A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

If the fluid flows through a noncircular duct, then the equivalent diameter, Deq, can be used in

equations (7.2-2, 3, 4, 5). The equivalent diameter is defined as Deq = 4rH = 4wet

cross

PA

where rH = hydraulic radiusAcross = cross sectional area of the flowPwet = wetted perimeter of the duct

Do Di

Figure 7.2-2 Flow through an annular tube.

For the flow through an annular tube, the equivalent diameter is given as

Deq = 4 = Do Di)(4/)( 22

io

io

DDDD

1 Churchill SW, Chem. Eng., Nov. 7, 1977, p. 91

7-5

Example 7.2-1. ----------------------------------------------------------------------------------

Water is pumped from the upper reservoir to the lower reservoir through the piping system shown. Determine the power required for the pump if the water flow rate is 60 kg/s. The fittings from pipe D1 to pipe D2 and from pipe D2 to pipe D3 can be considered to be standard 90o elbows with velocity in pipe 2 for the friction loss. Data:

h1 = 10 m, h2 = 3 m, L1 = 50 m, L2 = 300 m, L3 = 2 m, D1 = 0.2 m, D2 = 0.5 m, D3 = 0.03 m, water viscosity = 1 cP = 10-3 kg/ms, = 1000 kg/m3. The pipe roughness is 0.05 mm. The pump efficiency is 75%.

(1)

(2)

h1

D , L1 1

D , L2 2D , L3 3

h2

Globe valve

Solution ------------------------------------------------------------------------------------------

Applying the mechanical energy balance between (1) and (2) we have

+ gz1 + + wp = + gz2 + + ef 1P

2

211V

2P

2

222V

Let the reference level be at (2), the end of pipe 3, the energy equation becomes

+ g(h1 + L1 L3) + 0 + wp = + 0 + + ef atmP

2ghPatm

2

233V

g(h1 + L1 L3) + wp = gh2 + + ef2

233V

D(m) A(m2) V(m/s) NRe /D f.2.5.03

3.1410-2

1.9610-1

7.0710-4

1.910.30684.9

3.82105

1.53105

2.55106

2.5010-4

1.0010-4

0.0017

0.004060.004310.00600

7-6

ef = 4 + Kfitting,ji

ifi

i

DL

2

2iV

j

jV2

2

4 = 2 10-3[4.06 + 4.31 + 6 ] i

ifi

ii

DVL

2

2

2.091.150 2 2300 0.306

0.5 22 84.9

0.03

= 5.77 103 m2/s2

Kfitting,j = 0.51.9120.4 sudden contraction, Kfitting = 0.4j

jV2

2

+ 0.50.30620.7 standard 90o elbow, Kfitting = 0.7

+ 0.50.30627.5 open globe valve, Kfitting = 7.5

+ 0.584.920.7 standard 90o elbow, Kfitting = 0.7

Kfitting,j = 2.52 103 m2/s2j

jV2

2

Therefore ef = 5.77103 + 2.52103 = 8.29103 m2/s2

g(h1 + L1 L3) + wp = gh2 + + ef2

233V

9.81(10 + 50 2) + 0.75wp = 9.813 + + 8.2910329.84 2

wp = 1.51104 m2/s2

The power required for the pump is

= wp = 601.51104 = 9.08105 W = 1220 hppW m

Note: 1 hp = 746 W

---------------------------------------------------------------------------------------------------

7-7

Chapter 7

7.3 Pipe Flow Problems

There are three typical cases of calculation encountered in pipe flows with or without fittings. The problem depends on what is known and what is to be determined. Normally for the transport of fluid from location (1) to location (2), the following parameters are known

1. The elevation difference z = z2 z1.2. The total length of the piping system.3. The pipe wall roughness from the pipe materials.4. The physical properties of the fluid.5. The pipe fittings between location (1) and location (2).6. The work exchange between the fluid and the surroundings.

Three other variables are also relevant to the pipe flow problems. They are the volumetric flow rate Q, the internal pipe diameter D, and the pressure drop P = P1 P2. Specify any two of these three variables enables the remaining one to be determined. These three typical cases are the “unknown driving force,” “unknown flow rate,” and “unknown diameter” problems. You should realize that there exist many variations of the pipe flow problems. Almost any of the parameters and variables mentioned above could be the unknown(s). You need to find sufficient equations to solve for your unknown(s). The equations you might use are the conservation equations (mass, momentum, and energy balance) in the appropriate form, the relationship between friction factor f, Reynolds number NRe, and relative pipe roughness /D, and definitions that relate energy loss to friction factor, loss coefficients, and other relevant parameters. You will then obtain a set of nonlinear equations that might or might not be in the forms that can be solved for your unknown(s) explicitly. In case that you cannot solve for the unknown(s) directly you might be able to use a software package that solves a set of nonlinear equations provided appropriate first guesses for the unknowns are given. The procedure presented later in this section provides an algorithm to solve for the set of nonlinear equations iteratively. You might be able to find another method that could solve these equations more efficiently.

7.3a Unknown Driving Force

This is the easiest situation where no iterative steps are required for problems involving Newtonian fluids.

+ gz1 + + wp = + gz2 + + ef (6.2-1)

1P2

211V

2P

2

222V

From the energy equation (6.2-1), if (P1 P2) or wp is the unknown then the procedure is straight forward:

1. Calculate the mean velocities V1 and V2, the Reynolds numbers, and the pipe roughness ratio.

2. Determine the friction factors and then the friction loss ef.

7-8

3. Calculate (P1 P2) or wp from the energy equation.

Example 6.2-1 is a typical problem of this case. The following example provides another variation of the “unknown driving force” problems.

Example 7.3-1. ----------------------------------------------------------------------------------

If the flow rate through a 10 cm-diameter wrought iron pipe is 0.04 m3/s, find the difference in elevation H of the two reservoirs. The loss coefficients for the entrance and the exit are 0.5 and 1.0, respectively, based on the velocity in the pipe.

Standard elbow

20 m 10 m

20 m

Globe valve(fully open)Water

20 Co H

(1)

(2)

Figure 7.3-1 Two reservoirs systems.Solution ------------------------------------------------------------------------------------------

Applying the mechanical energy balance between (1) and (2) we have

+ gz1 + + wp = + gz2 + + ef 1P

2

211V

2P

2

222V

The equation is simplified to

H = z1 z2 = = (Kentrance + Kvalve+ 2Kelbow+ Kexit) + 4fge f

gV2

2

DL

gV2

2

H = = (Kentrance + Kvalve+ 2Kelbow+ Kexit + 4f ) ge f

DL

gV2

2

V = = = 5.09 m/s2RQ

)05.0(04.0

2

The kinematic viscosity of water at 20oC is = 10-6 m2/s, therefore

NRe = = = 5.09105

VD610

)1.0)(09.5(

= = 0.00045D

100045.0

7-9

The friction factor can be determined from the following equation

f = { 1.737 ln[0.269 ln (0.269 + )]}-2 = 0.0044D

Re

185.2N D

Re

14N

H = (Kentrance + Kvalve+ 2Kelbow+ Kexit + 4f )DL

gV2

2

The difference in elevation H of the two reservoirs is then calculated with values of loss coefficients given in Table 6.2-1.

H = (0.4 + 7.5 + 20.7 + 1 + 40.0044 ) = 25.4 m1.0

5081.92

09.5 2

--------------------------------------------------------------------------

7.3b Unknown Flow Rate

Since Q is unknown, you cannot determine NRe so the friction factor f is also unknown. We can proceed by a trial-and-error method with an initial guess of Reynolds number. We can then calculate the friction factor and solve the energy equation for the velocity that will be used to calculate the new Reynolds number. We repeat the process until agreement between calculated and guessed values is achieved. Using this approach, the following steps are needed:

1. Compute the roughness ratio .D

2. Assume NRe = 1105

3. Compute the friction factor from the following equation

f = 2 (6.2-5)

12/1

2/3

12

Re )(18

BAN

In this equation A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

4. Compute the mean velocity from the energy equation.

5. Compute NRe = . If this is within the accepted tolerance of the value used in

VD

step (3) then proceed to step (6). If not, return to step (3).

6. Compute the flow rate from Q = VD2/4.

7-10

Example 7.3-2.2 ----------------------------------------------------------------------------------

Oil is pumped from a tanker to a refinery storage tank as shown in Figure 7.3-2. Both free surfaces are open to the atmosphere. The friction loss for the piping system is equivalent to 6000 ft of commercial steel pipe Schedule 40 with a nominal diameter of 6 in. The discharge at point (4) is 200 ft above the pump exit, which is level with the free surface of oil in the tanker. The pressure at the pump exit is P2 = 132.7 psig. Determine the oil flow rate Q in gpm.

Oil density = 53 lb/ft3, oil viscosity = 13.2 cP. Note: cP = 0.000672 lb/fts.

Figure 7.3-2 Steady oil transferred to storage tank.

Solution ------------------------------------------------------------------------------------------

1. Compute the roughness ratio .D

For commercial steel pipe Schedule 40 with a nominal diameter of 6 in., we have = 0.00015 ft, D = 0.5054 ft. Therefore

= 0.000297D

2. Assume NRe = 1105

3. Compute the friction factor from the following equation

f = 2 (6.2-5)

12/1

2/3

12

Re )(18

BAN

In this equation A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

2 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 133

7-11

With = 0.000297 and NRe = 1105, f = 0.00487D

4. Compute the mean velocity from the energy equation.

Applying the mechanical energy balance between (2) and (4) we have

+ gz2 + + wp = + gz4 + + ef 2P

2

221V

4P

2

244V

Since there is no work and no change in kinetic energy, the equation is simplified to

P2 + gz2 = gz4 + ef = gz4 + 4 fDL

2

2V

The oil velocity inside the pipe is then

V = 2/1

242 )]([2

zzgPLf

D

V = = 6.684 ft2/1

)2002.32531442.327.132(60005300487.02

5054.0

5. Compute NRe = . If this is within the accepted tolerance of the value used

VD

in step (3) then proceed to step (6). If not, return to step (3).

NRe = = = 20,185

VD000672.02.13

5054.0684.653

Steps (3) to (5) are repeated until the difference between the guessed and calculated Reynolds numbers is less than 0.1 percent. The following values are obtained from the Matlab program listed in Table 7.3-1.

Re = 100000 , f = 0.00487, V(ft/s) = 6.684Re = 20185 , f = 0.00663, V(ft/s) = 5.732Re = 17310 , f = 0.00687, V(ft/s) = 5.630Re = 17001 , f = 0.00690, V(ft/s) = 5.618Re = 16965 , f = 0.00690, V(ft/s) = 5.617

6. Compute the flow rate from Q = VD2/4.

Q = VD2/4 = 5.617 7.48 60 = 506 gpm4

5054.0 2

sft3

3ft

gal

min

s

7-12

______ Table 7.3-1 Matlab program to iterate for the Reynolds number __________

% Program to iterate for Reynolds number.% den=53;L=6000;dz=200;pi=3.1416;g=32.2;vis=13.2*.000672;D=.5054eoD=.000297; R=1e5;ee=1;Rsave=R; while ee>0.001 A=(2.457*log(1/((7/R)^.9+0.27*eoD)))^16;B=(37530/R)^16; f=2*((8/R)^12+1/(A+B)^1.5)^(1/12); V=sqrt(D*(132.7*g*144-den*g*dz)/(2*f*53*L)); fprintf('Re = %8.0f , f = %8.5f, V(ft/s) = %7.3f\n',R,f,V) R=V*den*D/vis; ee=abs((R-Rsave)/R); Rsave=R; end

Example 7.3-3.3 ----------------------------------------------------------------------------------

Water is withdrawn from a reservoir as shown in Figure 7.3-3. The mass flow rate through the turbine is 3000 kg/s assuming constant water level in the reservoir. Calculate the mass flow rate through the pipeline if there is no turbine. You can assume the pipe is smooth.

Water density = 1000 kg/m3, water viscosity = 1 cP = 0.001 kg/ms.

Figure 7.3-3 Flow from a reservoir through a turbine.

Solution -------------------------------------------------------------------------------------

Applying the mechanical energy balance between (1) and (2) we have

+ gz1 + + wp = + gz2 + + ef 1P

2

211V

2P

2

222V

The equation is simplified to

3 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998, p. 460

7-13

g(hr + hp) 0.5V22 + 2f V2

2 DL

In this equation V2 is the water velocity inside the pipe. The procedure mentioned above can be used to determine the mass flow rate through the pipe.

9.81100 = (0.5 + 400f)V22

The expression (6.2-5) for friction factor can be used for the procedure.

f = 2 (6.2-5)

12/1

2/3

12

Re )(18

BAN

A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

However for a smooth pipe, a simpler equation predicts adequately the friction factor for turbulent flow.

f = 0.0791NRe-0.25 (7.3-1)

Table 7.3-2 lists the Matlab program and the result from the calculation.

______ Table 7.3-2 Matlab program to compute the mass flow rate __________

% Example 7.3-3, solve for mass flow with no turbine%L=200;D=1;vis=.001;den=1000;R=1e5;for i=1:20; f=0.0791*R^(-.25); V2=sqrt(981/(.5+400*f)); fprintf('V2(m/s) = %8.2f\n',V2) R=den*D*V2/vis; test=abs(R/Rsave-1); if test<.01, break, end Rsave=R;endmdot=den*V2*pi*D^2/4;fprintf('Mass flow rate (kg/s) = %g\n',mdot)

>> e6d3d3V2(m/s) = 20.75V2(m/s) = 31.82V2(m/s) = 32.63

7-14

V2(m/s) = 32.68Mass flow rate (kg/s) = 25665.8

Example 7.3-4. ----------------------------------------------------------------------------------

Water is withdrawn from a reservoir as shown in Figure 7.3-4. The pipe is 4-cm ID wrought iron pipe. Calculate the volumetric flow rate through the pipe system. The loss coefficients for the entrance and the screwed elbow are 0.5 and 0.95, respectively, based on the velocity in the pipe.

Water density = 1000 kg/m3, water viscosity = 1 cP = 0.001 kg/ms.

Screwed elbow

50 m 20 m

40 m

Globe valve(fully open)Water

20 Co

(1)

(2)

20 m

Figure 7.3-4 Flow from a reservoir through a piping system.

