Problem set #9
1) A centrifugal pump having the characteristics shown in the figure is used to pump water between two large
open tanks through 100 ft of 8-in.-diameter pipe.
The pipeline contains 4 regular flanged 90° elbows , a check valve, and a fully open globe valve. Other minor
losses are negligible. Assume the friction factor f=0.02 for the 100-ft section of pipe. If the static head
(difference in height of fluid surfaces in the two tanks) is 30 ft, what is the expected flowrate?
Solution
hp = 30 + 1.02×10-5 [Q(gal/min)]2
This curve intersect the pump curve at Q = 1740 gal/min. The efficiency at this flow rate is near peak efficiency
at about 83%. The pump should be satisfactory.
2) In a chemical processing plant a liquid is pumped from an open tank,
through a 0.1-m-diameter vertical pipe, and into another open tank as
shown in fig.a (assume elevation difference h = 30 m instead h = 30 m as
shown). A valve is located in the pipe, and the minor loss coefficient for
the valve as a function of the valve setting is shown in fig. b. The pump
head capacity relationship is given by the equation h a = 52.0 - 1.01x103 Q2
with ha in meters when Q is in m3/s . Assume the friction factor f = 0.02
for the pipe, and all minor losses, except for the valve, are negligible. The
fluid levels in the two tanks can be assumed to remain constant.
(a) Determine the flowrate with the valve wide open.
(b) Determine the required valve setting (percent open) to reduce the
flowrate by 50%.
Solution
hp = 33 + 5.78×103 [Q(m3/s)]2
Since the pump equation is ha = 52.0 - 1.01x103 Q2
hp = 33 + 5.78×103 Q2 = 52.0 - 1.01x103 Q2
Q = 0.0529 m3/s
(b) Determine the required valve setting (percent open) to reduce the flowrate by 50%.
ha = 52.0 - 1.01x103 Q2 = 52.0 - 1.01x103 (0.0529/2 )2 = 50.6 m
hp = 50.6 m = 33 m + (KL + 6)(826)(0.0529/2)2  KL = 24.3
From Figure 12.29b the valve should be 13% open to obtain this value.
3) A centrifugal pump having an impeller diameter of 1 m is to be constructed so that it will supply a head rise of
200 m at a flowrate of 4.1 m3/s of water when operating at a speed of 1200 rpm. To study the characteristics of this
pump, a 1/5 scale, geometrically similar model operated at the same speed is to be tested in the laboratory.
Determine (a) the required model discharge and (b) head rise. Assume that both model and prototype operate with
the same efficiency (and therefore the same flow coefficient).
Solution
For similarity, the model pump must operate at the same flow coefficient so that
 Q   Q 
  D3  =   D 3 
p

m 
m  Dm
 Qm =

 p  D p
3


 Qp


Qm = (1)(1/5)3(4.1) = 0.0328 m3
g p  m 
 gha   gha 
We also have
  2 D 2  =   2 D 2   ha,m = g   
p

m 
m  p
ha,m = (1)(1)2(1/5)2(200) = 8.0 m
2

 Dm

 Dp

2


 ha,p


4) A centrifugal pump has the performance characteristics of the pump with the 6-in.-diameter impeller
described in the Figure. Note that the pump in this figure is operating at 3500 rpm. What is the expected head
gained if the speed of this pump is reduced to 2800 rpm while operating at peak efficiency?
Solution
From the figure, for the 6-in diameter impeller operating at 3500 rpm, Q = 170 gpm and hL = 230 ft. If the pump
is still operated at peak efficiency with the speed reduced to 2800 rpm then
Q1
Q2
=
1
2
 Q2 =
2
1
Q1 = (2,800/3,500)(170) = 136 gpm
ha1 12
We also have
=
 ha2 =
2
ha 2  2

 2

 1
2

 ha1


ha2 = (2,800/3,500)2(230) = 147 ft
5) A centrifugal pump provides a flowrate of 500 gpm when operating at 1750 rpm against a 200-ft head.
Determine (a) the pump's flowrate and (b) developed head if the pump speed is increased to 3500 rpm.
Solution
Q1
Q2
=
1
2
 Q2 =
2
1
Q1 = (3,500/1,750)(500) = 100 gpm
ha1 12
We also have
=
 ha2 =
2
ha 2  2

 2

 1
2

 ha1


ha2 = (3,500/1,750)2(200) = 80 ft
6) In a certain application a pump is required to deliver 5000 gpm against a 300-ft head when operating at 1200
rpm. Find a specific speed Nsd of the pump and recommend an appropriate pump from the given figure.
Solution
Nsd =
 (rpm) Q( gpm) 1200 5000
=
= 1180
 ha ( ft )3 / 4
(300)3 / 4
From the above figure, at specific speed of 1180, a radial flow pump (centrifugal pump) would be
recommended.
7) A hydraulic turbine operating at 180 rpm with a head of 100 feet develops 20,000 horsepower. Estimate the
power if the same turbine were to operate under a head of 50 ft.
Solution
Head coefficient remains constant:
 gha   gha 
100 50
=

  2 D 2    2 D 2  1802 =  2

1 
2
2
2 = 127 rpm
Power coefficient is the same
since D1 = D2 and g1 = g2
 Wshaft   Wshaft  20,000 Wshaft ,2
=

 =
 
3
3
  3 D5    3 D5 
100
127

1 
2
since D1 = D2 and g1 = g2
Wshaft ,2 = 41,000 hp
8) Water for a Pelton wheel turbine flows from the headwater and through the penstock as shown in the given
Figure. The effective friction factor for the penstock, control valves, and the like is 0.032 and the diameter of
the jet is 0.20 m. Determine the maximum power output.
Solution
Maximum W
shaft
= 23,200 kW
12.70) Turbines are to be designed to develop 30000 horsepower while operating under a head of 70 ft and an
angular velocity of 60 rpm. Estimate the flowrate needed.
Solution
N’sd =
 (rpm) Wshaft (hp) 60 30,000
=
= 1180
 ht ( ft )5 / 4
(70)5 / 4
Wshaft = Qht  Q = Wshaft /ht = (30,000×550)/(62.4×70)
Q = 378 ft3/s