ANSWERS TO EXAMPLE SHEET FOR TOPIC 4
AUTUMN 2013
(Full worked answers follow on later pages)
Q1.
(a) H  12  0.08501Q 2 (H in m and Q in L s–1)
(b) 10.7 L s–1; 2.90 kW
(c) (i) 13.3 L s–1; 5.97 kW; (ii) 14.4 L s–1; 5.69 kW
Q2.
(a)
(b)
(c)
(d)
Q3.
(a) 77.1 L s–1; 38.5 m; 57.7 kW
(b) 71.8 L s–1
(c) 36.9 kW
Q4.
(a)
(b)
(c)
(d)
Q5.
(a) H  6  98320Q 2 (H in m and Q in m3 s–1)
(b) 410 s; 109 kJ
(c) 154 s
Q6.
(a) H  12  50430Q 2 (H in m and Q in m3 s–1)
(b) 8.93 L s–1; 1.89 kW
(c) 1440 rpm
Q7.
(a) H  12  952Q 2 (H in m and Q in m3 s–1)
(b) (i) 0.106 m3 s–1; (ii) 30.1 kW; (iii) –28.3 kPa gauge
(c) 0.158 m3 s–1; 75.4 kW
Q8.
(a)
(b)
(c)
(d)
H  3.2  0.003470Q 2 (H in m, Q in L s–1)
pump B on efficiency grounds (67% for B as opposed to 49% for A at duty point)
3610 W
3.50 m; no cavitation
Q9.
(a)
(b)
(c)
(d)
H  15  129.1Q 2 (H in m, Q in m3 s–1)
0.256 m3 s–1; 34.4 MJ
0.16 m3 s–1; 28.1 m; H  1098Q 2 (H in m, Q in m3 s–1)
1085 rpm; 21.0 MJ
Q10.
3170 rpm
Q11.
(a) 16.4 m
(b) 62%
hf = 3.4710–3 Q2 (hf in m and Q in L s–1)
28.1 L s–1; 5.38 kW
1910 rpm
(i) 4.80 kW; (ii) 1.15 kW; (iii) 0.47 kW; (iv) 0.47 kW; (v) 56%
H  80  0.01251Q 2 (H in m and Q in L s–1)
39.6 L s–1; 99.6 m; 0.706; 54.8 kW
13.9 rpm; centrifugal pump
2510 rpm
Hydraulics 2
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David Apsley
(c) 1.67 m s–1
(d) 11.0 m s–1
(e) 78%
Q12.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
20.5 MW
3
71.9 m s–1; 33.1 m s–1
1.58 m
3.42 MW
1.55 m3 s–1
0.166 m
85%
Hydraulics 2
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David Apsley
Q1.
(a)
System head = static lift + frictional head loss
Static lift:
hs  14  2  12 m
Frictional head loss:
L V2
Q
4Q
where
V

hf  λ
A
D 2g
πD 2
With Q in m3 s–1,
8λL
8  0.02  40
hf  2 5 Q2  2
Q 2  85010Q 2
5
π gD
π  9.81  0.06
It is more convenient here to have Q in L s–1. The system curve is then
H  12  0.08501Q 2 (H in m and Q in L s–1)
(b) Plot Hpump and Hsystem as functions of Q. The relevant data is given in the table below.
Discharge (L s–1):
Pump head (m)
System head (m)
0
30.0
12.00
3
29.5
12.77
6
27.6
15.06
9
24.4
18.89
12
19.7
24.24
15
13.5
31.13
18
5.9
39.54
60
pumps in series
50
H (m)
40
system characteristic
30
20
pumps in parallel
10
pump characteristic
0
0
5
10
15
20
Q (L
Hydraulics 2
25
30
35
s-1)
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David Apsley
100
90
80
efficiency (%)
70
60
50
40
30
20
10
0
0
5
10
15
20
Q (L s-1)
Duty point:
Q = 10.7 L s–1 = 0.0107 m3 s–1
H = 21.8 m
η = 0.79
Input power:
Pin 
ρgQH
η

