Chap.4. Control volume relations for fluid analysis We will present the analysis of the fluid flow based on a control volume ( not

differential element ) formulation.

Reynolds Transport Theorem

As a simple first example, consider a duct or streamtube with a nearly one-

dimensional flow as shown in Figure below. The selected control volume is a portion

of the duct which happens to be filled exactly by system 2 at a particular instant t. At

time t + dt, system 2 has begun to move out, and a sliver of system 1 has entered

from the left. The shaded areas show an outflow sliver of volume Ab Vb dt and an

inflow volume AaVa dt. Now let B be any property of the fluid (energy,

momentum, etc.), and let = dB/dm be the intensive value or the amount of B per

unit mass in any small portion of the fluid. The total amount of B in the control

volume is thus

Bsys could be total mass, total energy, total momentum, etc., of a system and Bsys per

unit mass is defined as βor βdB/dmThus βis the intensive equivalent of

Bsys.Applying a general control volume formulation to the time rate of change of Bsys ,

we obtain the following :

Conservation of Mass : For conservation of mass, we have that B = m and β =1 From the previous statement of conservation of mass and these definitions, Reynolds transport theorem becomes :

In steady flow, the mass flow per unit time passing through each section does not change , even if the pipe diameter changes. This is the law of conservation of mass. For the pipe shown in below Figure whose diameter decreases between sections 1 and 2, which have cross-sectional areas A1 and A2 respectively, and at which the mean velocities are υ1 and υ2 and the densities p1 and p2 respectively .

If the fluid is incompressible, e.g. water, with p being effectively constant, then

mass flow rate passing through any section is constant

Example :Water flows steadily through the Nozzle at 60 kg/s .The diameters

D1= 220 mm and D2=80 mm . Compute the average velocities at section 1 and

2 .

Q = m/ρ = 60/1000 = 0.06 m3 /s

V1 = Q / A1 = 1.58 m/s ,

V2 = Q / A2 = 12 m/s

Example: Three pipes steadily deliver water at to a large exit pipe

.The velocity

V2 =5 m/s, and the exit flow rate Q4= 0.0333 m3/s , if Q3 = 0.5 Q4 . Find V1 ,

V3 and V4

Solution: For steady flow we have

Q1 + Q2 + Q3 = Q4, or

V1 A1 + V2 A2 + V3 A3 = V4 A4

Q4 =(120 /3600 ) = 0.0333 m3 /s,

Substituting in above :

Q4= V4 A4 , V4 = 5.24 m/s

Conservation of Energy Bernoulli's equation

A streamline is chosen with the coordinates shown in below Figure. Around

this line , a cylindrical element of fluid having the cross - sectional area

dA and length ds is considered. Let p be the pressure acting on the lower

face, and pressure p + (dp/ds) ds acts on the upper face a distance ds

away. he gravitational force acting on thie element is its weight , pg dA ds.

Applying Newton’s second

∑F= m a , dm = ρ V = ρ dA ds , a = dv/dt , dw = g dm = g ρ dA ds

In the steady state, dv/dt = 0 and above equation would then become :

This equation is integrated with respect to s to obtain a relationship between points a finite distance apart along the streamline. This gives :

(1)

Multiplying each term by p,

If the streamline is horizontal, then the term ρgz can be omitted giving the following:

where ρv2/2 is called the dynamic pressure, ps the static pressure,

and pt the total pressure or stagnation pressure.Static pressure ps can be

detected, as shown in the Fig.a , by punching a small hole vertically

in the solid wall face parallel to the flow.

As Bernoulli's theorem applies to a flow line, it is also applicable to the flow in a pipe line as shown in Fig.b . Assume the pipe line is horizontal , and z1 = z2 . The following relative equation is obtained:

Fig.a Fig.b

Consequently, whenever A2 > A1 then v2 < v1 and p1 > p2 . As shown in below Figure , whenever water flows from point 1 to point 2, the energy equation for sections 1 and 2 are as follows from eq. (1) :

The line connecting the height of the pressure heads at respective points of

the pipe line is called the hydraulic grade line, while that connecting the heights of all the heads is called the energy line.

Application of Bernoulli’s equation Various problems on the one-dimensional flow of an ideal fluid can be solved by jointly using Bernoulli’s theorem and the continuity equation.

