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Chapter 6
6.3 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.
6.3a Unknown Driving Force
This is the easiest situation where no iterative steps are required for problems involving
Newtonian fluids.
P1

+ gz1 +
1V12
2
+ wp =
P2

+ gz2 +
 2V22
2
+ ef
(6.2-1)
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.
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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 6.3-1. ---------------------------------------------------------------------------------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.
(1)
Globe valve
(fully open)
Water
20oC
H
20 m
10 m
(2)
20 m
Standard
elbow
Figure 6.3-1 Two reservoirs systems.
Solution -----------------------------------------------------------------------------------------Applying the mechanical energy balance between (1) and (2) we have
P1

+ gz1 +
1V12
2
+ wp =
P2

+ gz2 +
 2V22
2
+ ef
The equation is simplified to
L V2
V2
H = z1  z2 =
= (Kentrance + Kvalve+ 2Kelbow+ Kexit)
+ 4f
D 2g
g
2g
2
e
L V
H = f = (Kentrance + Kvalve+ 2Kelbow+ Kexit + 4f )
D 2g
g
Q
0.04
V=
=
= 5.09 m/s
2
R
 (0.052 )
ef
The kinematic viscosity of water at 20oC is  = 10-6 m2/s, therefore
NRe =
VD

=
(5.09)( 0.1)
= 5.09105
6
10

0.045
=
= 0.00045
D
100
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The friction factor can be determined from the following equation
f = { 1.737 ln[0.269


2.185
14

ln (0.269
+
)]}-2 = 0.0044
D
D
N Re
N Re
H = (Kentrance + Kvalve+ 2Kelbow+ Kexit + 4f
L V2
)
D 2g
The difference in elevation H of the two reservoirs is then calculated with values of loss
coefficients given in Table 6.2-1.
50 5.09 2
H = (0.5 + 7.5 + 20.7 + 1 + 40.0044
)
= 25.4 m
0.1 2  9.81
--------------------------------------------------------------------------
6.3b 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 achieve. Using this approach, the following steps are
needed:
1. Compute the roughness ratio

.
D
2. Assume NRe = 1105
3. Compute the friction factor from the following equation
 8 12

1
 
f = 2 

3/ 2
( A  B) 
 N Re 

1 / 12
(6.2-5)



1

In this equation A = 2.457 ln 
0.9
 (7 / N Re )  0.27 / D 

16
 37,530 

and B = 
 N Re 
16
4. Compute the mean velocity from the energy equation.
5. Compute NRe =
VD
. If this is within the accepted tolerance of the value used in

step (3) then proceed to step (6). If not, return to step (3).
6. Compute the flow rate from Q = VD2/4.
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Example 6.3-2.2 ---------------------------------------------------------------------------------Oil is pumped from a tanker to a refinery storage tank as shown in Figure 6.3-2. 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/fts.
Figure 6.3-2 Steady oil transferred to storage tank.
Solution ------------------------------------------------------------------------------------------

.
D
For commercial steel pipe Schedule 40 with a nominal diameter of 6 in., we have  = 0.00015
ft, D = 0.5054 ft. Therefore

= 0.000297
D
1. Compute the roughness ratio
2. Assume NRe = 1105
3. Compute the friction factor from the following equation
 8 12

1
 
f = 2 

( A  B) 3 / 2 
 N Re 

1 / 12
(6.2-5)



1

In this equation A = 2.457 ln 
0.9
 (7 / N Re )  0.27 / D 

2
16
 37,530 

and B = 
 N Re 
Wilkes, James, Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 133
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16
With

= 0.000297 and NRe = 1105, f = 0.00487
D
4. Compute the mean velocity from the energy equation.
Applying the mechanical energy balance between (2) and (4) we have
P2

+ gz2 +
1V22
2
+ wp =
P4
+ gz4 +

 4V42
2
+ ef
Since there is no work and no change in kinetic energy, the equation is simplified to
P2 + gz2 = gz4 + ef = gz4 + 4 f
L V2
D 2
The oil velocity inside the pipe is then
1/ 2
 D

V= 
[ P2  g ( z 4  z 2 )]
 2 fL

1/ 2
0.5054


(132.7  32.2  144  53  32.2  200) 
V= 
 2  0.00487  53  6000

5. Compute NRe =
= 6.684 ft
VD
. If this is within the accepted tolerance of the value used

in step (3) then proceed to step (6). If not, return to step (3).
NRe =
VD 53  6.684  0.5054
=
= 20,185
13.2  0.000672

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 6.3-1.
Re =
Re =
Re =
Re =
Re =
100000 , f = 0.00487, V(ft/s) =
20185 , f = 0.00663, V(ft/s) =
17310 , f = 0.00687, V(ft/s) =
17001 , f = 0.00690, V(ft/s) =
16965 , f = 0.00690, V(ft/s) =
6.684
5.732
5.630
5.618
5.617
6. Compute the flow rate from Q = VD2/4.
 gal 
0.5054 2  ft 3 
 s 
Q = VD /4 = 5.617 
7.48  3  60 
= 506 gpm


4
 min 
 ft 
 s 
2
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______ Table
6.3-1 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=.5054
eoD=.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 6.3-3.3 ---------------------------------------------------------------------------------Water is withdrawn from a reservoir as shown in Figure 6.3-3. 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/ms.
Figure 6.3-3 Flow from a reservoir through a turbine.
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Middleman, Stanley, An Introduction to Fluid Mechanics, Wiley, 1998, p. 460
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