1000  9.81  0.0107  21.8
 2904 W
0.788
Answer: Q = 10.7 L s–1; Pin = 2.90 kW.
Hydraulics 2
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David Apsley
(c)
Pumps in parallel: same H, double Q.
Pumps in series: same Q, double H.
These are plotted in the H vs Q graph above. The duty points are as follows.
(i) Pumps in parallel:
For both pumps:
Q = 13.3 L s–1
H = 27.0 m
For each individual pump:
Q = 6.65 L s–1 = 0.00665 m3 s–1
H = 27.0 m
η = 0.590
ρgQH
1000  9.81  0.00665  27.0
Pin 

η
0.590
The total power consumption (two pumps) is then
2Pin  2  2985  5970 W
 2985 W
Answer: total discharge = 13.3 L s–1; total power consumption = 5.97 kW.
(ii) Pumps in series:
For both pumps:
Q = 14.4 L s–1
H = 29.7 m
For each individual pump:
Q = 14.4 L s–1 = 0.0144 m3 s–1
H = 14.85 m
η = 0.737
ρgQH
1000  9.81  0.0144  14.85
Pin 

 2846 W
η
0.737
The total power consumption is then
2Pin  2  2846  5692 W
Answer: total discharge = 14.4 L s–1; total power consumption = 5.69 kW.
Hydraulics 2
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David Apsley
Q2.
(a) Frictional head loss:
L V2
hf  λ
D 2g
Hence
8λLQ 2
hf  2 5
π gD
V
where
Q
A

4Q
πD 2
If hf is in m and Q in L s–1:
hf 
8  0.02  21  Q 


π 2  9.81  0.15  1000 
2
 0.003470Q 2
Answer: head loss due to friction is h f  0.00347Q 2 (hf in m and Q in L s–1).
(b) We require the system curve (static lift + losses).
The static lift is
H s  10  1.5  11.5 m
Hence the system characteristic is
H  11.5  0.003470Q 2 (hf in m and Q in L s–1)
(*)
The pump and system characteristics can be plotted graphically and the duty point
determined by their point of intersection.
20
system characteristic
15
H (m)
duty point
10
pump characteristic
5
0
0
Hydraulics 2
10
20
Q (L s-1)
30
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40
David Apsley
1.0
0.9
0.8
efficiency
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
10
20
Q (L s-1)
30
40
Q = 28.1 L s–1 = 0.0281 m3 s–1
H = 14.2 m
η  0.728
Input power:
Pin 
ρgQH
η

1000  9.81  0.0281 14.2
0.728
 5380 W
Answer: discharge = 28.1 L s–1; power consumption = 5.38 kW.
(c) Let N1 = 1750 rpm and the new speed be N2.
Q2 = 34.0 L s–1
From the system characteristic (*),
H 2  11.5  0.00347Q22  15.51 m
On the scaling curve through this new duty point,
2
Q 
H 1  N1 
   1 
 
H2  N2 
 Q2 
2
i.e.
H1
 Q 
 1 
15.51  34.0 
2
or
H1  0.01342Q12
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David Apsley
20
system characteristic
new duty point
15
H (m)
scaled duty point
10
pump characteristic
5
0
0
10
20
Q (L s-1)
30
40
This cuts the original pump characteristic at Q1 = 31.1 L s–1. Then
N 2 Q2
34.0


 1.093
N1 Q1
31.1
Hence,
N 2  1.093  N1  1.093 1750  1913 rpm
Answer: required pump speed = 1910 rpm.
(d) (i) When Q = 24 L s–1 the pump curves give H =15.5 m and η = 0.760.
The output power is
Pout  ρgQH  1000  9.81 (24 / 1000)  15.5  3646 W
The input power is
P
3646
Pin  out 
 4797 W
η
0.760
Answer: power consumption of the pump = 4.80 kW.
(ii) Power dissipated in the pump:
Pin  Pout  4797  3646  1151 W
Answer: power dissipated in the pump = 1.15 kW.
Hydraulics 2
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David Apsley
(iii) Head lost due to friction, with Q in L s–1:
h f  0.00347Q 2  0.00347  24 2  2.00 m
Power dissipated by friction:
Pfriction  ρgQh f  1000  9.81 (24 / 1000)  2.00  471 W
Answer: power dissipated by pipe friction = 0.47 kW.
(iv) Head lost at the control valve:
H valve  H pump  H system  15.5  11.5  2.00  2.00 m
Power dissipated in the control valve:
Pvalve  ρgQH valve  471 W
Answer: power dissipated in the control valve = 0.47 kW.
(v) Overall efficiency of the installation:
Powerin  Powerdissipated
4802  1153  471  471
η overall 

