Fluid Mechanics, 8th edition
In SI Units
Frank M. White
Solutions Manual
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Chapter 11 • Turbomachinery
P11.1 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: Clearly we are speaking of the human heart, driven periodically by traveling
compression of the heart walls. One ventricle serves the brain and the rest of one’s extremities,
while the other ventricle serves the lungs and promotes oxygenation of the blood. Ans.
P11.2
What would be the technical classification of the following turbomachines:
(a) a household fan = an axial flow fan. Ans. (a)
(b) a windmill = an axial flow turbine. Ans. (b)
(c) an aircraft propeller = an axial flow fan. Ans. (c)
(d) a fuel pump in a car = a positive displacement pump (PDP). Ans. (d)
(e) an eductor = a liquid-jet-pump (special purpose). Ans. (e)
(f) a fluid coupling transmission = a double-impeller energy transmission device. Ans. (f)
(g) a power plant steam turbine = an axial flow turbine. Ans. (g)
P11.3 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: High-viscosity fluids take a long time to enter and fill the inlet cavity of a
PDP. Thus, a PDP pumping high-viscosity liquid should be run slowly to ensure filling. Ans.
2
P11.4 Figure P11.4 shows the impeller on a common device which, in operation, turns
as fast as 300,000 r/min. Can you guess what it is and offer a description?
Fig. P11.4
Solution: This turbine, typically only 10 to 12 mm in diameter, drives a dental drill,
using high pressure air from tiny nozzles. Oy! When drilling against a tooth, it typically
runs at 360,000 r/min.
______________________________________________________________________
P11.5 What type of pump is shown in
Fig. P11.5? How does it operate?
Solution: This is a flexible-liner pump.
The rotating eccentric cylinder acts as a
“squeegee.” Ans.
Fig. P11.5
3
P11.6 Fig. P11.6 shows two points a
half-period apart in the operation of a
pump. What type of pump is this? How
does it work? Sketch your best guess of
flow rate versus time for a few cycles.
Solution: This is a diaphragm pump. As
the center rod moves to the right, opening A
and closing B, the check valves allow A to
fill and B to discharge. Then, when the rod
moves to the left, B fills and A discharges.
Depending upon the exact oscillatory motion
of the center rod, the flow rate is fairly steady,
being higher when the rod is faster. Ans.
Fig. P11.6
P11.7 A piston PDP has a 12.5-cm diameter and a 5-cm stroke and operates at 750 rpm
with 92% volumetric efficiency. (a) What is the delivery, in L/min? (b) If the pump
delivers SAE 10W oil at 20°C against a head of 15 m, what power is required when the
overall efficiency is 84%?
Solution: For SAE 10W oil, take r ≈ 870 kg/m3. The volume displaced is
=
υ
π
=
(12.5) 2 (5) 613.6 cm3 ,
4
3


cm  
1L
strokes 
∴ Q=
750
 613.6

 (0.92 efficiency)
3 
stroke   1,000 cm  
min 

or: Q ≈ 423 L/min Ans. (a)
 0.423 3 
870(9.81) 
m /s  (15)
r gQH
m⋅N
60


= =
=
Power
1, 074.5
η
0.84
s
4
Ans. (b)
P11.8 A 27.5 cm diameter Bell and Gossett pump at best efficiency, running at 1,750
r/min and 32.4 hp, delivers 4 m3/min of water against a head of 32 m. (a) What is its
efficiency? (b) What type of pump is this?
Solution: (a) For water take ρg = 9,790 N/m3. Pump efficiency is defined as
=
η
( ρ g )QH
=
P
(9, 790 N/ m3 )(4 / 60 m 3 / s)(32 m)
= 0.864
=
(32.4)(746) m − N/ s
86.4% Ans.( a )
(b) If you have only progressed as far as the beginning of this chapter, it would be
difficult to name this type of pump, which is centrifugal. You could rule out a PDP – to
pump water with a piston 27.5 cm in diameter at 1,750 r/min would result in a much
higher head than 32 m. Conversely, H = 32 m is too high, and Q = 4 m3/min is too low,
for an axial-flow pump. This pump has a specific speed (Section 11.4) of 260, about
right for a centrifugal pump.
5
P11.9 Figure P11.9 shows the measured performance of the Vickers Inc. Model PVQ40
piston pump when delivering SAE 10W oil at 180°F (r ≈ 910 kg/m3). Make some
general observations about these data vis-à-vis Fig. 11.2 and your intuition about PDP
behavior.
Solution: The following are observed:
(a) The discharge Q is almost linearly proportional to speed Ω and slightly less for the
higher heads (H or ∆p).
(b) The efficiency (volumetric or overall) is nearly independent of speed Ω and again
slightly less for high ∆p.
(c) The power required is linearly proportional to the speed Ω and also to the head H
(or ∆p). Ans.
Fig. P11.9
6
P11.10 Suppose that the pump of Fig. P11.9 is run at 1,100 r/min against a pressure rise of 210 bar.
(a) Using the measured displacement, estimate the theoretical delivery in m3/min. From the chart,
estimate (b) the actual delivery; and (c) the overall efficiency.
Solution: (a) From Fig. P11.9, the pump displacement is 41 cm3. The theoretical
delivery is
r 
cm3 
cm3
m3

=
Q 1,100
=
41
45,100
=
0.045



min  
r 
min
min

Ans. (a)
(b) From Fig. P11.9, at 1,100 r/min and ∆p = 210 bar, read
Q ≈ 0.047 m3/min Ans. (b)
(c) From Fig. P11.9, at 1,100 r/min and ∆p = 210 bar, read
ηoverall ≈ 87%. Ans. (c)
P11.11 A pump delivers 1,500 L/min of water at 20°C against a pressure rise of 270
kPa. Kinetic and potential energy changes are negligible. If the driving motor supplies 9
kW, what is the overall efficiency?
Solution: With pressure rise given, we don’t need density. Compute “water” power:
 1.5 m 3  
kN 
6.75
= 75% Ans.
Pwater = rgQH = Q ∆p = 
 270 2  = 6.75 kW, ∴ η =

9.0
m
 60 s 
P11.12 In a test of the pump in the figure,
the data are: p1 = 100 mmHg (vacuum),
p2 = 500 mmHg (gage), D1 = 12 cm, and
D2 = 5 cm. The flow rate is 0.68m3/min
of light oil (SG = 0.91). Estimate (a) the
head developed, and (b) the input power
at 75% efficiency.
Fig. P11.12
Solution: Convert 100 mmHg = 13,332 Pa, 500 mmHg = 66,661 Pa, 0.68 m3/min =
0.01133 m3/s. Compute V1 = Q/A1 = 0.01133/[(π /4)(0.12)2] = 1.00 m/s. Also, V2 = Q/A2
= 5.77 m/s. Calculate goil = 0.91(9,790) = 8,909 N/m3. Then the head is
H=
p2
g
+
V22
p V2
+ z2 − 1 − 1 − z1
2g
g 2g
7
=
66, 661 (5.52) 2
−13,332 (0.96) 2
+
+ 0.65 −
−
− 0, or: H = 11.3 m
8,909 2(9.81)
8,909
2(9.81)
Power
=
γ QH 8, 909(0.01133)(11.3)
=
= 1, 520 W
η
0.75
Ans. (a)
Ans. (b)
P11.13 A 3.5 hp pump delivers 5,000 N of ethylene glycol at 20°C in 12 seconds,
against a head of 5 m. Calculate the efficiency of the pump.
Solution: From Table A.3 for ethylene glycol, r = 1,117 kg/m3. The specific weight is
thus g = (1,117)(9.81) = 10,958 N/m3. We then may calculate the volume flow rate:
=
Q
w
=
γ
(5, 000 N/ 12 s)
=
10, 958 N/m3
0.038
m3
s
This gives us enough information to calculate the pump efficiency:
γ QH
746 W
(10, 958 N/m3 )(0.038 m3 /s)(5 m)
=
=
P
3.5 hp = 2, 611
W =
hh
hp
2, 082
=
h
Solve for =
0.797 or 79.7%
2, 611
8
Ans.
P11.14 A pump delivers gasoline at 20°C and 12 m3/h. At the inlet, p1 = 100 kPa, z1 = 1
m, and V1 = 2 m/s. At the exit p2 = 500 kPa, z2 = 4 m, and V2 = 3 m/s. How much power
is required if the motor efficiency is 75%?
Solution: For gasoline, take r g ≈ 680(9.81) = 6,671 N/m3. Compute head and power:
H=
p 2 V22
p
V2
500, 000
(3) 2
100, 000
(2) 2
+
+ z 2 − 1 − 1 − z1=
+
+4−
−
− 1,
g 2g
g 2g
6, 671 2(9.81)
6, 671 2(9.81)
ρρ
or:
 12 
6,671
(63.2)
r gQH
3,600 

≈ 1, 870 W
H ≈ 63.2 =
m, Power
=
η
0.75
Ans.
P11.15 A lawn sprinkler can be used as a simple turbine. As shown in Fig. P11.15, flow
enters normal to the paper in the center and splits evenly into Q/2 and Vrel leaving each
nozzle. The arms rotate at angular velocity ω and do work on a shaft. Draw the velocity
diagram for this turbine. Neglecting friction, find an expression for the power delivered to
the shaft. Find the rotation rate for which the power is a maximum.
Fig. P11.15
Solution: Utilizing the velocity diagram at right, we apply the Euler turbine formula:
=
P ρρ
Q(u2 Vt2 − u=
Q[u(W − u) − 0]
1Vt1 )
or: =
P r Qω R( Vrel − ω R) Ans.
9
Vrel
dP
= rQ(Vrel − 2u)
= 0 if =
ω
Ans.
du
2R
rQu(2u −=
u) rQ(w R)2
where P=
max
P11.16 The centrifugal pump in Fig. P11.16 has r1 = 15 cm, r2 = 25 cm, b1 = b2 = 6 cm,
and rotates counterclockwise at 600 r/min. A sample blade is shown. Assume α 1 = 90°.
Estimate the theoretical flow rate and head produced, for water at 20°C, and
comment.
Impeller
Fig. P11.16
30°
150°
40°
Solution: For water, take r = 998 kg/m3. For counterclockwise rotation, this is a
forward-facing impeller, that is, looking at the figure, b2 = 150°, not 30°. If α 1 = 90°,
then Vt1 = 0. Convert 600 r/min × 2π/60 = 62.8 rad/s. Evaluate u1 = ω r1 = (62.8)(0.15) =
9.425 m/s. Then
=
Vn1
u=
1 tan( b1 )
=
(9.425 m / s)
tan(40° )
7.91 m / s
m3
≈ Ans.
s
Evaluate u2 = ω r2 = (62.8)(0.25) = 15.71 m/s. From Eq. (3.18), the head produced is
=
And Q 2π=
r1 b1 | Vn1 | 2π (0.15 m)(0.06 m)(7.91
=
m / s)
H=
u22 u2 cot b 2 Q
−
=
2π r2 b2 g
g
0.447
(15.71) 2 (15.71) cot(150° )(0.447)
−
= 25.15 − (−13.15)= 38.3m Ans.
(9.81)
2π (0.25)(0.06)(9.81)
The water power is P = rgQH = (998)(9.81)(0.447)(38.3) = 168,000 m-N/s.
10
P11.17 A centrifugal pump has d1 = 17.5 cm, d2 = 33 cm, b1 = 10 cm, b2 = 7.5 cm, b1 = 25°,
and b2 = 40° and rotates at 1,160 r/min. If the fluid is gasoline at 20°C and the flow enters
the blades radially, estimate the theoretical (a) flow rate in L/min, (b) power in W, and (c)
head in m.
Solution: For gasoline, take r ≈ 680 kg/m3. Compute ω = 1,160 rpm = 121.5 rad/s.
=
u1 ω=
r1 121.5 ( 0.0875) ≈ 10.6 m/s
=
Vn1 u=
10.6 tan 25° ≈ 4.94 m/s
1 tan b1
=
π r1b1Vn1 2π ( 0.0875) ( 0.1) (5)
Q 2=
=0.275
=
Vn2
m3
= 16, 500 L / min
s
Ans. ( a )
Q
0.27
m
=
≈ 3.47
2π r2 b2 2π ( 0.165) ( 0.075)
s
=
u 2 ω=
r2 121.5(0.165) ≈ 20.0 m/s
Fig. P11.17
Vt2= u 2 − Vn2 cot 40° ≈ 15.9 m/s
Finally, Pideal = r Q u2Vt2 = 680(0.275)(20.0)(15.9) = 59,500 W. Ans. (b)
Theoretical head H = P/( r gQ) = 59,500/[680(9.81)(0.275)] ≈ 32.4 m. Ans. (c)
11
P11.18 A jet of velocity V strikes a vane
which moves to the right at speed Vc, as in
Fig. P11.18. The vane has a turning angle θ.
Derive an expression for the power delivered to the vane by the jet. For what vane
speed is the power maximum?
Fig. P11.18
Solution: The jet approaches the vane at relative velocity (V − Vc). Then the force is
F = rA(V − Vc)2(1 − cosθ ), and Power = FVc = rAVc(V − Vc)2(1 − cosθ) Ans. (a)
Maximum power occurs when
or: Vc =
dP
= 0,
dVc
1
Vjet
3
4


rAV3 [1 − cos θ ]
Ans.
(b)  P
=


27
P11.19 A centrifugal water pump has r2 = 23 cm, b2 = 5 cm, and b 2 = 35° and rotates at 1,060
r/min. If it generates a head of 55 m, determine the theoretical (a) flow rate in L/min and
(b) power in kW. Assume near-radial entry flow.
Solution: For water take r = 998 kg/m3. Convert ω = 1,060 rpm = 111 rad/s. Then
=
u 2 w=
r2 111 ( 0.23
=
) 25.5


Q
r Qu 2  u 2 −
cotb 2  , and
Power =
2π r2 b2


m
;
s
P
H=
=
55 m
r gQ
Q


or: P =9,790QH =998Q(25.5)  25.5 −
cot 35°
2π (0.23)(0.05)


Solve
=
for Q 0.220 m3 /s ≈ 13,180 L / min
Ans. (a)
With Q and H known, P = rgQH = 9,790(0.22)(55) = 118.5 kW.
12
with H =55 m
Ans. (b)
P11.20 Suppose that Prob. 11.19 is reversed into a statement of the theoretical power
P = 153 hp. Can you then compute the theoretical (a) flow rate; and (b) head? Explain
and resolve the difficulty that arises.
Solution: With power known, the basic theory becomes quadratic in flow rate:


m
Q
u 2 25.5
, P r Qu 2  u 2 −
cot b 2 
=
=
s
2π r2 b2


19.8Q] 153(746)
= 998Q (25.5)[25.5 −=
m3
m3
=
=
Two
roots: Q1 0.21
; Q2 1.08
s
s
m⋅N
s
Ans. (a)
These correspond
=
to H1 55.5
=
m; H 2 10.8 m
Ans. (b)
So the ideal pump theory admits to two valid combinations of Q and H which, for the
given geometry and speed, give the theoretical power of 114 kW. Prob. 11.19 was solution
1.
13
P11.21 The centrifugal pump of Fig.
P11.21 develops a flow rate of 16 m3/min
with gasoline at 20°C and near-radial
absolute inflow. Estimate the theoretical (a)
power; (b) head rise; and (c) appropriate
blade angle at the inner radius.
Solution: For
kg/m3.
gasoline
take
r ≈ 680
Convert Q = 16 m3/min = 0.267 m3/s and ω
= 1,750 rpm = 183 rad/s. Note r2 = 15 cm
and b 2 = 30°. The ideal power is computed
as
Fig. P11.21


Q
 15 
r Qu 2  u 2 −
w r2 =
P=
cot b 2  , where u 2 =
183 
 ≈ 27.5 m/s. Plug in:
2π r2 b 2
 100 




0.267
=
=
P 680(0.267)(27.5)  27.5 −
cot 30° 104.7 kW
2π (15 /100)(7.5/100)