Venturi tube

a device where the flow rate in a pipe line is measured by narrowing a part of the tube is called a Venturi tube. In the narrowed part of the tube, the flow velocity increases. By measuring the resultant decreasing pressure , the flow rate in the pipe line can be measured.

Fig.c

From the continuity equation,

However , since there is some loss of energy between sections A , and A , in actual cases, the above equation is amended as follows:

C is called the coefficient of discharge. It is determined through experiment.

The above equation is also applicable to the case where the tube is

inclined.

Pitot tube

The device incorporating that idea is shown in below Figure . This device is

called a Pitot tube, the tube is so designed that at the streamlined end a hole is

opened in the face of the flow ,while another hole in the direction vertical to

the flow is used in order to pick out separate pressures. Let pA and vA

respectively be the static pressure and the velocity at position A of the

undisturbed upstream flow. At opening B of the Pitot tube, the flow is

stopped, making the velocity zero and the

pressure pB . B is called the stagnation

point. Apply Bernoulli's equation between A

and B .

And, since (pB – pC)/pg = H, the following equation is obtained:

In the case where the flowing fluid is a gas, pB- pC is measured with a U tube.

However, with an actual Pitot tube, since some loss occurs due to its shape and

the fluid viscosity, the equation is modified as follows:

where Cv is called the coefficient of velocity.

Flow through a small hole 1: The case where water level does not change . As shown , we study here the case where water is discharging from a small hole on the side of a water tank. Such a hole is called an orifice.

Assume that fluid particle A on the water surface has flowed down to section B. Then, from Bernoulli's theorem,

Assuming that the water tank is large and the water level does not change, at point A , vA = 0 and zA = H, while at point B, zB = 0 . If pA is the atmospheric pressure, then

Coeficient of velocity: The velocity of spouting flow at the smallest section is less than the theoretical value vB .

, Cv = 0.95 , is the Coeficient of velocity

C= 0.60 , is the Coeficient of discharge

,

Flow through a small hole 2 : the case where water level changes.The theoretical flow velocity is :

Assume that dQ of water flows out in time dt with the water level falling by -dH . Then

The time needed for the water level to descend from H1 to H2 is

Flow through a small hole 3 : The section of water tank where the descending velocity of the water level is constant. Assume that the bottom has a small hole of area a, through which water flows then :

Example : Determine the velocity at point 1 in the devise as shown in below

Example : for the siphon , what are the pressure of the water in the tube at B and

at A .

Bernolli with Pump Example : The horizontal pump discharges water at 0.0158 m3/s . Neglecting losses , what power is delivered to the water by the pump. Take ρwater = 1000 kg/m3 .

Solution: First we need to compute the velocities at sections (1) and (2):

Bernolli with Turbine Example: The water flows from an upper reservoir to a lower one while passing through a turbine as shown . Fined the power generated by the turbine

Equation of momentum

+193.6 - 0

Elev 193.6

For linear momentum, we have that :

For Steady state :

F = movo – mi vi ∑

Application of equation of momentum

(a) Flow in a curved pipe

In the case where fluid flows in a curved pipe :

In this equation, m is the mass flow rate.

If Q is the volumetric flow rate , then

the following relation exists:

Fx

Fy

(b) Force of a jet

Applying the equation of momentum to the direction along the flat board :

-Fx = ( ρ Q1 v1 cosθ - ρ Q2 v2 cosθ ) - ρ Q v

Fy = ( ρ Q1 v1 sinθ - ρ Q2 v2 sinθ ) – 0

(c) Double Nozzle

-Rx + P1 A1 = ( ρ Q2 v2 cosθ + ρ Q3 v3 cosα ) - ρ Q1 v1

Ry = ( ρ Q2 v2 sinθ - ρ Q3 v3 sinα ) - 0

Ry

Rx

θ

α

Fy

Fx

(d) Moved and stationary blades

U: velocity of the blade relative to the earth

Vr : velocity of the fluid relative the blade

V1 : velocity of the fluid relative the earth

For multi-moving blades

Q = A V1

horizontal component for the fluid on the each moving blade is : The

-Rx = ρ A V1 ( -Vr cosβ - Vr cosα )