 0.56
Powerin
4802
Answer: overall efficiency of the system = 56%
Hydraulics 2
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David Apsley
Q3.
The hydraulic scaling laws give
2
Q2 N 2
H2  N2 

,


H 1  N1 
Q1 N1
Here, N1 = 1000 rpm, N2 = 1400 rpm, so that
N2
 1.4
N1
Hence, pump characteristics at 1400 rpm can be determined from those at 1000 rpm by
Q2  1.4Q1
H 2  1.96H1
This gives the following:
Pump characteristics at speed 1400 rpm:
Discharge (L s–1)
0
28
Head (m)
98
88.2
Efficiency (%)
–
60
56
64.68
69
Adding the static head (20 m) to the friction losses gives:
System characteristics:
Discharge (L s–1)
0
20
40
Head loss (m)
20
21.0
24.0
70
49
60
84
27.44
40
60
30.0
80
40.0
120
pump characteristic (1400 rpm)
100
H (m)
80
60
pumps in parallel (1000 rpm)
duty point (a)
40
system characteristic
duty point (b)
20
0
0
10
20
30
40
50
60
70
80
90
100
-1
Q (L s )
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David Apsley
80
70
single pump (1000 rpm)
in parallel operation
single pump (1400 rpm)
60
Efficiency (%)
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90
100
-1
Q (L s )
Duty point (see graphs):
Q = 77.1 L s–1 = 0.0771 m3 s–1
H = 38.5 m
η = 50.5 %
Then
ρgQH
1000  9.81  0.0771  38.5
Pin 

 57660 W
η
0.505
Answer: duty point is Q = 77.1 L s–1, H = 38.5 m; power consumption, Pin = 57.7 kW.
(b) For pumps in parallel, double the flow and keep the head the same. This gives the
following characteristics.
Pumps in parallel at speed 1000 rpm:
Discharge (L s–1)
0
40
80
100
120
Head (m)
50
45
33
25
14
This intersects the system curve at
Q = 71.8 L s–1
H = 35.7 m
Answer: maximum discharge rate for pumps in parallel, Q = 71.8 L s–1.
(c) For both pumps:
Q = 71.8 L s–1
H = 35.7 m
For each individual pump:
Hydraulics 2
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David Apsley
Q = 35.9 L s–1 = 0.0359 m3 s–1
H = 35.7 m
η = 0.681
ρgQH
1000  9.81  0.0359  35.7
Pin 

η
0.681
The total power consumption is then
2Pin  2  18460  36920 W
 18460 W
The discharge is similar to that of a single pump operating at a higher speed, but the power
consumption is substantially less (mainly because each pump operates closer to its maximumefficiency point). Assuming that the operating rather than capital and maintenance costs of
the system dominate, then this is the better option.
Answer: power consumption for pumps in parallel = 36.9 kW.
Hydraulics 2
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David Apsley
Q4.
(a)
System head = static lift + head losses
L V2
Q
where
V
H  Hs  λ
A
D 2g
8λL

H  H s  2 5 Q2
π gD
If H is measured in m and Q in L s–1 then
8  0.02  575
Q 2
H  80  2
(
)
5
π  9.81  0.15 1000

4Q
πD 2
Hence, the system curve is
H  80  0.01251Q 2
with H in m and Q in L s–1.
(b) Plot the system and pump characteristics as a graph of H against Q. The corresponding
efficiency is obtained from a graph of η against Q.
140
120
system characteristic
H (m)
100
duty point
80
pump characteristic
60
40
20
0
0
Hydraulics 2
10
20
30
Q (L s-1)
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40
50
60
David Apsley
100
90
80
efficiency (%)
70
60
50
40
30
20
10
0
0
10
20
30
Q (L s-1)
40
50
60
Duty point:
Q = 39.6 L s–1 = 0.0396 m3 s–1
H = 99.6 m
η = 0.706
Input power:
Pin 
ρgQH
η

