Essentials of Fluid Mechanics · Essentials of Fluid Mechanics The basic fundamentals of fluid...

1Essentials of Fluid Mechanics

The basic fundamentals of fluid mechanics are essential for understanding the fluid dynamicsof pumping machinery. This chapter aims to provide a quick revision of the definitions andbasic laws of fluid dynamics that are important for a thorough understanding of the materialpresented in this book. Of particular interest are the kinematics of fluid flow; the three conser-vation principles of mass, momentum, and energy; relevant dimensionless parameters; laminarand turbulent flows; and friction losses in piping systems. Some applications of relevance topumping machinery are also considered.

1.1 Kinematics of Fluid Flow

To fully describe the fluid motion in a flow field it is necessary to know the flow velocity andacceleration of fluid particles at every point in the field. This may be a simple task in laminarflows but may be difficult in turbulent flows. If we use the Eulerian method and utilizeCartesian coordinates, the velocity vector at any point in a flow field can be expressed as

V= ui + vj+wk ð1:1Þ

where V is the velocity vector; u, v, and w are the velocity components in the x, y, and z direc-tions; and i, j, and k are unit vectors in the respective directions. In general, each of the velocity

components can be a function of position and time, and accordingly we can write

u = u x,y,z, tð Þ, v= v x,y,z, tð Þ, w=w x,y,z, tð Þ ð1:2Þ

The components of acceleration in the three directions can be expressed as

Pumping Machinery Theory and Practice, First Edition. Hassan M. Badr and Wael H. Ahmed.© 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd.

COPYRIG

HTED M

ATERIAL

ax =∂u

∂t+ u

∂u

∂x+ v

∂u

∂y+w

∂u

∂zð1:3aÞ

ay =∂v

∂t+ u

∂v

∂x+ v

∂v

∂y+w

∂v

∂zð1:3bÞ

az =∂w

∂t+ u

∂w

∂x+ v

∂w

∂y+w

∂w

∂zð1:3cÞ

The acceleration vector becomes

a= axi+ ayj+ azk ð1:4Þ

This vector can be split into two components, the local component, alocal, and the convectivecomponent, aconv:, that can be expressed as

alocal =∂u

∂ti+

∂v

∂tj+

∂w

∂tk ð1:5aÞ

aconv: = u∂u

∂x+ v

∂u

∂y+w

∂u

∂z

� �i+ u

∂v

∂x+ v

∂v

∂y+w

∂v

∂z

� �j+ u

∂w

∂x+ v

∂w

∂y+w

∂w

∂z

� �k ð1:5bÞ

1.1.1 Types of Flows

The flow field can be described as steady or unsteady, uniform or non-uniform, compressible orincompressible, rotational or irrotational, one-, two-, or three-dimensional, and can also bedescribed as laminar or turbulent. The flow is said to be steady if the velocity vector at anypoint in the flow field does not change with time.Accordingly, the local component of acceleration (alocal) vanishes if the flow is steady. The

flow can also be described as uniform if the velocity vector does not change in the streamwisedirection. For example, the pipe flow shown in Figure 1.1 is uniform since the velocity vectordoes not change downstream, but the flow in the bend shown in Figure 1.2 is non-uniform.

Laminar flow in a pipe

ru

Figure 1.1 Laminar flow in a pipe as an example of uniform flow

2 Pumping Machinery Theory and Practice

The flow is described as incompressible if the density change within the flow field does notexceed 5%. Accordingly, most of the flows in engineering applications are incompressible as,for example, flow of different liquids in pipelines and flow of air over a building. However,compressible flows occur in various applications such as flow in the nozzles of gas and steamturbines and in high speed flow in centrifugal and axial compressors. In general, the flowbecomes compressible if the flow velocity is comparable to the local speed of sound.For example, the flow of air in any flow field can be assumed incompressible up to a Machnumber of 0.3.The flow is called one-dimensional (1-D) if the flow parameters are the same throughout any

cross-section. These parameters (such as the velocity) may change from one section to another.As an approximation, wemay call pipe or nozzle flows 1-D if we are interested in describing theaverage velocity and its variation along the flow passage. Figure 1.3 shows an example of 1-Dflow in a pipe with constriction. On the other hand, the flow is called 2-D if it is not 1-D and isidentical in parallel planes. For example, the viscous flow between the two diverging platesshown in Figure 1.4 is two-dimensional. In this case, two coordinates are needed to describethe velocity field.If the flow is not 1-D or 2-D, it is then three-dimensional. For example, flow of exhaust gases

out of a smoke stack is three-dimensional. Also, air flow over a car or over an airplane is three-dimensional.

