# Exam #1 Solutions

```CEE 345, Spring 2012, Midterm Exam Solutions: Pipe Networks and Pumps
l V2
. Although it is true
D 2g
that f declines with increasing V, the velocity head increases in proportion to V2. The
dependence of f on V is such that f varies with Vn, where n has a value between 1 (for
laminar flow) and 0 (for fully turbulent flow). Therefore, the product f*V2 is proportional to
Vm, where m varies between 1 and 2. As a result, hL always increases when V increases.
1. The frictional headloss is given by the D-W equation as hL  f
2. (a) Under fully turbulent flow conditions, f is independent of Re and therefore independent of
V for flow in a given pipe. Consequently, the D-W equation predicts that the headloss is
proportional to V2. The H-W equation predicts that the headloss is proportional to V1.85.
Therefore, for fully turbulent flow, the D-W equation predicts a stronger dependence of hL/l
on V than the H-W equation does.
(b) For laminar flow conditions, f equals 64/Re and is therefore proportional to V1. Under
these conditions, the D-W equation predicts that hL/l varies with V1. The H-W equation
predicts the same dependence of hL/l on V (i.e., hL/l is proportional to V1.85) for any value of
Re. Therefore, if the flow is laminar, the H-W equation predicts a stronger dependence of hL/l
on V than the D-W equation does (although the H-W equation was never intended to be used
for laminar flow, so its use under those conditions would be inappropriate).
3.
f 
hL D
0.1 m
  0.33
 0.033
V 2g l
1.0 m
2
From the Moody diagram, under fully turbulent flow conditions (high Re), a friction factor of
0.033 corresponds to a relative roughness of 0.0065. The relative roughness is /D, so
 
    D  0.0065  0.1 m   0.00065 m  0.65 mm
D
If the headloss is one-third of a velocity head for a travel distance of 1 m, it is 333.3 velocity
heads for a travel distance of a km. That is:
l V2
1000 m  0.5 m/s 
hL  f
  0.033
 4.20 m
D 2g
0.1 m 2  9.81 m/s 2 
2
Because the pipe is horizontal, the fluid’s elevation head remains constant, and because the
pipe has a fixed diameter, the velocity (and therefore the velocity head) remains constant. As
a result, the headloss is manifested entirely as a pressure loss. The headloss can be converted
to an equivalent pressure by multiplying by the specific weight of water, so:
p   hL   9.79 kN/m 2   4.20 m   41.1
kN
 41.1kPa
m2
1
4. (a) The total headloss is the sum of the frictional headloss in the pipe (i.e., the major
headloss) and the minor headlosses associated with the two bends and the loss of velocity
when the water enters the tank. The total headloss is:
hL ,tot  hL , f  2hL ,elbow  hL ,entrance
2
 l
V
  f  2 K elbow  K ent 
 D
 2g
2
45 ft

V
  0.023
 2  0.8  1.0 
0.333 ft

 2g
 5.71
V2
2g
The fraction of the total headloss attributed to the minor headlosses is therefore:
hL ,minor
hL ,tot

2  0.8  1.0
 0.455
45 ft
0.023
 2  0.8  1.0
0.333 ft
(b) The power lost in the two elbows, and therefore the power that could be saved by
eliminating the elbows, is:
Pelbows   QhL ,elbows
2
5.16 ft/s  

lb  
ft 3  
ft-lb

  18.6
  62.4 3   0.45   2  0.8
2


ft  
s 
s
2  32.2 ft/s  


ft-lb   1 hp 

 18.6

  0.034 hp
s   550 ft-lb/s 

5. The expression for headloss for each pipe individually can be written as hL = KQn, with n =
1.5, and K for each pipe equal to:
K  6.5
50
1.5x108   0.78
0.05
The value of K for a pipe that is equivalent to a group of real pipes in parallel is
K eq    K 1/ n  , so, for the combination of pipes A and B:
n
Keq ,A B   K A1/ n  K B1/ n 
n
  0.781/1.5  0.781/1.5 
1.5
 0.276
2
Pipes C and D are identical to pipes A and B, so Keq,C+D is also 0.276. The two pairs of pipes
are in series with one another, and the K value of a pipe that is equivalent to other pipes in
series is the sum of the K values of those pipes, so:
Keq,ABCD  2  0.276  0.552
Finally, we can use this value in conjunction with the definition of K to find the equivalent
length:
K eq ,A  B C D  0.552  6.5
leq
Deq4
C f ,eq
 0.552  0.05  35.4
leq 
 6.5 1.5x108 
4
Because all parameters are specified to have SI units, the length of the equivalent pipe is
35.4 m.
6. The system diagram indicates that the booster pump will be installed in series with the
existing pump. Therefore, all the flow will have to go through both pumps, and both pumps
be the sum of the heads added by the two pumps. Based on this logic, for any given flow rate
Q through the pumps, TDHtot will be TDHorig + TDHboost for that Q. This summation can be
carried out graphically.
For the pump and system to operate under steady conditions, TDHtot will have to equal the
headloss in the system, as characterized by the system curve. That equality is satisfied under
the conditions where the TDHtot curve intersects the system curve. The value of Q at the
operating point indicates the flow rate through both pumps and the piping. The head added
by the original pump is the value of TDHorig at that Q, and the head added by the booster
pump is the value of TDHboost at that Q. As seen on the following diagram, this condition
occurs around a Q of 80 cfs, at which TDHorig = 68 ft, and TDHboost = 42 ft, and TDHtot =
110 ft.
3
4
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