1000  9.81  0.0396  99.6
0.706
 54800 W
Answer: Q = 39.6 L s–1; H = 99.6 m; η = 0.706; Pin = 54.8 kW.
(c) Specific speed is
NQ1 / 2
Ns 
H 3/ 4
calculated at the maximum-efficiency point (Q = 31 L s–1 = 0.031 m3 s–1; H = 117 m). Thus
0.0311 / 2
N s  2800 rpm 
 13.9 rpm
117 3 / 4
Answer: 13.9 rpm; this is a centrifugal pump.
(d) At some unknown new rotational speed N2 the new discharge is
Q2 = 30 L s–1
The corresponding head is (from the system characteristic)
Hydraulics 2
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David Apsley
H2 = 91.26 m
The scaling curve through this point is
2
Q
H  N 
   
 
H2  N2 
 Q2 
2
or
H
H2 2
Q
Q22
 0.1014Q 2
140
120
system characteristic
scaled duty point
100
H (m)
new duty point
80
pump characteristic
60
40
20
0
0
10
20
30
Q (L s-1)
40
50
60
This scaling curve intersects the N1 = 2800 rpm curve at Q1 = 33.4 L s–1. Hence,
N 2 Q2

N1 Q1
Q
30

N 2  N1 2  2800 
 2510 rpm
Q1
33.4
Answer: N2 = 2510 rpm.
Hydraulics 2
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David Apsley
Q5.
(a)
L V2
H system  H s  λ
D 2g
8λLQ 2

H system  H s  2 5
π gD
Substituting parameter values:
H system  6  98320Q 2
where
V
Q
A

4Q
πD 2
(H in m, Q in m3 s–1)
(b) Plot:
H vs Q (for pump and system) to find the duty point.
η vs Q and read off efficiency.
12
10
H (m)
8
6
Hpump
Hsystem
4
2
0
0
Hydraulics 2
0.002
0.004
0.006
3
Q (m s-1)
A4-16
0.008
0.01
David Apsley
0.8
0.7
0.6
h
0.5
0.4
0.3
0.2
0.1
0
0
0.002
0.004
0.006
0.008
0.01
Q (m3 s-1)
From the graphs the duty point is:
Q = 0.00244 m3 s–1
H = 6.583 m
η = 0.593
Time taken:
1.0
T
Q

1
0.00244
 409.8 s
Power and energy:
ρgQH
1000  9.81  0.00244  6.583
Pin 

 265.7 W
η
0.593

E  PinT  265.7  409.8  108900 J
Answer: time taken = 410 s; energy used = 109 kJ.
(c) Scaling to from N1 = 1500 rpm to N2 = 2250 rpm:
N
Q2  2 Q1  1.5Q1
N1
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David Apsley
2
N 
H 2   2  H 1  2.25H 1
 N1 
The revised pump characteristics at 2250 rpm are given in the table below.
Discharge,
Q (L s–1)
Head,
H (m)
0
1.5
3
4.5
6
7.5
9
11.5
20.07
18.00
15.80
13.48
11.05
8.46
5.76
2.95
Plot new H vs Q (for pump and system) to find the duty point.
25
20
H (m)
15
Hpump
10
Hsystem
5
0
0
0.005
0.01
Q
(m3
0.015
s-1)
At the duty point:
Q = 0.00651 m3 s–1
The time taken is
1.0
1
T

 153.6 s
Q
0.00651
Answer: time taken = 154 s.
Hydraulics 2
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David Apsley
Q6.
(a) The system head requirements are:
L
V2
where
H  H s  (λ  K )
D
2g
V
Q
A

4Q
πD 2
L
8Q 2
 K) 2 4
D
π gD
Substituting the given parameters (Hs = 12 m, λ = 0.04, L = 30 m, D = 0.08 m, K = 10) gives,
(H in m and Q in m3 s–1)
H  12  50430Q 2
or, if preferred,
H  12  0.050430Q 2 (H in m and Q in L s–1)

H  H s  (λ
(b) Plot the graphs of H vs Q (pump and system characteristics) to determine the duty point.
Plot also η vs Q to find the efficiency. Either m3 s–1 or L s–1 may be used as units for Q.
30
25
Pump
H (m)
20
System
Duty point
15
10
5
0
0
0.005
0.01
0.015
0.02
Q (m3 s-1)
Hydraulics 2
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David Apsley
0.9
0.8
0.7
0.6
h
0.5
0.4
0.3
0.2
0.1
0
0
0.005
0.01
0.015
0.02
Q (m3 s-1)
The duty point is
Q = 0.00893 m3 s–1
H = 16.0 m
η = 0.743
The power input is given by
ρgQH
η
Pin
whence
ρgQH
1000  9.81  0.00893  16
Pin 

