Pipe Flow Problems-III
Example 1
A centrifugal pump has a 100 mm diameter suction pipe and a 75 mm diameter delivery
pipe. When discharging 15 l/s of water, the inlet water mercury manometer with one
limb exposed to the atmosphere recorded a vacuum deflection of 198 mm; the mercury
level on the suction side was 100 mm below the pipe centerline. The delivery pressure
gauge, 200 mm above the pump inlet, recorded a pressure of 0.95 bar. The measured in
put power was 3.2 kW. Calculate the pump efficiency. (See fig.1)
Solution;
Manometric head = rise in total head
 p1 V12 
p2 V22
 1bar  10.198m of water
Hm 

 z  

g 2 g

g
2
g


p2
 0.95 *10.198  9.65m of water
g
V2  3.39m / s;
V22
 0.588m
2g
V12
 0.186m
2g
Then Hm  9.69  0.588  0.2 - (-2.793  0.186)  13.09m
output power gQHm ( watts)
Efficiency ( ) 

input power
3200( watts)
3.2  0.015 13.09

 0.602 (60.2 percent )
3. 2
V1  1.91m / s;
Fig 1.
Example 2 (Pipeline selection in pumping system design)
As existing pump, having the tabulated characteristics is to be used to pump raw sewage
to a treatment plant through a static lift of 20 m. An uPVC pipeline 10 km long is to be
used. Allowing for minor losses totaling 10 V2/2g and taking an effective roughness of
0.15 mm because of sliming, select a suitable commercially available pipe size to
achieve a discharge of 60 l/s. Calculate this power consumption.
Discharge (l/s)
Total head (m)
Overall efficiency
(per cent)
0
45
10
44.7
20
43.7
30
42.5
40
40.6
50
38
60
35
70
31
35
50
57
60
60
53
40
Solution;
At 60 l/s, total head = 35.0 m, therefore the sum of the static lift and pipeline losses
must not exceed 35.0 m.
Try 300 mm diameter: A=0.0707 m2
Re = 2.25*100000 ; k/D = 0.0005
V=0.85 m/s
 =0.019= f
2
Friction head loss  0.019 10000  0.85  23 .32 m
0.3 19 .62
Hs + hf = 43.32 (> 35) pipe diameter too small
Try 300 mm diameter: A=0.0962 m2
Re = 1.93*100000 ; k/D = 0.00043
Hf=10.48m;
hm 
V=0.624 m/s;
 =0.0185
10  0.624 2
 0.2m
19 .612
Hs + hf + hm =30.68 (< 35 m) O.K
The pump would deliver approximately 70 l/s through the 350 mm pipe and to regulate
the flow to 60 l/s an additional head loss of 4.32 m by valve closure would be required.
Power consumption P  1000  9.81  0.06  35  38.85kW
0.55 1000
Example 3 (Pumps in parallel and series)
Two identical pumps having the tabulated characteristics are to be installed in a
pumping station to deliver sewage to a settling tank through a 200 mm uPVC pipeline
2.5 km long. The static lift is 15 m. Allowing for minor head losses of 10.0V2/2g and
assuming an effective roughness of 0.15 mm calculate the discharge and power
consumption if the pumps were to be connected: (a) in parallel, and (b) in series.
Pump Characteristics
discharge (l/s)
Total head (m)
Overall efficiency (per cent)
0
30
10
27.5
44
20
23.5
58
30
17
50
40
7.5
18
Solution;
The ‘system curve’ is computed as in the previous examples; this is, of course ,
independent of the pump characteristics. Calculated system heads (H) are tabulated
below for discrete discharges (Q)
H  HsT  h f  hm
Q (l/s)
H (m)
10
16.53
20
20.8
30
27.37
40
36.48
(a) Parallel operation
The predicted head v. discharge curve for dual pump operation in parallel mode is
obtained as described,.i.e. by doubling she discharge over the range of heads (since
the pumps are identical in this case). The system and efficiency curves are added as
shown in fig. 2. From the intersection of the characteristic and system curves the
following results are obtained:
Fig. 2 Parallel operation
Single pimp operation; Q = 22.5 l/S; Hm = 24 m ;  = 0.58
Power consumption = 9.13 kW
Parallel operation, Q = 28.5 l/S; Hm = 26 m ;  = 0.51
(Corresponding with 14.25 l/s per pump)
Power input = 14.11 kW
(b) Series operation
Using the method described in section 6.3 (b) and plotting the dual-pump
characteristic curve, intersection with the system curve yields (see Fig 3)
Fig 3. Series operation
Q=32.5 l/S; Hm = 28 m ;  = 0.41
Power input = 21.77kW
Note that for this particular pipe system, comparing the relative power consumptions the
parallel operation is more efficient in producing an increase in discharge than the series
operation.
Example 4
A laboratory test on a pump revealed that the onset of cavitations occurred at a
discharge of 35 l/s, when the total head at inlet was reduced to 2.5m and the total head
across the pump was 32 m. Barometric pressure was 760 mm Hg and the vapor pressure
17 mm Hg. Calculate the Thoma cavitation number. The pump is to be installed in a
situation where the atmospheric pressure is 650 mm Hg and water temperature 10C
(Vapor Pressure 9.22 mmHg) to give the same total head and discharge. The losses and
velocity head in the suction pipe are estimated to be 0.55 m of water. What is the
maximum height of the suction lift?
Solution;
H suc  2.5m  H suc 
pv