Solution -------------------------------------------------------------------------------------

Applying the mechanical energy balance between (1) and (2) we have

+ gz1 + + wp = + gz2 + + ef 1P

2

211V

2P

2

222V

The equation is simplified to

z1 z2 = + = (Kentrance + Kvalve+ 2Kelbow+ 1) + 4fg

V2

22

ge f

gV2

22

DL

gV2

22

z1 z2 = (Kentrance + Kvalve+ 2Kelbow+ 1 + 4f ) DL

gV2

22

40 = [0.5 + 7.5 + 20.95 + 1 + 4(110/0.04)f] g

V2

22

784.8 = (10.9 + 110,00f)V22

7-15

This equation and the friction factor expression, equation (6.2-5), are used in the Matlab program listed in Table 7.3-3. For wrought iron with a diameter of 40 mm., we have the roughness = 0.045 mm. Therefore

= 0.0011D

______ Table 7.3-3 Matlab program to compute the volumetric flow rate __________

% Example 7.3-4, solve for the volumetric flow rate%vis=.001;den=1000;D=.04;R=1e5;eoD=.0011;for i=1:20; Rsave=R; A=(2.457*log(1/((7/R)^.9+0.27*eoD)))^16;B=(37530/R)^16; f=2*((8/R)^12+1/(A+B)^1.5)^(1/12); V2=sqrt(784.8/(10.9+11000*f)); fprintf('V2(m/s) = %8.2f\n',V2) R=den*D*V2/vis; test=abs(R/Rsave-1); if test<.01, break, end endQ=V2*pi*D^2/4;fprintf('Volumetric flow rate (m3/s) = %g\n',Q)

>> e7d3d4V2(m/s) = 3.27V2(m/s) = 3.30Volumetric flow rate (m3/s) = 0.00415142

7.3c Unknown Diameter

The problem is to find the diameter D that will transport a given fluid at a specified flow rate Q and driving force. Since diameter is unknown, we cannot determine Reynolds number and the roughness ratio, therefore the friction factor is also unknown. We can proceed by a trial-and-error method with an initial guess of diameter. We can then calculate the friction factor and solve the energy equation for the diameter. We repeat the process until agreement between calculated and guessed diameters is achieve. Using this approach, the following steps are needed:

1. Guess the pipe diameter D and compute the mean velocity V = .24DQ

2. Compute the Reynolds number NRe = and the pipe roughness ratio .

VDD

7-16

3. Compute the friction factor from the following equation

f = 2 , where (6.2-5)

12/1

2/3

12

Re )(18

BAN

A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

4. Compute the diameter from the energy equation.

5. If the calculated D is within the accepted tolerance of the value used in step (2) then the problem is solved. If not, return to step (2).

Example 7.3-5 ----------------------------------------------------------------------------------

Oil is pumped from a tanker to a refinery storage tank as shown in Figure 7.3-2. Both free surfaces are open to the atmosphere. The friction loss for the piping system is equivalent to 6000 ft of commercial steel pipe. The discharge at point (4) is 200 ft above the pump exit, which is level with the free surface of oil in the tanker. The pressure at the pump exit is P2 = 132.7 psig. Determine the diameter required for an oil flow rate of 506 gpm.

Oil density = 53 lb/ft3, oil viscosity = 13.2 cP. Note: cP = 0.000672 lb/fts.

Figure 7.3-2 Steady oil transferred to storage tank.

Solution ------------------------------------------------------------------------------------------

1. Guess the pipe diameter D and compute the mean velocity V = .24DQ

Q = 506 gpm = 1.1275 ft3/s

Let D = 0.4 ft V = = = 8.97 ft24DQ

24.01275.14

7-17

2. Compute the Reynolds number NRe = and the pipe roughness ratio .

VDD

NRe = = = 21,443

VD000672.02.13

4.097.853

= = 0.000375D

4.000015.0

3. Compute the friction factor from the following equation

f = 2 , where (6.2-5)

12/1

2/3

12

Re )(18

BAN

A = and B = 16

9.0Re /27.0)/7(

1ln457.2

DN

16

Re

530,37

N

f = 0.00658

4. Compute the diameter from the energy equation.

Applying the mechanical energy balance between (2) and (4) we have

+ gz2 + + wp = + gz4 + + ef 2P

2

221V

4P

2

244V

Since there is no work and no change in kinetic energy, the equation is simplified to

P2 + gz2 = gz4 + ef = gz4 + 4 fDL

2

2V

P2 g(z4 z2) = 2 fDL 2

24

DQ

D = = f0.22.0

2422

2

)]([32

zzgPLQf

2.0

2422

2

)]([32

zzgPLQ

Let Con = then D = Con f0.22.0

2422

2

)]([32

zzgPLQ

The term Con does not change during iteration for the diameter.

7-18

Con = = 2.0

2422

2

)]([32

zzgPLQ

2.0

2

2

2002.32532.321447.132[1275.160005332

Con = 1.3676 D = 1.36760.006580.2 = 0.5008 ft

Steps (2) to (4) are repeated until the difference between the guessed and calculated diameters is less than 0.1 percent. The following values are obtained from the Matlab program listed in Table 7.3-4.

D(ft) = 0.4000 ,V(ft/s) = 8.9719 Re = 21442.70 ,e/D = 0.000375 , f = 0.00658

D(ft) = 0.5008 ,V(ft/s) = 5.7247 Re = 17128.20 ,e/D = 0.000300 , f = 0.00689

D(ft) = 0.5053 ,V(ft/s) = 5.6219 Re = 16973.73 ,e/D = 0.000297 , f = 0.00690

5. If the calculated D is within the accepted tolerance of the value used in step (2) then the problem is solved. If not, return to step (2).

The final value for the diameter is 0.5053 ft, which is essentially the value of the commercial steel pipe Schedule 40 with a nominal diameter of 6 in.

______ Table 7.3-4 Matlab program to iterate for the diameter __________

% Program to iterate for diameter.% Q=506/(60*7.48)den=53;L=6000;dz=200;pi=3.1416;g=32.2;vis=13.2*.000672;D=.4;ef=.00015; eD=1;Dsave=D; con=(32*53*Q*Q*L/(pi*pi*(132.7*g*144-den*g*dz)))^.2 while eD>0.001 V=4*Q/(pi*D*D); R=den*V*D/vis;eoD=ef/D; A=(2.457*log(1/((7/R)^.9+0.27*eoD)))^16;B=(37530/R)^16; f=2*((8/R)^12+1/(A+B)^1.5)^(1/12); fprintf('D(ft) = %8.4f ,V(ft/s) = %8.4f \n',D,V) fprintf('Re = %7.2f ,e/D = %8.6f , f = %8.5f \n\n',R,eoD,f) D=con*f^.2; eD=abs((D-Dsave)/D); Dsave=D; end

7-19

Chapter 7 7.4 Other Pipe Flow Problems

Example 7.4-1 ----------------------------------------------------------------------------------

A4 horizontal pipeline is designed for a given pipe roughness (), pipe length (L), volumetric flow rate (Q), density (), viscosity (), and pressure drop (P). The resulting diameter is calculated to be certain value D. It is then found that scaling or corrosion is likely to occur, and that may rise tenfold, giving the friction factor about twice as large as originally thought.

In order to maintain the same values for Q and P, by what ratio should the design diameter is increased over its original value to allow for scaling and corrosion?

Solution ------------------------------------------------------------------------------------------

Applying the energy over the horizontal pipe gives

= + ef P = (P2 P1) = ef 1P

2P

The frictional loss is given by

ef = 4 f = 2 f V2DL

2

2VDL

Since Q = we can obtain the expression for velocity in terms of flow rate4

2VD

V2 = 42

216DQ

We now solve for the pressure drop in terms of friction factor, flow rate, and diameter

P = ef

P = 2 f V2 = 2 fDL

DL

42

216DQ

Solving for the diameter in terms of the friction factor f gives

4 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 160

7-20

D5 = fP

LQ2

232

To maintain the same values for Q and P for this pipe we must have to be a P

LQ2

232

constant when f changes

D5 = Constantf

D = Cf0.2

In this expression C is another constant. We can then solve for the new required diameter

= old

new

DD

2.0

old

new

ff

= 20.2 = 1.15

old

new

DD

Example 7.4-2. 5----------------------------------------------------------------------------------

The simplest patient infusion system is that of gravity flow from an intraveous (IV) bag. A 500 ml IV bag containing an aqueous solution is connected to a vein in the forearm of a patient. Venous pressure in the forearm is 0 mm hg (gage pressure). The IV bag is placed on a stand such that the entrance to the tube leaving the IV bag is exactly one meter above the vein into which the IV fluid enters. The length of the IV bag is 30 cm. The IV is fed through an 18 gauge tube (internal diameter = 0.953 mm) and the total length of the tube is 2 meters. Calculate the flow rate of the IV fluid. Estimate the time needed to empty the bag.

Solution ------------------------------------------------------------------------------------------

Calculate the flow rate of the IV fluid.

Applying the mechanical energy balance between the surface of the fluid (1) and the entrance of the vein (2) we have

+ gZ1 + + Wp = + gZ2 + + hfriction 1P

2

211V

2P

2

222V

We have P1 = P2 = 0 (gage), Wp = 0, Z2 = 0, Z1 = 1.3 m, V1 << V2, and hfriction = 4f . LD

22

2V

Assume laminar flow, then 2 = 2. The mechanical energy equation becomes

5 Fournier R., Basic Transport Phenomena in Biomedical Engineering, Taylor & Francis, 2007, p. 148

7-21

gZ1 = V22 + 2f V2

2 (E-1)LD

For laminar flow, f = = , equation (E-1) becomes16Re 2

16DV

V22 + V2 gZ1 = 02

32 LD

This is a quadratic equation in terms of V2. Therefore

V2 = 1/ 22

12 2

16 16L L gZD D

Assuming the IV fluid has the same properties as water with = 0.001 kg/ms and = 1000 kg/m3.

V2 = = 0.1805 m/s1/ 22

2 2

(16)(0.001)(2) (16)(0.001)(2) (9.81)(1.3)(1000)(0.000953) (1000)(0.000953)

The flowrate of the IV fluid is then

Q = V2D2/4 = (18.05 cm/s)(0.0953 cm)2/4 = 0.1288 cm3/s = 7.72 ml/min

We need to check the assumption of laminar flow.

Re = = = 172 < 2000, laminar flow2DV

(1000)(0.000953)(0.1805)0.001

Estimate the time needed to empty the bag.

When the bag is empty, Z1 = 1.0 m, the exit velocity V2 is then

V2 = = 0.1389 m/s1/ 22

2 2

(16)(0.001)(2) (16)(0.001)(2) (9.81)(1)(1000)(0.000953) (1000)(0.000953)

We will use an average velocity to estimate the time needed to empty the bag

V2,ave = 0.5(0.1805 + 0.1389) = 0.1597 m/s

tempty = = = 4,390 s = 73 min22

4 bagVV D 2

(4)(500)(15.97) (0.0953)

7-22

Example 7.4-3. 6----------------------------------------------------------------------------------

The cardiac output in a human is normally about 6 liters/min. Blood enters the right side of the heart at a pressure of about 0 mm Hg gage and flows via the pulmonary arteries to the lungs at a mean pressure of 11 mm Hg gage. Blood returns to the left side of the heart through the pulmonary veins at a mean pressure of 8 mm Hg gage. The blood is then ejected from the heart through the aorta at a mean pressure of 90 mm Hg gage. Estimate the power delivered by the heart in W. Density of blood is 1.056 g/cm3.

Solution ------------------------------------------------------------------------------------------

Applying the mechanical energy balance between the inlet (1) and the outlet (2) we have

+ gZ1 + + Wp = + gZ2 + + hfriction1P

2

211V

2P

2

222V

Assume negligible potential and kinetic energy effects, = 1, and hfriction = 0.

Wp = 2 1P P

For both sides of the heart we have

Wp = = 11.7 J/kg(90 8) (11 0)1056

101,325760

Total work perform = Wp = (61.056/60)(11.7) = 1.24 Wm

6 Fournier R., Basic Transport Phenomena in Biomedical Engineering, Taylor & Francis, 2007, p. 153

8-1

Chapter 8 Pumps and Compressors

8.1 Introduction

A fluid may be transported from one location to another by gravity from elevated tanks or by passing it through a machine such as a pump or compressor that imparts energy to it. Pump is generally refers to a device for moving a fluid, while compressor denotes a device for moving a gas. However the terms air pump and vacuum pump are also used for devices that compress and move a gas.

There are many types of pumps and compressors. However they can be broadly classified into two categories: positive-displacement and centrifugal. Positive-displacement pumps apply pressure directly to the liquid by a reciprocating piston, or by rotating lobes (shown in Figure 8.1-11) that form chambers alternately filled by and emptied of the liquid. The volumetric flow rate is determined by the displacement per cycle of the moving member, times the cycle rate (e.g. rpm). In general, positive displacement pumps generate high pressure but have limited flow capacity. A positive displacement pump is usually equipped with a relief valve or a recycle line to avoid damage if a valve on the discharged line is inadvertently closed. Centrifugal pumps generate high rotational velocities. The resulting kinetic energy of the liquid is then converted to pressure energy that enables it to overcome the friction drag in the piping system. The discharge line in a centrifugal pump can be closed and the liquid will recirculate within the pump without causing damage unless excessive heating of the fluid occurs.

Rotating lobes

Inlet

Outlet

Fluid

Fluid

Fluid

Figure 8.1-1 Depiction of a rotary displacement pump.

Compressors may also be either positive displacement or centrifugal. Just as for pumps, the positive-displacement units are suitable for relatively high pressures and low flow rates, while the centrifugal units are designed for higher flow rates but lower pressures. A compressor may be considered as a pump for a compressible fluid at high enough pressure that the compressibility properties of the fluid must be considered. This normally occurs when the pressure change by more than 40 %. Compressors discharge pressures from 2 atm to several thousand atmospheres. Blowers are high-speed rotary devices that develop a maximum pressure of about 2 atm. Fans are low-speed machine that generate very low pressure, on the order of 0.04 atm.

1 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 176

8-2

The CD supplied with the text “Elementary Principles of Chemical Processes” by Felder and Rousseau provides pictures for many types of pumps and compressors. You should run the program 1_Main.exe within the folder encyclop to see the photographs with accompanying descriptions of equipment. Figure 8.1-2 shows the menu for pumps and compressors.

Figure 8.1-2 Visual Equipment Encyclopedia2.

2 CD Product by the Multimedia Education Laboratory, University of Michigan, Susan Montgomery (Director), 1997.

8-3

8.2 Centrifugal Pumps

Inlet

ExitHousing

Impeller

Vanes

Volute chamber

Figure 8.2-1 A simple centrifugal pump

Centrifugal pumps are widely used for transferring liquids of all types. These pumps are available in capacities from 2 GPM to 105 GPM, and for discharge pressures from a few feet to approximately 7000 psi3.

A centrifugal pump typically consists of an impeller rotating within a stationary housing. The impeller usually consists of two flat disks, separated by a number of curved vanes (blades), mounted on a shaft that project outside the housing. Power from an outside source is applied to the shaft, rotating the impeller within the stationary housing. The revolving vanes of the impeller produce a reduction in pressure at the inlet hole or eye of the impeller. Fluid then flows to the eye. From the eye the fluid is flung outwards by centrifugal force into the periphery of the housing and from there to the volute chamber and finally to the pump exit. The overall pressure increase across the pump at low flow rates is approximately4

P = V22 = 2D2N2 (8.2-1)where

= density of the fluidV2 = tangential impeller velocityD = diameter of impellerN = rotational speed of the impeller

so that the dimensionless group P/(D2N2) should be roughly constant at low flow rates. Let c1D be the gap between the disks of the impeller and c2V2 be the radial velocity outwards; then the volumetric flow rate is given by

Q = (D)(c1D)(c2V2) = (D)(c1D)(c2DN) = c1c22D3N (8.2-2)

3 Perry’s Chemical Engineers’ Handbook, Sixth Edition (Section 6-7)4 Wilkes, J. O., Fluid Mechanics for Chemical Engineers, Prentice Hall (1999), pg. 176-181.