η
0.743
 1886 W
Answer: discharge = 8.93 L s–1; power consumption = 1.89 kW
(c) For the new arrangement, need to change Hs (to 20 m) and L (to 38 m). The new system
curve is
(H in m and Q in m3 s–1)
H  20  58500Q 2
or, if preferred,
(H in m and Q in L s–1)
H  20  0.05850Q 2
It is not actually necessary to plot this curve because only the new duty point is required and
the discharge at the higher speed N2 is stated to be the same:
Q2 = 0.00893 m3 s–1
whence
H2 = 24.7 m (from the new system curve).
Hydraulics 2
A4-20
David Apsley
The old and new duty points do not scale onto each other. To determine where this came
from on the original curve use the hydraulic scaling laws:
H  N 


H 2  N 2 
Q
N

,
Q2 N 2
2
or
2
H Q
 
H 2  Q2 
With Q2 and H2 from above this gives (for H in m and Q in m3 s–1):
H  309700Q 2
Plot this to where it intersects with the original curve at speed N1.
30
(Q2,H2)
New duty point
25
Pump
System
H (m)
20
Scaling curve
(Q1,H1)
15
Old duty point
Pump (scaled)
10
5
0
0
0.005
0.01
3
0.015
0.02
-1
Q (m s )
This gives
Q1 = 0.00744 m3 s–1
Then,
N 2 Q2

N1 Q1

0.00893
 1.20
0.00744
Hence, the new rotation rate of the pump is
1.20 1200  1440 rpm
Answer: new rotation rate = 1440 rpm.
Hydraulics 2
A4-21
David Apsley
Q7.
(a) System head requirement is
H  static lift  losses
L V2
D 2g
Q
4Q
With V 
this is

A
πD 2
8λL
H  hs  2 5 Q 2
π gD
 hs  λ
Substituting numerical values (hs = 12 m, λ = 0.025, L = 35 m, D = 0.15 m) gives
H  12  952.1Q 2
when H is in m and Q is in m3 s–1. Note that L includes both length of pipe; alternatively add
the losses from separate parts – you will get the same answer.
(b) The system-head values may be calculated for the Q values in the table:
Q (m3 s–1) 0.00
Hsystem (m) 12.00
0.03
12.86
0.06
15.43
0.09
19.71
0.12
25.71
0.15
33.42
0.18
42.85
The H vs Q and η vs Q graphs can then be plotted as below.
50.0
45.0
40.0
pump
system
35.0
H (m)
30.0
25.0
20.0
15.0
10.0
5.0
0.0
0
0.05
0.1
0.15
0.2
0.25
0.3
Q (m3 s-1)
Hydraulics 2
A4-22
David Apsley
1.0
0.8
h
0.6
0.4
0.2
0.0
0
0.05
0.1
0.15
3
0.2
0.25
0.3
-1
Q (m s )
At the duty point,
Q = 0.106 m3 s–1
H = 22.70 m (from the system characteristic in part (a))
η = 0.785
(i) Answer: Q = 0.106 m3 s–1.
(ii) Since
η
ρgQH
P
one has
powerin 
ρgQH
η

1000  9.81  0.106  22.70
0.785
 30070 W
Answer: powerin = 30.1 kW.
(iii) At pump inlet,
H inlet  H sump  head losses over 10 m
Hence, in terms of gauge pressures
pinlet
L V2
V2
 z inlet 
 0  λ inlet
ρg
2g
D 2g
Hence,
Hydraulics 2
A4-23
David Apsley
pinlet  ρgz inlet  (1  λ
Linlet 1
) 2 ρV 2
D
Now,
zinlet  2.0 m
Linlet  10 m
4Q
4  0.106
V