1.1.2 Fluid Rotation and Vorticity

The rate of rotation of a fluid element represents the time rate of the angular displacementwith respect to a given axis. The relationship between the velocity components and the rateof rotation can be expressed as

Figure 1.2 Flow in a 90� bend as an example of non-uniform flow

V1

V2

V3

Figure 1.3 One-dimensional flow in a pipe with constriction


ωx =12

∂w

∂y−∂v

∂z

� �, ωy =

12

∂u

∂z−∂w

∂x

� �,ωz =

12

∂v

∂x−∂u

∂y

� �ð1:6Þ

where ωx, ωy, ωz represent the rate of rotation around the x, y, and z axes.The vorticity ζ is defined as twice the rate of rotation. Accordingly, the vorticity vector ζ can

be expressed as

ζ = ζxi+ ζyj+ ζzk= 2ωxi + 2ωyj+ 2ωzk ð1:7Þ

The flow is called irrotational when the rate of rotation around the three axes is zero. In thiscase, we must have ζx = ζy = ζz = 0 for irrotational flow. The components of the vorticity vectorin cylindrical coordinates can be written as

ζr =1r

∂vz∂θ

−∂vθ∂z

ð1:8aÞ

ζθ =∂vr∂z

−1r

∂

∂rrvzð Þ ð1:8bÞ

ζz =1r

∂

∂rrvθð Þ− 1

r

∂vr∂θ

ð1:8cÞ

1.2 Conservation Principles

1.2.1 Conservation of Mass

Considering the general case of a compressible flow through the control volume (c.v.) shownin Figure 1.5 and assuming that n is a unit vector normal to the elementary surface area dAand v is the flow velocity through this area, then the conservation of mass equation takesthe form

A

B

B

A

C

C

D

D

Figure 1.4 Two-dimensional flow between two diverging plates


∂

∂t

ðc:v:

ρdV +ðc:s:

ρv:ndA= 0 ð1:9Þ

where ρ is the fluid density, v is the fluid velocity, dV is an elementary volume, and t is the time.When the control volume tends to a point, the equation tends to the differential form,

∂ρ

∂t+∂ ρuð Þ∂x

+∂ ρvð Þ∂y

+∂ ρwð Þ∂z

= 0 ð1:10Þ

where u, v, and w are the velocity components in the x, y, and z directions. If the flow is incom-pressible, the above equation can be reduced to

∂u

∂x+∂v

∂y+∂w

∂z= 0 ð1:11Þ

In the special case of 1-D steady flow in a control volume with one inlet and one exit(Figure 1.6), the conservation of mass equation takes the simple form,

m• = ρ1A1V1 = ρ2A2V2 =Const: ð1:12aÞ

where m• is the mass flow rate, V is the flow velocity, and A is the cross-sectional area.

c.v.

vθ

c.s.

ndA

dV

Figure 1.5 A schematic of an arbitrary control volume showing the flow velocity through a smallelementary surface area

V2(1)

(2)

V1

Figure 1.6 One-dimensional flow in a diverging flow passage


1.2.2 Conservation of Momentum

1.2.2.1 Conservation of Linear Momentum

In the general case of unsteady flow of a compressible fluid, the linear momentum conservationequation (deduced from the Reynolds transport equation) can be expressed as

XF=

dM

dt=∂

∂t

ðc:v:

ρvdV +ðc:s:

ρv:v:dA ð1:12bÞ

where the term ΣF represents the vectorial summation of all forces acting on the fluid body andM is its linear momentum.In case of steady flow, the first term on the right-hand side of Eq. (1.12) vanishes and the

equation is reduced to

XF=

ðc:s:

ρv:v:dA ð1:13Þ

The right-hand side of the above equation represents the net rate of outflow of linear momentumthrough the control surface. In the special case of steady one-dimensional flow, the equationcan be written in the form

XF =

Xm•V

� �out

−X

m•V

� �in

ð1:14Þ

When the control volume is very small (tends to a point), the momentum equation tends to thefollowing differential form (known as the Navier–Stokes equation):

ρ∂V

∂t+ V:r� �

V

� �=F−rp+ μr2V ð1:15aÞ

If the flow is frictionless (μ = 0), the diffusion term, μr2V, vanishes and the equationbecomes

ρ∂V

∂t+ V:r� �

V

� �=F−rp ð1:15bÞ

The above equation is well-known as Euler’s equation. The equation can be applied along astreamline to yield the following 1-D Euler equation

∂

∂sp + γzð Þ = −ρas ð1:16Þ

where as is the acceleration in the streamwise direction. If the above equation is furthersimplified for the case of steady, incompressible, frictionless flow, it results in Bernoulli’sequation, which can be written as


p

γ+V2

2g+ z=Const: ð1:17Þ

The application of the momentum equation normal to the streamline results in an equationsimilar to (1.16) and can be written as

∂

∂np + γzð Þ = −ρan ð1:18Þ

where n is a coordinate normal to the streamline.

1.2.2.2 Conservation of Angular Momentum

In the general case of unsteady flow of a compressible fluid, the angular momentum conser-vation equation (deduced from the Reynolds transport equation) can be expressed as:

XM =

∂

∂t

ðc:v:

ρ r × v� �

dV +ðc:s:

ρ r× v� �

v :dA ð1:19Þ

where the termX

M represents the vectorial summation of all moment acting on the fluid body

within the control volume, v is the velocity vector and dV is the elementary volume.In the special case of steady one-dimensional flow, the first term in the right-hand side of

Eq. (1.19) will vanish and the equation can be written in the form

XM =

Xm•V× r

� �out

−X

m•V × r

� �in

ð1:20Þ

The termsX

m•V× r

� �inand

Xm•V× r

� �out

represent the rates of inflow and outflow of

angular momentum, respectively.

