CEE 345
Spring 2002
Problem set #2
Solutions
Problem 8.4
A pump has the characteristics given in Fig. 8-5. What discharge and head will be
produced at maximum efficiency if the pump size is 50 cm and the angular speed is 30 rps?
What power is required when pumping water under these conditions?
Solution:
At maximum efficiency, from Fig. 8-5,
C Q = 0.64,
C P = 0.60,
C H = 0.75
Q = C Q nD3 + 0.64 · 30 s−1 · (0.5 m)3 = 2.40 m3 /s
∆h =
C H n2 D2 0.75 · (30 s−1 )2 · (0.5 m)2
= 17.2 m.
=
g
9.81 m/s2
P = C P ρD5 n3 = 0.60 · 1000 kg/m3 · (0.5 m)5 · (30 s−1 )3 = 506 kW.
Problem 8.12
If the pump having the performance curve shown is operated at a speed of 1500 rpm,
what will be the maximum possible head developed?
Solution:
CH =
∆Hg
D2 n2
Since C H will be the same for the maximum head condition, then
!2
1500
2
∆H ∝ n or H1500 = H1000 ·
= 102 ft · 2.25 = 229.5 ft.
1000
Problem 8.19
What type of pump should be used to pump water at a rate of 12 cfs and under a head
of 25 ft? Assume N = 1500 rpm.
1
CEE 345
Spring 2002
Problem set #2
Solutions
Solution:
1500 rpm
= 25 rps
60 s/min
√
√
n Q
25 s−1 · 12 cfs
ns =
=
= 0.57
3
3
(gh) 4
(32.2 ft/s2 · 25 ft) 4
n=
Then from Fig. 8-15, n s < 0.60 so use mixed flow pump.
Problem 8.23
You want to pump water at a rate of 1.0 m3 /s from the lower to the upper reservoir
shown in the figure. What type of pump would you use for this operation if the impeller
speed is to be 600 rpm?
Solution:
!
L V2
20 m
(1.27 m/s)2
h = ∆z + 1.5 + f
= 18 m − 15 m + 1.5 + 0.01 ·
·
= 3.14 m
D 2g
1m
2 · 32.2 m/s2
p
√
10 s−1 · 1.0 m3 /s
n Q
=
= 0.76
ns =
3
3
(gh) 4
(9.81 m/s2 · 3.14 m) 4
From Fig. 8-15, use axial flow pump.
Problem 8.24
The pump used in the system shown has the characteristics given in Fig. 8-6, page 418.
What discharge will occur under the conditions shown, and what power is required?
Solution:
D = 35.6 cm = 0.356 m, n = 11.5 r/s.
Writing the energy equation from the reservoir surface to the center of the pipe at the outlet:
X
p2 V22
p1 V12
+
+ z1 + h p =
+
+ z2 +
hL
γ
2g
γ
2g
2
CEE 345
Spring 2002
Problem set #2
Solutions
or
Q2
L
h p = z2 − z1 = 1 + f + ke + kb
D
2gA2
(1)
Assuming L = 62.4 m, f = 0.014, kb = 0.35 and ke = 0.1.
Q
nD3
so we can get C H from Fig. 8-5.
CQ =
hp =
C H n2 D2
g
(2)
Going through this calculations with different values of Q gives us the following:
Q
m3 /s
0.1
0.15
0.2
0.25
0.3
0.35
CQ
CH
0.193 2.05
0.289 1.7
0.385 1.55
0.482 1.25
0.578 0.95
0.675 0.55
h p (1)
m
1.7
1.95
2.3
2.76
3.31
3.96
h p (2)
m
3.5
2.9
2.65
2.14
1.62
0.94
PSfrag replacementsThen plotting the system curve and the pump curve, we obtain the operating conditions:
Q = 0.22 m3 /s, Power: P = 6.7 kW (from Fig. 8-6).
5
h p [m]
4
Pump curve
3
2
1
0
0
Operating point
0.1
0.2
Q [m3 /s]
3
0.3
0.4
CEE 345
Spring 2002
Problem set #2
Solutions
Problem 8.27
What is the specific speed for the pump operating under the conditions given in Prob.
8-24? Is this a safe operation with respect to the susceptibility to cavitation?
Solution:
n=
690 rpm
= 11.5 rps
60 s/min
Assume temperature of 10◦ C,
Vapor pressure:
pv = 1.2 kPa so
pv
1.2 kPa
=
= 0.12 m.
