Problem 11.C1
The net head of a little aquarium pump is given by the manufacturer as a function of volume
flow rate as listed:
Q, m3/s:
0
1.0E−6
2.0E−6
3.0E−6
4.0E−6
5.0E−6
H, mm H2O:
1.10
1.00
0.80
0.60
0.35
0.0
What is the maximum achievable flow rate if you use this pump to pump water from the lower
reservoir to the upper reservoir as shown in Fig. C.11.1?
NOTE: The tubing is smooth, with an inner diameter of 5.0 mm and a total length of 29.8 m. The
water is at room temperature and pressure. Minor losses can be neglected.
Solution 11.C1
For water, take
= 998 kg/m3 and
= 0.001 kg/m·s. NOTE: The tubing is so small that the flow
is laminar, even at the highest pump flow rate:
Problem 11.C2
Reconsider Prob. 6.62 as an exercise in pump selection. Select an impeller size and rotational
speed from the Byron Jackson pump family of Prob. 11.28 which will deliver a flow rate of
3 ft3/s to the system of Fig. P6.62 at minimum input power. Calculate the horsepower required.
Problem 6.62
Water at 20C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as
shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient,
what horsepower pump is needed?
Neglect minor losses.
Problem 11.28
Tests by the Byron Jackson Co. of a 14.62-in-diameter centrifugal water pump at 2134 rpm yield
the following data.
Q, ft3/s:
0
2
4
6
8
10
H, ft:
340
340
340
330
300
220
bhp:
135
160
205
255
330
330
What is the BEP? What is the specific speed? Estimate the maximum discharge possible.
Solution 11.C2
For water take
= 1.94 slug/ft3 and
= 2.09E−5 slug/ft·s. For cast iron take
Problem 11.C3
Reconsider Prob. 6.77 as an exercise in turbine selection. Select an impeller size and rotational
speed from the Francis turbine family of Fig. 11.22d which will deliver maximum power
generated by the turbine. Calculate the water turbine power output and remark on the practicality
of your design.
Problem 6.77*
Modify Prob. 6.76 into an economic analysis, as follows. Let the 40 m of wrought-iron pipe have
a uniform diameter d. Let the steady water flow available be Q = 30 m3h. The cost of the turbine
is $4 per watt developed, and the cost of the piping is $75 per centimeter of diameter. The power
generated may be sold for $0.08 per kilowatt hour. Find the proper pipe diameter for minimum
payback time, i.e., minimum time for which the power sales will equal the initial cost of the
system.
Neglect minor losses.
Problem 6.76
The small turbine in Fig. P6.76 extracts 400 W of power from the water flow. Both pipes are
wrought iron. Compute the flow rate Q in m3/h. Why are there two solutions? Which is better?
Neglect minor losses.
Solution 11.C3
For water, take
= 998 kg/m3 and
= 0.001 kg/m·s. For wrought iron take
Problem 11.C4
The system of Fig. C11.4 is designed to deliver water at 20C from a sea-level reservoir to
another through new cast iron pipe of diameter 38 cm. Minor losses are K1 = 0.5 before the
pump entrance and K2 = 7.2 after the pump exit. (a) Select a pump from either Figs. 11.7a or
11.7b, running at the given speeds, which can perform this task at maximum efficiency.
Determine (b) the resulting flow rate; (c) the brake horsepower; and (d) whether the pump as
presently situated is safe from cavitation.
Solution 11.C4
For water take
= 998 kg/m3 and
= 0.001 kg/m·s. First establish the system curve of head loss
Problem 11.C5
For the 41.5-in water pump of Fig. 11.7b, at 710 r/min and 22,000 gal/min, estimate the
efficiency by (a) reading it directly from Fig. 11.7b; and (b) reading H and bhp and then
calculating efficiency from Eq. (11.5). Compare your results.
Solution 11.C5
For water, take the density to be 1.94 slug/ft3.
Problem 11.C6
An interesting turbomachine [58] is the fluid coupling of Fig. C11.6, which delivers fluid from a
primary pump rotor into a secondary turbine on a separate shaft. Both rotors have radial blades.