1.2.3 Conservation of Energy

Considering the case of steady 1-D flow, the application of the first law of thermodynamics fora control volume (Figure 1.7) results in a simplified form of the energy conservation equationthat can be expressed as

Q�

c:v: +X

m•

ihi +

V2i

2+ gzi

� �=X

m•

ehe +

V2e

2+ gze

� �+W

�

c:v: ð1:21Þ

where h is the enthalpy, Q�

c:v: is the rate of heat transfer to the c.v., andW�

c:v: is the rate of doingwork by the c.v. In the special case of one inlet and one exit, the above equation can beexpressed in the form


q + hi +V2i

2+ gzi = he +

V2e

2+ gze +w ð1:22Þ

where q is the heat transfer per unit mass and w is the work done per unit mass. The case ofsteady incompressible flow with no heat transfer has many applications in fluid mechanics.Now, by writing h = u + pv = u + p/ρ, where u is the specific internal energy, Eq. (1.22) canbe expressed in the form

p1γ+V21

2g+ z1 =

p2γ+V22

2g+ z2 +w + hL ð1:23Þ

where the term hL represents the amount of heat generation due to fluid friction per unit weight offluid. This term is normally referred to as the friction head loss. Every term in the above equationhas a unit of length and represents energy per unit weight of fluid. Equation (1.23) can besimplified to

H1 = H2 +w+ hL ð1:24Þ

where H = pγ +

V2

2g + z and is called the total head, and the terms w and hL are redefined in

Eqs. (1.23) and (1.24) to represent the work done and the energy loss per unit weight of fluid,respectively.

1.3 Some Important Applications

a. In the case of a pump, the work is done by the prime mover, and the total head developedby the pump can be obtained by applying Eq. (1.24) between the inlet (1) and exit (2)sections shown in Figure 1.8 as follows:

H1 + hp = H2 + hL ð1:25Þ

where hp is the head developed by the pump (hp = −w) and hL is the friction head lossbetween sections 1 and 2.The rate of doing work by the pump on the fluid, Pf, can be obtained from

Pf = γQhp ð1:26Þ

Exit

Inlet

Qc.v.

Wc.v. c.s.

c.v.

Figure 1.7 One-dimensional flow through a control volume


b. The application of Eq. (1.24) to the case of a hydraulic turbine (Figure 1.9) results in

H1 = H2 + ht + hL ð1:27Þ

where ht is the work produced by the turbine per unit weight of fluid (ht =w) and hL is thefriction head loss between sections 1 and 2. The power extracted from the fluid by the tur-bine will be

Pf = γQht ð1:28Þ

c. The application of Eq. (1.24) to the case of flow in a pipe (Figure 1.10) gives

hL =H1− H2 ð1:29Þ

The power loss in pipe friction can be obtained from

Pfriction = γQhL ð1:30Þ

Discharge nozzle (2)

Pump

Suction nozzle (1)

Input work

Figure 1.8 Schematic of a pump, showing the suction and discharge nozzles

Discharge nozzle (2)

Output work

Turbine

Inlet nozzle (1)

Figure 1.9 Schematic of a hydraulic turbine


The power loss in fluid friction represents a transformation of energy from a useful formto a less useful form (heat). Accordingly, the heat generation by fluid friction will cause anincrease in the fluid temperature. If we assume a thermally insulated pipe, the increase influid temperature (ΔT) can be obtained from the energy balance as follows:

Pfriction = γQhL = m•CpΔT = ρQCpΔT

This equation can be simplified to obtain

ΔT =ghLCp

ð1:31Þ

The above temperature increase is only appreciable in long pipelines. In long crude oilpipelines, heat exchangers are used at intermediate stations for cooling the pumped fluidespecially during the summer in hot areas.

d. The pressure variation in a rotating fluid (assuming solid body rotation or forced vortex) canbe obtained by applying the Euler’s equation as follows:

Equation 1:18ð Þ! ∂

∂np + γ zð Þ = −ρan

If we apply the above equation, considering the case of the rotating fluid shown inFigure 1.11, we obtain

p

γ−ω2r2

2g+ z = Const: ð1:32Þ

The pressure variation in section A-A can be obtained from Eq. (1.32) by equating z to aconstant and the pressure at any radius r can be expressed as

pr−po =γω2

2gr2 ð1:33Þ

where po is the pressure at the center O.e. The hydraulic and energy gradient lines (HGL and EGL) are used for graphical represen-

tation of the variation of piezometric head pγ + z

� �and the total head p

γ +V2

2g + z� �

along the

pipe respectively.

Pipe

Inflow section (1) Outflow section (2)

Figure 1.10 Schematic of flow in a pipe


As shown in Figure 1.12, the piezometric head at a point (or section) is the head that will bereached if a piezometer tube is attached to that point (or section). The energy gradient line isabove the hydraulic gradient line by a distance equal to the velocity head (V2/2 g) and representsthe variation of the total head (H) along the pipe.