8-4

where c1 and c2 are constants.

This model proposes that the flow rate is proportional to the tangential velocity of the impeller. Thus the dimensionless group Q/(ND3) is approximately constant at low flow rates. Even though the assumptions made for the two dimensionless groups ∆P/(D2N2) and Q/(ND3) are for low flow rates they are usually adequate to characterize all centrifugal pumps at any flow rate. A plot of ∆P/(D2N2) versus Q/(ND3) will have the general shape given by the curve in Figure. 8.2-2.

P D N/ 2 2

Q/ND3

Figure 8.2-2 General characteristic curve for centrifugal pumps.

The pump impeller is driven by a motor with an efficiency defined as

motor = /outmotorW ,

inmotorW ,

where = , is the angular velocity and is the torque applied to the outmotorW ,

impeller.

The efficiency of a pump depends on many factors: the pump and impeller design, the size and speed of the impeller, and the operating conditions. The efficiency must be determined from experiment data provided by the vendor. Pump efficiency is defined as

pump = /outpumpW ,

inpumpW ,

where = , andinpumpW ,

outmotorW ,

= P/, is the mass flow rate and P is the pressure rise across outpumpW , m m

the pump.

The overall pump efficiency is defined as

overall = /outpumpW ,

inmotorW ,

If the motor is 100% efficient then the brake horsepower HP ( = ) of the inpumpW ,

outmotorW ,

driving motor for the pump is given as

8-5

HP = = = inpumpW ,

pump

outpumpW

,

m

pump

P

The work put into the fluid by the pump goes to increasing the fluid pressure or the equivalent pump head Hp defined by

Hp = gP

The brake horsepower in terms of pump head is then

HP = = m

pump

pgH

pump

pgHm

The power delivered from the motor to the pump is also the product of the angular velocity of the shaft () and the torque on the shaft () driving the impeller given earlier

HP = = outmotorW ,

The pump head is then obtained by equating the above two expressions for the horsepower

= Hp = (8.2-3)pump

pgHm

gmpump

V2

r2r1 V1

Figure 8.2-3 Cross section of pump impeller.

The torque on the shaft can be obtained from an angular momentum balance. The impeller rotates with angular velocity , and its rotation causes the fluid to be thrown radially outwards between the vanes by centrifugal action. The fluid enters at a radial position r1 and leaves at a radial position r2; the corresponding tangential velocity V1 and V2 denote the inlet and exit liquid velocities relative to a stationary observer. Making an angular momentum balance on the impeller yields

8-6

= ( r1V1)in ( r2V2)out + dtId )( m m

For steady-state system, there is no accumulation of angular momentum, and the torque required to rotate the impeller is

= (r22 r1

2)m

Neglecting the angular momentum of the fluid entering the eye of the impeller and replacing r2 with the radius of the impeller Ri we obtain

= r22 = Ri

2m m

The expression for the pump head is then

Hp = = (8.2-4)gm

pump

g

Ripump22

The pressure rise across the pump can be evaluated

P = g Hp = pump2Ri2 (8.2-5)

You should note that equation (8.2-5) is equation (8.2-1) with the pumped efficiency included since

P|(8.2-1) = V22 = 2D2N2

P|(8.2-1) = 2(4Ri2) = 2Ri

22

2

Example 8.2-15 ----------------------------------------------------------------------------------

For a certain type of centrifugal pump, the head increase H (ft) is related to the volumetric flow rate Q (gpm) by the equation

H = a bQ2

In this equation, a and b are constants determined by tests on the pump. Two identical such pumps are now connected together, as shown in Figure 8.2-4, either in series or in parallel.

(a) Derive expressions for the head increase Hs and Hp for these two arrangements in terms of the total flow rate Q through them.

(b) Display the results graphically for a = 25 ft and b = 0.0025 ft/(gpm)2

5 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 180

8-7

Single pump Two in series Two in parallelQ Q

QH

Hs

Hp

Figure 8.2-4 Centrifugal pump arrangement.Solution ------------------------------------------------------------------------------------------

(a) Derive expressions for the head increase Hs and Hp for these two arrangements in terms of the total flow rate Q through them.

When the pumps are in series, the head increases are additive, and the total increase is twice that for the single pump:

Hs = 2(a bQ2)

When the pumps are in parallel, the flow through each pump is only Q/2. The overall head increase is the same as that for either pump singly:

Hp = a b2

2

Q

(b) Display the results graphically for a = 25 ft and b = 0.0025 ft/(gpm)2

Table 8.2-1 lists the Matlab program used to plot the performance curves for three different pump arrangements. The graphical results are displayed in Figure E8.2-1. From the performance curves it can be seen that the series arrangement can be used to increase pump head while the parallel arrangement can be used to increase flow rate.

______ Table 8.2-1 Matlab program plot pump performance curves __________

% Example 8.2-1: Pump performance curve%a=25;b=0.0025;Q=0:5:100;Qp=0:5:200;DH=a-b*Q.^2;DHs=2*DH;DHp=a-b*(Qp/2).^2;plot(Q,DH,Q,DHs,':',Qp,DHp,'--')xlabel('Q (gpm)');ylabel('Pump head (ft)')legend('Single','Series','Parallel')

8-8

0 20 40 60 80 100 120 140 160 180 2000

5

10

15

20

25

30

35

40

45

50

Q (gpm)

Pum

p he

ad (

ft)

SingleSeriesParallel

Figure E8.2-1 Performance curves for different pump arrangements.------------------------------------------------------------------------------------------

Pump Head

Pump head for a piping system with the necessary valves, fitting, and pipe size can be obtained by applying the energy equation from the up stream end (1) to the downstream end (2).

+ gz1 + + wp = + gz2 + + ef 1P

2

211V

2P

2

222V

Neglecting the change in kinetic energy and dividing the above equation by g, we have

+ z1 + = + z2 + g

P

1

gwp

gP

2

ge f

The pump head required is Hp = and is given asg

wp

Hp = + (z2 z1) + g

PP

12 ge f

The friction loss is the sum of all of the losses from point (1) to point (2), ef = . ii

fKV

2

2

Therefore

Hp = + (z2 z1) + g

PP

12 2

28gQ

i i

f

DK

4

8-9

Chapter 8 Net Positive Suction Head (NPSH)

The pressure at the pump inlet (suction) must be sufficiently high so that the minimum pressure anywhere in the pump will be above the vapor pressure. If this is not the case, the pressure at some point within the pump can drop below the vapor pressure of the liquid and cavitation will occur. In cavitation, vapor bubbles will form at the low pressure point and will move to another region in the fluid where the pressure is greater than the vapor pressure, at which point they will collapse. The formation and destruction of vapor bubbles occurs very rapidly and can cause erosion and serious damage to the impeller or pump. The minimum required net positive suction head is the minimum suction pressure in excess of the vapor pressure to prevent cavitation. A pump will not cavitate if the actual pressure head available at the pump suction (NPSHA) is greater than the net positive suction head plus the vapor pressure head.

NPSHA = > NPSH + g

Ps

gPv

In this expression, Ps is the actual pressure and Pv the vapor pressure of the fluid.

hmax

S

1

Figure 8.2-5 The maximum suction lift.

Consider Figure 8.2-5 where the pressure at the entrance to the upstream suction line is P1. The actual pressure at the pump suction, Ps, can be determined from the mechanical energy equation between points (1) and (S)

+ z1 + = + zs + + (8.2-6)g

P

1

gV2

21

gPs

gVs

2

2

ge f

In this equation, ef is the total friction loss in the suction line from the upstream entrance (point 1) to the pump inlet (point S) including all pipe and fittings.

The maximum suction lift is the distance hmax = zs z1 where = NPSH + . Solving g

Ps

gPv

equation (8.2-6) for hmax, we obtain

8-10

hmax = zs z1 = + gPP s

1

gVV s

2

221

ge f

In terms of NPSH, we have

hmax = zs z1 = NPSH + gPP v

1

gVV s

2

221

ge f

Example 8.2-26. ----------------------------------------------------------------------------------A centrifugal pump with the characteristics shown in Figure 8.2-6 is to be used to pump an organic liquid from a reboiler to a storage tank, through a 2 in. Schedule 40 line, at a rate of 200 gpm. The pressure in the reboiler is 2.5 atm, and the liquid has a vapor pressure of 230 mmHg, an SG of 0.85, and a viscosity of 0.5 cP at the working temperature. If the suction line upstream of the pump is also 2 in. Schedule 40 and has two elbows and one globe valve, what is the maximum height above the reboiler that the pump can be located without cavitating? The suction line is 30 ft long and the pump has a 73/4 in. impeller. Note: 1 cP = 0.000672 lb/fts.

Reboiler

Storage Tank

1

S

Figure 8.2-6 Flow from a reservoir through a piping system.

6 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 249

8-11

Curve forNPSH

Pumpcharacteristics

Figure 8.2-7 Typical pump characteristic curves.

Solution -------------------------------------------------------------------------------------

Figure 8.2-7 shows a typical set of pump characteristic curves as determined by the pump manufacturer. At the left top corner, “Size 2 3” means that the pump has a 2 in. discharge and a 3 in. suction port.”R&C” and “17/8” Pedestal” are the manufacturer’s designations, and 3500 rpm is the speed of the impeller. The pump should operate at the conditions on the left side of the efficiency contours (on the left side of the “Maximum Normal Capacity” curve) for stable operation. For the pump given in the problem, 200 gpm and 73/4 in. impeller, the NPSH is about 11 ft.

The maximum suction lift is given by

hmax = zs z1 = NPSH + gPP v

1

gVV s

2

221

ge f

The total friction loss in the suction line is

= (Kentrance + Kvalve+ 2Kelbow+ Kexpansion ) + 4fge f

gV2

2

DL

gV2

2

= (Kentrance + Kvalve+ 2Kelbow+ Kexpansion + 4f ) ge f

DL

gV2

2

Kentrance is the loss coefficient for the fluid entering the pipe from the reboiler and Kexpansion is the loss coefficient for the fluid entering the sudden expansion from the 2 in. pipe to the 3 in. suction port. From Table 7.2-1,

8-12

Kentrance = 0.4 0.4, Kvalve = 7.5 (gate valve, wide open), Kelbow = 0.7 (standard 90o 2

1

21

AA

elbow), and Kexpansion = = = 2

2

11

AA

2

2

2

321

8125

For 2 in. Schedule 40 pipe, D = 2.067 in. The velocity in the reboiler (V1) can be neglected, and the velocity in the pipe (Vs) is

V = Vs = = =19.12 ft2RQ

)48.7)(60)(0335.1(144200

2

NRe = = = 5.2105

VD000672.05.0

)12/067.2)(12.19)(4.6285.0(

= = 8.7110-4D

067.21200015.0

The friction factor can be determined from the following equation

f = { 1.737 ln[0.269 ln (0.269 + )]}-2 = 0.0049D

Re

185.2N D

Re

14N

= (Kentrance + Kvalve+ 2Kelbow+ Kexpansion + 4f )ge f

DL

gV2

2

The friction loss is then calculated with values of loss coefficients given in Table 7.2-1.

= (0.4 + 7.5 + 20.7 + + 40.0049 ) = 73.92 ftge f

8125

067.21230

2.32212.19 2

hmax = zs z1 = NPSH + gPP v

1

gVV s

2

221

ge f

hmax = 11 + 73.92 = 2.9 ft4.6285.0

)144)(7.14)(760/2305.2(

2.322

12.190 2

The pump must be located 3 ft below the liquid level in the reboiler.

8-13

Chapter 8

8.3 Compressor

The work supplied by the compressor to a compressible fluid can be determined from the energy equation. We start by reviewing the energy equation and its differential form given in section 5.3.

h + gz + V2 = q w (5.3-5)21

dh + gdz + VdV = q w (5.3-6)

We can use thermodynamics relations to convert the enthalpy term into a form that involves temperature, pressure, and density changes across the system.

du = Tds Pd(1/)

dh = du + d(P/) = Tds Pd(1/) + d(P/)

dh = Tds Pd(1/) + Pd(1/) + = Tds + (5.3-7)

dP

dP

For an idealized reversible process in which no energy dissipation occurs, the entropy change arises from heat transfer across the system boundaries

Tds = q

In any real system, the process is irreversible and there is dissipation of energy, therefore

Tds = q + ef = du + Pd(1/) (5.3-8)

dh = q + ef +

dP

The work of compression plus the energy dissipated (ef) due to friction and any heat transferred into the gas during compression all go to increasing the enthalpy of the gas that may be determined from integrating the above equation

h = = + + dh q fe 2

1

P

P

dP

h = q + ef + 2

1

P

P

dP

8-14

The compression work is given by

w = 2

1

P

P

dP

Assuming ideal gas properties, we can substitute the density as a function temperature and

pressure into the compression work to obtain

RTMP

w = = (8.3-1)2

1

P

P MPRTdP

MR

2

1

P

P PTdP

We cannot evaluate the compression work unless the temperature is a constant (isothermal condition) or a relation between temperature and pressure is known.

Isothermal Compression

For isothermal compression, equation (8.3-1) can be evaluated to give

w = = ln (8.3-2)MR

2

1

P

P PTdP

MRT

1

2

PP

The ratio is called the compression ratio (r), which is the ratio of the outlet to inlet 1

2

PP

pressures.

Isentropic Compression

For isentropic compression, the system is adiabatic and reversible. The change in internal energy is equal to the work supplied to the system

dU = dW CvdT = PdV

We now use ideal gas law to obtain an expression for PdV in terms of P, dP, T, and dT

PV = RT PdV + VdP = RdT PdV = VdP RdT

Therefore CvdT = VdP RdT = dP RdTP

RT

We can separate the variables to obtain

(Cv + R)dT = dP Cp = (Cp Cv)PRT

TdT

PdP

8-15

Since = = , where k = p

vp

CCC

vp

vp

CCCC/

1/ k

k 1

v

p

CC

= TdT

kk 1

PdP

Integrating the equation we obtain

= 1T

T kk

PP

1

1

Substituting the expression T = T1 into equation (8.3-1) yieldsk

k

PP

1

1

w = = MR

2

1

P

P PTdP

MRT1

kk

P

1

1

1

2

1

1

P

P

kk

PdPP

w = = MRT1

kk

P

1

1

1

2

1

11P

Pk

k

dPP)1(

1

kMkRT k

k

P

1

1

1

k

kk

kPP

1

1

1

2

w = (8.3-3))1(

1

kMkRT

1

1

1

2k

k

PP

The isothermal work required to compress a given gas to a given compression ratio is always less than the adiabatic work. At isothermal condition the gas is cooler than at adiabatic condition, therefore less volume of gas is being compressed which results in a lower amount of work. In reality, most compressor are somewhere between isothermal and adiabatic conditions. This can be accounted for in calculating the compression work by replacing the specific heat ratio k by an empirical polytropic constant in using equation (8.3-3). The constant must be determined from experiments and is between 1 and k (1 k).