2
πD
π  0.15 2
 5.998 m s 1 (quite quick!)
Hence,
pinlet  1000  9.81  (2)  (1  0.025 
10
)  12  1000  5.998 2
0.15
 28350 Pa
Answer: –28.3 kPa gauge.
(c) If all else (especially speed) is constant then the hydraulic similarity laws imply:
Q  D3
H  D2
η remains constant (because both ρgQH and the input power scale as D5)
For each value of Q in the original table there will be new Q values multiplied by
1.23 (= 1.728) and new H values multiplied by 1.22 (= 1.44). The new pump characteristics
are given below.
Q (m3 s–1)
H (m)
η (%)
0.00
43.2
–
0.052
42.9
29
0.104
40.5
54
0.156
36.0
73
0.207
29.4
80
0.259
20.3
70
0.311
8.9
38
The new characteristics may be plotted (or added to the original graphs). The system curve
might also need be extended in length (but is still the same function of H vs Q). Note that you
must plot a new efficiency graph.
Hydraulics 2
A4-24
David Apsley
50.0
45.0
40.0
pump (original)
system
pump (new)
35.0
H (m)
30.0
25.0
20.0
15.0
10.0
5.0
0.0
0
0.05
0.1
0.15
0.2
0.25
0.3
Q (m3 s-1)
1.0
0.8
h
0.6
0.4
0.2
0.0
0.00
0.05
0.10
0.15
3
0.20
0.25
0.30
-1
Q (m s )
The new duty point is:
Q = 0.158 m3 s–1
H = 35.77 m
η = 0.735
Hydraulics 2
A4-25
David Apsley
The new input power is
ρgQH
powerin 
η

1000  9.81  0.158  35.77
0.735
 75430 W
Answer: discharge = 0.158 m3 s–1; input power = 75.4 kW.
Hydraulics 2
A4-26
David Apsley
Q8.
(a) The system head requirements are:
L V2
Q
where V 
H  Hs  λ
A
D 2g

4Q
πD 2
8λLQ 2
π 2 gD 5
Substituting the given parameters (Hs = 3.2 m, L = 21 m, D = 0.1 m, λ = 0.02) gives, with
heads in m and discharges in L s–1 (for convenience, to be consistent with the data in the
tables):

H  Hs 

8  0.02  21  Q 


π 2  9.81  0.15  1000 
(H in m, Q in L s–1)
H  3.2  0.003470Q 2
2
H  3.2 
(b) Plot graphs of H vs Q and η vs Q (both pumps and, for head, the system requirements on
the same graphs) to determine the duty points.
25
Pump A
20
Pump B
System
H (m)
15
10
5
0
0
5
10
15
20
25
30
35
40
Q (L s-1)
Hydraulics 2
A4-27
David Apsley
100
90
80
70
h (%)
60
50
40
Pump A
30
Pump B
20
10
0
0
5
10
15
20
25
30
35
40
Q (L s-1)
Duty points:
Pump A:
Q = 32.9 L s–1; H = 6.96 m; η = 49%
Pump B:
Q = 34.1 L s–1; H = 7.23 m; η = 67%
The pumps provide almost exactly the same discharge (and the question tells you that both
are adequate), but pump B is considerably more efficient than pump A and should be
selected.
Answer: select pump B on efficiency grounds.
(c)
powerout
ρgQH

powerin
powerin
ρgQH
1000  9.81  0.0341 7.23
powerin 

 3610 W
η
0.67
η

Answer: input power = 3610 W.
(d) Considering the total head between the sump and the pump inlet:
Hydraulics 2
A4-28
David Apsley
head at pump = head at start – losses along half the pipe
1
LV2
p
p
V2

z
 ( atm  0  0)  λ 2
ρg
2g
ρg
D 2g
The velocity in the pipeline is
4Q
4  0.0341
V

 4.342 m s 1
2
πD
π  0.12
Hence,
2
1
p atm
p
10.5 4.342 2
2 L V

 z  (1  λ
)
 0  3.2  (1  0.02 
)
 6.179 m
ρg
ρg
D 2g
0.1 2  9.81
The NPSH is then
p  pcav
95000
 6.179 
 3.505 m
ρg
1000  9.81
This is well above zero and unlikely to cause cavitation.
Answer: net positive suction head = 3.50 m; no cavitation.
Hydraulics 2
A4-29
David Apsley
Q9.
(a) The system head requirements are:
L V2
Q
where V 
H  Hs  λ
A
D 2g