ω

A A

r

v = rω

Figure 1.11 Forced vortical motion in a cylindrical tank

Datum (z = 0)

Piezometer tube Stagnation tube

P/γ

HGLEGL

V2/2g

V

V

z

Figure 1.12 The hydraulic and energy gradient lines


Note:

1. The EGL is sometimes called the total energy line.2. The EGL has always a downward slope in the direction of flow because of friction losses.3. If the velocity is constant, the EGL and HGL are parallel lines.4. If the HGL is above the pipe centerline the pressure is above atmospheric and vice versa.5. The slope of the EGL represents the friction head loss/unit length.

1.4 Dimensionless Numbers

The important dimensionless numbers in fluid mechanics are the Reynolds number, Re; Machnumber, M; Froude number, Fr; and Weber number, We. The first two (Re andM) are of directrelevance to pumpingmachinery. The Reynolds number represents the ratio between inertia andviscous forces and it is important for achieving similarity in totally enclosed flows (such as flowin pipes and in air conditioning ducts). The Reynolds number is also important for achievingsimilarity for flow over fully submerged bodies (such as flow of air over a car or flow of waterover a submarine). On the other hand, theMach number represents the ratio between inertia andcompressibility forces and is important for achieving similarity in high-speed flows (such asflow of steam in a steam turbine nozzle or flow of air over a supersonic aircraft).

1.5 Laminar and Turbulent Flows

In laminar flows, the fluid moves in layers, every layer sliding over the adjacent one. There is nointerchange of momentum due to the movement of fluid particles between different layers. Theonly forces between fluid layers are the viscous shear forces in addition to pressure forces.On the other hand, turbulent flows are characterized by a high degree of mixing due to the

erratic movement of fluid particles between adjacent fluid layers, as shown in Figure 1.13. Inaddition to the viscous and pressure forces, there is a considerable turbulent shear force arisingfrom the strong interchange of momentum between adjacent layers. In fact, the turbulent shearstress is much greater than the viscous shear stress in turbulent flows.

u

yU

u

u

y

u

U

Laminar flow near a solid boundary Turbulent flow near a solid boundary

Figure 1.13 The velocity profiles in laminar and turbulent flows


1.6 Flow Separation

Themain streammay detach (separate) from the body surface as a result of the positive pressuregradient (∂p/∂x) due to the surface curvature (Figure 1.14). Separation may cause a transitionto turbulence. The location of the point of separation depends on the flow Reynolds number, thebody shape, and the surface roughness.

1.7 Cavitation

Cavitation is a phenomenon that is completely different from flow separation. It refers to theformation of vapor cavities in liquid flow as a result of the drop in liquid pressure belowthe vapor pressure. Cavitation always starts at the point of minimum pressure and it may causesevere damage in pump impellers and turbine runners due to the accompanying materialerosion. Figure 1.15 shows a typical p–T diagram for a pure substance. In thermodynamics,the vapor pressure is normally referred to as the saturation pressure, and it increases withthe increase in liquid temperature. At every temperature there is only one value for the vaporpressure.

Point of separation

Flow over an aerofoil with no separation Flow over an aerofoil with separation

Figure 1.14 Streamlines for flow over an aerofoil, showing the point of separation

T

p

Vaporization line

Solidification line

Sublimation line

Triple point

T1

Pv1

Liquid phase

Solid phase

Vapor phase

Figure 1.15 The p–T diagram for a pure substance showing the vaporization line


1.8 Friction Losses in Pipes and Pipe Fittings

Friction losses in piping systems are normally divided into two parts: major losses and minorlosses. The major losses represent the friction losses in straight pipes while the minor lossesrepresent the losses in various types of pipe fittings including bends, valves, filters, andflowmeters.

1.8.1 Major Losses

These losses represent the pipe friction losses and can be calculated using the Darcy–Weisbachformula which takes the form

hL =f LV2

2gDð1:34Þ

where f is the coefficient of friction,D is the pipe internal diameter, L is the pipe length, and V isthe average velocity in the pipe. Up to Reynolds number Re = 2000, the flow can be consideredlaminar in normal engineering applications and f can be calculated from f = 64/Re. For values ofRe > 2000, the flow can be considered turbulent and the friction coefficient can be obtainedfrom the Moody chart shown in Figure 1.16. In this case, f depends on the Reynolds numberand the pipe relative roughness and can be expressed as

f = f Re, ks=Dð Þ ð1:35Þ

where ks/D is the relative roughness. In the high Reynolds number range, the friction coefficientin rough pipes becomes more dependent on the relative roughness.The variation of the friction coefficient, f, with Re and ks/D is also given in a correlation

known as the Colebrook–White formula that can be written as:

f =0:25

log10ks

3:7D+5:74Re0:9

� �� 2 ð1:36Þ

The above formula can be used with reasonable accuracy in computational codes.

1.8.2 Minor Losses

These losses represent friction losses in various pipe fittings such as valves, bends, elbows,filters, and flowmeters. The minor losses can be calculated from

hLð Þminor =X

KV2

2gð1:37Þ

where K is a friction factor to be obtained experimentally for every pipe fitting. Tables forK areavailable for different pipe fittings.