Couplings are common in all types of vehicle and machine transmissions and drives. The slip of
the coupling is defined as the dimension-less difference between shaft rotation rates,
s = 1 −
s/
p. For a given volume of fluid, the torque T transmitted is a function of s,
,
p, and
impeller diameter D. (a) Non-dimensionalize this function into two pi groups, with one pi
proportional to T. Tests on a 1-ft-diameter coupling at 2500 r/min, filled with hydraulic fluid of
density 56 lbm/ft3, yield the following torque versus slip data:
Slip, s:
0%
5%
10%
15%
20%
25%
Torque T, ft·lbf:
0
90
275
440
580
680
(b) If this coupling is run at 3600 r/min, at what slip value will it transmit a torque of 900 ft·lbf?
(c) What is the proper diameter for a geometrically similar coupling to run at 3000 r/min and
5 percent slip and transmit 600 ft·lbf of torque?
Solution 11.C6
(a) List the dimensions of the five variables, from Table 5.1:
Variable:
T
s
p
D
Problem 11.C7
Report to the class on the Cordier method [63] for optimal design of turbomachinery. The
method is related to, and greatly expanded from, Prob. P11.46 and uses both software and charts
to develop an efficient design for any given pump or compressor application.
Problem 11.46
The answer to Prob. 11.40 is that the dimensionless “specific diameter” takes the form
Ds = D(gH*)1/4/Q*1/2, evaluated at the BEP. Data collected by the author for 30 different pumps
indicates, in Fig. P11.46, that Ds correlates well with specific speed Ns. Use this figure to
estimate the appropriate impeller diameter for a pump which delivers 20,000 gal/min of water
and a head of 400 ft when running at 1200 r/min. Suggest a curve-fitted formula to the data.
Hint: Use a hyperbolic formula.
Slip, s:
Problem 11.40
The specific speed Ns, as defined by Eq. (11.30), does not contain the impeller diameter. How
then should we size the pump for a given Ns? An alternate parameter is called the specific
diameter DS, which is a dimensionless combination of Q, (gH), and D. (a) If DS is proportional
to D, determine its form. (b) What is the relationship, if any, of DS to CQ*, CH*, and CP*?
(c) Estimate DS for the two pumps of Figs. 11.8 and 11.13.
Solution 11.C7
The Cordier method, developed in the textbook G. T. Csanady, Theory of Turbomachines,
Problem 11.C8
A pump-turbine is a reversible device that uses a reservoir to generate power in the daytime and
then pumps water back up to the reservoir at night. Let us reset Prob. P6.62 as a pump-turbine.
Recall that z = 120 ft, and the water flows through 2000 ft of 6-in-diameter cast iron pipe. For
simplicity, assume that the pump operates at BEP (92%) with H*p = 200 ft and the turbine
operates at BEP (89%) with H*t = 100 ft. Neglect minor losses. Estimate (a) the input power, in
watts, required by the pump; and (b) the power, in watts, generated by the turbine. For further
technical reading, consult the URL www.usbr.gov/pmts/hydraulics_lab/pubs/EM/EM39.pdf.
Problem 6.62
Water at 20C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as
shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient,
what horsepower pump is needed?
Neglect minor losses.
Solution 11.C8
For water, take
= 1.94 slug/ft3 and
= 2.09E-5 slug/ft-s. (a) Write the steady flow energy
equation for pump operation, neglecting minor losses:
Problem 11.W1
We know that an enclosed rotating bladed impeller will impart energy to a fluid, usually in the
form of a pressure rise, but how does it actually happen? Discuss, with sketches, the physical
mechanisms through which an impeller actually transfers energy to a fluid.
Solution 11.W1
Problem 11.W2
Dynamic pumps (as opposed to PDPs) have difficulty moving highly viscous fluids. Lobanoff
and Ross [15] suggest the following rule of thumb: D (in) > 0.015ν/νwater, where D is the
diameter of the discharge pipe. For example, SAE 30W oil (≈300νwater) should require at least a
4.5-in outlet. Can you explain some reasons for this limitation?
