Department of Civil Engineering and Applied Mechanics

McGill University

CIVE327 FLUID MECHANICS AND HYDRAULICS

Assignment No. 6

1. Consider the 371-mm-diameter pump curve of Fig. 12.9. If the pump is run at 1200 rpm to
deliver 0.8 m3/s of water, find the resulting head rise and fluid power. (Potter & Wiggert 12.8)

Solution:
1200 × 2π
Given: Q1200rpm = 0.8 m3/s, ω1200 rpm = 1200 rpm = = 125.66 rad/s , D1200rpm = 0.317 m
60
The speed of the pump in Figure 12.9, ω970rpm = 102 rad/s

H970rpm ≈ 11.75 m

NPSH970rpm ≈ 4.23 m
Q970rpm = 0.6494 m3/s

FIGURE 12.9 Mixed-flow pump performance curves for four different impellers and vane angles with N =
970 rpm (ω = 102 rad/s). Water at 20oC is the pumped liquid. (Courtesy of Sulzer Pumps Ltd.)

Q
Using discharge coefficient C Q = similarity between two different operating speeds,
ω D3
 Q1200 rpm Q970 rpm
3
=
ω1200 rpm D 1200 rpm ω970 rpm D9370 rpm
Since D1200rpm = D970rpm,
0 .8 Q970 rpm
=
125.66 102

Q970rpm = 0.6494 m3/s

From Figure 12.9, with Q970rpm = 0.6494 m3/s, H970rpm ≈ 11.75 m

gH
Using head coefficient C H = similarity,
ω 2 D2
gH 1200 rpm gH 970 rpm
= 2
ω1200 rpm D1200 rpm ω970 rpm D9270 rpm
2 2

Since g = constant and D1200rpm = D970rpm,

H 1200 rpm 11.75
=
125.66 2 102 2

Therefore, resulting head rise H1200rpm = 17.83 m

And the fluid power P1200rpm = γ H1200rpmQ1200rpm = 9810 × 17.83 × 0.8 = 139.93 kW

2. In Problem 12.8, how high above the suction reservoir can the pump running at 1200 rpm be
located if the water temperature is 50oC and the atmospheric pressure is 101 kPa? (Potter &
Wiggert 12.9). Assume that the pipe diameter is 0.4 m, friction factor f = 0.01, and that the length
of the pipe L = ∆z.

Solution:
From Figure 12.9,
with Q970rpm = 0.6494 m3/s (from Problem 12.8) the corresponding NPSH970rpm = 4.23 m

gNPSH
Using NPSH coefficient C NPSH = similarity,
ω 2 D2

gNPSH 1200 rpm gNPSH 970 rpm

2 2
= 2 2
ω 1200 rpm D 1200 rpm ω970 rpm D970 rpm

Since g = constant and D1200rpm = D970rpm,

NPSH 1200 rpm 4.23
=
125.66 2 102 2
NPSH1200rpm = 6.42 m
 From Table B.1 Properties of water (Potter & Wiggert, pg. 803), for water at 50oC,
pv = 12.3 kPa, and γ = 9693.26 N/m3

2 Q
• Pump
∆z
1
•

FIGURE Problem definition

Energy equation between 1 and 2 gives,

p1 V12 p V2
+ + z1 = 2 + 2 + z2 + hL
γ 2g γ 2g
p2 V22 pv
Since + = + NPSH
γ 2g γ
patmV12 0 p
++ z1 − z 2 = v + NPSH + hL
γ 2g γ
p p  ∆z  V 2
∆z = atm − v − NPSH −  K entrance + f 
γ γ  D  2g
2
 
 
 0 .8 
 π (0.4 )2 
101.3 × 10 3 12.3 × 10 3  ∆z   4 

∆z = − − 6.42 −  0.5 + 0.01 × 
9693.26 9693.26  0.4  2 × 9.81
∆z = 1.7288 − 0.0516∆z
∆z = 1.644 m

3. Prepare dimensionless performance curves for the 205-mm-diameter radial flow pump of Fig.
12.6. Compare these with the curves shown in Fig. 12.12. Can you explain why there may be
differences between the two sets of curves? (Potter & Wiggert 12.10). Prepare only the head-
discharge performance curve. Find the specific speed of the pump.

Solution:
Given: D = 0.205 m, ω = 304 rad/s

Since Q in Figure 12.6 is given in m3/h,

Q
Discharge coefficient CQ =
ω D3
Q
CQ =
304 × 3600 × 0.2053
 CQ = 1.0606 × 10 −4 Q

gH
Head coefficient CH =
ω 2 D2
9.81 × H
CH =
304 2 × 0.205 2
C H = 2.5259 × 10 −3 H
From Fig. 12.6,

• ηmax

FIGURE 12.6 Radial-flow pump performance curves for four different impellers with N = 2900 rpm (ω = 304
rad/s). Water at 20oC is the pumped liquid. (Courtesy of Sulzer Pumps Ltd.)

Using CQ and CH equations above,

TABLE: Summary of corresponding discharge and head coefficients

Q (m3/h) H (m) CQ × 102 CH × 10
0 53.4 0 1.349
50 53.2 0.530 1.344
100 52.6 1.061 1.329
150 50.2 1.591 1.268
200 46.6 2.121 1.177
250 40.9 2.652 1.033
300 32.6 3.182 0.823
 The dimensionless head-discharge curve,
1.5

CH
1
CH x 10

0.5

0
0 0.5 1 1.5 2 2.5 3 3.5
C Q x 100

The specific speed is evaluated at maximum efficiency.