Multistage Compressors

Multistage compressors can be arranged in series to increase the overall compression ratio. Since it is cheaper to cool a gas than to compress a larger volume, interstage coolers are usually used to cool the gas before it enters the next stage of the compression process. We now want to determine the work for a two stages compressor with interstage cooling. The inlet pressure P1 and the outlet pressure P3 are fixed. The gas enters stage 1 at (P1, T1), leaves stage 1 at (P2, T2) and is then cooled to T1. It then enters stage 2 at (P2, T1), and leaves at P3. The total isentropic work is given by

8-16

w = + )1(

1

kMkRT

1

1

1

2k

k

PP

)1(1

kMkRT

1

1

2

3k

k

PP

In this process, we want to determine the intermediate pressure P2 that will minimize the isentropic work required. We can take the derivative of w with respect to P2 and set it equal to zero to solve for P2.

= = 02dP

dwMRT1

kk

P

1

1

1

11

2

k

kP

MRT1 11

2

kk

P kk

P1

3

= = 11

2

k

kP 11

2

kk

P kk

P1

1

k

kP

1

3

kk

P1

22

kk

PP1

31

Finally we have

P22 = P1P3

or = = r = 1

2

PP

2

3

PP

2/1

1

3

PP

Thus, the total work is a minimum if the compression ratio of each stage is the same. This result can be generalized to any number (n) of stages with interstage cooling to the initial temperature as follows:

r = = = = = 1

2

PP

2

3

PP

n

n

PP 1

n

n

PP

/1

1

1

Compressor Performance

The performance of centrifugal compressors is somewhat similar to that of centrifugal pumps. However, compressor behavior is more complex than that for pumps because the fluid is compressible. Figure 8.3-1 shows the performance curve for a centrifugal compressor at two different rotation speeds. The y-axis represents the ratio of the outlet pressure to the inlet pressure unlike the pump curves, which have the pump head or the pressure difference between the outlet and the inlet. The compressor speed is often varied continuously to control the flow rate while the outlet valve is usually throttled to control flow rate for most pumps. It is economical to avoid throttling the outlet due to the higher power required in a compressor.

There is a maximum in the performance curve of a compressor. Starting a low flow rates, the pressure ratio increases with increasing flow rates, reaches a maximum, and then decreasing with increasing flow rates. The locus of the maxima in the performance curves is called the surge line. For safety reasons, compressors are operated far to the right of the surge line. The surge line is important for the following reason. Imagine that we start at a flow rate to the right of the surge line and then the flow rate is lowered continuously. The outlet pressure increases with decreasing flow rate until the surge line is reached. The outlet pressure is then lowered so that its value is less than the pressure downstream. This will create a backflow

8-17

that can cause the compressor to vibrate or surge. Severe surging can cause great damage to compressors since it can detach the compressors from the supports, causing them literally to fly apart. The surge line is considered a limiting operating condition, below which operation is prohibited.

P/ P

out

in

Flow rate

3500 rpm

2500 rpm

surge line

Figure 8.3-1 Centrifugal compressor performance curve.

Example 8.3-17 ----------------------------------------------------------------------------------Nitrogen is flowing under turbulent conditions at a constant mass flow rate through a long, straight, horizontal pipe. The pipe has a constant inside diameter of 2.067 in. At an upstream point (point 1), the temperature of the nitrogen is 70oF, the pressure is 15 psia, and the average velocity of the gas is 50 ft/s. At a given downstream point (point 2), the temperature of the gas is 140oF and the pressure is 50 psia. An amount of 10 Btu is transferred to each pound of the flowing gas between points 1 and 2. Nitrogen is an ideal gas with a mean heat capacity Cp = 7.0 Btu/lbmoloF. Estimate the total amount of energy (ftlbf/lb) supplied by the compressor located between points 1 and 2.Solution ------------------------------------------------------------------------------------------

From the energy equation (5.3-5)

h + gz + V2 = q w (5.3-5)21

We can solve for the work supplied by the compressor

w = h + gz + V2 q = h2 h1 + (V22 V1

2) q21

21

The mass balance gives

= V1A1 = V2A2 V2 = V1m2

1

7 Peters and Timmerhaus, Plant Design and Economics for Chemical Engineers, Fourth Edition, McGraw Hill

8-18

For ideal gas we have =

RTMP

2

1

2

1

PP

1

2

TT

= = 0.33962

1

5015

46070460140

Therefore V2 = V1 = (50)(0.3396) = 16.98 ft/s2

1

Since 1 Btu = 778 ftlbf, the change in enthalpy per pound of flowing gas is given by

h2 h1 = (T2 T1) = = 13,615 ftlbf/lbMC p

28)70140)(7787(

The energy supplied by the compressor is then

w = h2 h1 + (V22 V1

2) q21

w = 13,615 + 10778 = 5,800 ftlbf/lb21

2.32)5098.16( 22

-------------------------------------------------------------------------------------------------------

Example 8.3-28 ----------------------------------------------------------------------------------Calculate the horsepower for compressing 5,000 lbs/hr of ethylene from 100 psia, 40oF to 200 psia. The adiabatic efficiency of the compressor is 75%. Include your calculations and Mollier chart.Solution ------------------------------------------------------------------------------------------

The Mollier chart (Pressure-Enthalpy diagram) for ethylene can be obtained from the text by Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 511.

T=

-40

Fo

s =1.

64Bt

u/lb

Fo

Pres

sure

,psi

a

Enthalpy, Btu/lb1044 1064

200

100

8 Dr. Pang's Notes

8-19

Initial state

Final stage

Figure E1 Pressure-enthalpy diagram for ethylene9

The enthalpy of ethylene at 100 psia and 40oF can be located from the diagram to give

h1 = 1044 Btu/lb

For isentropic compression, s = 0, we follow the curve where s is a constant until it reaches the line where P = 200 psia to locate a value for the enthalpy

h2 = 1064 Btu/lb

Therefore (h2 h1)isentropic = 1064 1044 = 20 Btu/lb.

The actual increase in enthalpy of the gas is obtained from the adiabatic efficiency

(h2 h1)actual = (h2 h1)isentropic = 26.7 Btu/lb75.01

The horsepower supplied by the compressor to compress 5000 lb/hr of ethylene is

Power = 5000 26.7 = 133,500 Btu/hr

Power = (133,500 Btu/hr)(3.930110-4 hp/(Btu/hr)) = 52.5 hp

-------------------------------------------------------------------------------------------------------

9 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 511

8-20

Example 8.3-3 ----------------------------------------------------------------------------------Hydrogen at a temperature of 20oC and an absolute pressure of 1380 kPa enters a compressor where the absolute pressure is increased to 4140 kPa. If the mechanical efficiency of the compressor is 55 percent on the basis of an isothermal and reversible operation, calculate the pounds of hydrogen that can be handled per minutes when the power supplied to the pump is 224 kW. You can use ideal gas law and neglect kinetic energy effects.Solution ------------------------------------------------------------------------------------------

For isothermal compression, the work required to compress the gas is

w = = lnMR

2

1

P

P PTdP

MRT

1

2

PP

Using R = 1545 (lbf/ft2)ft3/lbmoloR yields

w = ln = 444,040 ftlbf/lbm016.2

)8.1293)(1545( 13804140

The actual power received by the gas in English unit is given by

Power = = 5,445,000 ftlbf/min746.0

)60)(550)(55.0)(224(

Note: We have use the conversion 1 hp = 0.746 kW = 550 ftlbf

Since Power = w, the pounds of hydrogen that can be handled per minute ism

= = = 12.2 lb/minmw

Power040,444000,445,5

9-1

Chapter 9 Flow Measurement and Control Valves

9.1 Introduction

We will study some of the more common devices for measuring flow rate in conduits, including the pitot tube, venturi meter, nozzle, orifice meter, and rotameter. These devices can be analyzed using the mass, energy, and momentum balances. We also study briefly control valves since they are frequently employed in conjunction with the measurement of flow rate to provide a means of controlling flow. More details treatment of control valves will be presented in the process control course CHE426.

9.2 The Pitot Tube

r

Stagnation point

v1

MovableAir

Liquid

h

(1) (2)

Figure 9.2-1 Pitot static tube.

The pitot-static tube is a device for measuring v(r), the local velocity at a given position in the conduit, as shown in Figure 9.2-1. The pitot tube offers negligible resistance loss in the conduit. Application of Bernoulli’s equation between points (1) and (2) gives

+ v12 = + v2

2 (9.2-1)

1P21

2P

21

In this equation, point 1 is in the stream just upstream of the probe and point 2 is just inside the open end of the probe (the stagnation point). Since the velocity at the stagnation point is zero, we can solve for the free stream velocity v1.

v1 = 2/1

12 )(2

PP

From hydrostatic, the pressure difference between point 2 and point 1 is

P2 P1 = gh

9-2

The velocity at the particular transverse location where the pitot tube is placed is given by

v1 = (2gh)1/2

Example 9.2-11 ----------------------------------------------------------------------------------The speed of a boat is measure by a Pitot tube. When traveling in seawater ( = 64 lb/ft3), the tube measures a pressure of 2.5 lbf/in2 due to the motion. What is the speed of the boat and the rise in level h of the seawater inside the tube.

h

d

(1)

(2)

v1

Solution ------------------------------------------------------------------------------------------

We choose a coordinate fixed with the boat, then the water is moving toward the boat at a velocity v1. Application of Bernoulli’s equation between points (1) and (2) gives

+ v12 = + g(h + d)

1P

21

2P

Since P1 = P2 + gd, we have

+ gd + v12 = + gh +gd

2P

21

2P

Solving for the velocity yields

v1 = (2gh)1/2

The pressure due to the motion causes a rise in the seawater within the tube: P = gh. Therefore

h = = = 5.625 ftgP

)2.32)(64()2.32)(144)(5.2(

The speed of the boat is

v1 = (232.25.625)1/2 = 19.0 ft/s

1 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, pg. 94

9-3

9.3 The Venturi Meter and the Flow Nozzle

1 2 1 2

Figure 9.3-1 The venturi meter and the flow nozzle.

The pitot tube provides the local or point velocity at a given location in the conduit. We need to integrate the velocity at various points over the cross sectional area of the conduit to obtain the volumetric flow rate. Other devices can provide the flow rate from a single measurement. These devices are sometimes referred to as obstruction meters, since they rely on the pressure drop across the obstruction introduced by the meters. Two such devices are the venturi meter and the flow nozzle shown in Figure 9.3-1. The flow rate can be obtained from the pressure drop between point 1 upstream of the constriction and point 2 at the position of the minimum area. From the conservation of mass, we have

1V1A1 = 2V2A2

For constant density: V1A1 = V2A2. Applying Bernoulli’s equation between points 1 and 2 yields

+ V12 = + V2

2 + ef1P

21

2P

21

In these equations, we have assumed uniform velocity over the cross-section area (plug flow). Let D and d be the conduit diameter and the minimum diameter at the throat of the venturi or nozzle respectively, we have

V1 = V2 = 2V2, where = 2

2

Dd

Dd

Neglecting friction loss and substituting V1 = 2V2 into Bernoulli’s equation yields

+ 4V22 = + V2

2

1P21

2P

21

Solving for the velocity at the throat gives

V2 = = (9.3-1))1()(2

412

PP

)1(2

4 P

9-4

The actual velocity is smaller than the ideal value given by equation (9.3-1) due to friction loss and assumption of plug flow. Therefore the actual throat velocity is usually calculated from

V2,actual = Cd (9.3-2))1(

24

P

In this equation, Cd is the discharge coefficient with a value of less than 1 (Cd < 1) that must be determined from experiment. For a given flow meter, the discharge coefficient is a function of Reynolds number in the conduit. For high Reynolds number (NRe > 105), the discharge coefficient is very close to 1 for both the venturi and the flow nozzle. You should note the definition used for the Reynolds number in reading the chart to obtain the discharge coefficient since the Reynolds number could be based on the pipe diameter D or the throat diameter d.

The discharge coefficient for the venturi, as well as for the nozzle and orifice, can be described as a function of NRe and by the general equation

Cd = C + (9.3-3)nNb

Re

Table2 9.3-1a lists the parameters C, b, and n for a venturi meter, a nozzle, and an orifice meter. Table 9.3-1b lists the applicable range of Reynolds number and the approximate accuracy for these meters.

Table 9.3-1a Values for Discharge Coefficient Parameters in Equation (9.3-3)Device C b n

Venturi (Machined inlet)Nozzle (ASME long radius)Orifice (corner taps)

0.9950.9975

0.5959 + 0.03122.1 0.1848

0 6.530.5

91.712.5

00.50.75

Table 9.3-1b Applicable Range and Accuracy of Equation (9.3-3)Device D, in (mm) NRe,D % error

Venturi (Machined inlet)Nozzle (ASME long radius)Orifice (corner taps)

2-10 (50-250)2-16 (50-400)2-36 (50-900)

0.4-0.750.25-0.750.2-0.6

2105-106

104-107

104-107

12.00.6

In Table 9.3-1, D is the nominal pipe diameter, is the ratio of the throat to pipe diameters, and NRe,D is the pipe Reynolds number. The error indicates the percentage accuracy of the discharge coefficient Cd.

2 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 297

9-5

Example 9.3-13 ----------------------------------------------------------------------------------

1 21 2

Figure E9.3-1 Venturi meter

A liquid of density flows through a pipe of cross-sectional area A, and then passes through the venturi meter shown in Figure E9.3-1, whose throat cross-sectional area is a. A manometer containing mercury (m) is connected between an upstream location (point 1) and the throat (point 2), and registers a difference in mercury levels of h. Derive an expression giving the volumetric flow rate Q in terms of A, a, g, h, , m, and Cd.

If the diameter of the pipe and the throat of the venturi are 6 in. and 3 in. respectively, what flow rate (gpm) of iso-pentane ( = 38.75 lb/ft3) would register a h of 20 in. on a mercury (specific gravity = 13.57) manometer? If the discharge coefficient is 0.98, determine the corresponding pressure drop in psi.

Solution ------------------------------------------------------------------------------------------

From the conservation of mass, we have

1V1A = 2V2a

For constant density: V1A = V2a. Neglecting friction loss and applying Bernoulli’s equation between points 1 and 2 yields

+ V12 = + V2

2

1P21

2P

21

Substituting V2 = V1 into Bernoulli’s equation yieldsaA

+ V12 = + V1

2

1P21

2P

21 2

aA

Solving for the velocity inside the pipe gives

3 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, pg. 94

9-6

V1 =

1

)(2

2

221

aA

PP

The volumetric flow rate is then

Q = CdAV1 = CdA = CdA

1

)(2

2

221

aA

PP

1

)(2

2

2

aA

hgm

The pressure drop is given by P = (m )gh

P = (13.5762.4 38.75)(32.2) = 4.34104 lbft/s2

1220

The pressure drop in psi is then

P = = 9.35 psi)144)(2.32(

1034.4 4

The volumetric flow rate of iso-pentane is

Q = CdA = CdA

1

)(2

2

2

aA

hgm

1

)(2

2

221

aA

PP

Q = 0.98 = 2.36 ft3/s = 1057 gpm4 2

126

)116)(57.38(1034.42 4

------------------------------------------------------------------------------------------------------

9-7

Chapter 9

9.4 The Orifice Meter

12

3

Vena contracta area

Figure 9.4-1 Orifice meter.