Example 1.1The water jet emerging from a circular pipe along the x-axis has a velocity of 60 m/s. The waterjet impacts a curved blade as shown in Figure 1.17.

a. Determine the x-component of the force exerted by the jet on the blade if the blade isstationary.

b. Determine the same force if the jet moves to the right at a speed of 20 m/s.

SolutionAssuming frictionless flow, the magnitude of the fluid velocity relative to the blade at exit willbe the same as at inlet. Now, we can apply Eq. (1.6) as follows:

a. Equation (1.6) )X

F=X

m•V

� �out

−X

m•V

� �in

) −Fx = m•V2x−V1xð Þ= ρAjVj −Vj cos45

�−Vj

� �)Fx = ρAjV

2j cos45� + 1ð Þ

)Fx = 103π

40:05ð Þ2 60ð Þ2 cos45� + 1ð Þ= 12:1 × 103N

( )

103 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 80.000,01

0.000,05

0.0001

0.0002

0.00040.00060.00080.001

0.002

0.004

0.0060.008

0.008

0.009

0.010

0.015

0.020Res

ista

nce

coef

ficie

nt, f

0.025

0.030

0.040

0.050

0.060

0.070

0.080

0.090

0.100

0.01

0.0150.02

0.03

0.040.05

104

103 2 4 6 8104 2 4 6 8105 2 4 6 8 2 4 6 8106

105 106 107 108

Re =VDv

Rel

ativ

e ro

ughn

ess,

Ks D

νD3/2

Re f1/2 =2ghf

L

1/2

Boundary material

Equivalent sandroughness, k2

In millimetersGlass, plastic Smooth (Smooth)

Copper or brass tubing 1.5 × 10–1 (5 × 10–6)Wrought iron, steel 4.6 × 10–2 (1.5 × 10–4)Asphalted cast iron 0.12 (4 × 10–4)Galvanized iron 0.15 (5 × 10–4)Cast iron 0.26 (8.5 × 10–4)Concrete 0.3 to 3.0 (10–3 × 10–2)

(In feet)

f = 64Re

Complete turbulence, rough pipes

Laminar flow

Smooth pipe

Figure 1.16 Friction factor versus Re (Reprinted with minor variations after Moody (1944) withpermission from ASME)


b. The relative velocity, Vr = 60 − 20 = 40 m/s Equation (1.6)

)m• = ρAjVr = 10

3 π

40:05ð Þ2 × 40 = 78:5kg=s

) −Fx = m•

−Vr cos45�−Vrð Þ

)Fx = m•Vr cos45� + 1ð Þ= 78:5 × 40 cos45� + 1ð Þ

)Fx = 78:5 × 40 cos45� + 1ð Þ = 5:36 × 103N

Example 1.2Figure 1.18 shows a lawn sprinkler having three arms.Water enters the sprinkler from a verticalcentral pipe at a flow rate of 3 × 10−3 m3/s. The water jet issuing from each arm has a velocity of12 m/s. Determine the torque required to hold the lawn sprinkler stationary.

Data R = 20 cm and θ = 30�

Solution

The mass flow rate from each nozzle, m•

n = 13mT

•. Now, apply the angular momentum conser-

vation equation, Eq. (1.12),

Equation 1:12ð Þ)X

M =X

m•V× r

� �out

−X

m•V× r

� �in

) T =X

m•V× r

� �out

−X

m•V× r

� �in

zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{= 0

= 3m•

n Vexit × rexit� �

x

y

45°V = 60 m/s

D = 5 cm

Stationary bladePipe

(a)

Fx

45°

(b) (c)

Fx

VB = 20 m/s

45°

Figure 1.17 Jet impingement on a stationary or a moving blade


) T =m•

T Vexit × rexit� �

= ρQ Vexit × rexit� �

= 103 × 3 × 10−3 12 × 0:2 sin 30�ð Þ

) T = 3:6N:m

Example 1.3A boat is powered by a water jet as shown in Figure 1.19. The pump sucks water through a10 cm diameter pipe and discharges it through a 5 cm diameter pipe at a rate of 0.047 m3/s.Knowing that the boat is moving at a constant speed of 10 m/s, determine the total resistanceto the motion of the boat.

Solution

Q=VjAj =Vj ×π

40:05ð Þ2 = 0:047m3=s

)Vj = 23:9m=s)m•= ρVjAj = 47kg=s

(a) Plan view

R

θ

12 m/s

(b) Photograph of the sprinkler

Figure 1.18 A plan view and a photograph of a water sprinkler

Pump

Vboat

Vjetx

Figure 1.19 A schematic of a boat powered by a water jet


Apply the momentum equation in the x-direction,

XFx =

Xm�

exVex ) −Fresist =m

�ex Vboat −Vj

� �) −Fresist = 47 10−23:9ð Þ ) F

resist= 653:3N

Example 1.4An explosion occurs in the atmosphere when an anti-aircraft missile meets its target as shown inFigure 1.20. A shock wave spreads out radially from the explosion. The pressure differenceacross the wave (ΔP) is a function of time (t), speed of sound (C), and the total amount ofenergy released by the explosion (E).