Solution 11.W2
Problem 11.W3
The concept of NPSH dictates that liquid dynamic pumps should generally be immersed below
the surface. Can you explain this? What is the effect of increasing the liquid temperature?
Solution 11.W3
Problem 11.W4
For nondimensional fan performance, Wallis [20] suggests that the head coefficient should be
replaced by FTP/(ρn2D2), where FTP is the fan total pressure change. Explain the usefulness of
this modification.
Solution 11.W4
Problem 11.W5
Performance data for centrifugal pumps, even if well scaled geometrically, show a decrease in
efficiency with decreasing impeller size. Discuss some physical reasons why this is so.
Solution 11.W5
Problem 11.W6
Consider a dimensionless pump performance chart such as Fig. 11.8. What additional
dimensionless parameters might modify or even destroy the similarity indicated in such data?
Solution 11.W6
Problem 11.W7
One parameter not discussed in this text is the number of blades on an impeller. Do some reading
on this subject, and report to the class about its effect on pump performance.
Solution 11.W7
Problem 11.W8
Explain why some pump performance curves may lead to unstable operating conditions.
Solution 11.W8
Problem 11.W9
Why are Francis and Kaplan turbines generally considered unsuitable for hydropower sites
where the available head exceeds 1000 ft?
Solution 11.W9
Problem 11.W10
Do some reading on the performance of the free propeller that is used on small, low-speed
aircraft. What dimensionless parameters are typically reported for the data? How do the
performance and efficiency compare with those for the axial-flow pump?
Solution 11.W10
Problem 11.1
Describe the geometry and operation of a human peristaltic PDP which is cherished by every
romantic person on earth. How do the two ventricles differ?
Solution 11.1
Clearly we are speaking of the human heart, driven periodically by travelling compression of the heart
Problem 11.2
What would be the technical classification of the following turbomachines: (a) a household fan,
(b) a windmill, (c) an aircraft propeller, (d ) a fuel pump in a car, (e) an eductor, (f ) a fluid-
coupling transmission, and (g) a power plant steam turbine?
Solution 11.2
What would be the technical classification of the following turbo machines:
(a) a household fan = an axial flow fan. Ans. (a)
Problem 11.3
A PDP can deliver almost any fluid, but there is always a limiting very high viscosity for which
performance will deteriorate. Can you explain the probable reason?
Solution 11.3
High-viscosity fluids take a long time to enter and fill the inlet cavity of a PDP. Thus a PDP
Problem 11.4
Figure P11.4 shows the impeller on a common device which, when operating, turns at up to
300,000 r/min. Can you guess what it is and offer a description?
Figure P11.4
Solution 11.4
This turbine, typically only 10 to 12 mm in diameter, drives a dental drill, using high pressure air
Problem 11.5
What type of pump is shown in Fig. P11.5? How does it operate?
Solution 11.5
Problem 11.6
Figure P11.6 shows two points a half-period apart in the operation of a pump. What type of
pump is this [13]? How does it work? Sketch your best guess of flow rate versus time for a few
cycles.
Solution 11.6
This is a diaphragm pump. As the center rod moves to the right, opening A and closing B, the
Problem 11.7
A piston PDP has a 5-in diameter and a 2-in stroke and operates at 750 rpm with 92 percent
volumetric efficiency. (a) What is the delivery, in gal/min? (b) If the pump delivers SAE 10W oil
at 20C against a head of 50 ft, what horsepower is required when the overall efficiency is
84 percent?
Solution 11.7
For SAE 10W oil, take
870 kg/m3 1.69 slug/ft3. The volume displaced is
Problem 11.8
A Bell and Gossett pump at best efficiency, running at 1750 r/min and a brake horsepower of
32.4, delivers 1050 gal/min against a head of 105 ft. (a) What is its efficiency? (b) What type of
pump is this?
Solution 11.8
(a) For water take ρg = 62.4 lbf/ft3. Pump efficiency is defined as
Problem 11.9
Figure P11.9 shows the measured performance of the Vickers Inc. Model PVQ40 piston pump
when delivering SAE 10W oil at 180F (
910 kg/m3). Make some general observations about
these data vis-à-vis Fig. 11.2 and your intuition about PDP behavior.