=
H
P
104, 700
=
≈ 59 m
ρ gQ 680(9.81)(0.267)
Ans. (a)
Ans. (b)
Compute Vn1 = Q/[2π r1b1] = 0.267/[2π (10/100)(7.5/100)] ≈ 5.7 m/s, u1 = ω r1 =
183(0.10) ≈ 18.3 m /s. For purely radial=
inflow, β 1 tan −1 (5.7/18.3) ≈ 17°. Ans. (c)
14
P11.22 A 37-cm-diameter centrifugal pump, running at 2,140 rev/min with water at
20°C produces the following performance data:
Q, m3/s:
0.0
0.05
0.10
0.15
0.20
0.25
0.30
H, m:
105
104
102
100
95
85
67
P, kW:
100
115
135
171
202
228
249
η:
0%
44%
74%
86%
92%
91%
79%
(a) Determine the best efficiency point. (b) Plot CH versus CQ. (c) If we desire to use this
same pump family to deliver 26.5 m3/min of kerosene at 20°C at an input power of 400 kW,
what pump speed (in rev/min) and impeller size (in cm) are needed? What head will be
developed?
Solution: Efficiencies, computed by η = rgQH/Power, are listed above. The best
efficiency point (BEP) is approximately 92% at Q = 0.2 m3/s. Ans. (a)
The dimensionless coefficients are CQ = Q/(nD3), where n = 2,160/60 = 36 rev/s and
D = 0.37 m, plus CH = gH/(n2D2) and CP = P/(r n3D5), where rwater = 998 kg/m3. BEP
and C*P 0.643. A plot of CH versus CQ is below.
111, C*H 5.35,=
values=
are C*Q 0.=
The values are similar to Fig. 11.8 of the text. Ans. (b)
(c) For kerosene, r k = 804 kg/m3. Convert 26.5 m3/min = 0.442 m3/s. At BEP, we
require the above values of dimensionless parameters:
Q
0.442
P
400, 000
0.111;
=
=
=
= 0.643;
3
3
3 5
nD
nD
ρn D
804n3 D5
rev
rev
=
n 26.1
= 1, 560 =
Solve
; D 0.534 m Ans. (c)
s
min
2
* (n 2 D 2 )/g 5.35(26.1)2 (0.534)
=
Also, H* C=
=
/9.81 106 m Ans. (c)
H
15
P11.23 When pumping water, (a) at what speed should the 27.5-cm Bell and Gossett
centrifugal pump of Prob. P11.8 be run, at best efficiency, to deliver 3 m3/min? Estimate
the resulting (b) head, and (c) power.
Solution: Recall the P11.8 data: n = 1,750 r/min, H = 32 m, Q = 4 m3/min, P = 24.2
kW. Scale the problem using our similarity rules, Eqs. (11.28):
Q2
(a ) =
Q1
n D 3
n2
3
ρ
= 2 ( 2=
)
(1.0)3 , solve n2 ≈ 1, 310
Ans.( a )
4
n1 D1
1, 750
min
H2
(b)=
H1
H2
n
D2 2
1, 310 2
H2
=
( 2 )2 ( =
)
(
) (1.0) 2 , solve=
n1
D1
32
1, 750
P
( c=
) 2
P1
P2
D2 5
ρ n
1, 310 3
P2 10.2 kW Ans.( c )
=
=
( 2 )( 2 )3 (=
)
(1.0)(
) (1.0)5 , solve
ρ1 n1 D1
24.2
1, 750
18 m
Ans.(b)
P11.24 Figure P11.24 shows performance data for the Taco, Inc., model 4013 pump.
Compute the ratios of measured shutoff head to the ideal value U2/g for all seven
impeller sizes. Determine the average and standard deviation of this ratio and compare it
to the average for the six impellers in Fig. 11.7.
Fig. P11.24 Performance data for a centrifugal pump.
(Courtesy of Taco. Inc., Cranston, Rhode Island.)
16
Solution: All seven pumps are run at 1,160 rpm = 121 rad/s. The 7 diameters are given.
Thus we can easily compute U = ω r = ω D/2 and construct the following table:
D, cm:
U, m/s:
Ho, m:
Ho /(U2/g):
25.4
15.43
13.41
0.553
The average ratio is 0.573
26.67
16.20
14.94
0.557
27.94
16.99
16.15
0.549
29.21
17.74
18.60
0.580
30.48
18.51
20.42
0.585
31.75
19.28
22.25
0.587
Ans. (a), and the standard deviation is 0.0196.
32.89
19.98
24.38
0.599
Ans. (b)
For the six pumps of Fig. 11.7, the average is 0.611, the standard deviation is 0.025.
These results are true of most centrifugal pumps: we can take an average of 0.60 ± 0.04.
P11.25 At what speed in rpm should the
0.889-m-diameter pump of Fig. 11.7(b) be
run to produce a head of 120 m at a
discharge of 75 m3/min? What power will
be required? Hint: Fit H(Q) to a formula.
Solution: A curve-fit formula for H(Q)
for this pump is H(m) ≈ 70.4 + 147.8Q −
4713Q2, with Q in km3/min. Then, for
constant diameter, the similarity rules
predict
Fig. 11.7b
2
 n1 
 n1 
H1 H=
=
2
 and Q1 Q2   ,
 n2 
 n2 
2
 75  710  
 710 
 75  710  
or: H1 =
− 4713 
120 
 ≈ 70.4 + 147.8 





 n2 
 1000  n 2  
 1000  n 2  
Solve for
n2
−11.09 ±
11.092 − 4 ( 0.09916 ) ( −104, 023)
2 ( 0.09916 )
≈ 970 rpm
2
Ans.
Now backtrack to compute H1 ≈ 64.3 m and Q1 ≈ 54.9 m3/min. From Fig. 11.7(b) we
may then read the power P1 ≈ 690 kW (or compute this from η1 ≈ 0.85). Then, by
similarity, P2 = (n2/n1)3 = 690(970/710)3 ≈ 1,760 kW.
17
Ans.
P11.26 Would the smallest, or the largest, of the seven Taco Inc. pumps in Fig. P11.24
be better (a) for producing, near best efficiency, a water flow rate of 2.3 m3/min and a
head of 29 m? (b) At what speed, in r/min, should this pump run? (c) What input power
is required?
Solution: First try the smallest pump, D = 25.4 cm. Read Fig. P11.24: Q* = 1.48
m3/min, H* = 12.5 m, ηmax = 65%. To scale up to 2.3 m3/min, use the similarity rules for
constant D:
Q2 2.3 n2
n2
,
=
= =
Q1 1.48 n1 1,160 rpm
solve n=
1,803 r / min
2
H2
H2
n2 2
1,803 2
Then =
(=
)
(
) , solve =
H 2 30.2 m
=
1,160
H1 12.5 m
n1
Power2
=
γ Q2 H 2 (9, 790)(2.3 / 60)(30.2)
=
= 17.4 kW
0.65
h
So the small pump will do the job, using 17.4 kW. Next try the largest pump, D = 32.89
cm. Read Fig. P11.24: Q* = 2 m3/min, H* = 21.95 m, ηmax = 80%. Again scale up to
2.3 m3/min, using the similarity rules at constant diameter:
Q2 2.3 n2
n2
=
= =
,
Q1
n1 1,160 rpm
2
solve n=
1, 334 r / min
2
H2
H2
n2 2
1, 334 2
= (=
H 2 29.0 m
Then =
)
(
) , solve =
H1 21.95 m
n1
1,160
=
Power2
γ Q2 H 2 (9, 790)(2.3 / 60)(29)
=
= 13.6 kW
h
0.80
So the largest pump is better to do the job, with slower speed and less power. Ans. (a)
The proper large-pump speed is 1,334 r/min. Ans. (b). The power is 13.6 kW. Ans. (c)
[NOTE that if we specify a different head, like 60 m, neither pump can do the job.]
18
P11.27 The 27.5-cm Bell and Gossett pump of Prob. P11.8 is to be scaled up to provide,
at best efficiency, a head of 75 m and a flow rate of 11.5 m3/min. Find the appropriate
(a) impeller diameter; (b) speed in r/min; and (c) power required.
Recall the P11.8 data: n = 1,750 r/min, H = 32 m, Q = 4 m3/min, P = 32.4 hp.
Solution: For water take ρ = 9,790 kg/m3. Compute the BEP dimensionless values and
apply them to the new head and flow rate to find the new n and D and P:
Q
4 / 60
11.5 / 60 0.192
=
=
=
=
0.11
3
3
nD
(1, 750 / 60)(27.5 / 100)
nD 3
nD 3
gH
(9.81)(32)
(9.81)(75)
735.75
=
=
=
=
=
(2) C
4.88
H*
2 2
2
2
2 2
n D
(1, 750 / 60) (27.5 / 100)
n D
n2 D2
Pnew
P
(32.4)(746)
=
=
=
(3) =
C P*
0.62
3 5
3
5
ρn D
(998)n 3 D 5
(998)(1, 750 / 60) (27.5 / 100)
(1) =
CQ*
Equations (1) and (2) can be solved simultaneously for n and D, after which the power P
can be computed from Eq. (3). This can be done by hand, or just throw them into Excel:
r
r
=
n 32.57
= 1, 954 =
; D 0.377
=
m 37.7=
cm; P 162.8 kW Ans.
s
min
19
P11.28 Tests by the Byron Jackson Co. of a 37-cm centrifugal water pump at 2,134
rpm yield the data below. What is the BEP? What is the specific speed? Estimate the
maximum discharge possible.
Q, m3/s:
0
0.057
0.113
0.17
0.227
0.283
H, m:
104
104
104
100
91
67
kW:
101
119
153
190
246
246
Solution: The efficiencies are computed from η = rgQH/P and are as follows:
Q:
0
0.057
0.113
0.17
0.227
0.283
η:
0
0.488
0.752
0.876
0.822
0.755
Thus the BEP is, even without a plot, close to Q ≈ 0.17 m3/s.
is
Ans.
The specific speed
nQ*1/2 2,134[(0.17)(60)]1/2
N=
≈
≈ 216 Ans.
s
H*3/4
(100)3/4
For estimating Qmax, the last three points fit a Power-law to within ±0.5%:
H ≈ 102.7 − 21,090Q5.055= 0 if Q ≈ 0.349
m3
= Q max
s
Ans.
P11.29 If the scaling laws are applied to the Byron Jackson pump of Prob. 11.28 for the
same impeller diameter, determine (a) the speed for which the shut-off head will be 84 m;
(b) the speed for which the BEP flow rate will be 0.225 m3/s; and (c) the speed for which
the BEP conditions will require 80 hp.
Solution: From the table in Prob. 11.28, the shut-off head at 2,134 rpm is 104 m. Thus
1/2
=
If D1 D=
n1 (H=
2,134(84/104)1/2 ≈ 1,918 rpm Ans. (a)
2 , n2
2 /H1 )
m3
* * 2,134(0.225 / 0.17) ≈ 2,824 rpm Ans. (b)
=
If Q*2 0.225=
, n 2 n=
1 (Q 2 / Q1 )
s
Finally, if r1 = r 2 (water) and the diameters are the same, then Power ∝ n3, or, at BEP,
n2 =
n1 (P2* /P1* )1/3 =
2,134(80 ⋅ 0.746/190)1/3 ≈ 1, 451 rpm Ans. (c)
20
P11.30 A pump, geometrically similar to the 33-cm model in Fig. P11.24, has a
diameter of 60 cm and is to develop 30 hp at BEP when pumping gasoline (not water).
Determine (a) the appropriate speed, in r/min; (b) the BEP head, in m; and (c) the BEP
flow rate, in m3/min.
Solution: For gasoline, from Table A.3, ρ = 680 kg/m3. Read Fig. P11.24 for the BEP
3
values:
=
H1* 21.95
=
m, Q1* 2 m
=
/ min, η1* 0.80. Compute the power from this
data:
=
P1
ρ gQH (998)(9.81)(2 / 60)(21.95)
=
= 8.95 kW
η
0.80
Use the scaling laws to find the new speed, head, and flow rate:
P2 30(0.746) ρ 2 n2 3 D2 5
680 n2 3 60 5
ρ
(=
) ( ) (
)(
) ( ) , solve n2 ≈ 660
Ans.(a )
=
=
8.95
998 1,160 33
P1
ρ1 n1 D1
min
H2
=
H1
Q2
=
Q1
H2
n
D2 2
( 2 ) 2 (=
)
=
21.95
n1
D1
Q2
=
2
(
n2 D2 3
)( =
)
n1 D1
(
(
660 2 60 2
) ( ) , solve H 2 ≈ 23.5 m
1,160 33
660 60 3
)( ) ,
1,160 33
solve Q2 ≈ 6.84
21
m3
min
Ans.(b)
Ans.(c)
P11.31 A centrifugal pump with backward-curved blades has the following measured
performance when tested with water at 20°C:
Q, m3/min:
0
1.5
3
4.5
6
7.5
9
H, m:
37.5
35
33
30.8
28.3
24.7
18.9
P, kW:
22.4
26.9
29.8
32.8
35.1
35.8
34.3
(a) Estimate the best efficiency point and the maximum efficiency. (b) Estimate the most
efficient flow rate, and the resulting head and power, if the diameter is doubled and the
rotation speed increased by 50%.
Solution: (a) Convert the data above into efficiency. For example, at Q = 1.51 m3/min,
=
η
γ QH (9, 790 N/m3 )(1.51/60 m3 /s)(35.05 m)
=
= 0.32
= 32%
P
(26,900 m⋅ N/s)
When converted, the efficiency table looks like this:
Q, m3/min:
0
1.5
3
4.5
6
7.5
9
η, %:
0
32%
54%
69%
79%
84%
81%
So maximum efficiency of 84% occurs at Q = 7.5 m3/min. Ans. (a)
(b) We don’t know the values of CQ* or C* or C* , but we can set them equal for
H
P
conditions 1 (the data above) and 2 (the performance when n and D are changed):
CQ*
=
Q1
Q2
Q2
,
=
=
3
3
n1D1 n2 D2 (1.5n1 )(2 D1 )3
or: =
Q2 12
=
Q1 12(7.5 m3 / min)
= 90 m3 /min
C*
=
H
gH1
gH 2
gH 2
=
=
,
2 2
2 2
n1 D1 n2 D2 (1.5n1 ) 2 (2 D1 ) 2
or: =
H 2 9=
H1 9(24.7 =
m) 222 m
C*P
=
Ans. (b)
Ans. (b)
P1
P2
P2
=
=
,
3 5
3 5
r n1 D1 r n2 D2 r (1.5n1 )3 (2 D1 )5
or:=
P2 108
=
P1 108(35.8 =
kW) 3, 870 kW
22
Ans. (b)
P11.32 The data of Prob. 11.31 correspond to a pump speed of 1,200 r/min. (Were you
able to solve Prob. 11.31 without this knowledge?) (a) Estimate the diameter of the
impeller [HINT: See Prob. 11.24 for a clue.]. (b) Using your estimate from part (a),
calculate the BEP parameters CQ* , C*H , and C*P and compare with Eq. (11.27). (c) For what
speed of this pump would the BEP head be 85 m?
Solution: Yes, we were able to solve Prob. 11.31 by simply using ratios.
(a) Prob. 11.24 showed that, for a wide range of centrifugal pumps, the shut-off head
Ho ≈ 0.6U2/g ± 6%, where U is the impeller blade tip velocity, U = ω D/2. Use this
estimate with the shut-off head and speed of the pump in Prob. 11.31:
m 0.6[(126 D/2) 2 /9.81 m/s 2 ]
ω 2p (1, 200 rpm)/60
=
= 126 rad/s, H o ≈ 37.5
=
Solve for D ≈ 0.39 m Ans. (a)
(b) With diameter D ≈ 0.39 m estimated and speed n = 1,200/60 = 20 r/s given, we can
calculate:
CQ*
(7.5/60) m3 /s
(9.81 m/s 2 )(24.7 m)
* =
C
≈ 3.98
≈
0.105;
H
(20 r/s)(0.39 m)3
(20 r/s) 2 (0.39 m) 2
C*P
=
(35,800 W)
(998 kg/m3 )(20 r/s)3 (0.39 m)5
≈ 0.497
Ans. (b)
(c) Use the estimate of C*
H to estimate the speed needed to produce 85 m of head:
C*H ≈=
3.98
(9.81 m/s 2 )(85 m)
=
, solve for
n 37 r/s ≈ 2, 227 r / min
n 2 (0.39 m) 2
23
Ans. (c)
P11.33 In Prob. P11.31, the pump BEP flow rate is 7.6 m3/min, the impeller diameter is
40 cm, and the speed is 1,200 r/min. Scale this pump with the similarity rules to find (a)
the diameter and (b) the speed that will deliver a BEP water flow rate of 15 m3/min and a
head of 54 m. (c) What power will be required for this new condition?
Solution: Take the specific weight of water to be g = 9790 N/m3. Find the efficiency:
n1 ≈ =
n2
γ Q1 H1
=
P1
(9,790 N/m3 )(7.6 / 60 m3 /s)(24.7 m)
= 0.856
(35,800 W)
or
85.6%
Meanwhile, use the scaling laws for BEP flow rate and head:
Q1
Q2
7.5
15
=
=
=
3
3
3
n1D1 (1, 200)(40)
n2 D2 n2 D23
gH1
gH 2 (9.81)(54)
(9.81)(24.7)
=
=
=
2 2
2
2
n1 D1 (1, 200) (40)
n22 D22
n22 D22
Solve simultaneously for D2 = 46.5 cm Ans.(a) and n2 = 1,528 r/min Ans.(b).
(c) Finally, determine the power (which we could have done above, using η1):
P2
=
γ Q2 H 2 (9, 790)(15 / 60)(54)
=
= 154 kW Ans.(c)
0.856
η2
24
P11.34 Consider a pump geometrically similar to the 23-cm-diameter pump of Fig.
P11.34 to deliver 4.5 m3/min of kerosene at 1,500 rpm. Determine the appropriate (a)
impeller diameter; (b) power; (c) shut-off head; and (d) maximum efficiency.
Fig. P11.34
Performance data for a family of centrifugal pump impellers.
(Courtesy of Taco, Inc., Cranston, Rhode Island.)
Solution: For kerosene, take r ≈ 790 kg/m3, whereas for water r ≈ 998 kg/m3. From
Fig. P11.34, at BEP, read Q* ≈ 2.56 m3/min, H* ≈ 23.16 m, and ηmax ≈ 0.77. Then
Q*
2.56/60
4.5/60
CQ* = 3 =
≈ 0.120 =
,
3
nD
(1, 760/60)(0.23)
(1,500/60)D3
Solve for Dimp ≈ 0.293 m
Ans. (a)
gH o
9.81H o
9.81(25.6 m)
=
Shut-off:
=
,
2 2
2
2
n D
(1,500/60) 2 (0.293) 2
(1, 760/60) (0.23)
Solve H o ≈ 30.2 m Ans. (c)
1 − η2
 0.23 
≈
78% Ans. (d) (crude estimate)
 , solve for η2 ≈ 0.784 =
1 − 0.77  0.293 
1/4
Moody:
25
Fig. P11.34: Read H* ≈ 23.16 m, whence
9.81 H*new
9.81(23.16)
=
,
(29.3) 2 (0.23) 2 (25) 2 (0.293) 2
or H*new ≈ 27.4 m
790(9.81)(4.5/60)(27.4)
*
=
Then Pnew
≈ 20.3 kW
0.784
Ans. (b)
P11.35 A 45-cm-diameter centrifugal pump, running at 880 rev/min with water at
20°C, generates the following performance data:
Q, m3/min:
0.0
7.5
15
22.5
30
37.5
H, m:
28.0
27.1
25.6
23.8
20.7
15.2
P, kW:
74.6
83.6
97
106.7
116.4
121.6
η:
0%
40%
65%
82%
87%
76%
Determine (a) the BEP; (b) the maximum efficiency; and (c) the specific speed. (d) Plot
the required input power versus the flow rate.
Solution: We have computed the efficiencies and listed them. The BEP is the nextto-last point: Q = 30 m3/min, ηmax = 87%. Ans. (a, b) The specific speed is
N′s = nQ*1/2/(gH*)3/4 = (880/60)(30/60)1/2/[9.81(20.7)]3/4 ≈ 0.193, or Ns = 497
(probably a centrifugal pump). Ans. (c)
The plot of input horsepower versus flow rate is shown below—there are no surprises
in this plot. Ans. (d)
26
P11.36 The pump of Prob. P11.35 has a maximum efficiency of 87% at 30 m3/min.
(a) Can we use this pump, at the same diameter but a different speed, to generate a BEP head
of 45 m and a BEP flow rate of 38 m3/min? (b) If not, what diameter is appropriate?
Solution: We are still pumping water, r = 998 kg/m3. Try scaling laws for head and then for flow:
m3
m3
r
30
; P1 116 kW=
; D1 0.457 m
; n1 880
rpm 14.7
= 0.5 =
=
=
min
s
s
H 2 45
n2 2
r
r
) , n=
21.5 = 1,300
(from the head)
D
=
= = (
1 D2 :
2
21 14.7
H1
s
min
H1
=
D
=
1 D2 :
21m=
; Q1
Q2 38
n
r
18.6 =
= = ( 2 ) , n=
2
Q1 30 14.7
s
1,100
r
(from the flow rate)
min
These rotation rates are not the same. Therefore we must change the diameter.
(b) Allow for a different diameter in both head and flow rate scaling:
H2
=
H1
Q2
=
Q1
45 m
=
21m
n
D2 2
( 2 ) 2 (=
)
n1
D1
38 m3 /m in
=
30 m3 /m in
n2 D2 3
=
( )
n1 D1
(
n2
D2
)2 (
)2 ,
14.67 r/s 0.457 m
(
n2
D2 3
)(
) ,
14.67 0.457m
2 2
or : n=
2 D2
or :=
n2 D23
Solve simultaneously to obtain
=
n2
=
22.6 r/s
1, 350 rpm =
; D2
0.43m
The power required increases to 338 kW.
27
Ans.(a)
Ans.(b)
96.7
1.8
P11.37 Consider the two pumps of Problems P11.28 and P11.35, respectively. If the
diameters are not changed, which is more efficient for delivering water at 11.5 m3/min
and a head of 120 m? What is the appropriate rotation speed for the better pump?
Solution: Unless we are brilliant, what can we do but try them both? The scaling law
for constant diameter states that Q is proportional to n, and H is proportional to n2. The
pump of Prob. P11.28 has BEP at Q* = 0.17 m3/s and H* = 100 m. Scale from there:
Q2 11.5 n2
n2
, Solve
=
=
−
Q1 10.2 n1 2,134 r/min
H2
H2
n2 2
2, 400 2
) (
) ,
=
= (=
2,134
H1 100 m
n1
n2 ≈ 2, 400 r/min
H 2 ≈ 127 m
Solve
It seems that the Prob. P11.28 pump can do the job quite well.
Answer.
Try the Prob. P11.35 pump, whose BEP is at Q* = 30 m3/min and H* = 20.7 m. Then
Q2 11.5 n2
n2
, Solve
=
=
−
30
Q1
n1 880 r/min
H2
H2
n2 2
335 2
) (
) ,
=
= (=
880
H1 20.7 m
n1
n2 ≈ 335 r/min
Solve
H2 ≈ 3 m
This is not just worse, it is ridiculous – too slow, almost no head, totally inefficient.
28
P11.38 A 17-cm pump, running at 3,500 rpm, has the measured performance at right
for water at 20°C. (a) Estimate the power at BEP. If this pump is rescaled in water to
provide 15 kW at 3,000 rpm, determine the appropriate (b) impeller diameter; (c) flow
rate; and (d) efficiency for this new condition.
Q , m3 /min:
H , m:
η , %:
0.189
61.3
29
0.379
61
50
0.568
60.4
64
0.757
59.1
72
0.946
57.6
77
1.136
55.2
80
1.325
51.5
81
1.514
47.5
79
1.703
42.4
74
Solution: The BEP of 81% is at about Q = 1.33 m3/min and H = 51.5 m. Hence the
power is
P*
=
ρ gQ*H* 9, 790(1.33/60)(51.5)
=
= 13.8 kW
η
0.81
If the new conditions are
coefficients:
=
C*P
Ans. (a)
15 kW at n = 3,000 rpm = 50 rps, we equate power
13,800
15, 000
= 0.437 ≡
,
998(50)3 D5
998(3,500/60)3 (17.4/100)5
Solve Dimp ≈ 0.194 m
Ans. (b)
With diameter known, the flow rate is computed from BEP flow coefficient:
Q*
1.33/60
Q*
=
C*Q =
= 0.0721 ≡
,
3
nD
50(0.194)3
(3,500/60)(17.4/100)3
=
Solve Q* 0.0263 m3 /s ≈ 1.58 m 3 / min Ans. (c)
Finally, since D1 ≈ D2, we can assume the same maximum efficiency: 81%. Ans. (d)
29
P11.39 The Allis-Chalmers D30LR centrifugal compressor delivers 950 m3/min of SO2
with a pressure change from 96.5 to 124 kPa absolute using an 800 hp motor at 3,550
r/min. What is the overall efficiency? What will the flow rate and ∆p be at 3,000 r/min?
Estimate the diameter of the impeller.
Solution: For SO2, take M = 64.06, hence R ≈ 130 J/(kg⋅°K). Then
 950 
∆p = (124 − 96.5) = 27.5 kPa, Power = Q∆p = 27.5 
 ≈ 435 kW delivered
 60 
Then
=
h Pdelivered=
/Pmotor 435/597 ≈ 73% Ans. (a)
 n2 
m3
 3, 000 
=
If n 2 3, 000 rpm,
=
Q 2 Q=
950
≈
803