Figure 9.4-1 shows one of the simplest flow measuring devices for a pipeline: the orifice meter. A plate with a hole of area A2 is inserted into a pipe of cross-sectional area A1 that is greater than the hole area. The major difference between the orifice meter and the venturi and nozzle meters is the fact that the fluid stream leaving the orifice contracts to an area called the vena contractor area, which is smaller than the orifice area. This happens because of the inward radial momentum possessed by the fluid. As it converges into the orifice hole it will continue to flow inward for a distance downstream of the orifice before the fluid starts to expand to fill the pipe. For highly turbulent flow, the vena contracta area is about 60 percent of the hole area.

Applying Bernoulli’s equation between points 1 and 2 yields

+ V12 = V2

2 + ef1

2

P

P

dP 2

121

In these equations, we have assumed uniform velocity over the cross-section area (plug flow). Let D and d be the pipe diameter and the minimum diameter at the vena contracta, we have

V1 = V2 = 2V2, where = 2

2

Dd

Dd

Neglecting friction loss and substituting V1 = 2V2 into Bernoulli’s equation yields

+ 4V22 = V2

21

2

P

P

dP 2

121

Solving for the velocity at the vena contracta gives

V2 = (9.4-1)2/14 )1(1

2/11

2

2

P

P

dP

9-8

The actual velocity is smaller than the ideal value given by equation (9.4-1) due to friction loss and assumption of plug flow. Therefore the actual velocity is usually calculated from

V2,actual = Cd (9.4-2)2/14 )1(1

2/11

2

2

P

P

dP

In this equation, Cd is the discharge coefficient with a value of less than 1 (Cd < 1) that must be determined from experiment. The discharge coefficient is a function of Reynolds number and beta ratio. If Ao is the hole area, the mass flow rate through the orifice is given by

= 2V2,actualAo = (9.4-3)m 2/142

)1(

od AC

2/11

2

2

P

P

dP

Incompressible Flow

For incompressible flow, equation (9.4-3) becomes

= CdAo (9.4-4)m2/1

421

1)(2

PP

Compressible Flow

Assuming ideal gas properties, we can substitute the density as a function temperature and

pressure into the integral to obtain

RTMP

1

2

P

P

dP

= = (9.4-5)1

2

P

P

dP

1

2

P

P MPRTdP

MR

1

2

P

P PTdP

If the flow is adiabatic and reversible, the change in internal energy is equal to the work supplied to the system

dU = dW CvdT = PdV

We now use ideal gas law to obtain an expression for PdV in terms of P, dP, T, and dT

PV = RT PdV + VdP = RdT PdV = VdP RdT

Therefore CvdT = VdP RdT = dP RdTP

RT

We can separate the variables to obtain

(Cv + R)dT = dP Cp = (Cp Cv)PRT

TdT

PdP

9-9

Since = = , where k = p

vp

CCC

vp

vp

CCCC/

1/ k

k 1

v

p

CC

= TdT

kk 1

PdP

Integrating the equation we obtain

= 2T

T kk

PP

1

2

Substituting the expression T = T2 into equation (9.4-5) yieldsk

k

PP

1

2

= = 1

2

P

P

dP M

R

1

2

P

P PTdP

MRT2

kk

P

1

2

1

1

2

1

P

P

kk

PdPP

= = 1

2

P

P

dP M

RT2k

k

P

1

2

1

1

2

11P

Pk

k

dPP)1(

2

kMkRT k

k

P

1

2

1

k

kk

kPP

1

2

1

1

= (9.4-6)1

2

P

P

dP )1(

2

kMkRT

1

1

2

1k

k

PP

The mass flow rate is then

= 2V2,actualAo = m 2/142

)1(

od AC

2/12

)1(2

kMkRT

2/11

2

1 1

k

k

PP

Since , = (9.4-7)

2

22 RT

MP m 2/142

)1(

od AC

2/1

2

2

)1(2

kkP

2/11

2

1 1

k

k

PP

However, the compressible mass flow rate is normally obtained from the equation for incompressible flow with a correction factor Y for the fluid expansion

9-10

= YCdAo (9.4-8)m2/1

421

1)(2

PP

The correction factor Y is found experimentally for radius taps to be

Y = 1 (0.41 + 0.352) (9.4-9)1

21

kPPP

Flange taps2.5 D 8 D

1 in.1 in.

Pipe taps

D D/2

Corner tapsFlow

Radius taps

Figure 9.4-2 Orifice pressure tap locations.

There are various pressure tap locations as shown in Figure 9.4-2. They are radius taps (1 D upstream and D/2 downstream); flange taps (1 in. upstream and downstream); pipe taps (2.5 D upstream and 8 D downstream); and corner taps. Radius taps are the most common while corner taps and flange taps are the most convenient since they can be installed in the flange that holds the orifice plate so that no additional taps through the pipe wall are required.

12

3

Vena contracta area

Controlvolume

Figure 9.4-3 Control volume for the calculation of loss coefficient.

The friction loss in an orifice meter causes a pressure drop P1 P3. Consider a control volume to be the region from a point just upstream of the orifice plate (P1) to a downstream position where the stream has filled the pipe (P3) as shown in Figure 9.4-3, the momentum balance becomes

0 = (V1 V3) + P1Ao + P2(A1 Ao) P3A1m

9-11

For incompressible flow V1 = V3, we have

P1Ao + P2(A1 Ao) P3A1 = 0 (9.4-10)

From the equation for the mass flow through the orifice

= V2Ao = CdAo (9.4-4)m2/1

421

1)(2

PP

We can solve for the pressure P2

P2 = P1 2

22V

2

41

dC

Substituting P2 into equation (9.4-10) yields

P1Ao + P1(A1 Ao) (A1 Ao) P3A1 = 02

22V

2

41

dC

A1(P1 P3) (A1 Ao) = 02

22V

2

41

dC

Dividing the equation by A1 and solving for (P1 P3) yields

P1 P3 = = Kf2

24 )1)(1(

dC

2

22V

2

22V

The loss coefficient for the flow through an orifice can be estimated as

Kf = 2

24 )1)(1(

dC

Example 9.4-1 ----------------------------------------------------------------------------------

A cylindrical tank of diameter 1 m has a sharp-edged orifice at the bottom so that the vena contracta area is 63 percent of the hole area If the hole diameter is 2 cm, how long will it take an initial dept of acetone to drain completely from the tank2?

Solution ------------------------------------------------------------------------------------------

Applying Bernoulli’s equation between points 1 and 2 yields

2 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, pg. 95

h

(1)

(2)

9-12

+ V12 + gh = + V2

2

1P21

2P

21

Since P1 = P2 = Patm and V1 0, we have

V2 = (2gh)1/2

Application of the mass balance on the liquid inside the tank gives

(Ath) = 0.63V2A2 = 0.63A2 (2gh)1/2dtd

Separation of variables yields

= 0.63 (2g)1/20

2 2/1hdh

2

2

tDD

tdt

0

Solving for the time we have

t = 2 = 221/22

0

2/1h2

2

DDt

2/1)2(63.01

g

2

02.01

2/1)81.92(63.01

t = 221/2 = 2,534 s2

02.01

2/1)81.92(63.01

------------------------------------------------------------------------------------------------------

Example 9.4-23 ----------------------------------------------------------------------------------An orifice meter was calibrated in air, and later put in service to monitor the flow of helium. The user of the meter was unaware that the calibration had been done in air. What would be the error in the reported flow rate? Solution ------------------------------------------------------------------------------------------

The volumetric flow rate is given by

Q = V2Ao = CdAo

2/1

421

)1()(2

PP

For a given meter

Q = C , where C = CdAo2/1

2/121 )(

PP

2/1

4 )1(2

3 Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998, p. 499

9-13

When the meter is calibrated in air we have

Q = C = Cair(P1 P2)1/22/1

2/121 )(

air

PP

When the meter is used with helium we have

Q = C = (P1 P2)1/2 = Cair (P1 P2)1/22/1

2/121 )(

He

PP

2/1air

C

2/1

He

air

2/1

He

air

If the meter is calibrated in air and we use it for helium without correcting the reading, we will find Q smaller by a factor

= = 2.72/1

He

air

2/1

429

Therefore, the calculated value Q = Cair(P1 P2)1/2 must be increased by a factor of 2.7

------------------------------------------------------------------------------------------------------

Example 9.4-3 ----------------------------------------------------------------------------------A horizontal 2-in ID pipe carries kerosene at 100oF with density of 50.5 lb/ft3 and viscosity of 3.18 lb/fthr. For a mass flow rate of 560 lb/min, specify the diameter of the corner taps orifice so that 15 in. of mercury is measured by the manometer. The discharge coefficient for the orifice can be estimated with Reynolds number based on the hole diameter by

Cd = C + nNb

Re

The parameters in this equation are given by the following table .

Device C b nOrifice (corner taps) 0.5959 + 0.03122.1 0.1848 91.712.5 0.75

1 2

hk

hmPe

9-14

Solution ------------------------------------------------------------------------------------------

The pressure difference across the orifice meter is given as

P1 P2 = hm(m k)g

The mass flow rate is given by equation (9.4-4)

= CdAo = CdA12 (9.4-4)m2/1

421

1)(2

PP2/1

41)(2

gh kmk

This equation can be rearranged to

= CdA1 = CdA1m2/1

4

4

1)(2

gh kmk2/1

4 1/1)(2

gh kmk

The pipe area A1 is = 0.0218 ft2. We have two unknowns and two equations: 2

122

4

equation (9.4-4) and Cd = C + . Assuming Cd = 0.62 yieldsnNb

Re

= (0.62)(0.0218)60

560

2/1

2

1 1

2.32)5.504.846(12155.502

oAA

= 2.79 = = 1.67 Do = = 1.2 in.oA

A1

2

1

oDD

oDD1

67.12

We need to check the assumed value of Cd

NRe = = = = 1.35105

oo DV oD

m4)3600/18.3)(12/2.1(

)60/560)(4(

= = = 0.6 b = 91.712.5 = 25.57381D

Do

67.11

C = 0.5959 + 0.03122.1 0.1848 = 0.6035 Cd = 0.61

Since the calculated Cd (0.61) is closed to the assumed value of 0.62, we accept 1.2 in. as the diameter of the orifice meter.

9-15

Chapter 9

9.5 The Rotameter

Scale

Flow

Mg

P1

P2

z1

z2

v1

v2

Float

Tapered tube Annulararea a

Tubearea A

+

Figure 9.5-1 Section of a rotameter

Figure 9.5-1 shows a section of a flow-measuring device called a rotameter. This meter consists of a gradually tapered glass tube that contains a solid float made of an inert material such as stainless steel, glass, or tantalum. The float is often stabilized by helical grooves incised into it. The grooves induce the float to rotate and that gives the meter the name. Other shapes of floats including spheres may be employed in the smaller rotameters. The venturi meter, flow nozzle, and orifice meter introduce a fixed reduction of flow area that results in pressure drop as a function of flow rate. In contrast, the rotameter depends on the change of an annular area a between the float and the tube, which is a function of the vertical location of the float, to yield an essentially fixed pressure drop at all flow rates. The annular area behaves like an orifice of variable area. We can determine the volumetric flow rate by applying the conservation laws to the control volume shown in Figure 9.5-1.

Mass balance: = V1A = V2a (9.5-1)m

Neglecting friction loss the energy balance (Bernoulli’s equation) yields

+ V12 + gz1 = + V2

2 + gz2 (9.5-2)

1P21

2P

21

Application of the momentum equation with the positive direction point upward gives

(P1 P2)A + V1 V2 [(z2 z1)A M/f]g Mg = 0 (9.5-3)m m

In this equation, M is the mass of the float and f is the density of the float. Therefore the term [(z2 z1)A M/f]g is just the weight of the fluid. We have assumed negligible change in the tube area across the float. Substituting (P1 P2) in terms of kinetic and potential energy from equation (9.5-2) into the momentum equation we have

9-16

g(z2 z1)A + (V22 V1

2)A + V1 V2 g(z2 z1)A Mg = 021 m m

f1

Canceling out the term g(z2 z1)A yields

(V22 V1

2)A + V1 V2 Mg = 021 m m

f1

Substituting V2 in terms of V1 from the mass balance gives

A(V12 V1

2) + AV12 AV1V1 = Mg

21

2

2

aA

aA

f1

AV12 = Mg

21

aA

aA 2212

2

f1

Solving for the velocity V1 yields

V1 = (9.5-4)

2/1

2

1

)/1(2

aAA

Mg f

The volumetric flow rate Q is

Q = AV1 = A (9.5-5)

2/1

2

1

)/1(2

aAA

Mg f

In the cases when a << A and << f, equation (9.5-5) is simplified to

Q = a (9.5-6)2/1

2

AMg

Since a and A depend on z, the flow rate can be determined from the position of the float. In practice, the rotameter is calibrated before use. Note that the reading on the rotameter is taken at the highest and widest point of the float.

9-17

Example 9.5-1 ----------------------------------------------------------------------------------A rotameter is used to measure air flow rate. The reading is 2.3 ft3/s for air at 100oF and 720 mmHg. However, the meter is calibrated with air at 60oF and 760 mmHg. Determine the corrected air flow rate.

Flow

Float

2.0

2.5

1.5

ft /s3

Solution ------------------------------------------------------------------------------------------

We use equation (9.5-6) since air density is much less than float density

Q = a (9.5-6)2/1

2

AMg

For air at the calibrated conditions of 70oF and 760 mmHg

Qcal = a = 2/1

2

AMgc 1/ 2

cal

cal

C

The reading is for air at 100oF and 720 mmHg, the corrected reading Q is given by

Q = = Qcal 2/1calC 1/ 2

1/ 2cal

In this expression, Qcal is the reading with meter calibrated at 60oF and 760 mmHg which is 2.3 ft3/s. The corrected reading is

Q = 2.3 = 2.3 = 2.45 ft3/s2/1

2/1

c

2/1

46060460100

720760

--------------------------------------------------------------------------------------------------------

9-18

9.6 Control Valves

Control valve is used to achieve a certain flow rate through a system. Let us consider a heat exchanger in which a process stream is heated by condensing steam as shown in Figure 9.6-1. A heat exchanger is a device that allows energy transfer between the hot and the cold streams. Heat exchangers can be classified as indirect contact type and direct contact type. Indirect contact type heat exchangers have no mixing between the hot and cold streams, only energy transfer is allowed. A control valve is installed on the hot or steam line.

Steam

Condensate

Processstream

Heatedstream

T(t)i T(t)

Control valve

Figure 9.6-1 An indirect contact heat exchanger.

The heat exchanger is used to heat the process fluid from some inlet temperature Ti(t) up to a certain desired outlet temperature T(t). The energy supplied to the process stream comes from the latent heat of condensation of the steam. In this process we want to maintain the outlet process temperature at its desired value regardless of any variation in other variables such as the process stream flow rate or the inlet process temperature. One way to accomplish this objective is by measuring the outlet temperature T(t), comparing it to the desired value call the set point. Any deviation from the set point can be corrected by adjusting the control valve to change the steam flow to the heat exchanger.