1. Using the above variables, obtain a dimensionless parameter for ΔP.2. Knowing that the radial distance of the blast wave (r) from the center of explosion depends

on the same variables. Obtain another dimensionless parameter for r.3. For a given explosion, if the time (t) elapsed after explosion doubles (while C and E are

unchanged), by what factor will ΔP decrease?

Solution

1.Variable Δp t E CDimensions ML−1 T−2 T ML2T−2 LT−1

π1 = ta EbCcΔp1 ) M0L0T0 = Tð Þa ML2T−2� �b

LT−1� �c

ML−1T−2� �1

M ) 0 = b + 1 ) b = −1

L ) 0 = 2b + c−1 ) c = + 3

T ) 0 = a−2b−c−2 ) a= + 3

9>=>; ) π1 = t3 E−1C3Δp ) π1 =

Δp t3C3

E

pp+Δp

C

r

Shock wave

Explosion

Figure 1.20 Schematic of a shock wave resulting from an explosion


2.Variable r t E CDimensions L T ML2T−2 LT−1

π2 = ta EbCc r1 ) M0L0T0 = Tð Þa ML2T−2� �b

LT−1� �c

Lð Þ1

M ) 0 = b ) b = 0

L ) 0 = 2b+ c + 1 ) c = −1

T ) 0 = a−2b−c ) a = −1

9>=>; ) π2 = t−1 E0C−1 r ) π2 =

r

Ct

3. π1 =Δp t3C3

E=Const: and since C and E are constants, thereforeΔp =

C

t3, accordingly, if t is

doubled, Δp will decrease by a factor of 8.

Example 1.5Amodel test is performed to study flow through a large valve having a 2 ft diameter at inlet andcarrying water at a flow rate of 30 ft3/s. The model and prototype are geometrically similar,and the model inlet diameter is 0.25 ft. Knowing that the working fluid is the same for modeland prototype, determine the flow rate in the model for dynamically similar conditions.

Solution

DataDm

Dp=0:252:0

=18, Qp = 30 ft

3=s, same fluid) ρm = ρp, and μm = μp

For dynamic similarity, the Reynolds number must be the same for model and prototype.

Therefore, ρ VDμ

� �m= ρ VD

μ

� �p

) VmVp

= Dp

Dm= 8

Knowing that Q = VA, therefore,

Qm

Qp=Vm

Vp

Am

Ap=Vm

Vp

Dm

Dp

� �2

= 818

� �2

=18

) Qm = 30 ×18= 3:75 ft3=s

Example 1.6Figure 1.21 shows a pumping system in which water (ρ = 103 kg/m3, μ = 1.31 × 10−3 N � s/m2)is pumped from reservoir A to reservoir B at a rate of 0.07 m3/s. The water surface elevation inreservoir A is 20 m, and the surface elevation in reservoir B is 80 m while the pump elevationis 10 m. All pipes are made of commercial steel.

a. Show whether the flow in the pipe is laminar or turbulent and determine the frictioncoefficient.


b. What is the maximum shear stress in the pipe?c. What is the pump power consumption assuming a pump efficiency of 70%? Consider minor

losses at the pipe entrance and exit sections, and assume Kent. = 0.5.d. Sketch the hydraulic and energy gradient lines.

Data L1 = 15 m, L2 = 120 m, D1 =D2 = 15 cm

Solution

a. Velocity of flow in the pipe, V =Q/A = 4Q/πD2 = 4 × 0.07/π(0.15)2 = 3.96 m/s

Re=ρVD

μ=103 × 3:96 × 0:15

1:31 × 10−3 = 4:53 × 105 ) Flow is turbulent

b. The friction coefficient f is a function of Re and ks/D and can be determined using theMoodychart. For the given steel pipe, ks/D = 0.046/150 = 3.07 × 10−4 and

Re =ρVD

μ=103 × 3:96 × 0:15

1:31 × 10−3 = 4:53 × 105

Now, using the Moody chart, we obtain f = 0.017.c. The maximum shear stress in the pipe occurs at the pipe wall and

τo =Cf12ρV2 and Cf = f =4 ) τo =

f

8ρV2

τo =0:0178

× 103 × 3:96ð Þ2 = 33:3N=m2

h1 PumpA

B

L1, D1

L2, D2

z = 20 m

z = 10 m

z = 80 m

h2

Figure 1.21 Pumping system


d. Apply the energy equation between points a and b and neglect all minor losses, thus

=paγ + =V2a

2g+ za + hp = =pbγ + =V

2b

2g+ zb +

XhL ) hp = zb−zað Þ +

XhL ð1Þ

In the above equation,P

hL can be determined using Darcy’s formula and consideringminor losses at pipe inlet (from reservoir A) and at pipe exit (to reservoir B),

)X

hL =X

fL

D

V2

2g+Kent:

V2

2g+Kexit

V2

2gð2Þ

Therefore, )X

hL = fL1 + L2ð Þ

DV2

2g + Kent: +Kexitð ÞV2

2g

)X

hL = 0:01715 + 120ð Þ0:15

+ 0:5 + 1

� �3:962

2 × 9:81= 13:4m:

Substitute in Eq. (1) ) hp = (80 − 20) + 13.4 = 73.4 m.