1
 3,550 
min


 n1 
Ans. (b)
2
 3, 000 
∆p 2 =
∆p1 (n 2 /n1 ) 2 =
(27.5) 
 ≈ 19.64 kPa
 3,550 
Ans. (c)
To estimate impeller diameter, we have little to go on except the specific speed:
ρavg ≈
∆p
110, 000
kg
27,500
≈ 2.93 3 , H =
=
≈ 957 m,
ρ g 2.93(9.81)
130(289)
m
rpm(m3 /s)1/2 3,550[950/60]1/2
Ns = 3/4 = 3/4
≈ 82 : Fig. P11.49: C*Q ≈ 0.45
(H-m)
(957)
=
Crudely, C*Q ≈ 0.45
950/60
, solve for Dimpeller ≈ 0.841m
(3,550/60)D3
Ans. (d)
Clearly this last part depends upon the ingenuity and resourcefulness of the student.
30
P11.40 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: If we combine CQ and CH in such a way as to eliminate speed n, and also to
make the result linearly proportional to D, we obtain Logan’s result:
1/4
D(gH*)1/4
C*H
Specific diameter D s =
Ans. (a) D s
=
1/2
1/2
Q*
C*Q
Ans. (b)
(c) For the pumps of Figs. 11.8 and 11.13, we obtain
(5.0)1/4
(1.07)1/4
D=
= 4.41; Ds=
= 1.37 Ans. (c)
s -Fig.11.8
-Fig.11.13
(0.115)1/2
(0.55)1/2
P11.41 It is desired to build a centrifugal pump geometrically similar to Prob. 11.28
(data below) to deliver 25 m3/min of gasoline at 1,060 rpm. Estimate the resulting (a)
impeller diameter; (b) head; (c) power; and (d) maximum efficiency.
Q, m3/s:
0
0.057
0.113
0.17
0.227
0.283
H, m:
104
104
104
100
91
67
kW:
101
119
153
190
246
246
Solution: For gasoline, take r ≈ 680 kg/m3. From Prob. 11.28, BEP occurs at Q* ≈
0.17 m3/s, ηmax ≈ 0.88. The data above are for n = 2,134 rpm = 35.6 rps and D = 37.14
cm.
Then=
C*Q
0.17
25/60
0.0933 ≡
,
=
3
35.6(37.14/100)
(1, 060/60)D3
Solve for
Dimp ≈ 0.63 m
Ans. (a)
9.81(100)
9.81H
C*H =
=
5.65 ≡
, solve for H ≈ 71 m
2
2
(35.6) (37.14/100)
(1, 060/60) 2 (0.63) 2
31
Ans. (b)
Step-up the efficiency with Moody’s correlation, Eq. (11.29a), for D1 = 37.14 cm ≈ .3714
m
1/4
D 
1 − η2
 0.3714 
≈  1 = 
= 0.877, solve for η2 ≈ 0.895
1 − 0.88  D2 
 0.63 
ρ gQ 2 H 2 680(9.81)(25/60)(71)
Then
P2 =
=
≈ 220 kW Ans. (c)
0.895
h2
1/4
P11.42 An 20-cm model pump delivering water at 80°C at 3 m3/min and 2,400 rpm
begins to cavitate when the inlet pressure and velocity are 82.7 kPa and 6 m/s,
respectively. Find the required NPSH of a prototype which is 4 times larger and runs at
1,000 rpm.
Solution: For water at 80°C, take r g ≈ 9,520 N/m3 and pv ≈ 76.6 kPa. From Eq. 11.19,
=
NPSH model
pi − p v Vi2 (82, 700 − 76, 600)
(6) 2
+ =
+
= 2.46 m
ρg
2g
9,520
2(9.81)
2
2
2
 n p   Dp 
 1, 000   4 
=
=
Similarity:
NPSH proto NPSH
 
 2.46 
m
   ≈7m
 2, 400   1 
 n m   Dm 
2
Ans.
P11.43 The 71-cm-diameter pump in Fig. 11.7a at 1,170 r/min is used to pump water at
20°C through a piping system at 53 m3/min. (a) Determine the required brake
horsepower. The average friction factor is 0.018. (b) If there is 20 m of 30-cm-diameter
pipe upstream of the pump, how far below the surface should the pump inlet be placed to
avoid cavitation?
Fig. 11.7a
32
Solution: For water at 20°C, take r g ≈ 9,790 N/m3 and pv ≈ 2,350 pa. From Fig. 11.7a
(above), at 71 cm and 53 m3/min, read H ≈ 97.5 m, η ≈ 0.81, and P ≈ 1,044 kW. Ans.
=
Or: Required
bhp
r gQH (9, 790)(53 /60)(97.5)
=
≈ 1, 040 kW Ans.
h
0.81
From the figure, at 53 m3/min, read NPSH ≈ 7.6 m. Assuming pa = 1 atm,
=
Eq.11.20: NPSH
pa − p v
101,325 − 2,350
L V2
− Zi −=
− Zi − h fi ≈ 7.6 m, =
h fi
h fi f
,
ρg
9, 790
D 2g
Q
53/60
m
V
= =
≈ 12.5 ,
2
A (π /4)(0.3 m)
s
 20   (12.5) 
so: =
Zi 10.1 − 7.6 − 0.018 
 ≈ −7 m

 0.3   2(9.81) 
2
Ans.
P11.44 The pump of Prob. 11.28 is scaled up to an 46-cm-diameter, operating in water
at BEP at 1,760 rpm. The measured NPSH is 4.8 m, and the friction loss between the
inlet and the pump is 6.6 m. Will it be sufficient to avoid cavitation if the pump inlet is
placed
2.7 m below the surface of a sea-level reservoir?
Solution: For water at 20°C, take r g = 9,790 N/m3 and pv = 2,350 pa. Since the NPSH
is given, there is no need to use the similarity laws. Merely apply Eq. 11.20:
NPSH ≤
pa − p v
101,325 − 2,350
− Zi − h fi , or: Zi ≤
− 6.5 − 5 =−1.4 m, OK,
9, 790
rg
Z actual = −2.7 m
Ans.
This works. Putting the inlet 2.7 m below the surface gives 1.3 m of margin against
cavitation.
P11.45 Determine the specific speeds of the seven Taco, Inc. pump impellers in Fig.
P11.24. Are they appropriate for centrifugal designs? Are they approximately equal
within experimental uncertainty? If not, why not?
Solution: Read the BEP values for each impeller and make a little table for 1,160 rpm:
D, cm:
25.4
26.7
27.9
29.2
30.5
31.8
32.9
Q*,
m3/min:
1.48
1.59
1.67
1.74
1.82
1.93
2.00
H*, m:
12.5
13.41
14.94
17.07
18.29
20.12
21.95
Specific
speed NS:
212
209
197
182
177
170
162
These are well within the centrifugal-pump range (NS < 600) but they are not equal because
they are not geometrically similar (7 different impellers within a single housing). Ans.
P11.46 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 writer 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
76 m3/min of water and a head of 120 m running at 1,200 rev/min. Suggest a curve-fit
formula to the data (Hint: a hyperbola).
Fig. P11.46
34
Solution: We see that the data are very well correlated by a single curve. (NOTE: These
are all centrifugal pumps—a slightly different correlation holds for mixed- and axial-flow
pumps.) The data are well fit by a hyperbola:
Figure P11.46: Ds ≈
Const
, where Const ≈ 850 ± 50
Ns
Ans.
For the given pump-data example, we compute
=
Ns
Hence Ds ≈
(rpm)(m3 / min)1/2 1, 200(76)1/2
289
= =
( Head −m)3/4
(120)3/4
D[9.81(120)]1/4
850
≈ 2.94 =
, solve D = 0.57 ± 0.04 m
289
(1.27)1/2
Ans.
P11.47 A pump must be designed to deliver 6 m3/s of water against a head of 28 m. The
specified shaft speed is 20 r/s. What type of pump do you recommend?
Solution: Then the specific speed is
=
Ns
(rpm)(m3 / min)1/2
=
( H − m)3/4
(1200 rpm)[(6 m3 /s)(60 s / min)]1/2
=
(28 m)3/4
From Fig. 11.11a, an axial-flow pump is appropriate for this job.
1,870
Ans.
P11.48 Using the data for the pump in Prob. P11.8, (a) determine its type: PDP,
centrifugal, mixed-flow, or axial-flow. (b) Estimate the shutoff head at 1,750 r/min. (c)
Does this data fit on Fig. 11.14? (d) What speed and flow rate would result if the head
were increased to 48 m?
Solution: Recall the P11.8 data: n = 1,750 r/min, H = 32 m, Q = 4 m3/min,
P = 24.2 kW.
(a) The specific speed can be calculated immediately:
=
Ns
rpm m3 / min
=
( H − m)0.75
(1, 750) 4
≈ 260 Centrifugal pump Ans.(a )
(32 m)0.75
35
(b) From the discussion below Eq. (11.18), the shutoff head is about 60% of the theory:
2π 27.5 / 2
=
)][
] 25.2 m / s
60
100
U2
(25.2) 2
≈ 39 m Ans.(b)
H 0 ≈ 0.6 = 0.6
g
9.81
= ω=
U
r [1, 750(
The actual pump has a shutoff head of only 37 m, or 57% of the theoretical value.
(c) In Fig. 11.14, at 4 m3/min and Ns = 260, read η ≈ 84%. The actual pump has a best
efficiency, from Prob. P11.8, of 86.4%.
(d) To get a higher head for the same pump diameter, we have to increase the speed:
H2
=
H1
48
=
32
n
D2 2
( 2 )2 ( =
)
n1
D1
Q2
=
Q1
Q2
=
4
n D 3
m3
48
( 2 )( 2=
)
(
)(1.0)3 , solve Q2 ≈ 4.90
min
n1 D1
32
(
n2 2
r
) (1.0) 2 , solve n2 ≈ 2,143
min
1750
Ans.(d )
P11.49 Data collected by the writer for flow coefficient at BEP for 30 different pumps are
plotted at right in Fig. P11.49. Determine if the values of C*Q fit this correlation for the
pumps of Problems P11.28, P11.35, and P11.38. If so, suggest a curve fit formula.
Solution: Make a table of these values:
Fig. P11.49 Flow coefficient at BEP for
30 commercial pumps.
36
Prob. 11.28:
Prob. 11.35:
Prob. 11.38:
Q*,
m3/min
10.19
30.28
1.32
D, cm
n, rpm
NS
C*Q = Q * /(nD 3 )
37.13
45.72
17.40
2134
880
3500
216
497
210
0.0933
0.360
0.0719
When added to the plot shown below, all three data points seem to fit quite well. The data
are useful for predicting general centrifugal-pump behavior and are well fit to either a
2nd-order polynomial or a single-term Power-law slightly less than parabolic:
Polynomial: C*Q ≈ −0.0146 + 2.597 E − 4N S + 9.021E − 7N S2 :(Correlation r 2 ≈ 0.97)
37
Ans.
P11.50 Data collected by the writer for power coefficient at BEP for 30 different pumps
are plotted at right in Fig. P11.50. Determine if the values of C*P for the three pumps of
Prob. 11.49 above fit this correlation.
Solution: Make a table of these values, similar to Prob. 11.49:
Fig. P11.50 Power coefficient at BEP for
30 commercial pumps.
*P P */( ρ n 3 D5 )
C=
P*, kW
D, cm n, rpm
NS
Prob. 11.28:
Prob. 11.35:
Prob. 11.38:
190.2
116.4
13.8
37.13
45.72
17.40
2134
880
3500
38
216
497
210
0.600
1.85
0.435
When added to the plot shown below, all three of them seem to fit reasonably well. The
data are moderately useful for predicting general centrifugal-pump behavior and can be
fit to either a 2nd-order polynomial or a single-term Power-law:
Polynomial: C*Q ≈ −0.0886 + 2.399 E − 3N S + 2.628E − 6N S2 :(Correlation r 2 ≈ 0.99)
P11.51 An axial-flow pump delivers 1 m3/s of air which enters at 20°C and 1 atm. The
flow passage has a 25-cm outer radius and a 20-cm inner radius. Blade angles are α1 =
60° and b 2 = 70°, and the rotor runs at 1,800 rpm. For the first stage, compute (a) the
head rise; and (b) the power required.
Solution: Assume an average radius of (20 + 25)/2 = 22.5 cm and compute the blade
speed:
2π   22.5 
m
Q
1 m3 /s
m

