345 S 02 P 2

 

CEE 345 Spring 2002

Problem set #2 Solutions

Problem 8.4

A pump has the character characterist istics ics given given in Fig. 8-5. 8-5. What discha discharg rgee and head will will be produced at maximum efficienc ciency y 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   H   =  0 .75 Q  = C Q nD3 + 0.64 30 s

·

g

· (0.5 m)

3

=  2 .40 m3 /s

  0.75 (30 s 1 )2 (0.5 m)2

  C  H n2 D2 ∆h  =

1

−

−

·

=

9.81 m/s2

·

 

=  17 .2 m.

P  = C P ρ D  ρ D5 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:

C   H  H   =

  ∆ H g

 D2 n2

Since C   H  H  will be the same for the maximum head condition, then ∆ H   n

 ∝  ∝

2

or   H 1500 1500   =   H 1000 1000

·

  1500 1000

2

=  102 ft 2.25  =  229 .5 ft.

·

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:

n  =

 1500 rpm   =  25 rps 60 s/min

ns   =

√ 

 n Q

(gh)

3 4

  25 s

−

=

1

·

√ 

12 cfs

(32.2 ft/s2 25 ft)

·

3 4

=  0 .57

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 show n in the figure. What type of pump would you use for this operati operation on if the impeller speed is to be 600 rpm? Solution:

  L  V 2 =  18 m h  = ∆ z + 1.5 +   f   D 2g





ns   =

√ 

 n Q

(gh)

3 4

=

10 s

1

−

·

 

(9.81 m/s

− 15 m +

1.0 m3 /s

2

· 3.14 m)

3 4



1.5 + 0.01

·

20 m 1m



(1.27 m/s)2

 · 2 · 32.2 m/s

2

  =  3 .14 m

=  0 .76

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:  p1 γ 

+

V 12

2g

+  z1  + h p   =

  p2 γ 

+

V 22

2g

+  z2  +

h L

 2

 

CEE 345 Spring 2002

Problem set #2 Solutions

or 2

h p   =  z 2

 − z   = 1





1 +   f   L + k e  + k b   Q 2 2g A  D

 

(1)

Assuming  L  =  62 .4 m,   f   =  0 .014,  k b   =  0 .35 and  k e   =  0 .1. C Q   =

 Q nD3

so we can get C   H  H  from Fig. 8-5. h p   =

2 2   C   H  H n  D

(2)

g

Going through this calculations with di ff erent erent values of  Q  Q  gives us the following: Q m3 / s 0.1 0.15 0.2 0.25 0.3 0.35

C Q   C  H    h p  (1)   h p  (2) m m 0.193 2. 2.05 1.7 3.5 0.289 1.7 1.95 2.9 0.385 1.55 2.3 2.65 0.482 1.25 2.76 2.14 0.578 0.95 3.31 1.62 0.675 0.55 3.96 0.94

Then 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 4

Pump curve

     ] 3     m      [      p

        h

2 1 0

0

Operating point

0.1

0.2 s] Q  [m 3 / 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 60 s/min

  =  11 .5 rps

Assume temperature of 10 C, ◦

Vapor pressure:   p   =  1 .2 kPa v

so

  p

v

γ 

=

  1.2 kPa   =  0 .12 m. 9.81 kN

Assume atmospheric pressure head of 10.3 m. Negl Neglectin ecting g head loss and velocit velocity y head, the gauge pressure head on the suction side of the impell impellor or will be approx approximate imately ly 1 m. Then NPSH  =  10 .3 m + 1 m 0.12 m  =  11 .2 m n ss   =

 

(g

nQ

−

1 2

× NPSH)

  11.5 s

3 4

=

1

3

· (0.21 m /s) (9.81 m/s · 11.2 m) −

1 2

3 4

2

=  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 diame ter steel pipe. When both are operating, operating, estimate the time to fill the tank from the 150-ft level level to to the 200-ft level. level. Estimate Estimate the maximum pressure in the pipe during the filling filli ng phase. Where will this maximum pressure 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:  p1 γ 

+

V 12

2g

+  z1  + h p   =

  p2 γ 

+

V 22

2g

+  z2  +

 4

h L

 

CEE 345 Spring 2002

Problem set #2 Solutions

  L  V 2 0 + 0 + 95 ft + h p   =  0 + 0 +  z2  + K e  + 2 K b  + K  E  +   f   D 2g





300 ft 95 ft + 0.1 + 2 0.2 + 1.0 + 0.014 1.5 ft



(Fig. 5-5) and   f   =   0.013 (Fig. Assuming   K e   =   0.1,   K b   =   0.2,   K  E    =   1.0,   k s / D   =   0.0001 (Fig. 5-4). Then h p   =   z2

 −



·

 − 95 ft + 4.10 ·

=   z2

·

Q2

 · 2g A

2

Q2

2 32.2 ft/s2

·

2

2

5

 − 95 ft + 0.0204 s /ft · Q

=   z2

2

 π

·  · (1.5 ft)

4

16

The performance performance curve for the two pumps in series is given below below. The initia initiall discharg dischargee 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

·

Plot this on the same graph and find the intersection intersection Qi   =  25 .3 cfs.