A possible control system for the heat exchanger is shown in Figure 9.6-2. A sensor such as a thermocouple, a thermistor, or any resistance temperature device can measure the outlet process stream temperature. This sensor is usually connected to transmitter, which amplifies the output from the sensor and sends it to a controller. The controller compares the signal with the set point and decides the action necessary to maintain the desired temperature. The controller then sends a pneumatic or electrical signal to the final control element, which is the control valve in this case, to adjust the steam flow rate accordingly. The control valves acts as a variable resistance in the steam line since the flow rate depends on the valve stem or plug position. To regulate flow, the flow capacity of the control valve varies from zero when the valve is closed to a maximum when the valve is fully opened. Part of the job of a control engineer is to size control valves for a given service.

9-19

Steam

Condensate

Processstream

Heatedstream

T(t)i T(t)

Control valve

TT Temperature transmitter

TCTemperature controller

Valve stemSet pointSP

Figure 9.6-2 Heat exchanger control system using control valve.

Incompressible Flow

A control valve is simply an orifice with a variable area of flow. The volumetric flow rate for incompressible fluid through an orifice is given by

Q = CdAo (9.6-1)2/1

4 )1(2

P

In this equation P is the pressure drop across the orifice. For a control valve, the flow area and geometric factors, the density of the reference fluid, and the friction loss coefficient are combined into a single coefficient Cv to provide the following formula for the liquid flow through the valve

Q = Cv = Cv(wghv)1/2 (9.6-2)SGPv

In this equation, Pv is the pressure drop across the valve, SG is the fluid specific gravity, and hv is the head loss across the valve. The reference fluid for the density is water for liquids and air for gases. Although equation (9.6-2) is similar to equation (9.6-1), the flow coefficient Cv is not dimensionless like the discharge coefficient Cd, but has dimensions of [L3][L/M]1/2. The normal engineering units of Cv are gpm/(psi)1/2. If Q is in gpm and is hv in ft, equation (9.6-2) becomes

Q = 0.658Cv(hv)1/2 (9.6-3)

The value of the flow coefficient Cv is different for each valve and also varies with the valve opening or stem position for a given valve. Figure 9.6-3 shows the flow coefficients for Masoneilan’s valves that are provided by the manufacturer from measurements. Different valve plugs are usually available for a given valve, each providing a different flow response or “trim” characteristic when the stem position is changed.

9-20

Nominal Trim size 1/4 3/8 ½ 3/4 1 1.5 2 3 4 6 8 10Orifice Dia. (in.) .250 .375 .500 .750 .812 1.250 1.625 2.625 3.500 5.000 6.250 8.000

Valve size (in.) Reduced Trim Full Capacity Trim

3/41

1½2346810

1.7 3.7 6.4 111.71.71.7

3.7 6.4 11 123.8 6.6 12 13 253.8 6.7 13 19 26 46

14 31 47 11032 49 113 195

53 126 208 400133 224 415 640

233 442 648 1000Figure 9.6-3 Flow coefficient for Masoneilan’s valve Schedule 40.

Example 9.6-12 ----------------------------------------------------------------------------------A process for transferring oil from a storage tank to a separation tower is shown in Figure E9.6-1. The tank is at atmospheric pressure, and the tower works at 12.7 psia. Nominal oil flow is 700 gpm, its specific gravity is 0.94, and its vapor pressure at the flowing temperature of 90oF is 13.85 psia. The pipe is 8-in. Schedule 40 commercial steel pipe, and the pump efficiency is 75%. Size a valve to control the flow of oil if the frictional pressure drop in the line is found to be 6 psi. Use a pressure drop of 5 psi across the valve and estimate the annual cost if the electricity price is $0.10/kW-hr and the pump operates 8200 hr per year.

Crude tank

P = 1 atm

8 ft

4 ft

64 ft

Separation towerP = 12.7 psia

Pump

Figure E9.6-1 An oil transferring system

Solution ------------------------------------------------------------------------------------------

The flow coefficient for the valve can be determined from equation (9.6-2)

2 Smith and Corripio, Principles and Practice of Automatic Process Control, Wiley, 1997, pg. 209

9-21

Chapter 9

Q = Cv (9.6-2)SGPv

Since the nominal flow is 700 gpm, we want 2700 or 1400 gpm of flow when the pump is fully opened. The maximum valve coefficient is

Cv = Q = 1400 = 607 gpm/psi1/2

vPSG 5

94.0

From Figure 9.6-3 an 8-in. Masoneilan’s valve Schedule 40 has a Cv of 640 gpm/psi1/2. This valve is suitable for the service. Now we need to decide where to place the valve to prevent liquid flashing due to the pressure drop through it. Liquid will flash (vaporize) when the pressure in the line is less than its vapor pressure. A good place for the valve is at the discharge of the pump, where the exit pressure is higher as a result of the hydrostatic pressure due to the 60 ft of elevation plus most of the 6 psi of friction drop. The minimum pressure at the valve exit is

Pmin = (62.3 lb/ft3)(0.94)(60 ft)/(144 in2/ft2) + 12.7 psa = 37.1 psia.

This pressure is much higher than the vapor pressure of the oil at 13.85 psia.

The annual cost attributable to the 5-psi drop across the valve is

= $ 1,665/yearmin

700galgal

ft48.71 3

75.0)/144)(5( 2ftlbf

yearhr200,8

hrhW 10.0$

To estimate the extra work required by the pump in the above equation we have used the equation

Energy = (Power)(time)

Energy = (time)

pump

PQ

It is important to note that a typical process may require several hundreds control valves. An optimum control valve sizing also requires the determination of the pressure drop across the valve that might save thousand of dollars on the energy cost annually.

------------------------------------------------------------------------------------------------------

9-22

Compressible Flow

Different manufactures have developed different formulas to model the flow of compressible fluids - gases, vapors, and steam - through their control valves. Among several manufactures that produce good valves, including the Crane Company, DeZurik, Foxbora, Fisher Controls, Honeywell, and Masoneilan, we only present the compressible formulas of Masoneilan and Fisher Controls. Their equations and methods are typical of the industry.

The equations for compressible flow derive from the equation for liquids. However the expressions look quite different from the equation for liquids since they contain the unit conversion factors and density corrections for temperature and pressure. The flow coefficient Cv of a valve is the same whether the valve is used for liquid or gas services.

Masoneilan uses the following set of equations.

Qscfh = 836CvCf (y 0.148y3) (9.6-3)1

1

TSGP

In this equation Qscfh is the volumetric flow rate in standard cubic feet per hour at the standard conditions of 1 atm and 60oF. Cf is the critical flow factor with a numerical value between 0.6 and 0.95 depending on the valve types. SG is the gas specific gravity with respect to air and is calculated by dividing the molecular weight of the gas by 29, the average molecular weight of air. T1 is the absolute temperature at the valve inlet in oR. P1 is the pressure at the valve inlet in psia. y represents the compressibility effects on the flow and is defined by

y = (9.6-4)fC

63.1

1PPv

For gas or vapor mass flow rate in lb/hr,

= 2.8CvCfP1 (y 0.148y3) (9.6-5)m1

520T

SG

For steam flow in lb/hr,

= 1.83CvCf (y 0.148y3) (9.6-6)m)0007.01(

1

SHTP

In this equation TSH is the degree of superheat in oF.

Fisher Controls uses the following set of equations.

Qscfh = Cg P1sin (9.6-7a)1

520TSG rad

PP

C

1

1

1

64.59

9-23

Qscfh = Cg P1sin (9.6-7b)1

520TSG deg1

1

1

3417

PP

C

In equations (9.6-7a) and (9.6-7b) all the symbols are the same as for the Masoneilan formulas except for the two new coefficients Cg and C1. The coefficient Cg determines the gas flow capacity of the valve, where as the coefficient C1, defined as Cg /Cv, is functionally the same as the Cf factor in the equations used by Masoneilan. The values of C1 can be found on Table 10-3, page 318, in Darby’s text for various valves. The argument of the sine function must be limited to /2 radians or 90 degrees since the flow has reached critical flow conditions and cannot increase above this value without increasing P1. Let 1 be the density of the gas at P1 in lb/ft3, the mass flow rate in lb/hr for steam or vapor at any pressure is given by

= 1.06Cg sin (9.6-8a)m 11Prad

PP

C

1

1

1

64.59

= 1.06Cg sin (9.6-8b)m 11Pdeg1

1

1

3417

PP

C

Example 9.6-23 ----------------------------------------------------------------------------------A 3-in Masoneilan valve with full trim has a capacity factor of 110 gpm/psi1/2 when fully opened. The pressure drop across the valve is 10 psi.

(a) Calculate the flow of a liquid solution with density 0.8 g/cm3.(b) Calculate the flow of gas with average molecular weight of 35 when the valve inlet conditions are 100 psig and 100oF.(c) Calculate the flow of the gas from part (b) when the inlet pressure is 5 psig. Calculate the flow both in volumetric and in mass rate units, and compare the results for a 3-in. Fisher Controls valve (Cv = 120, Cg = 4280, C1 = 35.7).Solution ------------------------------------------------------------------------------------------

(a) Calculate the flow of a liquid solution with density 0.8 g/cm3.

For the liquid solution, the volumetric flow rate is given by

Q = Cv = 110 = 389 gpmSGPv

8.010

The mass flow rate is

= (389 gpm)(60 min/hr)(8.330.8 lb/gal) = 155,600 lb/hrm

3 Smith and Corripio, Principles and Practice of Automatic Process Control, Wiley, 1997, pg. 206

9-24

(b) Calculate the flow of gas with average molecular weight of 35 when the valve inlet conditions are 100 psig and 100oF.

Let Cf = 0.9, we have

y = = = 0.535fC

63.1

1PPv

9.063.1

7.1410010

The volumetric flow rate is given by

Qscfh = 836CvCf (y 0.148y3)1

1

TSGP

y 0.148y3 = 0.535 0.148(0.535)3 = 0.512

Qscfh = 836(110)(0.9) (0.512) = 187,000 scfh)560)(29/35(

7.114

The mass flow rate is determined from

= 2.8CvCfP1 (y 0.148y3) (9.6-5)m1

520T

SG

= 2.8(110)(0.9)(114.7) (0.512) = 17,240 lb/hrm560520

2935

(c) Calculate the flow of the gas from part (b) when the inlet pressure is 5 psig. Calculate the flow both in volumetric and in mass rate units, and compare the results for a 3-in. Fisher Controls valve (Cv = 120, Cg = 4280, C1 = 35.7).

y = = = 1.290fC

63.1

1PPv

9.063.1

7.14510

The volumetric flow rate is given by

Qscfh = 836CvCf (y 0.148y3)1

1

TSGP

y 0.148y3 = 1.290 0.148(1.290)3 = 0.972

Qscfh = 836(110)(0.9) (0.972) = 61,000 scfh)560)(29/35(

7.19

9-25

The mass flow rate is determined from

= 2.8CvCfP1 (y 0.148y3) (9.6-5)m1

520T

SG

= 2.8(110)(0.9)(19.7) (0.972) = 5,620 lb/hrm560520

2935

Using formulas from Fisher Controls we have

Qscfh = Cg P1sin1

520TSG rad

PP

C

1

1

1

64.59

Qscfh = 4280 (19.7) sin = 68,700 scfh)560)(29/35(

520

rad

7.19

107.35

64.59

= 1.06Cg sinm 11Prad

PP

C

1

1

1

64.59

The density is determined from ideal gas law: 1 = = = 0.115 lb/ft3

1

1

RTMP

)560)(73.10()7.19)(35(

= 1.06(4280) sin = 6,340 lb/hrm )7.19)(115.0(rad

7.19

107.35

64.59

------------------------------------------------------------------------------------------------------

Example 9.6-34 ----------------------------------------------------------------------------------A control valve is to regulate the flow of steam into a heat exchanger with a design heat transfer rate of 15 million Btu/hr. The supply steam is saturated at 20 psig. Size the control valve for a pressure drop of 5 psi and 100% overcapacity.Solution ------------------------------------------------------------------------------------------

At 20 psig, the steam latent heat of condensation is Hevap = 930 Btu/lb, the nominal steam flow is

= = 16,130 lb/hrm930

105.1 7

The steam is saturated, therefore the degrees of superheat, TSH = 0, is zero. Assuming a Masoneilan valve with Cf = 0.8, we have

4 Smith and Corripio, Principles and Practice of Automatic Process Control, Wiley, 1997, pg. 208

9-26

y = = = 0.773 (9.6-4)fC

63.1

1PPv

8.063.1

7.14205

y 0.148y3 = 0.773 0.148(0.7733) = 0.705

For steam flow in lb/hr,

= 1.83CvCf (y 0.148y3) (9.6-6)m)0007.01(

1

SHTP

16,130 = 1.83Cv(0.8)(34.7)(0.705) Cv = 450 gpm/psi1/2

To design for 100% overcapacity, we need a maximum Cv,max = 2Cv = 950 gpm/psi1/2. From Figure 9.6-3, a 10-in. Masoneilan valve, with a coefficient of 1000, is the smallest valve that can provide the service.

Using Fisher Controls equation, we have

= 1.06Cg sinm 11Prad

PP

C

1

1

1

64.59

Saturated steam at 20 psig, Tsat = 259oF, 1 = 0.0833 lb/ft3. Let C1 = 35, we have

16,130 = 1.06Cg sin Cg = 15,000)7.34)(0833.0(rad

7.34

535

64.59

For 100% overcapacity Cg ,max = 2(15,000) = 30,000 . The corresponding Cv is then

Cv = Cg/C1 = 30,000/35 = 856 gpm/psi1/2

We obtain similar value for the valve coefficient Cv from both methods.

10-1

Chapter 10

Flow in Chemical Engineering Application

10.1 Drag Force on Solid Particles in Fluids

Turbulent eddies at high NRe

Vh

DFD

Figure 10.1-1 Flow past a sphere.

When a fluid flows past a spherical particle it will exert a net total drag force FD on the sphere in the direction of the flow as shown in Figure 10.1-1. This total drag force consists of the viscous drag and the form drag. The viscous or frictional drag is the results of the momentum transfer between the fluid and the sphere due to the difference in their velocities. The form drag is the net force exerted on the sphere due to the difference in pressure upstream and downstream of the particle. The total drag force FD can be determined from the following equation

CD = (10.1-1)2

21

/

V

AF p

This equation is the definition of the drag coefficient CD where Ap is the projected area of the particle in the direction of motion. For a sphere Ap = D2/4. The definition of the drag coefficient from equation (10.1-1) is analogous to that of the friction factor for flow in a pipe

f = (10.1-1)2

21 V

w

In this equation, w is the force exerted on the wall by the fluid per unit area. For laminar flow (NRe = DV/ < 1) the differential momentum equation can be solved for the flow over the sphere to obtain the pressure and local stress distributions. The integration of the stress distribution over the sphere surface gives the viscous drag and the integration of the pressure over the sphere surface gives the form drag. For this case the drag coefficient is given as

CD = Re

24N

10-2

For NRe > 1, the drag coefficient can be obtained from experiment. A simple correlation, which represents the entire range of CD vs. NRe reasonably well up to about NRe = 2105, is given by Dallavalle1 for fluids flow over a sphere:

CD = (10.1-3)2

Re

8.4632.0

N

For flow past a circular cylinder with L/D >>1 normal to the cylinder axis, the flow is similar to that over a sphere. The drag coefficient over the entire range of NRe (up to 2105) is given by

CD = 2

Re

9.105.1

N

10.2 Separations by Free Settling

Many engineering processes require the separation of solid particles from fluids. Separation might be achieved by free settling under gravity or other forces since particles of different size and density will move through the fluid at a different rate. As a first step, we will determine the velocity of the particle in free settling or the travel time of a particle through a given distance. Consider a spherical particle with diameter D and density s shown in Figure 10.2-1, which is settling under gravity in a fluid of density and viscosity .

