Chapter 6
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
g(hr + hp)  0.5V22 + 2f
L
V22
D
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.
9.81100 = (0.5 + 400f)V22
The expression (6.2-5) for friction factor can be used for the procedure.
 8 12

1
 
f = 2 

( A  B) 3 / 2 
 N Re 

1 / 12
(6.2-5)



1

A = 2.457 ln 
0.9
(
7
/
N
)

0
.
27

/
D
Re



16
 37,530 

and B = 
 N Re 
16
However for a smooth pipe, a simpler equation predicts adequately the friction factor for
turbulent flow.
f = 0.0791NRe-0.25
(6.3-1)
Table 6.3-2 lists the Matlab program and the result from the calculation.
______ Table
6.3-2 Matlab program to compute the mass flow rate __________
% Example 6.3-3, 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)
6-13
R=den*D*V2/vis;
test=abs(R/Rsave-1);
if test<.01, break, end
Rsave=R;
end
mdot=den*V2*pi*D^2/4;
fprintf('Mass flow rate (kg/s) = %g\n',mdot)
>> e6d3d3
V2(m/s) = 20.75
V2(m/s) = 31.82
V2(m/s) = 32.63
V2(m/s) = 32.68
Mass flow rate (kg/s) = 25665.8
Example 6.3-4. ---------------------------------------------------------------------------------Water is withdrawn from a reservoir as shown in Figure 6.3-4. 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/ms.
(1)
Water
20oC
20 m
Globe valve
(fully open)
50 m
20 m
40 m
Screwed
elbow
Figure 6.3-4 Flow from a reservoir through a piping system.
Solution ------------------------------------------------------------------------------------Applying the mechanical energy balance between (1) and (2) we have
P1

+ gz1 +
1V12
2
+ wp =
P2

+ gz2 +
The equation is simplified to
6-14
 2V22
2
+ ef
(2)
z1  z2 =
2
2
2
e
L V2
V2
V
+ f = (Kentrance + Kvalve+ 2Kelbow+ 0.5) 2 + 4f
D 2g
2g
2g
g
z1  z2 = (Kentrance + Kvalve+ 2Kelbow+ 0.5 + 4f
2
L V2
)
D 2g
2
40 = [0.5 + 7.5 + 20.95 + 0.5 + 4(110/0.04)f]
V2
2g
392.4 = (27.5 + 110,00f)V22
This equation and the friction factor expression, equation (6.2-5), are used in the Matlab
program listed in Table 6.3-3. For wrought iron with a diameter of 40 mm., we have the
roughness  = 0.045 mm. Therefore

= 0.0011
D
______ Table
6.3-3 Matlab program to compute the volumetric flow rate __________
% Example 6.3-4, solve for the volumetric flow rate
%
vis=.001;den=1000;D=.04;
R=1e5;eoD=.0011;
for i=1:20;
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(392.4/(27.5+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
Rsave=R;
end
Q=V2*pi*D^2/4;
fprintf('Volumetric flow rate (m3/s) = %g\n',Q)
>> e6d3d4
V2(m/s) = 2.09
V2(m/s) = 2.08
Volumetric flow rate (m3/s) = 0.00261025
6-15
6.3c 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 trialand-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 =
2. Compute the Reynolds number NRe =
4Q
.
D 2

VD
and the pipe roughness ratio
.
D

3. Compute the friction factor from the following equation
 8 12

1
 
f = 2 

( A  B) 3 / 2 
 N Re 

1 / 12
, where



1

A = 2.457 ln 
0.9
 (7 / N Re )  0.27 / D 

(6.2-5)
16
 37,530 

and B = 
 N Re 
16
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 6.3-5 ---------------------------------------------------------------------------------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. 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/fts.
6-16
Figure 6.3-2 Steady oil transferred to storage tank.
Solution -----------------------------------------------------------------------------------------4Q
1. Guess the pipe diameter D and compute the mean velocity V =
.
D 2
Q = 506 gpm = 1.1275 ft3/s
Let D = 0.4 ft  V =
4Q
4  1.1275
=
= 8.97 ft
2
D
  0 .4 2
2. Compute the Reynolds number NRe =
NRe =

VD
and the pipe roughness ratio
.
D

53  8.97  0.4
VD
=
= 21,443
13.2  0.000672


0.00015
=
= 0.000375
D
0 .4
3. Compute the friction factor from the following equation
 8 12

1
 
f = 2 

( A  B) 3 / 2 
 N Re 

1 / 12
, where



1

A = 2.457 ln 
0.9
 (7 / N Re )  0.27 / D 

(6.2-5)
16
 37,530 

and B = 
 N Re 
f = 0.00658
4. Compute the diameter from the energy equation.
Applying the mechanical energy balance between (2) and (4) we have
6-17
16
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  4Q 
P2  g(z4  z2) = 2 f  2 
D  D 


32 fLQ 2
D=  2

 [ P2  g ( z 4  z 2 )] 


32 LQ 2
Let Con =  2

 [ P2  g ( z 4  z 2 )] 
0.2
L V2
D 2
2


32 LQ 2
=  2

 [ P2  g ( z 4  z 2 )] 
0.2
f0.2
0.2
then D = Con f0.2
The term Con does not change during iteration for the diameter.


32 LQ 2
Con =  2

 [ P2  g ( z 4  z 2 )] 
0.2


32  53  6000  1.12752
=  2

  [132.7  144  32.2  53  32.2  200 
0.2
Con = 1.3676  D = 1.36760.006580.2 = 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 6.3-4.
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.
6-18
______ Table
6.3-4 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
6-19