≈
=
=
≈ 14.1
u avg =
ω ravg =
1,800
42.4
;
V



n
2
2
60   100 
s
A π [(0.25) − (0.20) ]
s

2
2
Theory: gH = u − uVn(cot α1 + cot b 2) = (42.4) − 42.4(14.1)(cot 60° + cot 70°),
H ≈ 126 m Ans. (a)
 101, 325 
=
Ptheory r=
gQH 
= 1, 490 W
 (9.81)(1)(126)
 287(293) 
39
Ans. (b)
Ans.
P11.52 An axial-flow fan operates in sea-level air at 1,200 r/min and has a blade-tip
diameter of 1 m and a root diameter of 80 cm. The inlet angles are α1 = 55° and b1 = 30°,
while at the outlet b2 = 60°. Estimate the theoretical values of the (a) flow rate,
(b) power required, and (c) outlet angle α2.
Solution: For air, take r ≈ 1.205 kg/m3. The average radius is 0.45 m. Thus
Fig. P11.52
2π 
m

=
u=
ωR =
Vn (cot α1 + cot β1 ) =
Vn 2 (cot α 2 + cot β 2 )
1, 200
 (0.45) ≈ 56.6
60 
s

56.6
m
Solve Vn1 =
Vn2 = ≈ 23.2
and a 2 ≈ 28.3° Ans. (c)
cot 55° + cot 30°
s
m3
Then Q =
Vn A =
(23.2)[π{(0.5)2 − (0.4)2}] ≈ 6.56
Ans. (a)
s
gH= u 2 − uVn (cot α1 + cot β 2=
) (56.6) 2 − 56.6(23.2)(cot 55° + cot 60°=
) 1, 520 m 2 /s2
Finally,
=
P ρ=
QgH 1.205(6.56)(1,520)
= 12,000 W
40
Ans. (b)
P11.53 Figure P11.46 is an example of a centrifugal pump correlation, where Ds
is defined in the problem. From data in the literature, we can suggest the following
correlation for axial-flow pumps and fans:
873
Ds ≈ 0.946
for N s > 100
Ns
where Ns is the dimensional specific speed, Eq. (11.30b). Use this correlation to
find the appropriate size for a fan that delivers 680 m3/min of air at sea-level
conditions when running at 1,620 r/min with a pressure rise of 5 cm of water. Hint:
Express the fan head in meters of air, not meters of water.
Solution: For air, take r ≈ 1.205 kg/m3. ∆p of 5 cm of water =
(998)(9.81)(0.05)=490Pa.
rpm(Q : m 3 / min)0.5
1, 620(680)0.5
=
= 2, 586
Ns =
H ( m )0.75
[490 / (1.205)(9.81)]0.75
D=
s
D ( gH *)1/4 D[490 / 1.205]1/4
873
=
=
=
→D
= 0.387 m Ans.
0.516
Q *1/2
2, 5860.946
(680 / 60)1/2
P11.54 It is desired to pump 1.5 m3/s of water at a speed of 22 r/s, against a
head of 25 m. (a) What type of pump would you recommend? Estimate (b) the
power in kW.
Solution: Evaluate the specific speed:
m3
m3
r
=
Q 1.5= 90
=
, n 22
= 1, 320rpm,=
H 25m
s
min
s
rpm(Q : m 3 / min)0.5 1, 320(90)0.5
=
Ns
= =
1,120
H ( m )0.75
(25)0.75
From Fig. 11.11a, this pump should have a mixed-flow impeller.
41
Ans.(a)
(b) According to Fig. 11.11a, a really good mixed-flow impeller, Ns = 1,000, would
have an efficiency of about 92 per cent. Thus we can estimate the power required:
=
Pinput
γ QH (9, 790 N / m 3 )(1.5m 3 / s )(25m )
=
= 399 kW
η
0.92
Ans.(b)
P11.55 Suppose that the axial-flow pump of Fig. 11.13, with D = 45 cm, runs at
1,800 r/min. (a) Could it efficiently pump 95 m3/min of water? (b) If so, what
head would result? (c) If a head of 36 m is desired, what values of D and n would
be better?
Solution: Convert n = 1,800 r/min = 30 r/s. The flow coefficient at BEP is
approximately 0.55. Check this against the data:
=
CQ*
Q
(95 / 60 m 3 / s )
=
= 0.579 (Yes, this is at BEP )
nD 3 (30r / s )(0.45m )3
Ans.( a )
(b) Since we are at BEP, we can use the BEP head coefficient to get H:
CH * ≈ 1.07
=
gH
(9.81m / s 2 ) H
=
→H
= 19.9 m
n 2 D 2 (30r / s ) 2 (0.45m ) 2
Ans.(b)
(c) For D = 45 cm, H = 36 m is far off BEP, CH ≈ 1.9, efficiency only 40 per cent. So
let diameter be unknown and set both head and flow coefficients equal to BEP values:
CQ *
=
0.579
=
(95 / 60 m3 /s)
nD
3
,
C
=
H*
r
Solve simultaneously for
=
n 49.0
=
s
1.07
=
2, 940
42
(9.81)(36 m)
n2 D2
r
, D 0.38 m
=
min
Ans.( c)
P11.56 Determine if the Bell and Gossett pump of Prob. P11.8 (a) fits the three
correlations in Figs. P11.46, P11.49, and P11.50. (b) If so, use these correlations
to find the flow rate and horsepower that would result if the pump is scaled up to
D = 60 cm but still runs at 1,750 r/min.
Solution: Recall the P11.8 data: n = 1,750 r/min, H* = 32 m, Q* = 4 m3/min, P*
= 24.2 kW. Calculate the specific speed and the specific diameter (from Prob.
P11.46):
=
Ns
rpm m3 / min
=
( H − m)0.75
=
Ds
D ( gH *)1/4
=
Q *1/2
(1, 750) 4
≈ 260
(32)0.75
(0.275)[9.81(32)]1/4
=
(4 / 60)1/2
Read Fig. P11.46 at a specific speed of 260: Ds ≈ 4.4
4.48
(reasonable agreement).
Read Fig. P11.49 at a specific speed of 260: CQ* ≈ 0.12
(fair agreement).
Read Fig. P11.50 at a specific speed of 260: CP* ≈ 0.68
(fair agreement)
Ans.(a)
(b) Apply the last two numbers to the new D2 = 0.60 m:
Q2
Q2
m3
m3
C =3 =
≈ 0.12 , Q2 ≈ 0.756
≈ 45.4
nD2
(1, 750 / 60)(0.6)3
s
min
*
Q
CP*
=
P2
P2
=
= 0.68, P2 ≈ 1, 309 kW
3 5
ρ n D2
(998)(1, 750 / 60)3 (0.6)5
43
Ans.(b)
P11.57 Performance data for a 53-cm-diameter air blower running at 3,550 rpm
are shown below. What is the specific speed? How does the performance compare
with Fig. 11.13? What are C*Q , C*H , C*P ?
∆p, cm H2O:
Q, m3/min:
kW:
74
14
4.48
76
28
5.97
71
56.5
8.95
53
85
13.43
25
113
18.65
Solution: Assume 1-atm air, r ≈ 1.2 kg/m3. Convert the data to dimensionless
form and put the results into a table:
∆p, pa:
7245
7441
6951
5189
2448
3
Q, m /s:
0.233
0.467
0.942
1.417
1.883
H, m (of
air):
615.4
632.1
590.5
440.8
208.0
CQ :
0.0265
0.0530
0.107
0.161
0.214
CH :
6.14
6.31
5.89
4.40
2.08
CP:
0.431
0.574
0.861
1.292
1.794
η:
0.377
0.582
0.731
0.547
0.247
Close enough without plotting: C*Q ≈ 0.107 , C*H ≈ 5.89 , C*P ≈ 0.861
=
Specific speed N s
Ans.
(3,550 rpm)(56.5 m3 /min)1/2
≈ 223 Ans.
(590.5 m)3/4
This centrifugal pump is very similar to the dimensionless data of Fig. 11.8.
Ans.
44
P11.58 Aircraft propeller specialists claim that dimensionless propeller data,
when plotted as (CT/J2) versus (CP/J2), form a nearly straight line, y = mx + b. (a)
Test this hypothesis for the data of Fig. 11.16, in the high-efficiency range J =
V/(nD) equal to 0.6, 0.7, and 0.8. (b) If successful, use this straight line to predict
the rotation rate n, in r/min, for a propeller with D = 1.5 m, P = 30 hp, T = 420 N,
and V = 153 km/h, for sea-level standard conditions. Comment.
Solution: For sea-level air, take r = 1.22 kg/m3and P = 30 hp = 22 kW. The
writer read Fig. 11.16 as best he could and came up with the following data,
rewritten in the desired form:
J
0.6
0.7
0.8
CT
0.055
0.040
0.023
CP
0.042
0.035
0.023
CP/J^2
0.1167
0.0714
0.0359
CT/J^2
0.1528
0.0816
0.0359
The plot is, sure enough, reasonably linear:
CT/J
2
CP/J
2
The specialists were correct; this propeller data fits the simple formula
CT
J
=
2
T
≈ 1.453(
CP
) − 0.0184=
2
1.453 (
P
ρρ
D2 V 2
J
n D3 V 2
45
) − 0.0184
We have all the data except n. Convert to BG units: 153 km/h = 42.5 m/s. Put
these values into our straight-line formula:
420
22, 000
] − 0.0184 ,
(1.22) n (1.5)3 (42.5) 2
4.298
4.298
r
r
or : 0.0853 ≈
− 0.0184 , n ≈
≈ 41.7
≈ 2500
n
0.1037
s
min
(1.22)(1.5) 2 (42.5) 2
≈ 1.453[
Ans.
This is reasonably accurate and close to maximum efficiency, V/nD ≈ 0.68.
However, beware! This clever correlation has reduced the propeller data from
three variables (CT, CP, and V/nD) to only two (CT/J2 and CP/J2), without any
theoretical justification. Thus, information is lost. Do not use this straight line as
a general problem solver. It will fail miserably if fed with data that does not
match the conditions in Fig. 11.16.
P11.59 Suppose it is desired to deliver 20 m3/min of propane gas (molecular
weight = 44.06) at 1 atm and 20°C with a single-stage pressure rise of 20 cm
H2O. Determine the appropriate size and speed for using the pump families of (a)
Prob. P11.57 and (b) Fig. 11.13. Which is the better design?
Solution: For propane, with M = 44.06, the gas constant R = 189 m⋅N/(kg⋅°K).
Convert ∆p = 20 cm H2O = (9,790)(0.2) = 1,958 N/m2. The propane density and
head rise are
p 101,325 pa
kg
r=
=
≈ 1.83 3 ,
gas
RT
189(293)
m
1,958
Hence=
H pump
≈ 109 m propane
1.83(9.81)
*
(a) Pr ob.11.57 : C=
Q ≈ 0.105
20/60
9.81(109)
*
and C
=
H ≈ 5.83
3
nD
n 2 D2
Solve for n =
28.5 rps ≈ 1, 710 rpm and D ≈ 0.48 m
Ans. (a) (centrifugal pump)
(b) Fig. 11.13: C*Q ≈ 0.55 and C*H ≈ 1.07 yield
n ≈ 14, 000 rpm, D ≈ 0.14 m
46
Ans. (b) (axial flow)
The centrifugal pump (a) is the better design—nice size, nice speed. The axial
flow pump is much smaller but runs too fast. Ans.
P11.60 Performance curves for a certain free propeller, comparable to Fig.
11.16, can be plotted as shown in Fig. P11.60, for thrust T versus speed V for
constant power P.
(a) What is striking, at least to the writer, about these curves? (b) Can you deduce
this behavior by rearranging, or replotting, the data of Fig. 11.16?
1400
250 kW
1200
Thrust, N
350 kW
Fig. P11.60
1000
450 kW
800
600
400
200
0
0
50
100
150
200
Speed, kilometer per hour
Solution: (a) The three curves look exactly alike! In fact, we deduce, for this
propeller, that the thrust, at constant speed, is linearly proportional to the
horsepower! This implies that the dimensionless thrust, CT, is proportional to the
dimensionless power, CP, for this propeller. We can go back and see that this is
not the case for Fig. 11.16.
47
250
(b) Here is a plot of thrust versus speed for the actual data of Fig. 11.16. We see
that the curves are similar in shape, but the thrust is not proportional to the
horsepower. Ans.(b)
Thrust - N
Data plotted from Figure 11.16
1800
125 kW
1500
250 kW
375 kW
1200
900
600
300
0
0
100
200
300
400
Speed - m/s
P11.61 A mine ventilation fan delivers 500 m3/s of sea-level air at 295 rpm and
∆p = 1,100 Pa. Is this fan axial, centrifugal, or mixed? Estimate its diameter in
meter. If the flow rate is increased 50% for the same diameter, by what percent
will ∆p change?
Solution: For sea-level air, take r g ≈ 11.8 N/m3, hence H = ∆p/r g = 1,100/11.8
≈ 93 m. Calculate the specific speed:
rpm(m3 /min)1/2 295(500 ⋅ 60)1/2
=
Ns
=
≈ 1, 700 (axial-flow pump ) Ans. (a)
(head)3/4
(93)3/4
=
Estimate C*Q ≈ 0.55
500 m3 /s
, solve Dimpeller ≈ 5.7 m
(295/60)D3
Ans. (b)
At constant D, Q ∝ n and ∆p (or H) ∝ n2. Therefore, if Q increases 50%, so does
n, and therefore ∆p increases as (1.5)2 = 2.25, or a 125% increase. Ans. (c)
48
P11.62 The actual mine-ventilation fan in Prob. 11.61 had a diameter of 6 m [Ref.
20, p. 339]. What would be the proper diameter for the pump family of Fig. 11.14
to provide 500 m3/s at 295 rpm and BEP? What would be the resulting pressure
rise in Pa?
Solution: For sea-level air, take r g ≈ 11.8 N/m3. As in Prob. 11.61 above, the
specific speed of this fan is 1,700, hence an axial-flow fan. Figure 11.14 indicates
an efficiency of about 90%, and the only values we know for performance are
from Fig. 11.13:
=
Ns ≈ 1,800: C*Q ≈ 0.5
500
, solve D impeller ≈ 5.88 m
(295/60)D3
2
Ans. (a)
2
(295/60) (5.88)
= 106.5 m,
C*H ≈=
1.25, H 1.25
9.81
=
∆p (11.8)(106.5) ≈ 1, 257 Pa Ans. (b)
P11.63 A good curve-fit to the head vs. flow for the 81 cm pump in Fig. 11.7a is
H (in m) ≈ 150-(0.00617) Q2 Q in m3/min
Assume the same rotation rate, 1,170 r/min, and estimate the flow rate this pump
will provide to deliver water from a reservoir, through 270 m of 30 cm pipe, to a
point 45 m above the reservoir surface. Assume a friction factor f = 0.019.
Solution: Apply Bernoulli eqn.
p1 V12
p2 V22
+
+z +h = +
+ z + h f , p1 = p2 = 0, V1 =V2 , z1 = 0, z2 = 45
g 2g 1 p g 2g 2
L V2
270 m [4Qm / (60 ⋅ p ⋅ 0.32 )]2
=
hp =
150 − 0.00617Qm 2 , h f =
f
(0.019)(
)[
]
D 2g
0.3 m
2(9.81 m/s 2 )
Qm : flow rate in m3 / min
150 − 0.00617Qm 2 =
45 + (0.019)(
2
1922 → Q
Qm=
= 43.8
270 m [4Qm / (60 ⋅ π ⋅ 0.32 )]2
)[
]
0.3 m
2(9.81 m/s 2 )
m3
m3
= 0.731
min
s
49
Ans.
P11.64
A leaf blower is essentially a centrifugal impeller exiting to a tube.
Suppose that the tube is smooth PVC pipe, 1.2 m long, with a diameter of 6.35
cm. The desired exit velocity is 117 km/h in sea-level standard air. If we use the
pump family of Eq. (11.27) to drive the blower, what approximate (a) diameter
and (b) rotation speed are appropriate? (c) Is this a good design?
Solution: Recall that Eq. (11.27) gave BEP coefficients for the pumps of Fig.
11.7:
CQ* ≈ 0.115; C*H ≈ 5.0; C*P ≈ 0.65
Apply these coefficients to the leaf-blower data. Neglect minor losses, that is, let
the pump head match the pipe friction loss. For air, take r = 1.2 kg/m3 and m =
1.8E−5
kg/m⋅s.
Convert 117 km/h = 32.5 m/s.
2
 1.2 m   (32.5 m/s) 2 
n2 D 2
L V pipe
=
h f H=
= f
= f
=
5.0
, f f (Re d )