Rx = ρ A V1 Vr ( cosβ + cosα )

The vertical component for the fluid

on the each moving blade is :

Ry = ρ A V1 (Vr sinβ –

Vr sinα )

Ry = ρ A V1 Vr ( sinβ –

sinα )

β

α

v1

vr

u

vr

u

v2

u

β

α Rx

Ry

β

α

β

α Rx

Ry

Vr

For the stationary single blade V1=Vr

Rx =ρ A Vr2 ( cosβ + cosα )

( Ry =ρ A Vr2 ( sinβ - sinα

(e) Jet Pump If vo is the velocity of the jet discharging at section 1 and v , the

velocity of the surrounding water, and assuming that mixing finishes at section 2 and the flow is then at uniform velocity v2 , then we have the following:

By the low of momentum

(f) Efficiency of a propeller

From the changes in momentum and kinetic energy across the revolving face of the propeller, the thrust T is given by :

Example: Water flows through the elbow and exits to the atmosphere. The pipe diameter is D1 = 10 cm , while D2 = 3 cm. The flow rate is 0.0135 m3/s, the pressure p1 = 233.105 kN/m2. Neglecting the weight of water and elbow, estimate the force on the flange bolts at section 1.

m = ρ A1 V1 = (998) (π×0.12 /4) (1.95) = 15.25 kg/s.

Fbolt = 233105 ×( π× 0.12 /4) + 15.25 × ( 21.7× cos 40 + 1.95 ) = 2100 N

Example A water jet 4 cm in diameter with a velocity

of 7 m/s is directed to a stationary turning vane with θ = 40o. Determine the force Fb necessary to hold the vane stationary when : (a)the cart moving to the right with a velocity Uc = 2 m/s (b) the cart moving to the right with a velocity Uc = 2 m/s

Fbolt

Solution :

Problems 1. Find the flow velocities v1 , v2 and v3 in the conduit shown in Fig.1. The flow

rate Q is 800 L/min and the diameters d1 , d2 and d3 at sections 1, 2 and 3 are 50, 60 and 100 mm respectively. 2. Water is flowing in the conduit shown in Fig.1 .If the pressure p1 at section 1 is 24.5 kPa, what are the pressures p2 and p3 at sections 2 and 3 respectively ? 3. In Fig. 2, if water flows at rate Q = 0.013 m3 /s radially between two discs of radius r1 = 30cm each from a pipe of radius r1 = 7cm, obtain the pressure and the flow velocity at r2 =12 cm. Assume that h = 0.3 cm and neglect the frictional l oss.

(a)

(b)

Fig.1 Fig.2

4.As shown in Fig. 3, a tank has a hole and a << A. Find the time necessary for the tank to empty . 5.As shown in Fig. 4, water flows out of a vessel through a small hole in the bottom. What is a suitable section shape to keep the velocity of descent of the water surface constant? Assume the volume of water in thevessel is 21, R/d =100 (where R is the radius of the initial water surface in the vessel, d the small hole on the bottom ) , and the flow discharge coefficient of the small hole is C = 0.6. What should R and d be in order to manufacture a water clock for measuring 1 hour? 6.In the case shown in Fig.5, water at a flow rate Q = 0.2m3 /s is supplied to the cylindrical water tank of diameter 1 m discharging through a round pipe of length 4 m and diameter 15 cm. How deep will the water in the tank be ?

7.As shown in Fig.6, a jet of water of flow rate Q and diameter d strikes the stationary plate at angle 8. Calculate the force on this stationary plate and its direction. Furthermore, if θ= 60o, d = 25 mm and Q = 0.12 m3 /s, obtain Q1 , Q2 and F.

Fig.3 Fig.4 Fig.5

8.As shown in Fig. 7 , if water flows out of the tank of head 50cm through the throttle, obtain the pressure at the throat.

9.A jet-propelled boat as shown in Fig.8 is moving at a velocity of 10 m/s. The river is flowing against the boat at 5m/s. Assuming the jet flow rate is 0.15 m3 /s and its discharge velocity is 20 m/s, what is the propelling power of this boat? ( Jet boats like this are actually in use.)