4Q
πD 2
8λLQ 2
π 2 gD 5
Substituting the given parameters (Hs = 15 m, L = 800 m, D = 0.4 m, λ = 0.02) gives, with
heads in m and discharges in m3 s–1:
8  0.02  800 2
H  15  2
Q
π  9.81  0.4 5

(H in m, Q in m3 s–1)
H  15  129.1Q 2

H  Hs 
(b) Plot graphs of H vs Q and η vs Q to determine the duty point.
45
40
35
30
H (m)
25
H (pump)
20
H (system)
15
10
5
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Q (m3 s-1)
Hydraulics 2
A4-30
David Apsley
Efficiency
90
80
70
60
h (%)
50
40
Efficiency
30
20
10
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Q (m3 s-1)
Duty point:
Q = 0.256 m3 s–1; H = 23.5 m; η = 67%
Input power:
powerin 
ρgQH
η

1000  9.81  0.256  23.5
 88090 W
0.67
Time taken to pump 100 m3 is
100
 390.6 s
0.256
The energy used is then
88090  390.6  3.44  10 7 J
Answer: discharge = 0.256 m3 s–1; energy consumed = 34.4 MJ.
(c) From the efficiency graph the most efficient operation (η = 79.5%) is for Q = 0.16 m3 s–1,
H = 28.1 m.
Since the hydraulic scaling laws give
2
Q
N
H
 N 
and



0.16 1400
28.1  1400 
we have, eliminating the speed ratio,
H
 Q 


28.1  0.16 
Hydraulics 2
2
A4-31
David Apsley
or
H  1098Q 2
(H in m, Q in m3 s–1)
Answer: Q = 0.16 m3 s–1; H = 28.1 m; H  1098Q 2 (H in m, Q in m3 s–1).
(d) Since efficiency is unchanged by hydraulic scaling, the scaling line from part (c) will
always indicate the maximum-efficiency conditions as the rotational speed changes. This line
is plotted below. It will form the duty point when it meets the operating requirements (i.e. the
system curve) at a rotational speed such that
Q = 0.124 m3 s–1
H = 17.0 m
45
40
35
30
H (m)
25
H (pump)
20
H (system)
Scaling line
15
10
5
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Q (m3 s-1)
*** Alternative method
In this instance the intersection point can also be found analytically as the intersection of two
quadratics:
H  15  129.1Q 2 (system curve)
H  1098Q 2 (scaling line)
Equating gives (in metre-second units throughout):
1098Q 2  15  129.1Q 2

968.9Q 2  15

Q  0.124 (as above)
*** End of alternative method
The speed ratio is
Hydraulics 2
A4-32
David Apsley

N
0.124

 0.775
1400 0.16
N  0.775 1400  1085 rpm
The power consumption is (noting that efficiency is unchanged by scaling):
ρgQH
1000  9.81  0.124  17.0
powerin 

 26010 W
η
0.795
Time taken to pump 100 m3 is
100
 806.5 s
0.124
The energy used is then
26010  806.5  2.10  10 7 J
Answer: operating speed = 1085 rpm; energy consumed = 21.0 MJ.
Hydraulics 2
A4-33
David Apsley
Q10.
The pump and system characteristics are initially given in different units. Convert them both
to metres of water.
The pump characteristic at N1 = 2950 rpm is then
(H in m of water, Q in m3 s–1)
H1  0.24(1.0  Q1 )
The system characteristic in the same units is
18600
H system 
Q1.75  1.896Q1.75
1000  9.81
(*)
Pump characteristics at other speeds can be determined using the hydraulic scaling laws:
2
Q 
H2  N2 
   2 
 
H 1  N1 
 Q1 
2
(**)
From the system characteristic, the duty point at the higher speed is
Q2  0.28 m 3 s 1  H 2  1.896  0.281.75  0.2043 m
Substituting in (**),
0.2043  0.28 

 
H1
 Q1 
2
or
H1  2.606Q12
(***)
The scaled duty point at the lower speed may be found by solving (*) and (**)
simultaneously – either graphically or, here, by solving a quadratic:
2.606Q12  0.24(1.0  Q1 )
0.30
whence
3 1
system
Q1  0.2609 m s
Then
0.28