V

FD

Fb

Fg

+

Figure 10.2-1 Settling of a spherical particle under gravity.

Application of a downward momentum balance on the particle gives

= Fg FD Fb (10.2-1)dtd

VD s

6

3

where

Fg = gravity force = g6

3sD

1 Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 341

10-3

FD = drag force = CD V2Ap21

Fb = buoyant force = g6

3D

If there is no change in particle velocity, equation (10.2-1) becomes

0 = g CD V2Ap g (10.2-2)6

3sD

21

6

3D

The velocity obtained from solving equation (10.2-2) is called the terminal velocity Vt when

the gravity force exactly balances the buoyant and drag forces and there is no acceleration.

Example 10.2-12 ----------------------------------------------------------------------------------A stainless steel sphere of diameter D = 1 mm and density s = 7,870 kg/m3 falls steadily under gravity through a polymeric fluid of density = 1,052 kg/m3 and viscosity = 0.1 kg/ms. Determine the downwards terminal velocity of the sphere.Solution ------------------------------------------------------------------------------------------

At terminal velocity, there is no acceleration of the particle

0 = Fg FD Fb

0 = g CD V2Ap g6

3sD

21

6

3D

Solving for the velocity yields

V2 = = (E-1)

43

)(2

3

DC

gD

D

s

34

D

s

CDg

)(

We need an expression for the drag coefficient CD to solve for the velocity, we could use equation (10.1-3)

CD = (10.1-3)2

Re

8.4632.0

N

However, using this expression will result in a nonlinear equation. A simpler method that might work is to assume laminar flow so that

2 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, pg. 215

10-4

CD = = Re

24N VD

24

Substituting this expression into equation (E-1) gives

V = = = 0.0372 m/s

18

)(2 sgD)1.0)(18(

)052,1870,7)(81.9()10( 23

We now need to check the assumption of laminar flow:

NRe = DV/ = (1,052)(0.0372)(0.001)/0.1 = 0.39 < 1

The flow is laminar hence the terminal velocity is 3.72 cm/s.

------------------------------------------------------------------------------------------------------

Example 10.2-23 ----------------------------------------------------------------------------------A spherical hot air balloon of diameter 40 ft and deflated mass 500 lb is released from rest in still air at 50oF and 1 atm. The gas inside the balloon is effectively air at 200oF and 1 atm. Assuming a constant drag coefficient of CD = 0.60, estimate

(a) The steady upward terminal velocity of the balloon.(b) The time in seconds it takes to attain 99% of this velocity.

Solution ------------------------------------------------------------------------------------------

Air density can be estimated from ideal gas law: = RT

PM w

At 50oF and 1 atm, 1 = = 0.078 lb/ft3)510)(73.10(

)7.14)(29(

At 200oF and 1 atm, 2 = = 0.060 lb/ft3)660)(73.10(

)7.14)(29(

Application of the momentum balance on the balloon gives

M = Fb Mg FD dtdV

In this equation M is the total mass of the balloon. The buoyant force is evaluated as

Fb = g = (32.2) = 84,165 lbft/s26

13D

6)078.0)(40( 3

The total mass of the balloon is

3 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, pg. 216

V

FD

Fb

Fg+

10-5

M = 500 + = 500 + = 2,511 lb 6

23D

6)060.0)(40( 3

The gravity force is

Mg = (2,511)(32.2) = 80,842 lbft/s2

The drag force is

FD = CD V2Ap = (0.6)(0.078)V2()(202) = 29.4V221

21

(a) The steady upward terminal velocity of the balloon.

29.4Vt2 = Fb Mg = 84,165 80,842 = 3,323 lbft/s2

Vt = = 10.6 ft/s2/1

4.29323,3

(b) The time in seconds it takes to attain 99% of this velocity.

2,511 = Fb Mg FD = 3,323 29.4V2dtdV

85.4 = 113 V2dtdV

t = 85.4 tV

VdV99.0

0 2113

0.99Vt = 10.5

Using the integration formula = ln , we have 2xadx

a21

xaxa

2/1

2/1

t = 85.4 ln = 20.4 s11321

5.101135.10113

2/1

2/1

------------------------------------------------------------------------------------------------------

Solid particles can be removed from a dilute suspension by passing the suspension through a large vessel as shown in Figure 10.2-2. In the vessel, if the upward velocity of the fluid (Q/A) is less than the terminal velocity (Vt) of the particles and the residence time is long enough, the particles will settle out.

10-6

A

Vt

Overflow Overflow

Feed (Q)

Underflow

Q/A

Figure 10.2-2 Gravity settling tank.

For laminar flow, the terminal velocity of the particles is given by

V = (10.1-4)

18

)(2 sgD

This velocity must be greater than the upward velocity of the liquid. Therefore

>

18

)(2 sgDAQ

The diameter of the smallest particle that will settle out is

D = (10.1-5)2/1

)(18

AgQ

s

If the flow is not laminar, we can divide the correlation for the drag coefficient (10.1-3), by NRe to obtain

= = ReN

CD

Re

1N

2

Re

8.4632.0

N

2/1

Re

NCD

Re

1N

Re

8.4632.0N

This equation can be rearranged to a quadratic equation in term of Re

1N

+ = 0Re

8.4N Re

632.0N

2/1

Re

NCD

10-7

Solving for yieldsRe

1N

= 0.0658Re

1N

2/1

Re

208.000433.0

NCD

Once the Reynolds number is determined, the diameter of the smallest particle that will settle out can be evaluated

NRe = D = =

VDV

N

Re

QAN

Re

We first need to determine the ratio . From the force balance on the spherical particle at ReN

CD

terminal velocity

0 = Fg FD Fb

0 = g CD V2Ap g6

3sD

21

6

3D

CD = = = p

s

AVgD

2

3

3)(

4/3)(

22

3

DVgD s

23)(4

VgD s

We can divide the above equation by Reynolds number to eliminate the unknown D

= (10.1-6)ReN

CD323

)(4V

gs

The procedure to determine the smallest particle that will settle out is as follows.

1) Calculate the terminal velocity V = AQ

2) Calculate = ReN

CD323

)(4V

gs

3) Calculate = 0.0658Re

1N

2/1

Re

208.000433.0

NCD

4) Calculate D = V

N

Re

10-8

If the problem is to determine the maximum flow rate Q so that particles of a given size D will settle out we can follows the following procedure:

1) Solve for CD from the force balance on the particle

0 = g CD V2Ap g6

3sD

21

6

3D

CD = = = p

s

AVgD

2

3

3)(

4/3)(

22

3

DVgD s

23)(4

VgD s

2) The unknown in this case is the terminal velocity that can be eliminated by multiplying CD with NRe

2

CDNRe2 = = = NAr23

)(4V

gD s

2

VD2

3

3)(4

gD s

34

In this equation NAr is the Archimedes number = . Reynolds number can be 2

3 )(

gD s

evaluated using the correlation for the drag coefficient (10.1-3), by NRe to obtain

CD = CDNRe2 = NRe

2 = NAr

2

Re

8.4632.0

N

2

Re

8.4632.0

N 34

NRe = NAr1/2

Re

8.4632.0N

2/1

34

3) This is a quadratic equation in term of NRe1/2, therefore

NRe1/2 = 3.798 2/1

827.142.14 ArN

4) Calculate the terminal velocity from Reynolds number

NRe = V =

VDD

N

Re

5) The maximum volumetric flow rate so that particles of a given size D will settle out is then

Q = VA = AD

N

Re

10-9

Chapter 10

10.3 Flow Through Packed Beds

Air flowFigure 10.3-1 Flow through a packed bed.

A packed bed is simply a column or tube packed with solid particles as shown in Figure 10.3-1. Flow through packed beds occurs in many engineering application for example the fluid flow through a tubular reactor containing catalyst particles or the water flow through cylinders packed with ion-exchange resin in order to produce deionized water. Flow through a packed bed is a sub area of the flow through a porous media in which a solid, or collection of solid particles, has sufficient open space in or around the solids to enable a fluid to pass through or around them.

Since the fluid in a packed bed follows a tortuous path through the interstices or pores between particles, one method of modeling the flow behavior is to consider the flow path as a noncircular conduit with a hydraulic diameter defined as

Dh = 4 = 4 = 4p

i

WA

LWLA

p

i

areasurfaceWettedvoidsofVolume

Dh = 4 (10.3-1))/)(.( ParticleareaSurfaceparticlesofNo

volumeBed

10-10

The bed porosity, , which is the fraction of total volume that is void is defined as

= (10.3-2)bedentireofvolume

voidsvolume

= bedentireofvolume

particlesofvolumebedentireofvolume

Ai

Wp

Fluid path

Particles

Idealized pore

L

Figure 10.3-2 A model of the flow through pores.

Figure 10.3-2 shows a model for the flow through the pores of the particles in a packed bed. All the empty volume within the bed is considered as the volume AiL of a noncircular conduit with surface area WpL representing the surface of contact between the fluid and the solid particles. The number of particles Np in the bed may be estimated from the average particle volume

Np = ParticleVolume

bedinsolidsofFractionvolumeBed/

))((

Np = (10.3-3)ParticleVolume

volumeBed/

)1)((

Substituting equation (10.3-3) into equation (10.3-1) yields

Dh = 4 (10.3-4)

1

sa1

In this equation as = . If the particles are spherical with diameter D, volumeparticle

areasurfaceparticle

then

as = =

6

3

2

DD

D6

10-11

For a bed composed of uniform spherical particles with diameter D

Dh = (10.3-5))1(3

2

D

When the packing has a shape different from spherical, an effective particle diameter is defined

Dp = 6 = = Ds (10.3-6)areasurfaceparticle

volumeparticlesa

6

In this equation Ds is the diameter of a sphere with the same volume as the particle and is the sphericity or shape factor defined by

= particletheofareaSurfaceparticletheasvolumesamewithsphereaofareaSurface

Vs

Vi

A

Ai

P1 P2

Figure 10.3-3 A horizontal packed bed with idealized pores.

Applying the energy equation over a horizontal packed bed shown in Figure 10.3-3 yields

= + ef (10.3-7)

1P

2P

In this equation, the friction loss ef is given by

ef = 2fporeVi2 (10.3-8)

hDL

where fpore is the friction factor in the pores and Vi is the interstitial or actual velocity within the pores. The actual velocity relates to the approach or superficial velocity Vs by the expression

Vi = = = (10.3-9)iA

QA

Q

sV

Substituting the hydraulic diameter, equation (10.3-5), and the interstitial velocity, equation (10.3-9), into equation (10.3-8) yields

ef = 2fporeVi2 = 2fpore L = 3fporeVs

2

hDL 2

sV

pD2)1(3

3)1(

pDL

10-12

From equation (10.3-7), the pressure drop across the bed is given by

P = (P2 P1) = ef = 3fporeVs2

3)1(

pDL

Rearranging the equation so that the friction factor in the pore is on the right-hand side, we have

= 3fpore2sV

P

LDp

1

3

The numerical factor 3 is normally absorbed in the fiction factor fpore. There for the above equation is usually written as

= fpore2sV

P

LDp

1

3

For turbulent flow, the friction factor fpore was found experimentally to be a constant so that the above equation becomes

= 1.752sV

P

LDp

1

3

For laminar flow, the friction factor fpore, also found experimentally, is expressed in terms of Reynolds number by an equation similar to that in laminar flow in a pipe

fpore = poreN Re,

180

The pore Reynolds number is defined as NRe,pore = . An expression that adequately

)1(

ps DV

represents the pressure drop in both laminar and turbulent flow is

= + 1.752sV

P

LDp

1

3

poreN Re,

180

This equation with a value of 150 instead of 180 is called the Ergun equation.

10-13

Example 10.3-1 ----------------------------------------------------------------------------------A packed bed is composed of cylinders having a diameter D = 0.02 m and a length h = D. The bulk density of the overall packed bed is 962 kg/m3 and the density of the solid cylinder is 1600 kg/m3. Calculate the void fraction and the effective diameter Dp.Solution ------------------------------------------------------------------------------------------

Let the volume of the packed bed = 1 m3. The total mass of the bed is then (962 kg/m3)(1 m3) or 962 kg. This is also the mass of the solid cylinders. The volume of the cylinders or particles contained in the bed is

Volcyl = = 0.601 m33/1600

962mkg

kg

The void fraction is then

= = = 0.399bedentireofvolume

particlesofvolumebedentireofvolume 1

601.01

Dp = 6 = 6 = 6 = D = 0.02 mareasurfaceparticle

volumeparticle

DDD

DD

2

2

42

415.0

25.0D

------------------------------------------------------------------------------------------------------Example 10.3-2 ----------------------------------------------------------------------------------Air at 311oK is flowing through a packed bed of spheres having a diameter of 12.7 mm. The void fraction of the bed is 0.38. Find the pressure drop across the bed.

H

P1

P2

Packed bed

Bed diameter = 0.61 m P1 = 1.10 atmMass flow rate = 0.358 kg/s

H = 2.44 m

Solution ------------------------------------------------------------------------------------------

We can use the Ergun equation to determine the pressure drop through the packed bed

= + 1.75 NRe,pore = 2sV

P

LDp

1

3

poreN Re,

150

)1(

ps DV

10-14

P = + 32

2)1(150

p

s

DLV

3

2 )1(75.1

p

s

DLV

In this equation, the density should be evaluated at the 0.5(P1 + P2). However since P2 is not known, we will evaluate the density at the inlet pressure

= 1 = = = 1.248 kg/m3RT

PM w

)311)(8314()10013.11.1)(97.28( 5

Note: R = 8314 m3Pa/kmoloK

The superficial velocity is

Vs = = = = 0.981 m/sA

m

24

Dm

2)61.0)()(248.1()358.0)(4(

For air at 311oK and 1.1 atm: = 1.910-5 kg/ms

P = 3)1(

p

s

DLV

sp

VD

75.1)1(150

P = 3)38.0)(0127.0()38.01)(44.2)(981.0(

)981.0)(248.1)(75.1(0127.0

)38.01)(109.1)(150( 5

P = 4.86103 Pa

P = (P2 P1) = 4.86103 Pa

P2 = P1 4.86103 = 1.101.013105 4.86103 = 1.114105 Pa

Since P2 is close to P1 (less than 1% difference), we do not need to reevaluate the air density. The pressure drop across the bed is

P = 4.86103 Pa

10-15

Chapter 10

10.4 Laminar Flow Through Porous Medium

The Ergun equation can also be applied to consolidated medium such as a brick or a sandstone rock formation through which oil is flowing. For laminar flow

= = 2sV

P

LDp

1

3

poreN Re,

150

ps DV )1(150

Solving for the superficial velocity yields

Vs = = (10.4-1)LP

2

32

)1(150

pD

LP

Equation (10.4-1) is called d’Arcy’s law with the permeability defined by

= 2

32

)1(150

pD

The most common unit for the permeability is the darcy, which is defined as the flow rate in cm3/s that results when a pressure drop of 1 atm is applied to a porous medium that is 1 cm2 in cross-sectional area and 1 cm long, for a fluid with viscosity of 1 cP. For this case the superficial velocity is 1 cm/s, therefore

1 darcy = = 0.98610-8 cm2cmatmcPscm

/))(/(

The differential form of d’Arcy’s law in one-dimensional flow is

vx = (10.4-2)

dxdP

In this expression, vx is a flux (a volumetric flow rate per unit area), is a conductance, and

is a driving force so that the equation has the general form:dxdP

Flux = = Conductance Driving forceResistance

forceDriving

For three-dimension flow, equation (10.4-2) becomes v = P (10.4-3)

Substituting this equation into the continuity equation for incompressible flow yields

10-16

v = P = 0

For constant , the equation is simplified to

P = 2P = 0 (10.4-4)

Numerical Solution of the Laplace equation using COMSOL Multiphysics

Flow in two-dimensional porous medium is governed by Laplace equation in terms of pressure as:

2P = = 02

2

2

2

yP

xP

Once the pressure is known, the velocity vector can be evaluated from

v = P or vx = , vy = where is the permeability

xP

yP

= 0nP

A B

CD

Ep=10 p=0

= 0nP

Solve the problem4 of flow in a porous medium shown in the above figure. The flow region is a rectangle ABCD with a central circular region E of zero permeability that allows no

flow through it. The boundaries AB and CD are also impervious to flow so that = 0. The nP

4 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice-Hall, 1999, p.566

10-17

pressure along the two ends are P = 10 psig and P = 0 psig. Let = 1.0 lbft/s throughout

the region.

