fluid-mechanics-4-5

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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 βor βdB/dmThus βis 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 ρv
2
/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 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 smallestsection 
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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Equat
ion of 
mom
entum 
+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 anon-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/m
3 
Glycerin (model): μm = 1.5 kg/(m.s) , ρm = 1263 kg/m
3 
 
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 relationshipbetween 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 Vm
2 Dm
 2= FP/ ρp Vp
2 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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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