pump
2 
d pipe 2 g
9.81 m/s 2
 0.0635 m   2(9.81 m/s ) 
=
Q
pp
2
m3
(0.0635 m) 2 (32.5
m/s) 0.103
=
d pipeV
=
= 0.115nD3
4
4
s
We know the Reynolds number, Red = rVd/m = (1.2)(32.6)(0.0635)/(1.8E−5) =
130,000, and for a smooth pipe, from the Moody chart, calculate fsmooth = 0.0168.
Then H = hf = 17.1 m, and the two previous equations can then be solved for
Dpump ≈ 0.39 m ; n ≈ 15.1 r/s =
906 r/min
Ans. (a, b)
(c) This blower is too slow and too large, a better (mixed or axial flow) pump
can be designed. Ans. (c)
50
P11.65 An 29-cm diameter centrifugal pump, running at 1,750 rev/min, delivers
3.2 m3/min and a head of 32 m at best efficiency (82%). (a) Can this pump
operate efficiently when delivering water at 20°C through 200 m of 10-cmdiameter smooth pipe? Neglect minor losses. (b) If your answer to (a) is
negative, can the speed n be changed to operate efficiently? (c) If your answer to
(b) is also negative, can the impeller diameter be changed to operate efficiently
and still run at 1,750 rev/min?
Solution: For water at 20°C, take r = 998 kg/m3 and m = 0.001 kg/m-s.
Convert to SI units: Q = 3.2 m3/min = 0.0533 m3/s, D = 0.29 m, H 32 m.
Compute ReD and f:
V=
Q
0.0533
m
ρVD 998(6.79)(0.1)
=
=
6.79 ; Re D=
=
= 678, 000; f smooth= 0.0125
2
A (π / 4)(0.1)
s
m
0.001
(a) Now compute the friction head loss at 3.2 m3/min and see if it matches the
pump head:
L V2
200 m (6.79 m/s) 2
=
h f f= (0.0125)(
=
)[
] 58.7 m
D 2g
0.1 m 2(9.81 m/s 2 )
The answer to part (a) is No, the pump will operate far off-design to lower the
flow rate and hf.
(b) Can we just change the pump speed? This answer is also No. At BEP,
(gH/n2D2) is constant, therefore pump head is proportional to n2. But the pipe
friction loss is also approximately proportional to n2. Thus, for any speed n, the
pipe head will always be about twice as large as the pump head and not operate
efficiently. Ans.(b)
(c) Can we change the impeller diameter, still at 1750 rev/min, and operate
efficiently? This time the answer is Yes. First note the BEP dimensionless
parameters, n = 1,750/60 = 29.2 rps:
=
CH *
gH * (9.81 m/s 2 )(32.0 m)
Q*
(0.0533 m3 /s)
=
=
4.38;
=
C
=
= 0.0748
Q*
n 2 D p2
(29.2/s) 2 (0.29 m) 2
nD po (29.2/s)(0.29 m)3
Now combine these with the fact that hf is approximately proportional to n2. In SI
units,
n 2 D p2
h f (m) ≈ 20, 660Q 2 (in m3 /s) =
H pump =
4.38
≈ 20, 660[0.0748 n D3p ]2
g
n cancels, solve for D p4 ≈ 0.00386 m 4 , D p ≈ 0.25 m Ans. (c)
51
This is approximately correct. Check: the pump head is 23.8 m and flow rate is
2.04 m3/min. For this flow rate, the pipe head loss is 23.8 m. Good match!
Ans.(c)
NOTE: Part (c) is almost independent of n. For example, if n = 20 rev/s, the best
efficiency converges to an impeller diameter of 0.24 m, with Q ≈ 1.22 m3/min and
H = hf ≈ 10.2 m.
P11.66 It is proposed to run the pump of Prob. 11.35 at 880 rpm to pump water
at 20°C through the system of Fig. P11.66. The pipe is 20-cm diameter
Is this an efficient
commercial steel. What flow rate in m3/min results?
operation?
Fig. P11.66
Solution: For water, take r = 998 kg/m3 and m = 0.0010 kg/m⋅s. For
commercial steel, take e = 0.046 mm. Write the energy equation for the system:

40 [Q/(p (0.2) 2 /4)]2
m3 
LV2
=11 − 4 + f ( )
=7 + 10,328 fQ 2 ,  Q in

0.2
2(9.81)
s 
d 2g


e
Meanwhile, H p = fcn(Q)Prob. 11.35 and f = fcn  Red , 

d Moodychart
H pump =∆z + f
where e /d = 0.046/200 = 0.00023. If we guess f as the fully rough value of 0.0141,
we find that Hp is about 25 m and Q is about 0.35 m3/s. To do better would
require some very careful plotting and interpolating, or: iteration with Excel leads
to the more accurate solution:
m
; Ppump 77 =
kW; H p 23.6 m
=
f 0.0145;
=
V 10.6 =
s
m3
Q ≈ 20
Ans.
min
52
The efficiency is 74%, considerably off the maximum of 88% - not a very good
system fit. Ans.
Here is a plot of the pump and system curves:
P11.67 The pump of Prob. 11.35, running at 880 r/min, is to pump water at
20°C through 75 m of horizontal galvanized-iron pipe (e = 0.15 mm). All other
system losses are neglected. Determine the flow rate and input power for (a) pipe
diameter = 20 cm; and (b) the pipe diameter yielding maximum pump efficiency.
Solution: (a) There is no elevation change, so the pump head matches the
friction:
=
Hp
f
LV2
=
d 2g
f
75 [Q/ (pρ
(0.2) 2 /4)]2
4 Q e 0.15
= 19,366 fQ 2, =
Red
, = = 0.00075
pµ d
d 200
0.2
2(9.81)
But also Hp = fcn(Q) from the data in Prob. 11.35. Guessing f equal to the fully
rough value of 0.0183 yields Q of about 26.5 m3/min. Use Excel to get closer:
m
=
f 0.0185;
=
V 14.4=
; Red 2.87=
E 6; H p 73 m
s
m3
Q = 0.452
Ans. (a)
s
The efficiency is 87%, not bad! (b) If we vary the diameter but hold the pump at
maximum efficiency (Q* = 30 m3/min), we obtain a best d = 0.211 m. Ans.
(b)
53
P11.68 A popular small aircraft cruises at 230 km/h at 2,500 m altitude. It
weighs 10 kN, has a 180-hp engine, a 193-cm diameter propeller, and a drag-area
CDA ≈ 0.52 m2. The propeller data in Fig. P11.68 is proposed to drive this
aircraft. Estimate the required rotation rate, in r/min, and power delivered, in hp.
[NOTE: Simply use the coefficient pairs. The actual advance ratio is too high.]
Fig. P11.68
Solution: V = 230 km/hr = 63.9 m/s, ρ at 2,500 m ≈ 0.95 kg/m3, D = 193/100 =
1.93 m. The weight is immaterial because we know the drag-area. The two
coefficients are
=
=
T 0.5
ρ (CD A)V 2 0.5(0.95)(0.52)(63.9) 2 ≈ 1010 N
T
1, 010
=
=
2 4
ρn D
(0.95)n 2 (1.93) 4
P
η[(180)(746]
=
=
=
CP
3 5
ρn D
(0.95)n3 (1.93)5
=
CT
76.6
n2
5, 280η
n3
From the chart, at max efficiency, read, approximately,
η ≈ 0.58, CT ≈ 0.024, and CP ≈ 0.012. This yields n = 56 r/s for the CT relation
and n = 63 for the CP relation, not equal. Try other advance ratios. We find the
54
best agreement at V/nD ≈ 0.32, where η ≈ 0.56, CT ≈ 0.0185, and CP ≈ 0.012.
This yields
n ≈ 63.5 r / s ≈ 3, 800 r / min, P = 0.56(134) ≈ 75kW
Ans.
P11.69 The pump of Prob. 11.38, running at 3,500 rpm, is used to deliver water at
20°C through 180 m of cast-iron pipe to an elevation 30 m higher. Find (a) the proper
pipe diameter for BEP operation; and (b) the flow rate that results if the pipe diameter
is 7.5 cm.
Q, m3/min:
0.189 0.379 0.568 0.757 0.946 1.136 1.325 1.514 1.703
H, m:
61.3
61.0
60.4
59.1
57.6
55.2
51.5
47.5
42.4
η, %:
29
50
64
72
77
80
81
79
74
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. For cast
iron, take e ≈ 0.00026 m. (a) The data above are for 3,500 rpm, with BEP at 1.325
m3/min:
LV2
180 [Q/ (π d 2 /4)]2
0.00725 f
1.325 m3
,
*
Q
Q
= 30 + f
= 30 +
=
=
d 2g
d
2(9.81)
60 s
d5
e
Iterate, converges
to Red 2.87E5,
f 0.0258,
=
= 0.00265,
=
d
d ≈ 0.098 m =
9.8 cm Ans. (a)
H *= 51.5 m= ∆z + f
(b) If d = 7.5 cm, the above solution changes to a new flow rate:
Curve-fit: H ≈ 60 + 6.7Q - 9.94Q 2 =
30 + f
183 [Q/ (π (0.075) 2 /4)]2
=
30 + 6,371,857 fQ 2
0.075
2(9.81)
Iterate=
or EES: f 0.0277,
=
Re 2.21E5,
=
H 60 m,
Q = 0.013
m3
s
Ans. (b)
55
P11.70 The pump of Prob. 11.28, operating at 2,134 rpm, is used with water at
20°C in the system of Fig. P11.70. The diameter is 20 cm. (a) If it is operating at
BEP, what is the proper elevation z2? (b) If z2 = 68 m, what is the flow rate?
Fig. P11.70
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. For cast
iron, take e ≈ 0.00026 m, hence e /d ≈ 0.0013. (a) At BEP from Prob. 11.28, Q* ≈
0.17 m3/s and H* ≈ 100 m. Then the pipe head loss can be determined:
Q
0.17
m
998 × 5.4(0.2)
= =
=
= 1.07E6, f Moody ≈ 0.0212
V
5.4
, Re=
d
2
A (π /4)(0.2)
s
1E−3
Hsyst = ∆z + f
2
L V2
 450  (5.4)
= z 2 − 30 + 0.0212 
= z 2 − 30 + 71 ≡ H* = 100 m

d 2g
 0.2  2(9.81)
Solve for z 2 ≈ 59 m Ans. (a)
(b) If z2 = 68 m, the flow rate will be slightly lower and we will be barely offdesign:
L V2
450 V 2
Q
, V=
=68 − 30 + f
d 2g
0.2 2(9.81)
(π /4)(0.2) 2
m
Converges to f ≈ 0.0212, V ≈ 5.1 , H ≈ 103 m, Q ≈ 0.16 m 3 / s Ans. (b)
s
H =H(Q)Table =∆z + f
56
P11.71 The pump of Prob. 11.38, running at 3,500 r/min, delivers water at 20°C
through 2,200 m of horizontal 12.5-cm-diameter commercial-steel pipe. There are
a sharp entrance, sharp exit, four 90° elbows, and a gate valve. Estimate (a) the
flow rate if the valve is wide open and (b) the valve closing percentage which
causes the pump to operate at BEP. (c) If the latter condition holds continuously for
1 year, estimate the energy cost at 10 ¢/kWh.
Data at 3,500 rpm:
Q, m3/min:
0.189 0.379 0.568 0.757 0.946 1.136 1.325 1.514
1.703
H, m:
61.3
61.0
60.4
59.1
57.6
55.2
51.5
47.5
42.4
η, %:
29
50
64
72
77
80
81
79
74
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. For
commercial steel, take e ≈ 0.000046 m, or e /d ≈ 0.00037. The data above show BEP
at 1.325 m3/min. The minor losses are a sharp entrance (K = 0.5), sharp exit (K
= 1.0), 4 elbows (4 × 0.28), and an open gate valve (K = 0.1), or ∑K ≈ 2.72. Pump
and systems heads are equal:
=
H p H(Q)Table
= H=
syst
V2  L
L

V Q/[(p /4)(0.125) 2 ], ≈ 17,560
 f + ∑ K  , where=
2g  d
d

The friction factor f ≥ 0.0155 depends slightly upon Q through the Reynolds
number.
(a) Iterate on Q until both heads are equal. The result is
f ≈ 0.0175, Red ≈ 287,000, H p =
Hsyst ≈ 51.8 m, Q ≈ 1.35 m3 /min Ans. (a)
(b) Bring Q down to BEP, ≈ 1.32 m3/min, by increasing the gate-valve loss. The
result
is
f ≈ 0.0179, Red ≈ 221,000, H ≈ 51.5 m, Q ≈ 1.32 m3/min, Kvalve ≈ 21 (25%
open) Ans. (b)
(c) Continue case (b) for 1 year. What does it cost at 10¢ per kWh? Well, we
know the power level is exactly BEP, so just figure the energy:
ρ gQH 9,790(1.32/60)(51.5)
m⋅N
P= =
≈ 13,700
=
13.7 kW
η
0.81
s
Annual cost 13.7(365 days*24 hours)($0.1/kWh) ≈ $12, 000 Ans. (c)
57
P11.72 Performance data for a small commercial pump are shown below. The
pump supplies 20°C water to a horizontal 1.6-cm-diameter garden hose (e ≈ 0.025
cm), which is 15 m long. Estimate (a) the flow rate; and (b) the hose diameter
which would cause the pump to operate at BEP.
Q, m3/min:
0
H, m:
22.86 22.86 22.56 21.95 20.73 18.90 14.33 7.32
0.038 0.076 0.114 0.151 0.189 0.227 0.265
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. Given
e /d = 0.025/(1.6) ≈ 0.016, so f (fully rough) ≈ 0.045. Pump and hose heads must
equate, including the velocity head in the outlet jet:
V2  L 
Hp =
H(Q)Table =
Hsyst =  f + 1 ,
2g  d 
L
15
Q
=
≈ 970, V =
d 0.0155
(p /4)(0.0155) 2
(a) Iterate on Q until both heads are equal. The result is:
f ≈ 0.0456, Red ≈ 50,900, V ≈ 3.3
m
, H ≈ 24.9 m, Q ≈ 0.037 m3 /min
s
Ans. (a)
The hose is too small.
(b) We don’t know exactly where the BEP is, but it typically lies at about 60% of
maximum flow rate (0.6 × 0.29 ≈ 0.175 m3/min). Iterate on hose diameter D to
make the flow rate lie between 0.151 and 0.19 m3/min. The results are:
Q = 0.15 m3/min,
H = 20.73 m, f ≈ 0.0362,
Re ≈ 118,600,
Dhose ≈ 2.7 cm
0.17
19.80
0.0354
124,500
2.9 cm
0.19
18.90
0.0348
134,800
3 cm
58
Ans. (b)
P11.73 The Bell and Gossett pump of Prob. P11.8, running at the exact same
conditions, delivers water at 20ºC through a long, smooth, 20-cm-diameter pipe.
Neglect minor losses. How long is the pipe?
Solution: For water at 20ºC, from Table A.3, take ρ = 998 kg/m3 and μ = 0.0010
kg/m·s. “Exact same” means Q = 4 m3/min and H = 32 m. For a 20-cm
π
2
=
V Q=
=
/ A (4 / 60) m3 / s/ [ (0.2 m)
] 2.1m/ s. Find ReD and fsmooth:
diameter,
4
f
Re
ρVD (998)(2.1)(0.2)
1
Re D =
419, 200 ; Eq.(6.48) :
) yields f ≈ 0.0136
=
=
≈ 2 log( D
µ
1E − 3
2.51
f
The pump head must balance the friction loss in the pump:
H
= 32
=
f
L V2
L (2.1) 2
= (0.0136)( )[
];
0.2 2(9.81)
D 2g
L
solve =
2, 095 m
Ans.
P11.74 The 81 cm pump in Fig. 11.7a is used at 1,170 rpm in a system whose
head curve is Hs(m) ≈ 30 + 0.105Q2, with Q in c/min. Find the discharge and
brake horsepower required for (a) one pump; (b) two pumps in parallel; and (c)
two pumps in series. Which configuration is best?
Solution: Assume plain old water, rg ≈ 9,790 N/m3. A reasonable curve-fit to
the pump head is taken from Fig. 11.7a: Hp(m) ≈ 150 − 0.021Q2, with Q in
m3/min. Try each case:
(a) One pump: Hp = 150 − 0.021Q2 = Hs = 30 + 0.105Q2: Q ≈ 30.9 m3/min
Ans. (a)
(b) Two pumps in parallel: 150 − 0.021 (Q/2)2 = 30 + 0.105Q2, Q ≈ 33.0 m3/min
Ans. (b)
(c) Two pumps in series: 2(150 − 0.021Q2) = 30 + 0.105Q2: Q ≈ 42.9 m3/min
Ans. (c)
Clearly case(c) is best, because it is very near the BEP of the pump. Ans.
59
P11.75 Two 89 cm pumps from Fig. 11.7b are installed in parallel for the
system of Fig. P11.75. Neglect minor losses. For water at 20°C, estimate the flow
rate and power required if (a) both pumps are running; and (b) one pump is shut
off and isolated.
Fig. P11.75
Solution: For water at 20°C, take r = 998 kg/m3 and m = 0.0010 kg/m·s. For
cast iron, e ≈ 0.00025 m, or e /d ≈ 0.000417. The 89 cm pump has the curve-fit
head relation Hp(m) ≈ 72 − 1.106E-4Q3, with Q in m3/min. In parallel, each pump
takes Q/2:
L V2
L
1 
=Hsyst =∆z + f
=2667
H p =fcn  Q 
,
d 2g
d
 2 curve-fit
Q (m3 /s)
=
=
and V
, ∆z 30 m
(p /4)(0.6 m) 2
(a) Two pumps: Iterate on Q (for Q/2 each) until both heads are equal. The results
are:
f ≈ 0.0164, Red ≈ 1.23E6, H p =
Hs =
67.4 m
m3
Q total ≈ 69.5
(34.7 for each pump) Ans. (a)
min
 9,790(34.7/60)(67.4) 
=
P 2P
=
2
each
 ≈ 1.05 M W Ans. (a)
0.73