Power required to drive the pump = γQhpηpump

= 9:81 × 0:07 × 73:40:7 = 72kW

e. The hydraulic and energy gradient lines are shown in Figure 1.22.

References1. Elger, D.F., Williams, B.C., Crowe, C.T., and Roberson, J.A. (2012) Engineering Fluid Mechanics, 10th edn.

John Wiley & Sons, Inc., Hoboken, NJ.2. White, F.M. (2011) Fluid Mechanics, 7th edn. McGraw Hill, New York.3. Fox, R.W. and McDonald, A.T. (2010) Introduction to Fluid Mechanics, 8th edn. John Wiley & Sons, Inc.,

Hoboken, NJ.4. Douglas, J.F., Gasiorek, J.M., Swaffield, J.A., and Jack, L.B. (2011) Fluid Mechanics, 6th edn. Prentice Hall

Publishers, New York.

HGL

a

b

Pump

EGL

HGL

EGL

L1, D1

L2, D2

z = 10 m

Figure 1.22 Hydraulic and energy gradient lines


5. Munson, B.R., Young, D.F., Okiishi, T.H., and Huebsch, W.W. (2012) Fundamentals of Fluid Mechanics, 7th edn.John Wiley & Sons, Inc., Hoboken, NJ.

6. Street, R.L., Watters, G.Z., and Vennard, J.K. (1995) Elementary Fluid Mechanics, 7th edn. John Wiley & Sons,Inc., New York.

7. Moody, L.F. (1944) Friction factors for pipe flow. Trans ASME, 66, 671.

Problems

1.1 Figure 1.23 shows a series-parallel piping system in which all pipes are 8 cm diameter. Ifthe flow rate at section 2 is 35 L/s calculate the total pressure drop (p1 − p2) in kPa, assum-ing that the fluid is water at 20 �C. Neglect minor losses and consider a friction coefficientf = 0.025 for all pipes.

1.2 Figure 1.24 shows two water reservoirs connected by a pipe (A) branching to two pipes(B and C). Knowing that all pipes are 8 cm diameter, determine the flow rate from reservoir1 to reservoir 2 if the valve in branch C is fully open. Consider Kv = 0.5 and assume afriction coefficient f = 0.02 for all pipes.

1.3 Water at 20 �C is to be pumped through 2000 ft of pipe from reservoir 1 to reservoir 2 at arate of 3 ft3/s, as shown in Figure 1.25. If the pipe is cast iron of diameter is 6 in. and thepump is 75% efficient, what is the pump power consumption? Neglect minor losses.

L = 150 m

L = 250 m

L = 100 m1

2

Figure 1.23 Series-parallel piping system

1

2LA = 100 m

LC = 70 m

LB = 50 m

z = 25 m

z = 0

10 m

30 m

Valve

Figure 1.24 Two water reservoirs connected by a branching pipe


1.4 Figure 1.26 shows a U-tube which is filled with water at 20 �C. The tube is sealed at A andis open to the atmosphere at D. If the tube is rotated about the vertical axis AC, identify thepoint of minimum pressure and determine the maximum speed of rotation if cavitation isto be avoided.

1.5 A fluid (ρ = 900 kg/m3 and ν = 10−5 m2 /s) flows at a rate of 0.2 m3/s through 500 m castiron pipe of diameter 20 cm. Knowing that the pipe has a 10� downward slope, determine(a) the friction head loss and (b) the pressure drop.

1.6 Water flows in a horizontal pipe with velocity 4 m/s and accelerating at a rate of 1.25 m/s2.The pipe is 5 cm diameter and 15 m long. Assuming a friction coefficient f = 0.01, deter-mine the pressure drop through the pipe.

1.7 A compressed air drill requires 0.2 kg/s of air at 500 kPa gage at the drill. The hose from theair compressor to the drill is 25 mm inside diameter. The maximum compressor discharge

A B

CD

75 mm

300 mm

Figure 1.26 A U-tube rotating about axis AC

1

2

Pump

L = 2000 ft

120 ft

Figure 1.25 Simple pumping system


pressure is 550 kPa gage at which air leaves the compressor at 40 �C. Neglect changes indensity and any effects due to hose curvature. Calculate the longest hose that may be used.Assume the hose roughness to be negligibly small.

1.8 The pipes shown in Figure 1.27 are all concrete pipes (ks = 1 mm). Water flows from A toB at a rate of 0.8 m3/s, find the friction head loss and the division of flow in the pipes fromA to B. Assume that f = 0.03 for all pipes. Determine the total power loss in fluid friction.

1.9 Fuel oil (sp. gr. = 0 .9 and ν = 5 × 10−5 m2/s) is pumped from reservoir A to reservoir B ata rate of 0.05 m3/s as shown in Figure 1.28. All pipes used in the system are commercialsteel pipes (ks = 0.046 mm) of diameter 15 cm. The lengths of the suction and deliverypipes are 20 and 180 m, respectively. Determine the total head developed by thepump. Assuming a pump overall efficiency of 75%, determine the pump powerconsumption.