A partial differential equation solver such as COMSOL Multiphysics developed by COMSOL Inc could solve the set of differential and algebraic equations encountered in fluid mechanics. The ability of this software can be found from the COMSOL User’s Guide:

“COMSOL Multiphysics is designed to make it as easy as possible to model and simulate physical phenomena. With COMSOL Multiphysics, you can perform free-form entry of custom partial differential equations (PDEs) or use specialized physics application modes. These physics modes consist of predefined templates and user interfaces that are already set up with equations and variables for specific area of physics. Further, by combining any number of these application modes into a single problem description, you can model a multiphysics problem-such as one involving simultaneous mass and momentum transfer.”

COMSOL Multiphysics can solved a variety of problems beside fluid mechanics, including those in heat transfer, chemical reactor design, electromagnetics, fuel cells, transport phenomena, polymer processing and so on. COMSOL is very flexible with a friendly interface. The Graphical User Interface consists of the following steps:

1. Model Navigator: Use this dialog box to choose the model equation, its dimensions, stationary or time-dependent analysis.

2. Geometry Modeling: Use this interface to draw or specify object and its dimensions and coordinates.

3. Physics Settings: The boundary and initial conditions are specified in the Boundary Settings. The physical properties in the model equations are specified in the Subdomain Settings.

4. Mesh Generation and Solution: The mesh can be initialized and refined if necessary. The problem is then solved with various options for solver like: stationary linear, stationary nonlinear, time dependent and so on. The solver can also be auto selected.

5. Postprocessing and Visualization: The results can be displayed in different graphical or numerical formats.

Getting Started

1) Start the program by going through the Start Menu and double-clicking on Comsol Multiphysics v3.x. Doing so opens the Model Navigator Dialog box. This is the starting point for the definition of the certain types of physics phenomena of interest. Note: The software provides default values for each step therefore you might not have to choose a value for some of the steps explained in the following intructions.

2) Click on the New Tab in the Model Navigator.3) In the Space Dimension, select (2D) to simulate the two-dimensional flow.4) Under the Application Modes folder, single-click on the + sign for PDE Modes

folder. Single-click + sign for the Classical PDEs. Single-click the + sign again for Laplace’s Equation.

10-18

5) The Model Navigator shows the dependent variable, which is to be solved for. Enter P as the dependent variable.

6) Click OK.

Geometry Modeling

7) The next step in the modeling process is to create the model geometry. In this case, simulate the reactor as a rectangle. To do this, select Specify Object on Draw menu. Select Rectangle.

8) On the dialog box, enter 2.5 for width, and 1.8 for height. Choose Center for the Base Postion. Click OK.

9) Select Specify Object on Draw menu. Select Circle. On the dialog box, enter 0.5 for radius. Click OK.

10-19

10) Since we want the domain to be a rectangle with a circular hole in it, Select Create Composite Object on Draw menu. Edit the Set formula field to contain: R1 – C1

11) Click OK and click on the Zoom Extents button.

Zoom Extents

10-20

12) The following screen will appear.

Boundary Conditions

13) Select Boundary Settings on Physics menu. Choose Dirichlet boundary condition for Boundary 1 and a value of 10 for r (since P = 10). Choose Dirichlet boundary condition for Boundary 4 and a value of 0 for r (since P = 0). Choose Newmann

boundary condition for Boundaries 2, 3, 5, 6, 7, and 8 since = 0 for these nP

boundaries.

10-21

14) Select Plot Parameters on Postprocessing menu. On the General tab, check Surface, Contour, Arrow, and Geometry edges. Click on the Arrow tab and edit “ Px” for x component and “ Py” for y component since velocity is defined by

vx = = Px, vy = = Py

xP

yP

This will give the correct direction for the arrow.

15) Choose Initialize Mesh from the Mesh menu. Next choose Refine Mesh to improve the first rough mesh. This option can be repeated to reduce the mesh size further.

Everything is now ready for generating a solution. Choose Solve Problem from the Solve menu. The results will be generated automatically and the values of the dependent variable will be color-coded in the solution domain. A color bar at the right assigns numerical solution values to each color. The isobar and velocity vector are also plotted.

10-22

To summarize, Comsol Multiphysics allows you to use drawing tools to create solution domains. You can then choose the PDE to be solved, assign PDE parameters appropriate for the domain, assign boundary conditions to boundary segments, and specify initial conditions for the PDE if necessary. You can then generate triangular meshes of different refinements, compute discrete solutions at the nodes of the mesh, and display high-quality plots of the continuous approximation to the PDE solution over the domain and even over times.

10-23

Chapter 10

10.5 Flow Through Fluidized Beds

The upward flow of fluid through a bed of solid particles is an important process occurring in nature and in industrial operations. At low velocities, the pressure drop in a horizontal bed increases with the fluid velocity according to the Ergun equation

= + 1.75 (10.5-1)2sV

P

hDp

1

3

poreN Re,

150

whereP = pressure drop through the packed bedh = bed l(height)Dp = particle diameter = fluid densityVs = superficial velocity at a density averaged between inlet and outlet conditions = bed porosity

NRe,pore = = average Reynolds number based upon actual velocity

)1(

ps DV

The overall energy balance over the horizontal bed is

+ ef = 0P

Therefore the frictional dissipation term per unit mass flowing is

ef = = + (10.5-2)

P

2150

pDVh

3

21

pDhV 275.1

3

1 mf

When the packing has a shape different from spherical, an effective particle diameter is defined as

Dp = 6 = = Ds (10.5-3)areasurfaceparticle

volumeparticlesa

6

In this equation Ds is the diameter of a sphere with the same volume as the particle and is the sphericity or shape factor defined by

= particletheofareaSurfaceparticletheasvolumesamewithsphereaofareaSurface

10-24

Equation (10.5-3) can also be written as

Dp = 6 = 6)(

)(areasurfaceparticleN

volumeparticleN

p

p

areasurfaceparticletotalvolumeparticletotal

In this equation, Np is the number of particles in the bed.

Dp = 6 )1)((

)1)((

areasurfaceparticletotal

volumeparticletotal

Dp = 6 = (10.5-4))(

)1)((areasurfaceparticletotal

volumebed

sA)1(6

In this equation As is the interfacial area of packing per unit of packing volume, ft2/ft3 or m2/m3. The bed porosity is defined as

= bedentireofvolume

particlesofvolumebedentireofvolume

= (10.5-5)hR

densityparticleparticlesallofweighthR

2

2

where R = inside radius of column, As and are characteristics of the packing. Experimental values of can easily be determined from Eq. (10.5-5) but As for non-spherical particles is usually more difficult to obtain. Values of As and are available for the common commercial packing in various references. As for spheres can be computed from the volume and surface area of a sphere.

As the gas velocity increases, conditions finally occur where the force of the pressure drop times the cross-sectional area just equals the weight of the particles in the bed. A slight increase in gas velocity, to increase the pressure drop, is required to unlock the intermeshed fixed-bed particles. Once the particles disengage from each other, they begin to move. The pressure drops to the point where the upward force on the bed is balanced by the downward force due to the weight of the bed particles. Further increases in gas velocity fluidize the bed, the pressure drop rises slightly until slugging and entrainment occurs.

The point of maximum pressure drop shown in Figure 10.5-1 is the point of minimum fluidization. Applying the energy equation for a vertical bed at this point yields

P1 P2 = ef + ghmf = hmf[(1 mf)p + mf)] g (10.5-6)

where hmf = bed height at the minimum fluidization point.

10-25

Equation (10.5-6) is rearranged to

ef = g(1 mf)(p ) (10.5-7)mfh

Log Re

Inches ofwater

Pressuredrop P,

point of fluidization

Figure 10.5-1 Pressure drop in a bed of particulate solids.

Eq. (10.5-2) can be rearranged with Dp substituted by Ds, Vs substituted by Vmf, and h substituted by hmf to obtain

ef = + (10.5-8)

mfh

22

150

s

mf

DV

3

21

mf

mf

s

mf

DV

275.1

3

1

mf

mf

The minimum fluid velocity Vmf at which fluidization begins can be calculated by combining Eqs. (10.5-7) and (10.5-8) to obtain the quadratic equation in terms of Vmf:

+ = g(1 mf)(p ) (10.5-9a)s

mf

DV

275.1

3

1

mf

mf

22

150

s

mf

DV

3

21

mf

mf

+ = g(p ) (10.5-9b)3

275.1

mfs

mf

DV

22

150

s

mf

DV

3

1

mf

mf

For many systems, Equation (10.5-9b) simplifies to:

Vmf = for NRe,mf = < 20 (10.5-10a)

1650

2 gD ps

mfs VD

and

10-26

Vmf = for NRe,mf > 1000 (10.5-10b)

5.24

2 gD ps

At high fluid velocities, when the expansion of the bed is large, the behavior of fluidization depends on whether the fluid is a liquid or a gas. With a liquid, fluidization is smooth and uniform without large bubbling. This kind of fluidization is known as “particulate” fluidization. With a gas, uniform fluidization is frequently observed only at low velocities. At high velocities, non-uniform or “aggregative” fluidization will be observed with large bubbling and the bed is then often referred to as a “boiling” bed. In long, narrow fluidized beds, coalescence of the bubbles might be large enough to cover the entire cross-section of the column. These slugs of gas alternate with slugs of fluidized solids are carried upwards and subsequently collapse, causing the solids to fall back again. Slugging can cause severe entrainment problems and hence is undesirable.

0 ln Vmf ln Vs

ln Pbedheight h

h

Packed bed

Fluidized bed

P

Figure 10.5-2 Variation of bed height with superficial velocity

Figure 10.5-2 shows the relation between the actual bed height h and the superficial velocity Vs. For low velocity, the bed height h remains constant and the bed functions like a packed bed. However, as V is increased, the pressure drop will eventually high enough to counter balance the downward weight of the particles. This is the point of minimum fluidization where the velocity has reached the incipient fluidizing velocity Vmf. From this point on, the bed behave as a fluidized bed where the bed continues to expand with increasing velocity while the pressure drop increases only slightly.

Example 10.5-1 ----------------------------------------------------------------------------------Solid particles having a size of 0.12 mm, a sphericity shape factor of 0.88, and a density of 1000 kg/m3 are to be fluidized using air at 2 atm and 25oC. The voidage (void fraction) at minimum fluidizing conditions is 0.42.

(1) If the cross sectional area of the empty bed is 0.30 m2 and the bed contains 300 kg of solid, calculate the minimum height hmf of the fluidized bed.(2) Determine the pressure drop at the minimum fluidizing conditions.(3) Determine the incipient fluidizing velocity Vmf.

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Solution ------------------------------------------------------------------------------------------

(1) Calculate the minimum height hmf of the fluidized bed.

The mass of the solid particles in the bed is given by

ms = pAhmf(1 - mf) hmf = )1( mfp

s

Am

hmf = = 1.724 m)42.01)(3.0)(1000(

300

(2) Determine the pressure drop at the minimum fluidizing conditions.

The pressure drop is balanced by the gravity and buoyant forces on the particles

(P)(A) = (Ahmf) [(1 mf)p + mf)] g

Air at 2 atm and 25oC, = 2.374 kg/m3

P = (9.81)(1.724)[(1 0.42)(1000) + (0.42)(2.374)] = 9,827 Pa

(3) Determine the incipient fluidizing velocity Vmf.

We can solve equation (10.5-b) for the incipient fluidizing velocity Vmf

+ = g(p ) (10.5-9b)3

275.1

mfs

mf

DV

3

1

mf

mf

22

150

s

mf

DV

For air at 2 atm and 25oC, = 1.8510-5 Pas. We also have

Ds = (0.88)(0.1210-3) = 1.05610-4 m

Using numerical values in equation (10.5-9b), we have

+ = 9.81(1000 2.374)34

2

)42.0)(10056.1()374.2)(75.1(

mfV

3

1 0.420.42

24

5

)10056.1()1085.1(150

mfV

5.31105(Vmf)2 + 1.9481106Vmf 9.7867103 = 0

Vmf = 5.0210-3 m/s

10-28

Example 10.5-2 ----------------------------------------------------------------------------------A5 horizontal water purification unit consists of a hollow cylinder that is packed with ion-exchange resin particles. Tests with water flowing through the unit gave the following results:

Flow rate Q (gallons/hr) 10.0 20.0Pressure drop, P (psi) 4.0 10.0

If the available pump limits the pressure drop over the unit to a maximum of 54 psi, what is the maximum flow rate of water that can be pumped through it?Solution ------------------------------------------------------------------------------------------

The pressure drop across the packed bed can be determined by the Ergun equation

= + 1.75 NRe,pore = 2sV

P

LDp

1

3

poreN Re,

150

)1(

ps DV

For a given packed bed, the only two variables are the pressure drop ( P) and the flow rate (Q). The Ergun equation can be rearranged to

P = Vs + Vs2 = aQ + bQ2

In this equation, a and b are parameters that depend on the characteristics of the packed bed. They can be obtained from the data of pressure drop versus flow rate:

4 = 10a + 100b

10 = 20a + 400b

Solving for a and b we obtain

a = 0.3 psi/(gallons/hr) and a = 0.01 psi/(gallons/hr)2

Therefore the pressure drop for this particular bed can be correlated by

P = 0.3Q + 0.01Q2

For a maximum pressure drop of 54 psi, we have

54 = 0.3Qmax + 0.01Qmax2 Qmax

2 + 30Qmax 5,400 = 0

This is a quadratic equation in terms of Qmax, hence

Qmax = 15 + (152 + 5,400)0.5 = 60 gpm

5 Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice-Hall, 1999, p.218