(b) One pump: Iterate on Q alone until both heads are equal. The results are:
f ≈ 0.0164, Re ≈ 1.94E6, H ≈ 53.5 m, Q ≈ 55 m3 /min
=
P 9, 790(55/60)(53.5)/0.85 ≈ 565 kW
Ans. (b)
Ans. (b)
The pumps in parallel give 26% more flow at the expense of 85% more power.
60
P11.76 Two 81-cm pumps from Fig. 11.7a are combined in parallel to deliver
water at 20°C through 450 m of horizontal pipe. If f = 0.025, what pipe diameter
will ensure a flow rate of 130 m3/min at 1,170 rpm?
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. As in
Prob. 11.75, a reasonable curve-fit to the pump head is taken from Fig. 11.7a:
Hp(m) ≈ 150 − 0.07857Q − 0.005132Q2, with Q in m3/min. Each pump takes half
the flow, 65 m3/min, for which
H p ≈ 150 − 0.07857(65) − 0.005132(65)2 ≈ 123 m.
Then Q=
pipe
130
m3
= 2.17
and the pipe loss is
60
s
2
2
4.38
 450  [2.17/(π d /4)]
=
=
H=
0.025


syst
2(9.81)
d5
 d 
123 m, solve for=
d 0.51 m
Ans.
P11.77 Two pumps of the type tested in Prob. 11.22 are to be used at 2,140
r/min to pump water at 20°C vertically upward through 100 m of commercialsteel pipe. Should they be in series or in parallel? What is the proper pipe
diameter for most efficient operation?
Solution: For water take r = 998 kg/m3 and m = 0.001 kg/m⋅s. For commercial
steel take e = 0.046 mm. Parallel operation is not feasible, as the pump can
barely generate 100 m of head and the friction loss must be added to this. For series
operation, assume BEP operation, Q* = 0.2 m3/s, H* = 95 m:
H 2 pumps= 2(95)= ∆z + f
L V2
100 [0.20/(p d 2 /4)]2
0.3305 f
= 100 + f
= 100 +
d 2g
d
2(9.81)
d5
Given Red = 4r Q/(πm d) = 4(998)(0.2)/[π(0.001)d] = 254,000/d and e /d =
0.000046/d, we can iterate on f until d is obtained. The Excel result is:
f = 0.0156; Red = 1.8E6; V = 12.7 m/s, dbest = 0.142 m Ans.
61
P11.78 Consider the axial-flow pump of Fig. 11.13, running at 4,200 r/min,
with an impeller diameter of 90 cm. The fluid is propane gas (molecular weight
44.06). (a) How many pumps in series are needed to increase the gas pressure
from 1 atm to 2 atm? (b) Estimate the mass flow of gas.
Solution: Find the gas constant and initial density of the propane. Assume T =
293°K.
=
R propane
p
8330
J
= 189
r =
;=
°
44.06
RT
kg K
101325 N/m 2
=
(189)(293°K)
1.83
kg
m3
From Fig. 11.13, at BEP, CQ* = 0.55 and CH* = 1.07. Calculate the head rise for
one stage:
gH
=
n2 D 2
=
1.07
(9.81) H
(70 r/s) 2 (0.9m) 2
or =
: H
,
433m
Dp1st stage = r g H = (1.83)(9.81)(433) = 7775
N
m2
(a) Crudely, then, to increase pressure by 1 atm = 101,325 pa, we need
101,325/7,775 = 13 stages. However, the pressure rises about 8 per cent per stage,
so a better estimate would be ln(2.0)/ln(1.08) =
10 stages.
Ans.(a)
(b) The mass flow follows from the flow coefficient:
=
CQ*
0.55
=
Q
=
n D3
Q
(70)(0.9)
m propane
=
, or :=
Q
3
r
=
Q (1.83
28
m3
s
m3
)(28
=
)
s
m3
kg
62
51.2
kg
s
Ans.(b)
P11.79 Two 81- cm pumps from Fig. 11.7a are to be used in series at 1,170 rpm
to lift water through 150 m of vertical cast-iron pipe. What should the pipe
diameter be for most efficient operation? Neglect minor losses.
Solution: For water at 20°C, take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. For cast
iron, e ≈ 0.00026 m. From Fig. 11.7a, read H* ≈ 117.5 m at Q* ≈ 75.5 m3/min.
Equate
L V2
H p =2H* =2(117.5) =H syst =∆z + f
d 2g
2
 75

(p d 2 /4) 

150
19.37f

  60
 =
=
+
150 + f  d 
150
2(9.81)
d5


Iterate, guessing f ≈ 0.02 to get d, then get Red and e /d and repeat. The final result
is
f ≈ 0.0185, V ≈ 15.6 m/s, Red ≈ 4.98E6, d ≈ 0.32 m Ans.
P11.80 Determine if either (a) the smallest, or (b) the largest of the seven Taco
pumps in Fig.P11.24, running in series at 1,160 r/min, can efficiently pump water
at 20°C through 1 km of horizontal 12-cm-diameter commercial steel pipe.
Solution: For water at 20°C, take r = 998 kg/m3 and m = 0.001 kg/m-s. For
commercial steel pipe, from Table 6.1, e = 0.046 mm. Then e/D = 0.046 mm/120
mm = 0.000383.
(a) For the smallest pump in Fig. P11.24, the BEP is at about 1.5 m3/min, with a
head of about 12.5 m. See what friction head loss results from this flow rate:
V=
Q
=
A
ρVD (998)(2.2)(0.12)
(0.025) m3 / s
m
= 2.21 ; Re D=
=
= 265, 000
2
m
s
0.001
(π / 4)(0.12 m)
e
= 0.000383; Eq.(6.48) : f Moody ≈ 0.0177 ,
D
L V2
1, 000 (2.21)2
=
=
hf
f
(0.0177)(
)= 36.7 m
D 2g
0.12 2(9.81)
So three small pumps in series, each with 12.5 m of head, would be an efficient
system. Ans. (a)
63
(b) For the largest pump in Fig. P11.24, the BEP is at about 2 m3/min, with a
head of about 22 m. See what friction head loss results from this flow rate:
V
=
(2 / 60) m3 / s
m
Q
ρVD (998)(2.95)(0.12)
2.95 ; Re=
=
=
=
= 353, 000
D
2
s
0.001
A (π / 4)(0.12 m)
m
e
=
D
0.000383;
hf
=
Eq. (6.48) :
f Moody ≈ 0.0173,
1000 (2.95)2
L V2
) = 63.9 m
f= (0.0173)(
0.12 2(9.81)
D 2g
So three large pumps in series, each with 22 m of head, would be an efficient
system. Ans. (b) The largest pump is a better solution because of its higher
efficiency (80% compared to 65%).
P11.81 Reconsider the system of Fig. P6.62. Use the Byron Jackson pump of
Prob. 11.28 running at 2,134 r/min, no scaling, to drive the flow. Determine the
resulting flow rate between the reservoirs. What is the pump efficiency?
Solution: For water take r = 998 kg/m3 and m = 1E−3 kg/m⋅s. For cast iron take e
= 0.00026 m, or e /d = 0.00026/0.15 = 0.00173. The energy equation, written
between reservoirs, is the same as in Prob. 6.62:
H p =∆z + f
LV2
610 [Q / (p (0.15) 2 /4)]2
=36 m + f
=36 + 663, 700 fQ 2
d 2g
0.15
2(9.81)
where

4r Q
e
f = f Moody = fcn  Red ,  with Red =

πµ d
d
64
From the data in Prob. 11.28 for the pump, Hp = fcn(Q) and is of the order of 90
m. Get a close result using Excel:
=
H p 105
=
m; f 0.0229;
=
Red 573, 600;
m
m3
=
V 3.83
=
; bhp 125 kW; Q = 0.068
s
s
Ans.
Interpolating, the pump efficiency is η ≈ 56%. Ans. The flow rate is too low for
this particular pump.
P11.82 The S-shaped head-versus-flow curve in Fig. P11.82 occurs in some axial-flow
pumps. Explain how a fairly flat system-loss curve might cause instabilities in the operation of
the pump. How might we avoid instability?
a
b
c
Fig. P11.82
Solution: The stability of pump operation is nicely covered in the review article by
Greitzer (Ref. 41 of Chap. 11). Generally speaking, there is little danger of instability if the
slope of the pump-head curve, dH/dQ, is negative, unless there are two such points. In
Fig. P11.82 above, a flat system curve may cross the pump curve at three points (a, b, c).
Of these 3, point b is statically unstable and cannot be maintained. Consider a small
disturbance near point b: Suppose the flow rate drops slightly—then the system head
decreases, but the pump head decreases even more. Then the flow rate will drop still
more, etc., and we move away from the operating point, which therefore is unstable. The
general rule is:
A pump operating point is statically unstable if the (positive) slope of the
pump-head curve is greater than the (positive) slope of the system curve.
By this criterion, both points a and c above are statically stable. However, if the points
are close together or there are large disturbances, a pump can “hunt” or oscillate between
points a and c, so this could also be considered unstable to large disturbances.
65
Finally, even a steep system curve (not shown above) which crosses at only a single
point b on the positive-slope part of the pump-head curve can be dynamically unstable,
that is, it can trigger an energy-feeding oscillation which diverges from point b. See
Greitzer’s article for further details of this and other turbomachine instabilities.
____________________________________________________________________
P11.83 The low-shutoff head-versus-flow curve in Fig. P11.83 occurs in some centrifugal pumps. Explain how a fairly flat system-loss curve might cause instabilities in the
operation of the pump. What additional vexation occurs when two of these pumps are in
parallel? How might we avoid instability?
Fig. P11.80
Solution: As discussed, for one pump with a flat system curve, point a is statically unstable,
point b is stable. A ‘better’ system curve only passes through b.
For two pumps in parallel, both points a and c are unstable (see above). Points b and d are
stable but for large disturbances the system can ‘hunt’ between the two points.
P11.84 Turbines are to be installed where the net head is 120 m and the flow rate is 950
m3/min. Discuss the type, number, and size of turbine which might be selected if the
generator selected is (a) 48-pole, 60-cycle (n =150 rpm); or (b) 8-pole (n = 900 rpm).
Why are at least two turbines desirable from a planning point of view?
Solution: We select two turbines, of about half-flow each, so that one is still available
for power generation if the other is shut down for maintenance or repairs. Ans.
Assume η ≈ 90%:
 950 
Ptotal= hr gQH ≈ 0.9(9, 790) 
 (120) ÷ 1, 000 ≈ 16, 740 kW (each turbine ≈ 8, 370 kW)
 60 
(a) n = 150 rpm:
1/2
nP1/2 150(8, 370) ≈ (select two impulse turbines)
=
N sp = 5/4
35
(120)5/4
H
66
Ans. (a)
Estimate φ ≈ 0.47=
=
(b) n 900
=
rpm: N sp
π nD
2gH
=
π (150/60)D
2(9.81)(120)
, or: D ≈ 2.9 m
Ans. (a)
900(8, 370)1/2
≈ 207 (select two Francis turbines)
(120)5/4
Fig. 11.21: C*p ≈ 2.6
=
P*
8, 370 ×1, 000
, solve D ≈ 1.0 m
=
3 5
 900  3 5
ρn D
998 
 D
 60 
Ans. (b)
Ans. (b)
P11.85 For a high-flow site with a head of 13 m, it is desired to design a single 2-mdiameter turbine that develops 4,000 bhp at a speed of 360 r/min and 88% efficiency. It
is decided first to test a geometrically similar model of diameter 0.3 m, running at 1,180
r/min. (a) What likely type of turbine is the prototype? What are the appropriate (b)
head, and (c) flow rate for the model test? (d) Estimate the power expected to be
delivered by the model turbine.
Solution: For water, take r = 998 kg/m3. (a) We have enough information to determine
power specific speed:
=
N sp
(rpm)(kW)1/2 (360 rpm)[(4, 000)(0.746)]1/2
=
=
[ H (m)]5/4
(13 m)5/4
800
For a water-flow application, this number is characteristic of a propeller turbine.
Ans.(a)
(b) The similarity rules give us the required model head by equating the head
coefficients:
CH
=
(9.81) H 2
gH1
gH 2
(9.81)(13)
, Solve H 2 ≈ 3.14 m Ans.(b)
=
=
=
2
2
2
2
2
2
n1 D1 (360) (2)
n2 D2 (1,180)2 (0.3)2
[We didn’t bother changing n to r/s because the factors of “60” would have cancelled out.]
(c) To get the model flow rate, first calculate the prototype flow rate from the known power:
=
P hr
=
gHQ (4, 000)(746)
= (0.88)(998 kg / m3 )(9.81 m / s2 )(13 m)Q1
Solve for Q1 = 26.6 m3 / s
Then=
CQ
Q1
Q2
Q2
(26.6)
m3
0.294
=
=
=
Solve
=
Q
2
s
n2 D13 (360)(2)3 n2 D23 (1,180)(0.3)3
67
Ans.( c )
If we assume the efficiency is unchanged, the model power output will be
=
P2 ηρ
=
gH 2Q2 (0.88)(998)(9.81)(3.14 m)(0.294
=
m3 / s) 7, 950
m−N
Ans.( d )
s
[The Moody size-effect formula, Eq. (11.29a), suggests η2 ≈ 0.805, or P2 ≈ 7,275 W.]
P11.86 The Tupperware hydroelectric plant on the Blackstone River has four 90-cmdiameter turbines, each providing 447 kW at 200 rpm and 5.8 m3/s for a head of 9 m.
What type of turbines are these? How does their performance compare with Fig. 11.21?
Solution:
nP1/2 200(447)1/2
=
Nsp =
≈ 270 (These are Francis turbines)
H5/4
(9)5/4
Ans.
Use the given data to compute the dimensionless BEP coefficients:
6
447 × 103
*Q
*P
C=
≈
2.47;
C
=
≈ 20.5 (both are quite different!)
(200/60)(0.9)3
998(200/60)3 (0.9)5
9.81(9)
C*H = 2
≈ 9.8 (OK );
(200/60) (0.9) 2
447(1, 000)
≈ 85% (OK )
9, 790(6)(9)
ηmax =
68
Ans.
P11.87 An idealized radial turbine is
shown in Fig. P11.87. The absolute flow
enters at 30° and leaves radially inward.
The flow rate is 3.5 m3/s of water at 20°C.
The blade thickness is constant at 10 cm.
Compute the theoretical power developed
at 100% efficiency.
Fig. P11.87
Solution: For water, take r ≈ 998 kg/m3. With reference to Fig. 11.22 and Eq. 11.35,
m
3.5
m
 2π 
u2 =ω r2 =
135 
, α 2 =30°, α1 =90°, Vn2 =
≈ 7.96
 (0.7) =9.90
s
2π (0.7)(0.1)
s
 60 
Vn2
7.96
m
Vn1
=
= 13.8
=
Vt2 =
and=
Vt1
0
tan a 2 tan 30°
s
tan 90°
=
= 477, 000 W ≈ 477 kW
Thus
Ptheory r=
Qu 2 Vt2 998(3.5)(9.90)(13.8)
69
Ans.
P11.88 Performance data for a very small (D = 8.25 cm) model water turbine, operating
with an available head of 15 m, are as follows:
Q, m3/h:
18.7
18.7
18.5
rpm:
0
500
1,000 1,500 2,000 2,500 3,000 3,500
η:
0
14%
27%
18.3
38%
17.6
50%
16.7
65%
15.1
61%
11.5
11%
(a) What type of turbine is this likely to be? (b) What is so different about this data
compared to the dimensionless performance plot in Fig. 11.21d? Suppose it is desired to
use a geometrically similar turbine to serve where the available head and flow are 45 m
and 0.2 m3/s, respectively. Estimate the most efficient (c) turbine diameter; (d) rotation
speed; and (e) power.
Solution: (a) Convert Q = 16.7 m3/h = 0.00464 m3/s. Use BEP data to calculate power
specific speed:
=
P ρ gQH
=
η (9, 790 N/m3 )(0.00464 m3 /s)(15 m)(0.65)
= 443 m⋅N/s
rpm(kW)1/2 (2, 500 rpm)(0.384 kW)1/2
=
N sp
=
≈ 52.5 (Francis turbine)
( H -m)5/4
(15 m)5/4
Ans. (a)
(b) This data is different because it has variable speed. Our other data is at constant
speed. Ans. (b)
(c, d, e) First establish the BEP coefficients from the small-turbine data:
3
Q
(16.7/3, 600 m /s)
=
= 0.198
nD 3 (2, 500/60 r/s)(0.0825 m)3
gH
(9.81)(15 m)
=
=
= 12.5
C*H
2 2
n D
(2, 500/60) 2 (0.0825)2
*
=
CQ
=
C*
P
(384 W)
P
=
= 1.40
3 5
3
rn D
(998 kg/m )(2, 500/60 r/s)3 (0.0825 m)5
0.2 m3 /s
(9.81 m/s2 )(45 m)
*
*
=
CQ =
0.198;
=
CH
= 12.5
n2 D23
n22 D22
Solve for
=
n2 14.5
=
r/s 870 r / min=
; D2 0.41 m
Ans. (c, d)
3 5
*
=
P2 C=
Then
1.40(998 kg/m3 )(14.5 r/s)3 (0.41 m)5
P r n2 D2
P2 = 49.3 kW
Ans. (e)
70
Actually, since the new turbine is 5 times larger, we could use the Moody step-up formula,
Eq. (11.29a), to predict (1 − η2 ) ≈ (1 − 0.65)/(5)1/4 = 0.234, or η2 = 0.766 = 76.6%. We thus
expect more power from the larger turbine:
P2 = (49.3kW)(76%/65%) ≈ 57.6 kW Better
Ans. (e)
P11.89 A Pelton wheel of 3.6-m pitch diameter operates under a new head of 600 m.
Estimate the speed, power output, and flow rate for best efficiency if the nozzle exit
diameter is 10 cm.
Solution: First get the jet velocity and then assume BEP at φ ≈ 0.47:
C v ≈ 0.94, so: Vjet =
C v 2gH =
0.94 2(9.81)(610) ≈ 102 m/s
π nD
π n(3.6)
=
BEP at φ ≈ 0.47=
,
2gH
Solve =
n 4.51
2(9.81)(600)
r
× 60 ≈ 270 rπm
s
Ans.
m3
 π   10 
Q V=
102  =
=