1.10 Figure 1.29 shows a cylindrical tank full of water at 20 �C. The tank is completely closed,except for a vent at the center of its top plate. The tank is forced to rotate about its verticalaxis at a speed of 300 rpm. Determine the pressure distribution on the top plate. Deter-mine also the location and magnitude of the maximum pressure in the tank.

A

L1 = 500 m, D1 = 50 cm L5 = 800 m, D5 = 50 cm

L2 = 800 m, D2 = 25 cm

L3 = 600 m, D3 = 20 cm

L4 = 800 m, D4 = 25 cm B

Figure 1.27 Series-parallel piping arrangement

Pump

A

20 m

B

L1 = 20 m

L2 = 180 mTwo 90° bends, Kb = 0.5

Figure 1.28 Schematic of the pumping system


1.11 Figure 1.30 shows a closed tank 1 m diameter and 2 m long that is completely filled withwater. If the tank rotates about its own horizontal axis at a speed of 120 rpm, determinethe maximum pressure difference in the tank and identify the points of maximum andminimum pressures. Assume forced vortex motion.

1.12 Study the system shown in Figure 1.31 and answer the following questions:

a. What is the flow direction?b. What kind of machine is at B?c. Do the pipes AB and BC have the same diameter?

2 m

1 m

Figure 1.30 A cylindrical tank rotating about its own horizontal axis

30 cm

20 cm

Figure 1.29 A cylindrical tank rotating about its own vertical axis

A

C

B

EGLEGL

Figure 1.31 Schematic of a hydraulic system showing the energy gradient line


d. Where is the point of minimum pressure?

e. Sketch the HGL for the system.

1.13 Determine the elevation of the water surface in the upstream reservoir shown inFigure 1.32 if the system flow rate is 0.09 m3/s. Sketch the HGL and EGL, showingthe relative magnitudes and slopes. Calculate the major and minor losses in the pipingsystem.

Data L1 = 20 m, D1 = 25 cm, f1 = 0.02 and L2 = 15 m, D2 = 15 cm, f2 = 0.015,Kelbow = 0.5

1.14 Consider the water sprinkler shown in Figure 1.33, which has two arms rotating in a hor-izontal plane. Each arm has a radius r = 20 cm and ends with a nozzle of c.s. area 10 mm2.Knowing that θ = 20� and the flow rate delivered by the sprinkler is 0.30 L/s, what torque(T) will be required to hold the sprinkler arms stationary?

1.15 The total shear force, F, acting on the walls of a pipe due to fluid flow depends on thetube diameter, D; tube length, L; surface roughness, k; fluid density, ρ; viscosity, μ; andthe average flow velocity, V. Show that the relationship between these variables can bewritten in the following dimensionless form:

F

ρV2D2= f

ρVD

μ,L

D,k

D

� �

1.16 The power required to drive a fan, P, depends on the impeller diameter, D; speed ofrotation, N; flow rate, Q; and the fluid properties (ρ, μ). Show that the relationshipbetween P and the rest of the variables can be expressed in dimensionless form as

P

ρN3D5= f

Q

ND3,ρND2

μ

� �

How does the term ρND2/μ relate to the dimensionless parameters that you know?

A

B

Z = 10 mPipe 1

Pipe 2

Z = ?

Figure 1.32 Two reservoirs connected by pipes of different diameters


1.17 A three-nozzle water sprinkler discharges water while the sprinkler head rotates. Theangular velocity, ω, depends on the water density, ρ; volume flow rate, Q; water velocityat the nozzle exit, V; nozzle area, A; and the resisting frictional torque, Tf. Use dimensionalanalysis to develop an expression for the angular velocity.

1.18 The valve at the end of the inclined pipe shown in Figure 1.34 is gradually closed. Theresulting fluid deceleration during valve closure is 1.2 m/s2. Neglecting viscous effects,determine the difference in pressure between points 1 and 2.

Data L = 20 m, D = 10 cm, V = 4 m/s, θ = 30�

Angle = θRadius r

V2

T

ω

V2

V1

Figure 1.33 Schematic of a two-arm sprinkler

Valve

L

θ

Water

V

2

1

Figure 1.34 Decelerating flow in an inclined pipe


1.19 The power P generated by a certain wind turbine, depends on its diameter, D; the airdensity, ρ; the wind velocity, V; the speed of rotation, Ω; and the number of blades,N. Write the relationship between these variables in dimensionless form.

1.20 A model of the wind turbine mentioned in Problem 1.17 was built and tested in a windtunnel. The model which has a diameter of 0.5 m was tested at a wind speed of 40 m/s anddeveloped an output power of 2.7 kW when rotating at 4800 rpm.

a. Determine the power developed by a similar prototype of diameter 5 m when operatingat a wind speed of 12m/s.

b. What is the prototype speed for dynamically similar conditions?


Essentials of Fluid Mechanics · Essentials of Fluid Mechanics The basic fundamentals of fluid...

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