Answer 1. v 1 = 6.79 m/s , v2 = 4.02 m/s, v3 = 1.70 m/s

2. p2 = 39.5 kPa , p3 = 46.1 kPa

3. vr = 5.75 m/s , pr - po = -1.38 x 104 Pa

5. condition of the section shape

6. H = 2.53 m

Fig.6 Fig.7

Fig.8

4.

7.

8. -7.49 mH,O 9. F=749N

Chap 5. Dimensional Analysis , Similarity and Modeling

The method of dimensional analysis is used in every field of engineering, especially in such fields as fluid dynamics and thermodynamics where problems with many variables are handled. This method derives from the condition that each term summed in an equation depicting a physical relationship must have same dimension. By constructing non-dimensional quantities expressing the relationship among the variables, it is possible to summaries the experimental results and to determine their functional relationship. Next, in order to determine the characteristics of a full-scale device through model tests , besides geometrical similarity, similarity of dynamical conditions between the two is also necessary. When the above dimensional analysis is employed, if the appropriate non-dimensional quantities such as Reynolds number and Froude number are the same for both devices, the results of the model device tests are applicable to the full-scale device.

Dimensionless Analysis When the dimensions of all terms of an equation are equal the equation is dimensionally correct. In this case, whatever unit system is used, that equation holds its physical meaning. If the dimensions of all terms of an equation are not equal, dimensions must be hidden in coefficients, so only the designated units can be used. Such an equation would be void of physical interpretation. Utilizing this principle that the terms of physically meaningful equations have equal dimensions , the method of obtaining dimensionless groups of which the physical phenomenon is a function is called dimensional analysis. If a phenomenon is too complicated to derive a formula describing it , dimensional analysis can be employed to identify groups of variables which would appear in such a formula. By supplementing this knowledge with experimental data, an analytic relationship between the groups can be constructed allowing numerical calculations to be conducted.

Buckingham,s π theorm In order to perform the dimensional analysis, it is convenient to use the π

theorem.Consider a physical phenomenon having n physical variables υ1 , υ2

, υ3 , . . ., υn , and k basic dimensions' (L, M, T or L, F, T or such) used to describe them. The phenomenon can be expressed by the relationship among n

- k = m non-dimensional groups π1 , π2 , π3 , . . . πm . In other words, the

equation expressing the phenomenon as a function f of the physical .

can be substituted by the following equation expressing it as a function φ of a smaller number of non-dimensional groups:

This is called Buckingham's x theorem. In order to produce π1 , π2 , π3 , . . . πm . k core physical variables are selected which do not form a π themselves.

Each π group will be a power product of these with each one of the m remaining variables. The powers of the physical variables in each x group are determined algebraically by the condition that the powers of each basic dimension must sum to zero. Example Let us study the resistance of a sphere placed in a uniform flow . .In this case the effect of gravitational and buoyancy forces will be neglected. First of all, as the physical quantities influencing the drag F of a sphere, sphere diameter d, flow velocity U, fluid density ρ and fluid viscosity μ , are candidates. In this case n = 5, k = 3 and m = 5 – 3 = 2 , so the number of necessary non-dimensional groups is two. Select ρ , U and d as the k core physical quantities, and the first non-dimensional group n, formed with D, is

Solving the above simultaneously gives

Next, select ,u with the three core physical variables in another group, and

= F

F

Example : As the quantities influencing pressure loss ∆p/L per unit length due to pipe

friction, flow velocity v, pipe diameter d , fluid density ρ , fluid viscosity μ and pipe wall roughness ε , are candidates. In this case, n = 6, k = 3, m = 6 - 3 =

3. Obtain π1 , π2 , π3 by the same method as in the previous case, with ρ , v and d as core variables: Therefore, from the π theorem, the following functional relationship is obtained

F μ

Similarity : When the characteristics of a water wheel, pump, boat or aircraft are obtained by means of a model, unless the flow conditions are similar in addition to the shape, the characteristics of the prototype cannot be assumed from the model test result. In order to make the flow conditions similar, the respective ratios of the corresponding forces acting on the prototype and the model should be equal. Similarity generally includes three basic classifications in fluid mechanics: (1) Geometric similarity (2) Kinematic similarity (3) Dynamic similarity

List the dimensions of each variable according to {MLT} or {FLT} is given in this Table

Geometric similarity : Al l linear dimensions of the model are related to the corresponding dimensions of the prototype by a constant scale factor SFG . Consider the following airfoil section in this Figure :

For this case, geometric similarity requires the following:

Kinematic Similarity :The velocities at corresponding points on the model and prototype are in the same direction and differ by a constant scale factor SFk. Therefore , the flows must have similar streamline patterns , Flow regimes must be the same. These conditions are shown in the following kinematically similar flows.