 0.8
jet Anozzle
s
 4   100 
2
Ans.
=
u best Vjet /2 ≈ 51 m/s
Ptheory
= r Qu(Vj − u)(1 − cos β )
≈ 998(0.8)(51)(102 − 51)(1 − cos165°) ≈ 4, 080 kW100%
Pturb ≈ Ptheory ⋅ 80% = 3,264 kW,
N sp ≈
270(3, 264)1/2
≈ 5.2 :
(600)5/4
Fig. 11.27, read ηmax < 80%, say, 75%
Actual power output ≈ η P100%
= 0.75(4, 080) ≈ 3,060 kW
71
Ans.
P11.90 An idealized radial turbine is
shown in Fig. P11.90. The absolute flow
enters at 25° with the blade angles as
shown. The flow rate is 8 m3/s of water
at 20°C. The blade thickness is constant
at 20 cm. Compute the theoretical power
developed at 100% efficiency.
Fig. P11.90
Solution: The inlet (2) and outlet (1)
velocity vector diagrams are shown at
right. The normal velocities are
8.0
=
=
= 5.31 m/s
Vn2 Q/A
2
2π (1.2)(0.2)
8.0
=
Vn1 Q/A
=
= 7.96 m/s
1
2π (0.8)(0.2)
From these we can compute the
tangential velocities at each section:
 2π 
=
= 10.1 m/s;
u2 ω=
r2 80 
 (1.2)
 60 
 2π 
u1 ω=
r1 80 
=
= 6.70 m/s
 (0.8)
 60 
Vt2= Vn2 cot 25°= 11.4 m/s; Vt1= u1 − Vn1 tan 30°= 2.11 m/s
Ptheoρy
= ρ Q(u2 Vt2 − u1Vt1=
) 998(8)[10.1(11.4) − 6.7(2.11)] ≈ 800, 000 W
72
Ans.
P11.91 The flow through an axial-flow turbine can be idealized by modifying
the stator-rotor diagrams of Fig. 11.12 for energy absorption. Sketch a suitable
blade and flow arrangement and the associated velocity vector diagrams. For
further details, see Chap. 8 of Ref. 25.
Solution: Some typical velocity diagrams are shown below, where u =
ω r = blade speed. The power delivered to the turbine, at 100% ideal shock-free
flow, is
Pideal ρ Q(u1Vt1 − u2 Vt2 )
=
wheρe Vt1,2 aρe the tangential components of V1,2
Fig. P11.88
P11.92 A dam on a river is being sited for a hydraulic turbine. The flow rate is
1,500 m3/h, the available head is 24 m, and the turbine speed is to be 480 r/min.
Discuss the estimated turbine size and feasibility for (a) a Francis turbine; and (b)
a Pelton wheel.
Solution: Assume η ≈ 89%, as in Fig. 11.21d. The power generated by the
turbine would be P = ηg QH = (0.89)(9790 N/m3)(0.417 m3/s)(24 m) = 87,200 mN/s = 87.2 kW. Now compute Nsp = (480 rpm)(87.2 kW)1/2/(24 m)5/4 ≈ 84,
appropriate for a Francis turbine. (a) A Francis turbine, similar to Fig. 11.21d,
would have CQ* ≈ 0.34 = (0.417 m3/s)/ [(480/60 r/s)D3]. Solve for a turbine
diameter of about 0.54 m, which would be excellent for the task. Ans. (a)
73
(b) A Pelton wheel at best efficiency (half the jet velocity) would only be 45 cm
in diameter, with a huge nozzle, d ≈ 15 cm, which is too large for the wheel. We
conclude that a Pelton wheel would be a poor design. Ans. (b)
P11.93 Figure P11.93, shown below, is a crossflow or “Banki” turbine [Ref. 55],
which resembles a squirrel cage with slotted curved blades. The flow enters at
about 2 o’clock, passes through the center and then again through the blades,
leaving at about 8 o’clock.
Report to the class on the operation and advantages of this design, including
idealized velocity vector diagrams.
Brief Discussion (not a “Solution”):
The crossflow turbine is ideal for small dam owners, because of its simple,
inexpensive design. It can easily be constructed by a novice (such as the writer)
from wood and plastic.
It is not especially efficient (≈ 60%) but
makes good, inexpensive use of a small
stream to produce electric power. For
details, see Ref. 55 or the paper “Design
and Testing of an Inexpensive Crossflow
Turbine,” by W. Johnson et al., ASME
Symposium on Small Hydropower Fluid
Machinery, Phoenix, AZ, Nov. 1982,
ASME vol. H00233, pp. 129−133.
Fig. P11.93
74
P11.94 A simple crossflow turbine, Fig. P11.93 above, was constructed and tested at
the University of Rhode Island. The blades were made of PVC pipe cut lengthwise into
three 120°-arc pieces. When tested in water at a heads of 1.65 m and a flow rate of 2.4
m3/min, the measured power output was 0.6 hp. Estimate (a) the efficiency; and (b) the
power specific speed if n ≈ 200 rpm.
Solution: We have sufficient information to compute the available water power:
0.448
 2.4 
= ρ gQH
= (9790) 
= 646 W
= 0.646 kW, ∴ =
≈ 69%
Pavail
η
 (1.65)
0.646
 60 
At 200 =
rpm, N sp
rpm(kW)1/2 200(0.448)1/2
=
≈ 71.6
(head)5/4
(1.65)5/4
Ans. (a)
Ans. (b)
P11.95 One can make a theoretical estimate of the proper diameter for a penstock in an
impulse turbine installation, as in Fig. P11.95. Let L and H be known, and let the turbine
performance be idealized by Eqs. (11.38) and (11.39). Account for friction loss hf in the
penstock, but neglect minor losses. Show that (a) the maximum power is generated when
hf = H/3, (b) the optimum jet velocity is (4gH/3)1/2, and (c) the best nozzle diameter is
Dj = [D5/(2fL)]1/4, where f is the pipe-friction factor.
Fig. P11.92
Solution: From Eqs. 11.38 and 39, maximum power is obtained when u = Vj /2, or:
Vj 
Vj 
 1 − cos β  3
Pmax
Q  Vj −  (1 − cos =
)
Aj 
CVj2 Vj , =
C constant
= ρβρ
=
j
V
2 
2 
4


75
Now apply the steady-flow energy equation between the reservoir and the outlet jet:
4
2
2
L Vpipe Vj
L  Dj 
pp
Dz = H = f
+
, or: Vj2 = 2gH − f   Vj2 since Vj D2j = Vp D2
D 2g
2g
D D 
4
4
4


D


dP
L
L
j
Thus Pmax =
C 2gHVj − f   Vj3  ; Differentiate: max =
0 if 2gH =
3f Vp2


D D 
dVj
D


=
=
or if: H 3h
f,pipe! The pipe head loss H/3
Ans. (a)
Continuing, Vj2 |optimum= 2g(H − h f )= 2g(H − H/3), or: Vj,optimum =
4
gH
3
Ans. (b)
Then the correct pipe flow speed is obtained by back-substitution:
2
L Vpipe H
=
f
=
, or: Vpipe
D 2g
3
2gH
3fL/D
 D5 1/4
2gH
, solve for D jet =  p 
=
 D  3fL / D
 2fL 

4
D
Continuing, Vp2 = Vj2  j 
Ans. (c)
P11.96 Apply the results of Prob. P11.95 to determine the optimum (a) penstock
diameter and (b) nozzle diameter for a head of 330 m and a flow rate of 5,400 m3/h with a
cast iron penstock of length 600 m.
Solution: For water take ρ = 998 kg/m3 and μ = 0.0010 kg/(m·s). For cast iron take ε =
0.26 mm. Prob. P11.95 showed that the optimum friction head loss is H/3:
H pipe
=
H 330
=
= 110 m
=
3
3
2
f
L Vp
,=
L 600 m , V=
p
Dp 2 g
4Q
m3
m3
,
=
Q
5,
400
=
1.5
p D p2
h
s
These are two equations in the two unknowns Dp and f. We use Eq. (8.48) to calculate f :
e / Dp
ρV p D p
1
2.51
) , Re Dp =
≈ − 2.0 log(
+
3.7
µ
f
Re Dp f
76
There is no direct solution. We have to iterate or use Excel to obtain the solutions:
m
=
=
=
=
Re Dp 4.27
; V p 9.60
; D p 0.446m Ans.(a )
E 6 ; f 0.0174
s
D 5p 1/4
(0.446)5
[=
]
[
]
=
D jet
=
0.17 m
Ans.(b)
2f L
2(0.0174)(600)
P11.97 Consider the following non-optimum version of Prob. 11.95: H = 450 m, L = 5 km,
D = 1.2 m, Dj = 20 cm. The penstock is concrete, e = 1 mm. The impulse wheel diameter
is 3.2 m. Estimate (a) the power generated by the wheel at 80% efficiency; and (b) the
best speed of the wheel in r/min. Neglect minor losses.
Solution: For water take r = 998 kg/m3 and m = 0.001 kg/m·s. This is a non-optimum
condition, so we simply make a standard energy and continuity analysis. Refer to the
figure on the next page for the notation:
∆z = H = h f +
V j2
2g
, V j ∆ 2j = Vpipe ∆ 2 , combine and solve for jet velocity:
4

L  D j  
2
V j 1+ f   = 2gH ,

D  D  
 rVD e 
where f = fcn 
, ,
 µ D
e 0.001
=
= 8.33E−4
D 1.2
For example, guessing f ≈ 0.02, we estimate Vj ≈ 91.2 m/s. Using Excel yields
m3
m
, Q = 2.87
s
s
The power generated (at 80% efficiency) and best wheel speed are
D
1
Poweρ =
Quwheel (V j − uwheel )(1 − cos )(0.8), uβest =Vjet =
ρββ
Ω wheel wheel ,
≈ 165°
2
2
rev
rad
If Dwheel
= 3.2 m, solve for Ω wheel
= 28.5 = 272
,
min
s
Power = 9.36 MΩ Ans. (a, b)
As shown on the figure below, which varies Dj, the optimum jet diameter is 34 cm, not
20 cm, and the optimum power would be 16 MW, or 70% more!
=
f 0.0189,
=
Red 3.03E6,
=
V jet 91.23
77
P11.98 Francis and Kaplan (enclosed) turbines are often provided with draft tubes,
which lead the exit flow into the tailwater region, as in Fig. P11.98. Explain at least two
advantages to using a draft tube.
Solution: Draft tubes have two big advantages:
(1) They reduce the exit loss, since a draft tube is essentially a diffuser, as in Fig. 6.23 of
the text, so more of the water head is converted to power.
Fig. P11.95
(2) They reduce total losses downstream of the turbine, so that the turbine runner can be
placed higher up without the danger of cavitation.
78
P11.99 Turbines can also cavitate when the pressure at point 1 in Fig. P11.98 drops too
low. With NPSH defined by Eq. 11.20, the empirical criterion given by Wislicenus [Ref. 4]
for cavitation is
(rpm)(m3 / min)1/2
=
N ss
≥ 1, 650
[NPSH(m)]3/4
Use this criterion to compute how high (z1 − z2) the impeller eye in Fig. P11.98 can be
placed for a Francis turbine, with a head of 90 m, Nsp = 180, and pa = 96.5 kPa absolute
before cavitation occurs in 15°C water.
Solution: For water at 15°C, take rg = 9,790 N/m3 and pv ≈ 1.72 kPa. Then
Eq. 11.20: NPSH
=
pa − p v
(96.5 − 1.72)(1, 000)
− ∆z=
− h fi
− ∆=
z − 0 9.68 m − ∆z
ρg
9, 790
Now we need the NPSH, which we find between the two specific-speed criteria:
N=
180
=
sp
N ss 1,=
650
=
n(P in hp)1/2
(90 m )5/4
with =
P hρ g(Q m3 / s)(90 m) ÷ 746
rpm(Q in m3 / min)1/2
. Iterate to solve by assuming η ≈ 90%
(NPSH in m)3/4
(We can’t find n or Q, only nQ1/2). Result: NPSH ≈ 13.86 m, ∆z = 9.68 – 13.86 ≈ −4.18 m.
Ans.
P11.100 The manufacturer of the 21-m-diameter wind turbine in the chapter-opener
photo claims that it develops exactly 100 kW at a wind speed of 15 m/s. Compare this
with an estimate from the correlations in Fig. 11.32.
Solution: It is near sea level, so take ρ = 1.2 kg/m3. There is a curve for a 3-blade
HAWT in Fig. 11.32, with a maximum CP ≈ 0.43 at ωr/V ≈ 5.0. The predicted maximum
power is thus
P
P
CP ,max =
0.43 = 3 =
, solve Pmax ≈ 300, 000 W =
300kW Ans.
0.5 r AV
0.5(1.2)[p (10.5) 2 ](15)3
ω r ω (10.5m)
rad
r
=
≈ 5.0 , solve ω � 7.1
≈ 68
Rotation speed :
V
15 m / s
s
min
79
This correlation indicates the turbine is much more powerful than stated. The
manufacturer is constraining it to spin off-maximum, in this case at ωr/V ≈ 7, or about 95
r/min. The power coefficient at this point is about 0.14 << 0.43. The writer observed ω
≈ 90 r/min when taking the photo. Here is the manufacturer’s (unexpected) data for
power developed versus wind speed:
This data shows a maximum power coefficient of about 0.44 at a wind speed of 6.5 m/s ≈
15 mi/h.
P11.101 A Darrieus VAWT in operation in Lumsden, Saskatchewan, that is 10 m high
and 6 m in diameter sweeps out an area of 40 m2. Estimate (a) the maximum power and
(b) the rotor speed if it is operating in 25 km/h winds.
Solution: For air in Saskatchewan (?), take r ≈ 1.19 kg/m3. Convert 25 km/h = 7.0 m/s.
From Fig. 11.34 for the Darrieus VAWT, read optimum CP and speed ratio:
C P,max ≈=
0.42
Pmax
P
=
3
(1/2) ρ AV1
(1/2)(1.19)(40)(7.0)3
Solve for
Pmax ≈ 3.4 kW
Ans. (a)
ωr
ω (6/2)
rad 60
≈ 4.1
×
≈ 91 rpm
At Pmax=
,
, or: ω ≈ 9.57
V1
7.0
s
2p
80
Ans. (b)
P11.102 An American 1.8-m diameter multi-blade HAWT is used to pump water to a
height of 3 m through 7.5-cm-diameter cast-iron pipe. If the winds are 20 km/h, estimate
the rate of water flow in m3/min.
Solution: For air, take r ≈ 1.19 kg/m3. Convert 20 km/h = 5.6 m/s. For water, take r =
998 kg/m3 and m = 1E−3 kg/(m⋅s). From Fig. 11.34 for the American multi-blade HAWT,
read optimum CP and speed ratio:
C P,max ≈ 0.29 at
ωr
π
m⋅N
1
≈ 0.9: Pmax ≈ 0.29   (1.19) (1.8)2 (5.6)3 ≈ 77
V1
4
s
2
2