 2.5  0.23  2.27
2.27
 0.071
32
p5 V52
p1 V12

 z 5  hL 

 z1

2g
 2g


p5


pv


p1

V52

  z 5  z1  
 hL  NPSH

2g



pv
0.55
 z 5  z1 
p1


pv

 0.55  NPSH  8.84  0.1254  0.55  2.27  5.89m
Example 4
The pump of Fig 4a.is placed in a 10-in- diameter pipe (f=0.020), 1300 ft long, which is
used to lift water from one reservoir to another. The difference in water surface
elevations between the reservoirs fluctuates from 20ft to 100ft. Plot a curve showing
delivery rate versus water surface elevation difference. Plot also the corresponding
efficiencies. The pump is operated at a constant speed of 1450 rpm. Neglect minor
losses.
Solution;
Plot the pump characteristic curve and pipe system curves for a variety of z’s.
 z,ft
20
40
60
80
100
110
See plotted curves in fig 4b
Q, gpm
2500
2320
2100
1860
1350
940
efficiency, %
68
76
81
84
77
66
Example 5
Suppose a pump is to pump water at a head of 130ft, the water temperature being
100F and the barometric pressure being 14.3 psia. At intake the pressures is a vacuum
of 17 in Hg and the velocity is 12 fps. What are the values of NPSH and ?
Solution;
NPSH   p s abs /   Vs2 / 2 g  p v / 
 p s abs 
p 0 /   p s /   14.3144  / 62.0   17 / 12 847.3 / 62.0   13.85 ft
Vs2 / 2 g  12 2 / 2 * 32.2  2.24 ft
p v /   135 / 62  2.18 ft
NPSH  13.85  2.24  2.18  13.91 ft
  13.91 / 130  0.107
Example 6
A pump is delivering 7500 gpm of water at 140 F at a head of 240 ft, and the
barometric pressure is 13.8 psia. Determine the reading on a pressure gage in inches of
mercury vacuum at the suction flange when cavitation is incipient. Assume the suction
pipe diameter equals 2 ft and neglect the effects of prerotation. Take c=0.085
Solution;
   p s abs /   Vs2 / 2 g  pv /  / h
let p  gage pressure at suction flange
 p s abs  p atm  p  13.8  p


Vs  Q / As  75000.002228 /  2 2 / 4  5.319 ft
p v /   416 / 61.4  6.775 ft


0.085  13.8  p 144 / 61.4  5.319 2 /( 2 * 32.2)  6.775 / 240
p   2.400 lb / in    2.40029.9 / 14.7    4.88 inHg
2
or 4.88 in Hg vacuum
Example 7
Water is being pumped from a reservoir to the top of a hill, where it is discharged, as
shown in Fig. The pump, which is 70 % efficient, is rated at 150kW. Find the flow rate
at which water is being discharged from the pipe. Neglect minor losses
Solution;
p1 /   V12 / 2 g  z1  E p  p 2 /   V22 / 2 g  z 2  hL
E P  P / Q  1500.7  / Q9.79  10.725 / Q
V Q/ A
  500  2 
V 2  Q /   
 / 4  5.093Q
  1000 


hL  h f   f L / D  V 2 / 2 g

Assume

 500  
2
2
f  0.018, hL  0.018 975 / 
  5.093Q) / 2 * 9.807   46.42Q
 1000  



. 0  0  111.0  10.725 / Q  0  5.093Q) 2 / 2 * 9.807   150.2  46.42Q 2
Q  0.254 m 3 / s (by trial and error)
This solution was based on the assumed value of f of 0.018. However, the value of f is
dependent on the Reynolds number and relative roughness. Therefore a new value of f
should be determined based on the computed value of Q of 0.254 m3/s.
 / D  0.00030 / 500 / 1000  0.00060


V  Q / A  0.254 /  500 / 1000 / 4  1.294 m / s
2


N R  DV / v  500 / 10001.294 / 1.02 *10 6  6.34 *10 5
From Fig, f=0.018. Hence, the computed flow rate of 0.254 m3/s is correct.
Example 8
Oil with a specific gravity of 0.87 is being pumped from a lower reservoir to an
elevated tank as shown in Fig. The pump in the system is 78% efficient and is rated at
185 KW. Determine the flow rate of the oil in the pipe if the total head loss from point 1
to 2 is 12 of oil.
Solution;
p1 /   v12 / 2 g  z1  E p  p 2 /   v 22 / 2 g  z 2  hL
p1 /   p 2 /   v 22 / 2 g  0
z1  150m
0.78185  Q * 0.87 * 9.79E P 
EP
2
2
2
v 22 / 2 g  Q / A / 2 g  Q /  0.160  / 4 /( 2 * 9.807)  126.12Q 2
P  QE P
z 2  200m
hL  12m
 16.94 / Q
0  0  150  16.94 / Q  0  126.12Q 2  200  12
126.12Q 2  16.94 / Q  62  0
By trial and error solution, Q= 0.244 m3/s