γ
9.81 kN
Assume atmospheric pressure head of 10.3 m. Neglecting head loss and velocity head, the
gauge pressure head on the suction side of the impellor will be approximately 1 m.
Then NPSH = 10.3 m + 1 m − 0.12 m = 11.2 m
1
1
n ss =
nQ 2
3
(g × NPSH) 4
=
11.5 s−1 · (0.21 m3 /s) 2
3
(9.81 m/s2 · 11.2 m) 4
= 0.155
The n ss value of 0.155 is much less than the critical value of 0.494, therefore, the pump is
in the safe operating range.
Problem 8.32
Two pumps having the performance curve shown are operated in series in the 18-in.
diameter steel pipe. When both are operating, estimate the time to fill the tank from the
150-ft level to to the 200-ft level. Estimate the maximum pressure in the pipe during the
filling phase. Where will this maximum pressure occur? What would have been the initial
discharge if the pumps had been installed in parallel?
Solution:
First write the energy equation from the lower to upper reservoir:
X
p2 V22
p1 V12
+
+ z1 + h p =
+
+ z2 +
hL
γ
2g
γ
2g
4
CEE 345
Spring 2002
Problem set #2
Solutions
L V2
0 + 0 + 95 ft + h p = 0 + 0 + z2 + Ke + 2Kb + KE + f
D 2g
Assuming Ke = 0.1, Kb = 0.2, KE = 1.0, k s /D = 0.0001 (Fig. 5-5) and f = 0.013 (Fig.
5-4). Then
!
Q2
300 ft
·
h p = z2 − 95 ft + 0.1 + 2 · 0.2 + 1.0 + 0.014 ·
1.5 ft 2gA2
Q2
= z2 − 95 ft + 4.10 ·
π2
2 · 32.2 ft/s2 · 16
· (1.5 ft)4
= z2 − 95 ft + 0.0204 s2 /ft5 · Q2
The performance curve for the two pumps in series is given below. The initial discharge
will be obtained by solving the performance curve and the energy equation (for z 2 = 150 ft)
h p = 55 ft + 0.0204 s2 /ft5 · Q2
PSfrag replacements
Plot this on the same graph and find the intersection
Qi = 25.3 cfs.
300
250
Performance Curve
h p [ft]
200
150
100
System curve
50
0
20
30
15
25
Q [cfs]
To calculate the time to fill the reservoir consider increments of filling in 5 ft steps.
0
5
10
5
CEE 345
Spring 2002
Problem set #2
Solutions
z2
[ft]
150
z̄2
[ft]
Q̄
[cfs]
∆−
V
[ft3 ]
∆t
[s]
152.5
25.2
25133
997
157.5
25
25133
1007
162.5
24.7
25133
1018
167.5
24.4
25133
1028
172.5
24.2
25133
1039
177.5
23.9
25133
1051
182.5
23.6
25133
1065
187.5
23.3
25133
1079
192.5
23
25133
1093
197.5
22.7
25133
1108
155
160
165
170
175
180
185
190
195
200
P
∆t = 10485 s
P
So the total time will be t = ∆t = 10485 s or 2 hours 55 minutes. The discharge Q
was obtained by solving the system equation with with the performance curve as done for
obtaining Qi .
Check f :
Vi =
Qi
25.3 cfs
=
= 14.34 ft/s
A
1.767 ft2
for the initial values and V f = 12.82 ft/s in the final value in our table. So
Rei =
Vi D 14.32 ft/s · 1.5 ft
=
= 1.79 · 106
−5
2
ν
1.2 · 10 ft /s
and Re f = 1.60 · 106 . For either of this value we find our initial assumption of f = 0.013 to
be valid.
6
CEE 345
Spring 2002
Problem set #2
Solutions
The maximum pressure will occur immediately downstream of the pumps when the 200
foot level is reached in the tank. Write the energy equation from the maximum pressure
point to the water surface in the reservoir.
X
pr Vr2
pmax V 2
+
+ zp =
+
+ zr +
hL
γ
2g
γ
2g
!
pmax
Q2
280 ft Q2
+
+ z p = zr + K E + f
γ
2gA2
D
2gA2
or
!
!
280 ft Q2
= γ zr − z p + −1 + 1.0 + f
D
2gA2
!