L Vpipe 
If ηpump ≈ 80%, Ppump ≈ 0.8(77)
9,
790Q
z
f
= r water gQH
=
D
+


syst


D
2g


where =
Vpipe
e 0.00026
Q
L
3
=
= 40,
=
≈ 0.00347
,
2
D 7.5 /100
D 7.5/100
(p /4)(7.5/100)
Clean up to: 0.0063
= Q(3 + 104, 500f Moody Q2 ), with Q in m3 /s. Iterate to obtain
f ≈ 0.0305, Vpipe ≈ 0.48
m
m3
, Re ≈ 35, 600, Q ≈ 0.0021
≈ 0.126 m3 / min
s
s
81
Ans.
P11.103 At a distance of 1.5 km from the wind turbine in the chapter-opener photo is a
30-m-high, 7-m-diameter HAWT, in Fig. P11.103. It is rated at 10 kW and provides half
of the electricity for the Salty Brine State Beach bathhouse. From the data in Fig. 11.32,
at a wind velocity of 32 km/h, estimate (a) the maximum power developed, and (b) the
rotation speed, in r/min?
Fig. P11.103 – Official RI State photo.
Solution: Assume sea-level, ρ ≈ 1.22 kg/m3. Convert 32 km/h to 8.9 m/s. The curve for
a 3-blade HAWT in Fig. 11.32 shows a maximum CP ≈ 0.43 at ωr/V ≈ 5.0. (a) The
predicted maximum power is
P
P
m⋅ N
=
≈ 7.1kW Ans.( a )
0.43 =
, solve P ≈ 7,120
3
π
0.5ρ AV
s
0.5(1.22)[ (7) 2 ](8.9)3
4
(b) At this maximum-power condition, the rotation speed is
ωr
V
=
ω (3.5 m)
8.9 m/ s
≈ 5.0 , solve ω ≈ 12.7
rad
r
≈ 120
Ans.(b)
s
min
__________________________________________________________________
82
P11.104 The controversial Cape Wind project proposes 130 large wind turbines in
Nantucket Sound, intended to provide 75% of the electric power needs of Cape Cod and
the Islands. The turbine diameter is 100 m. For an average wind velocity of 22 km/h,
what are the best rotation rate (r/min) and total power output (MW) estimates for (a) a
HAWT; and (b) a VAWT?
Solution: Convert 22 km/h = 6.1 m/s. Use Fig. 11.32 of the text to make your
estimates. (a) HAWT geometry. Best efficiency is at (ωr/V) ≈ 5.8, where CP ≈ 0.45:
ωr
= 5.8
=
V
ω (100 / 2 m)
6.1 m/ s
, Solve =
ω 0.708
rad
60(0.708)
r
, RPM
=
= 6.8
s
2π
min
Ans.(a )
r
1.22 π
m− N
2
=
Power C=
AV 3 0.45(
) (100)
=
(6.1)3 489, 400
= 489 kW
P
2
2 4
s
(b) Darrieus VAWT geometry. Best efficiency is at (ωr/V) ≈ 4.0, where CP ≈ 0.40:
(100 / 2 m)
r
ww
rad
60(0.488)
r
, RPM
Ans.(b)
=
= 4.7
6.1m/ s
s
2p
min
1.22 p
m− N
r
2
) (100)
(6.1)3 435, 000
Power C=
AV 3 0.40(
=
=
= 435 kW
P
2
2 4
s
Total power (0.435
MW)(130 units) 57 MW
Ans.(b)
=
=
= 4.0
=
V
, Solve =
w 0.488
Primarily because of the variability of Cape winds, both of these are overestimates. The
official power estimate for the Cape Wind (HAWT) project is 42 MW.
83
P11.105 In 2007, a wind-powered-vehicle contest, held in North Holland [64], was won
with a design by students at the University of Stuttgart. A schematic of the winning 3wheeler is shown in Fig. P11.101. It is powered by a shrouded wind turbine, not a
propeller, and, unlike a sailboat, can move directly into the wind. (a) How does it work?
(b) What if the wind is off to the side? (c) Cite some design questions you might have.
Fig. P11.105
Solution: (a) The wind turns the turbine. The turbine turns a shaft that is geared to the
wheels and drives the vehicle.
(b) If the wind is from the side, the entire turbine swivels to face into the wind.
(c) Design questions might include:
1. How large to make the turbine? Answer: D = 1.8 m.
2. How to design the drive train? Answer: Two sets of bicycle gears and a chain
drive.
3. How to balance the weight to resist the overturning moment of the turbine thrust?
Answer: More weight in the front – total weight was about 1,290 N.
4. How to manage variable winds? Answer: The turbine blades could change pitch
if the wind speed changed.
5. How to make light but strong blades? Answer: The students molded the blades
from composite materials.
________________________________________________________________________
84
P11.106
Analyze the wind-powered-vehicle of Fig. P11.101 with the following data:
turbine diameter D = 1.8 m, power coefficient (Fig. 11.32) = 0.3, vehicle CDA = 0.4 m2,
and turbine rotation 240 r/min. The vehicle moves directly into a head wind, W = 40
km/h. The wind backward thrust on the turbine is approximately T ≈ CT(r /2)Vrel2 Aturbine,
where Vrel is the air velocity relative to the turbine and CT ≈ 0.7. Eighty per cent of the
turbine power is delivered by gears to the wheels, to propel the vehicle. Estimate the sealevel vehicle velocity V, in km/h.
Solution: At sea level take r = 1.22 kg/m3. Recall the definition of power coefficient from
Eq. (11.46). The air velocity relative to the moving turbine is (W+V). Hence
=
Pturbine
CP
r
3
Aturbine =
Vrel
r π
( ) D 2 (W + V )3 , CP ≈ 0.4 ( Fig .11.32)
2
2 4
=
=
≈ 0.8
Pturbine
Pvehicle
Fvehicle V , where
Also, hh
CP
The retarding force on the vehicle is the sum of drag on the vehicle and thrust on the
turbine:
F vehicle =
Fdrag + Fthrust =
r
2
(CD A)(W + V ) 2 + CT
r π
2 4
D 2 (W + V ) 2
where CT ≈ 0.7. Combine these two equations and we find that both (W+V)2 and r cancel,
and the turbine area can be divided out. The result is
h CP (=
W +V ) V[
=
or
: V
CD A
(π / 4) D
2
W
(analytic answer)
CD A
1
+
−
C
(
)[
]
1
T
h CP (π / 4) D 2
+ CT ] ,=
or : V
W
=
1
0.42
+ 0.7] −1
[
][
0.8(0.4) (π / 4)(1.8) 2
W
=
2.7 − 1
40 km/ h
=
1.7
23.5 km / h
Ans.
This is a simplified analysis, independent of density and RPM, in reasonable agreement with the
actual performance of the Stuttgart student design, which averaged 24 km/h during the contest.
85
P11.107
Figure 11.32 showed the typical power performance of a wind turbine. The
wind also causes a thrust force that must be resisted by the structure. The thrust
coefficient CT of a wind turbine may be defined as follows:
CT
=
Thrust force
=
( r / 2) A V 2
T
( r / 2)[(π / 4) D 2 ] V 2
Values of CT for a typical horizontal-axis wind turbine are shown in Fig. P11.107. The
abscissa is the same as in Fig. 11.32. Consider the turbine of Prob. P11.103. If the wind is
32 km/h and the rotation rate 115 r/min, estimate the bending moment about the tower base.
Fig. P11.107
ω r/V1
Solution: Assume sea-level density of 1.22 kg/m3 – all of Rhode Island is pretty much
sea-level. Convert 115 r/min x 2π/60 = ω = 12.04 rad/s. Convert wind speed = 32 km/h =
8.9 m/s. Now find where we are on the graph:
ωr
=
V
(12.04 / s )(3.5 m)
=
8.9 m / s
4.7
From Fig. P11.107, read CT ≈ 0.66. Then the thrust force is, approximately,
1
1
π
2
=
Thrust C=
(0.66)( )(1.2)[ (7m ) 2 ](8.9) 2 ≈ 1, 230 N
T ( ) r AV
2
2
4
Moment about the base
= T=
h (1, 230 N )(30 m
=) 36, 900 Nm Ans.
_______________________________________________________________________
86
P11.108 To avoid the bulky tower and impeller and generator in the HAWT of the
chapter-opener photo, we could instead build a number of Darrieus turbines of height 4 m
and diameter 3 m. (a) How many of these would we need to match the HAWT’s 100
kW output for 15 m/s wind speed and maximum power? (b) How fast would they
rotate? Assume the area swept out by a Darrieus turbine is two-thirds the height times
the diameter.
Solution: Again take ρ = 1.2 kg/m3. From Fig. 11.32, for a Darrieus turbine, CP,max =
0.42 at ωr/V = 4.2. The total power generated by one turbine is thus
1
1
2
3
3
(0.42)( )(1.2)[ (4)(3)](15)
PDarrieus C=
=
=
P ( ) r AV
2
2
3
6,800 W
To match the big 3-blade HAWT, we would need 100,000/6,800 ≈ 15 Darrieus
turbines. Ans.
The rotation speed at this maximum-power condition is
ωr
ω (1.5m)
rad
r
=
4.2 =
, solve ω ≈ 42
≈ 400
V
15 m / s
s
min
Ans.
__________________________________________________________________
87
COMPREHENSIVE PROBLEMS
C11.1 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
1E−6
2E−6
3E−6
4E−6
5E−6
H, m 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 the figure?
NOTE: The tubing is smooth, with an inner diameter of 5 mm and a total length of 29.8 m.
The water is at room temperature and pressure, and minor losses are neglected.
Fig. C11.1
Solution: For water, take r = 998 kg/m3 and m = 0.001 kg/m·s. NOTE: The tubing is so
small that the flow is laminar, even at the highest pump flow rate:
4ρQ
4(998)(5E−6)
=
Remax = max =
1, 270 < 2,000 ∴ Laminaρ
πm d
π (0.001)(0.005)
The energy equation shows that the pump must fight both friction and elevation:
L V2
128 µ LQ
128(.001)(29.8)Q
0.8
,
=∆z + h f ,laµ =∆z +
=
+
d 2g
π d4ρg
π (0.005)4 (998)(9.81)
or: H p =
0.8 + 198, 400Q =
H p (Q ) from the pump data above
H πuµπ =∆z + f
88
One can plot the two relations, as at right,
or use Excel to get a final result for flow
rate and head:
H p = 1.00 m
=
Q 1.0E− 6 m 3 /s Ans.
The results are Red = 255, H = 0.999 m, Q =
1.004E−6 m3/s.
C11.2 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 0.085 m3/s
to the system of Fig. P6.62 at minimum
input power. Calculate the power required.
Solution: For water take r = 998 kg/m3
Fig. C11.2
and m = 1E−3 kg/m·s. For cast iron take
e = 0.00026 m, or e /d = 0.00026/0.15 = 0.00173. The energy equation, written between
reservoirs, is the same as in Prob. 6.62:
LV2
610 [Q/ (p /4)(0.15) 2 ]2
=36 m + f
=36 + 663, 735 fQ 2
H p =∆z + f
d 2g
0.15
2(9.81)

e
4 rQ
where f = fMoody = fcn  Red ,  with Red =

d
πµ d
If, as given, Q = 0.085 m3/s, then, from above, f = 0.0228 and Hp = 145.3 m.
Now we have to optimize: From Prob. 11.28, Q* = 0.17 m3/s, H* = 100.58 m, and P* =
255 bhp when n = 2,134 rpm (35.57 rps) and D = 37.135 cm. Thus, at BEP:
=
CQ*
Q*
gH*
P*
=
0.0933; =
C*H
= 5.66; =
C*P
= 0.599
3
2 2
nD
n D
ρn3 D 5
For the system above, (0.085)/[nD3] = 0.0932, or nD3 = 0.912, and (9.81)(145.3)/n2D2 =
5.66, or n2D2 = 251.84. Solve simultaneously:
=
n 66
=
rps 3,960
rev
;=
D p 0.24 m; Power
= 0.599 r=
n3 D 5p 136.5 kW
min
89
Ans.
C11.3 Reconsider Prob. 6.77 as an exercise in turbine selection. Select an impeller
size and rotational speed from the Francis
turbine family of Fig. 11.21d which will
deliver maximum power generated by the
turbine. Calculate the water turbine power
output and remark on the practicality of
your design.
Fig. C11.3
3
Solution: For water, take r = 998 kg/m and m = 0.001 kg/m·s. For wrought iron take
e = 0.046 m, or e /d1 = 0.046/60 = 0.000767 and e /d2 = 0.046/40 = 0.00115. The energy
and continuity equations yields
2
d 
V22
L1 V12
L2 V22
∆z= 20 m=
+ f1
+ f2
+ hturbine ; V2= V1  1  = 2.25V1
2g
d1 2 g
d2 2 g
 d2 
with the friction factors to be determined by the respective Reynolds numbers and
roughness ratios. We use the BEP coefficients from Eq. (11.36) for this turbine:
CQ*
=
Q*
gH *
P*
0.34; =
C*H
C*
=
= 9.03; =
= 2.70
P
3
2 2
ρ n3 Dt5
nDt
n Dt
We know from Prob. 6.76 that the friction factors are approximately 0.022. With only one
energy equation (above) and two unknowns (n and Dt ), we vary, say, n from 0 to 100 rev/s
and find the resulting turbine diameter and power extracted. The power is shown in the
graph below, with a maximum of 381 watts at n = 70 rev/s, with a resulting turbine
diameter Dt = 5.3 cm. This is a fast, small turbine! Ans.
90
C11.4 The system of Fig. C11.4 is designed to deliver water at 20°C 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.
Fig. C11.4
Solution: For water take r = 998 kg/m3 and m = 0.001 kg/m·s. First establish the system
curve of head loss versus flow rate. For cast iron take e = 0.26 mm, hence e /d = 0.26/380 =
0.000684. The pumps of Figs. 11.7a,b deliver flows of 15.14 to 106 m3/min, no doubt
91
turbulent flow. The energy equation, written from the lower free surface to the upper
surface, gives
2
Vpipe
p1 V12
p2 V22
 L

+
+ z1 = 0 + 0 + 0 =
+
+ z2 + h f − H pump = 0 + 0 + 11 m +
 f ∑K
2g  d
g 2g
g 2g
ρρ

Whερε
 ε /d
ρVπiπε d
2.51 
Q
f −1/2 =
−2.0 log10 
+
and Vπiπε =
 , Rε d =
 3.7 Rε
µ
(π /4)d 2
f 
d

Take, for example, Q = 1.39 m3/s. Then Vpipe = 12.24 m/s, Red = 4.64E6,
f = 0.0180, hence the head loss becomes
(12.24) 2 

 28 m 
=
(0.0180)

 + 0.5 + 7.2  68.9 m,

2(9.81) 
 0.38 m 

Thus a match requires H = h f + z2 = 68.9 m + 11 m = 79.9 m
=
hf
No pump in Fig. 11.7 exactly matches this, but the 71-cm pump in Fig. 11.7a and the
105-cm pump in Fig. 11.7b are pretty close, especially the latter. We can continue the
calculations:
Q, m3/min:
hf, m:
15.14
13.41
30.82
20.12
45.42
31.39
60.57
47.55
75.71
67.97
83.28
79.86
90.85
106
92.96 122.53
(a) The best match seems to be the 96.5-cm pump of Fig. 11.7b at a flow rate of
75.71 m3/min, near maximum efficiency of 88%. Ans. (a)
(b, c) The appropriate flow rate is 75.71 m3/min.
The power is 955 kW. Ans. (c)
92
Ans. (b)
(d) Use Eq. (11.20) to check the NPSH for cavitation. For water at 20°C, pv = 2,337 Pa.
The velocity in the pipe is V = Q/A = 11.13 m/s. The theoretical minimum net positive
suction head is:
NPSH =
pa
− zi − h fi −
pv
γγ
=
101,350 Pa
(11.13 m/s)2
2,337 Pa
−
−
1
m
(1 + 0.5) −
3
2
9,790 N/m
2(9.81 m/s )
9,790 N/m3
NPSH = −0.36 m
In Fig. 11.7b, at Q = 75.71 m3/min, read NPSH ≈ 4.88 m >> −0.36 m! So this pump,
when placed in its present position, will surely cavitate! Ans. (d) A new pump
placement is needed.
C11.5 For the 105-cm water pump of Fig. 11.7b, at 710 r/min and 83 m3/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: For water, take the density to be 998 kg/m3.
(a) From Fig. 11.7b, we are close to maximum efficiency: 89%
Ans.(a)
(b) From Fig. 11.7b, at 83 m3/min, read H ≈ 82.3 m and bhp ≈ 1,700. Then
Efficiency
=
r gQH
=
bhp
(998)(9.81)(83 / 60)(82.3)
= 0.878 ≈ 88% Ans.(b)
(1, 700)(746)
This estimates are quite close, considering the difficulty of reading Fig. 11.7b.
93
C11.6 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 dimensionless difference between shaft rotation rates, s = 1 − ω s /ω p. For a given percentage of
fluid filling, the torque T transmitted is a function of s, r, ωp, and impeller diameter D.
(a) Non-dimensionalize this function into two pi groups, with one pi proportional to T.
Tests on a 30-cm-diameter coupling at 2,500 r/min, filled with hydraulic fluid of density 896
kg/m3, yield the following torque versus slip data:
Slip, s:
0%
5%
10%
15%
20%
25%
Torque T, m·N:
0
122
373
597
786
922
(b) If this coupling is run at 3,600 r/min, at what slip value will it transmit a torque of 1,220
m·N?
(c) What is the proper diameter for a geometrically similar coupling to run at 3,000 r/min
and 5% slip and transmit 815 m·N of torque?
Fig. C11.6
Solution: (a) List the dimensions of the five variables, from Table 5.1:
Variable:
Dimension:
T
{ML2/T2}
s
{1}
94
r
ωp
3
{M/L }
{1/T}
D
{L}
Since s is already dimensionless, the other four must form a single pi group. The result is:
T
ρω p2 D5
= fcn( s) Ans. (a)
Now, to work parts (b) and (c), put the data above into this dimensionless form.
Convert ωp = 2,500 r/min = 41.7 r/s. The results are:
Slip, s:
(
)
T ρω p2 D 5 :
0%
0
5%
10%
15%
20%
25%
0.0297 0.0.909 0.145
0.192
0.225
(b) With D = 0.3048 m and ω p = 3,500 r/min = 58.3 r/s and T = 1,220 m·N,
T
rω 2p D5
1, 220 m⋅N
=
0.152, Estimate s ≈ 15%
3
(897 kg/m )(58.3 r/s) 2 (0.305 m)5
Ans. (b)
(c) With D unknown and s = 5% and ω p = 3,000 r/min = 50 r/s and T = 815 m·N,
T
815 m⋅N
=
=
0.0297, Solve D ≈ 0.415 m
2 5
rω p D
(897 kg/m3 )(50 r/s) 2 ( D)5
Ans. (c)
C11.7 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.
Solution: The Cordier method, developed in the textbook G. T. Csanady, Theory of
Turbomachines, McGraw-Hill, New York, 1964, is an extensive correlation of the best
features of various pumps (or turbines) to design an optimal machine for a given set of
conditions. Its modern use, with software, is discussed in Ref. 63.
95
C11.8 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 = 36 m, and the water flows through 600 m of
15-cm-diameter cast iron pipe. For simplicity, assume that the pump operates at BEP
(92%) with H*p = 60 m and the turbine operates at BEP (89%) with H*t = 30 m. 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.
Solution: For water, take r = 998 kg/m3 and m = 1E-3 kg/m-s. (a) Write the steady flow
energy equation for pump operation, neglecting minor losses:
h p =60 m =Dz + f
Qp
L V2
V2
600 m
=36 m + f (
)[
], V =
D 2g
0.15 m 2(9.81)
(p / 4)(0.15 m)2
For cast iron, e = 0.00026 m, hence e /D = 0.00026 m/0.15 m = 0.00173. Iterate to find
the friction factor and complete the pump head equation. Excel quickly finds the result:
Re D
=
m
ρVD (998)V (0.15)
f 0.0230;=
V 2.26 ;
=
= 338,300;
=
1E − 3
s
m
m3
; H 60 m
Q 0.04
=
=
s
(b) Write the steady flow energy equation for turbine operation, neglecting minor losses:
ht =30 m =Dz − f
L V2
600 m
V2
Qt
=36 m − f (
)[
], V =
D 2g
0.15 m 2(9.81)
(π / 4)(0.15 m) 2
Iterate again and obtain a considerably lower flow rate:
=
Re D
m
m3
rVD (998)V (0.15)
=
= 164,700;
=
f 0.0236;
=
V 1.1 =
; Q 0.019 =
; H 30 m
1E − 3
s
s
m
Hence Power
QH (0.88)(9,790)(0.04)(60)
=
ηγ
=
= 21,000
pump
m−N
Ans.(b)
s
The pumping power is much greater than the power generated by the turbine, but the
pump’s required electricity is very cheap because usage is so small at night.
96