Dynamic Similarity:This is basically met if model and prototype forces differ by a constant scale factor at similar points.This is illustrated in the following figure for flow through a sluice gate .

The forces acting on the flow element are due to gravity FG , pressure FP , viscosity Fv , surface tension FT , inertia F, and elasticity FE . The forces can be expressed as shown below.

Non-dimensional groups which determine flow similarity

1-Reynolds number

2-Froude number

3- Weber number

4-Mach number

Example : A copepod is a water crustacean approximately 1 mm in diameter. We want to know the drag force on the copepod when it moves slowly in fresh water. A scale model 100 times larger is made and tested in glycerin at V =30 cm/s. The measured drag on t he model is 1.3 N. For similar conditions, what are the velocity and drag of the actual copepod in water .

Water (prototype): μP = 0.001 kg/(m.s) , ρP = 998 kg/m3

Glycerin (model): μm = 1.5 kg/(m.s) , ρm = 1263 kg/m3

The length scales are Lm= 100 mm and LP =1 mm. Assume that Equation below is applies and the fluid properties are :

F= ρ V2 D2 f( Re )

Solution : We are given enough model data to compute the Reynolds number and force coefficient

Problems 1. Obtain the drag on a sphere of diameter d placed in a slow flow of velocity U. 2. Assuming that the traveling velocity a of a pressure wave in liquid depends

upon the density ρ and the bulk modulus k of the liquid, derive a relationship

for a by dimensional analysis.

3. Assuming that the wave resistance D of a boat is determined by thevelocity v

of the boat, the density p of fluid and the acceleration of gravity g, derive the

relationship between them by dimensional analysis.

4. When fluid of viscosity p is flowing in a laminar state in a circular pipe of

length L and diameter d with a pressure drop ∆p, obtain by dimensional

analysis a relationship between the discharge Q and d , ∆p/L and p.

5. Obtain by dimensional analysis the thickness δ of the boundary layer distance

x along a flat plane placed in a uniform flow of velocity U ( density p, viscosity

v ).

6. Fluid of density p and viscosity μ is flowing through an orifice of diameter d bringing about a pressure difference ∆p . For discharge Q , the discharge

coefficient is given by :

Fm/ ρm Vm2 Dm

2= FP/ ρp Vp2 Dp

2

Show by dimensional analysis that there is a relationship C = f( Re ). 7.An aircraft wing , chord length 1.2 m, is moving through calm air at 20°C and 1 bar at a velocity of 200 km/h. If a model wing of scale 1:3 is placed in a wind tunnel, assuming that the dynamical similarity conditions are satisfied by Re, then: (a) If the temperature and the pressure in the wind tunnel are respectively equal to the above, what is the correct wind velocity in the tunnel? (b) If the air temperature in the tunnel is the same but the pressure is increased by five times, what is the correct wind velocity ? Assume that the

viscosity p is constant.

(c)If the model is tested in a water tank of the same temperature, what is the correct velocity of the model ? For a pump of head H, representative size I and discharge Q, assume that

the following similarity rule is appropriate:

8.Obtain the Froude number when a container ship of length 245m is sailing at 28 knots. Also, when a model of scale 1:25 is t ested under similarity conditions where the Froude numbers are equal, what is the proper towing velocity for the model in the water tank? Take 1 knot = 0.514 m/s. 9. For a pump of head H , representative size L and discharge Q , assume that the following similarity rule is appropriate: where, for the model, subscript m is used. If a pump of Q = 0.1 m3 /s and H = 40m is model tested using this relationship in t he situation Q, = 0.02m3/s and H, = 50m , what is the model scale necessary for dynamical similarity?

Answer

1

2

3

4

5

6

7

8

9