(22.7 ft3 /s)2
280 ft
3
= 62.4 lbs/ft · 200 ft − 90 ft + 0.013 ·
1.5 ft 2 · 32.2 ft/s2 · (1.767 ft2 )2
!2
1 ft
= 7252 lbs/ft2 ·
= 50.4 psi
12 in
pmax
Consider parallel pump installation. The performance curve for the two pumps would
PSfrag replacements
be as shown below. Solving the initial system equation with this performance curve yields
Qi = 36.4 cfs.
150
Performance Curve
h p [ft]
100
System curve
50
0
0
10
20
30
Q [cfs]
7
40
50
60
CEE 345
Spring 2002
Problem set #2
Solutions
Problem 8.33
The pump of Prob. 8-12 is used to pump water from reservoir A to reservoir B. The
pump is installed in a 2-mi long, 12-in pipe joining the two reservoirs. There are two bends
in the pipe (r/D = 1.0), and two gate valves are open when pumping. When the water
surface elevation in reservoir B is 30 ft above the water surface in reservoir A at what rate
will water be pumped?
Solution:
Write the energy equation from the water surface of reservoir A to the water surface in
reservoir B:
X
p1 V12
p2 V22
+
+ z1 + h p =
+
+ z2 +
hL
γ
2g
γ
2g
L V2
0 + 0 + 0 + h p = 0 + 0 + 30 ft + Ke + KE + 2Kb + 2KV + f
D 2g
where Ke = 0.5, KE = 1.0, Kb = 0.35 and KV = 0.20 (Table 5-3). Also k s /D = 0.00015
(Fig. 5-5), assume f = 0.013.
!
2 mi · 5280 ft/mi V 2
h p = 30 ft + 2.6 + 0.013 ·
1 ft
2g
2
Q
Q2
=
30
ft
+
139.9
= 30 ft + 139.9
2
2gA2
2 · 32.2 ft/s2 · π · (1 ft)4
16
3.52 s2 /ft5
= 30 ft + 3.52 s2 /ft5 · Q2 = 30 ft +
Q2
(448.8 gpm/cfs)2
= 30 ft + 1.75 · 10−5 ft/(gpm)2 · Q2
Plotting the above equation (system curve) on the performance curve for problem 8-12
yields a discharge of 1500 gpm or 3.34 cfs
V=
Q 3.34 ft3 /s
=
= 4.26 ft/s
A
0.785 ft2
Re =
VD
4.26 ft/s · 1 ft
=
= 3.5 · 105
−5
2
ν
1.2 · 10 ft /s
giving f = 0.016. With this larger f the system equation becomes
h p = 30 ft + 2.14 · 10−5 ft/(gpm)2 · Q2 .
8
CEE 345
Spring 2002
Problem set #2
Solutions
Plotting this new system curve, etc. yields Q = 1450 gpm = 3.23 cfs.
Problem 8.34
Work Prob. 8-33 but have two pumps like that of Prob. 8-12 operating in parallel.
Solution:
Assume same system curve as for the solution to Prob. 8-33:
h p = 30 ft + 2.14 · 10−5 ft/(gpm)2 · Q2 .
PSfrag replacements
The parallel pump performance curve is given below: Plotting the system curve on the
performance curve yields a solution of Q = 1650 gpm or 3.67 cfs
120
System curve, 1200 pipe
100
Performance Curve
h p [ft]
80
60
40
System curve, 1800 pipe
20
0
0
1000
2000
Q [gpm]
9
3000
4000
CEE 345
Spring 2002
Problem set #2
Solutions
Problem 8.35
Work Prob. 8-33 but have two pumps like that of Prob. 8-12 operating in parallel and
have an 18-in pipe instead of a 12-in pipe.
Solution:
For this pipe k s /D = 0.0001, assume f = 0.014. Then the energy equation reduces to
!
2 mi · 5280 ft/mi
Q2
h p = 30 ft + 2.6 + 0.014 ·
π2
1.5 ft
· (1.5 ft)4
2 · 32.2 ft/s2 · 16
= 30 ft + 0.503 s2 /ft5 · Q2 = 30 ft + 2.50 · 10−6 ft/(gpm)2 · Q2
Plotting the above equation on the graph of solution for problem 8-34 yields Q = 3300 gpm
or 7.35 cfs.
10