0.2609
 1.073
Hence,
duty point
0.20
H (m)
N 2 Q2

N1 Q1
characteristic
pump characteristic
at N2
0.25
N 2  1.073N1
scaled
duty point
0.15
0.10
 1.073  2950  3170 rpm
0.05
Answer: estimated fan speed = 3170 rpm.
0.00
0.0
0.1
0.2
0.3
0.4
Q (m3 s-1)
Hydraulics 2
A4-34
David Apsley
0.5
Q11.
(a)
Total head = piezometric head + velocity head
Need velocities at suction (inlet) and discharge (outlet):
Q
4Q
4  0.025
Vin 


 1.415 m s 1
2
Ain
πDin
π  0.15 2
Q
4Q
4  0.025
Vout 


 3.183 m s 1
2
2
Aout
πDout
π  0.1
Hence,
p *in Vin2
1.415 2
H in 

 4 
 3.898 m
ρg
2g
2  9.81
2
p *out Vout
3.1832
H out 

 12 
 12.52 m
ρg
2g
2  9.81
Hence the difference in total head across the pump is
H out  H in  16.42 m
Answer: head difference = 16.4 m.
(b) Overall efficiency,
ρgQ( H out  H in )
η
Pin

1000  9.81 0.025  (12.52  3.898)
6500
 0.6195
Answer: η = 62%.
(c)
Vr 
Q
A

0.025
 1.667 m s 1
0.015
Answer: Vr = 1.67 m s–1.
(d)
Ideal whirl velocity (i.e. assuming leaves parallel to blades) is comprised of forward motion
Rω and backward component Vr / tan β. The actual whirl velocity is 80% of this. Hence,
Vt  0.8( Rω  Vr / tan β)
But,
R  D / 2  0.25 / 2  0.125 m
ω  N  2π / 60  1400  2π / 60  146.6 rad s 1
Hence,
Vt  0.8  (0.125  146.6  1.667 / tan 20)  11.00 m s 1
Answer: Vt = 11.0 m s–1.
Hydraulics 2
A4-35
David Apsley
(e)
Manometric efficiency 
piezometric head
Euler head

(h *out h *in )  g
Vt Rω

9.81  (12  (4))
11.00  0.125  146.6
 0.7787
Answer: manometric efficiency = 78%.
Hydraulics 2
A4-36
David Apsley
Q12.
(a) Total available head is
H  H gross  H lost  300  20  280 m
Specific speed:
NP1 / 2
Ns  5/ 4
H
Inverting for the power (per wheel):
2
N 
P   s  H 5/ 2
 N 
2
 50 
5/ 2

  280
 400 
 20500 kW  20.50 MW
Answer: output power per wheel = 20.5 MW.
(b) Number of wheels required:
Total power
60

Power per wheel 20.50
 2.9
Answer: 3 wheels required.
(c) Jet velocity:
v  cv 2 gH
 0.97  2  9.81 280  71.90 m s 1
Bucket velocity:
u  0.46  v  0.46  71.90  33.07 m s 1
Answer: jet velocity = 71.9 m s–1; bucket velocity = 33.1 m s–1.
(d) Use u = Rω.
2π  N
ω
 41.89 rad s 1
60
u
33.07
R

 0.7894 m

ω
41.89

D  2R  2  0.7894  1.579 m
Answer: wheel diameter = 1.58 m.
(e)
Power per jet 
power per wheel
number of jets

20.5  10 6
6
 3.417  10 6 W
Answer: power per jet = 3.42 MW.
Hydraulics 2
A4-37
David Apsley
(f) Inverting P = ηρgQH:
P
3.417  10 6
Q

ηρgH
0.8  1000  9.81  280
 1.555 m 3 s 1
Answer: quantity of flow per jet = 1.55 m3 s–1.
(g) Since
Q  vA  v
πD 2
4
the jet diameter is
4Q
D

πv
4  1.555
π  71.90
 0.1659 m
Answer: jet diameter = 0.166 m.
(h) Hydraulic efficiency is
ρQ(v  u )u (1  k cos θ)
v2
 (1  0.46)  0.46  (1  0.85 cos 165) 
ρgQH
gH
 0.4523  2  0.97 2
 0.851
Answer: hydraulic efficiency = 85%.
Hydraulics 2
A4-38
David Apsley