From Fig. 12.6, at the maximum efficiency, Q = 213.08 m3/h and H = 45.19 m

Thus, specific speed of the pump

1 1
CQ 2 ωQ 2
ωs = 3
= 3
CH 4 (gH ) 4
1
 213.08  2
304 ×  
ωs =  3600 
3
(9.81 × 45.19)4
ωs = 0.7655 ( < 1.0, therefore radial-flow pump)

4. The following performance data were obtained from a test on a 216-mm double-entry centrifugal
pump moving water at a constant speed of 1350 rev/min:

Q (m3/min) H (m) ηP
0 12.2 0
0.454 12.8 0.26
0.905 13.1 0.46
1.36 13.4 0.59
1.81 13.4 0.70
2.27 13.1 0.78
2.72 12.2 0.78
3.80 11.0 0.74
Plot H versus Q, ηP versus Q, and W& P versus Q. If the pump operates in a system whose demand
curve is given by 5 + Q2, find the discharge, head, and power required. In the demand curve, Q is
given in cubic meters per minute. (Potter & Wiggert 12.22)

Solution:
γ HQ
The pump power W& P =
ηP

Q (m3/min) H (m) ηP W& P (kW)

0 12.2 0 0
0.454 12.8 0.26 3.654
0.905 13.1 0.46 4.214
1.36 13.4 0.59 5.050
1.81 13.4 0.70 5.665
2.27 13.1 0.78 6.233
2.72 12.2 0.78 6.956
3.80 11.0 0.74 9.236

20 1.5

HD
15
HP
1
HP (m)

10

ηP
η
0.5
5

0 0
10
W& P
W P (kW)

Q = 2.682 m3/s
0
0 0.5 1 1.5 2 2.5 3 3.5 4
3
Q (m /min)
FIGURE H versus Q, ηP versus Q, and W& P versus Q
 Q (m3/min) HD = 5 + Q2 (m)
0 5.000
0.454 5.206
0.905 5.819
1.36 6.850
1.81 8.276
2.27 10.153
2.72 12.398
3.80 19.440

In the figure above, at the intersection where demand curve meets the system curve,
Q = 2.682 m3/min, H = 12.2 m, W& P = 6.926 kW

5. An oil (S = 0.85) is to be pumped through a pipe as shown in Fig. P12.23a. Using the
characteristic curves of Fig. P12.23b, determine the required impeller diameter, speed, and input
power to deliver the oil at a rate of 14.2 L/s. Would you determine the actual speed and size of
the pump to operate at peak efficiency. (Potter & Wiggert 12.23). Correction to the typo in
Figure P12.23a: The sum of the minor loss coefficients is ΣK = 0.5 + 1.0, not fifteen.

FIGURE P12.23a

FIGURE P12.23b
Solution:
Given: Q = 14.2 L/s = 0.0142 m3/s , ΣK = 1.5 , Soil = 0.85

•2
1
• 1.5

FIGURE P12.23a

Energy equation between 1 and 2 gives

0 20 0 0
p1 V p V2
+ 1 + z1 + H P = 2 + 2 + z 2 + hL
γ 2g γ 2g
 L V 2
H P = ∆z +  ∑ K + f 
 D  2g
 61  0.0142 2
H P = 12.2 + 1.5 + 0.01 ×  2
 0.075   π × 0.075 2 
2 × 9.81 ×  
 4 
H P = 17.273 m

FIGURE P12.23b

From Fig. P12.23b, ηmax = 0.745, CQ = 0.0165, and CH = 0.1239.

 Q
Discharge coefficient CQ =
ω D3
0.0142
0.0165 =
ωD 3
ωD 3 = 0.8606

gH
Head coefficient CH =
ω 2 D2
9.81 × 17.273
0.1239 =
ω 2D2
ω D = 36.981

Substituting ω D = 36.981 into discharge coefficient,

0.8606
D2 =
36.981
D = 0.1525 m

36.981
Therefore, ω=
0.1525
ω = 242.5 rad/s = 2316 rpm

γ HQ
Power required W& P = oil
ηP
0.85 × 9810 × 17.273 × 0.0142
W& P =
0.745
&
W P = 2.745 kW

6. A model pump (Fig. P12.32) delivers 80oC water at a speed of 2400 rpm. It begins to cavitate
when the inlet pressure pi = 83 kPa absolute and inlet velocity Vi = 6 m/s.
(a) Compute the NPSH of the model pump. Neglect losses and elevation changes between the
inlet section where pi is recorded and the cavitation region in the pump.
(b) What is the required NPSH of a prototype pump which is four times larger and runs at 1000
rpm? (Potter & Wiggert 12.32)

pi, Vi Q
+ +
Location of
∆z cavitation in
Inlet section
pump

FIGURE P12.32
Solution:

(a) From Table B.1 Properties of water (Potter & Wiggert, pg. 803), for water at 80oC,
pv = 47.3 kPa, and γ = 9533.36 N/m3

pi, Vi Q
+ +
Location of
∆z cavitation in
Inlet section
pump

FIGURE P12.32

Energy equation between inlet and cavitation location gives,

pi Vi 2 p V 2
+ = +
γ 2g γ 2g
p i Vi 2 p v
+ = + NPSH
γ 2g γ
83 × 10 3 62 47.3 × 10 3
+ = + NPSH
9533.36 2 × 9.81 9533.36

NPSH = 5.58 m

g NPSH
(b) Using NPSH coefficient C NPSH = similarity,
ω 2D2

g NPSH model g NPSH prototype

2 2
= 2 2
ω model Dmodel ω prototype Dprototype

Since g = constant and Dprototype = 4Dmodel,

5.58 NPSH prototype

=
2400 × Dmodel 1000 2 × (4 Dmodel )2
2 2

NPSHprototype = 15.5 m