300 250 Performance Curve

200      ]     t      f 150      [      p         h

100

System curve

50 0

0

 

5

 

10

 

  15 Q [cfs]

20

 

25

 

30

To calculate the time to fill the reservoir consider increments of filling in 5 ft steps.

5

 

CEE 345 Spring 2002

Problem set #2 Solutions  z2    z¯2   Q¯   ∆V  V    ∆t  [ft] [ft] [cfs] [ft3 ] [s] 150 152.5 25.2 25133 997 155 157.5 25 25133 1007 160 162.5 24.7 25133 1018 165 167.5 24.4 25133 1028 170 172.5 24.2 25133 1039 175 177.5 23.9 25133 1051 180 182.5 23.6 25133 1065 185 187.5 23.3 25133 1079 190 192.5 23 25133 1093 195 197.5 22.7 25133 1108 200 ∆t  =  =   10485 s

−



So the total time will be   t   =

hours 55 minute minutes. s. The discha discharge rge   Q ∆t   =   10485 s or 2 hours



was obtained by solving the system equation with with the performance curve as done for obtaining Q i . Check   f : V i   =

  Qi  A

=

  25.3 cfs 1.767 ft2

  =  14 .34 ft/s

for the initial values and  V   f f   =  12 .82 ft/s in the final value in our table. So Rei   =

  V i D ν

=

  14.32 ft/s 1.5 ft   =  1 .79 106 5 2 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 level is reached in the tank. Write the energy equation equation from the maximum pressure pressure point to the water surface in the reservoir.  pmax γ 

+

  V 2

2g

+  z p   =

  pr 

+

γ 

V r 2

2g

  Q2

+  zr  +



h L

 280 ft   Q2   +  z p   =  z r  + K  E   +   f  + 2g A2  D γ  2g A2

 pmax





or





 pmax   =   γ   zr   z p  +

 −

−

 280 ft   Q2 1 + 1.0 +   f  2g A2  D

 

3

=   62.4 lbs/ft =   7252 lbs/ft

· 2

 −   200 ft

·

  1 ft

12 in

280 ft (22.7 ft3 /s)2 90 ft + 0.013 1.5 ft 2 32.2 ft/s2 (1.767 ft2 )2

·

·

2

·



=  50 .4 psi

Consider parallel pump installation. Consider installation. The performance curve for the two pumps would be as shown below. Solving the initial system equation with this performance curve yields Qi   =  36 .4 cfs. 150 Performance Curve

     ]     t100      f      [

System curve

     p

        h

50

0

0

10

20

  30 Q [cfs]

7

40

 

50 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 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 two gate gate valv valves es are open open whe when n pum pumpin ping. g. Whe When n the water 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 water surface surface of reservoir A  to the water surface in reservoir  B :  p1 γ 

+

V 12

2g

+  z1  + h p   =

  p2 γ 

V 22

+

2g

+  z2  +

h L

 

  L  V 2 0 + 0 + 0 + h p   =  0 + 0 + 30 ft + K e  + K  E   + 2K b  + 2 K V  V   +   f   D 2g



where   K e   =   0.5,   K  E   =   1.0,   K b   =   0.35 and   K V  (Table 5-3). Also k s / D   =   0.00015 V    =   0.20 (Table (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



·

  Q2

=   30 ft + 139.9



·

  =  30 ft + 139.9 2

2g A 2

=   30 ft + 3.52 s /ft =   30 ft + 1.75 10

·

2

5

5

−

Q =  30 ft +

·

ft/(gpm)2 Q2

·

 

Q2

2 32.2 ft/s2

·

  3.52 s2 /ft5

2

 π

·  · (1 ft)

4

16

2

Q2

(448.8 gpm/cfs)

Plotting the above equation (system curve) on the performance curve for problem 8-12 yields a discharge of 1500 gpm or 3.34 cfs   3.34 ft3 /s V   = =   =  4 .26 ft/s 0.785 ft2  A  Q

Re  =

 VD ν

=

  4.26 ft/s 1 ft 1.2 10

·

·

5

2

ft /s

−

  =  3 .5 105

·

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 curve as for the solution to Prob. Prob. 8-33: h p   =  30 ft + 2.14 10

·

5

−

ft/(gpm)2 Q2 .

·

The parallel pump performance performance curve is given given below: Plotting Plotting the system curve on the performance curve yields a solution of  Q  Q  =  1650 gpm or 3 .67 cfs 120 System curve, 12 pipe 

100 Performance Curve 80      ]     t      f      [ 60      p         h

40 System curve, 18 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 1.5 ft 2 32.2 ft/s2 16 (1.5 ft)4



·



·

2

·

=   30 ft + 0.503 s2 /ft5 Q2 =  30 ft + 2.50 10

·

·

 π

6

−

·  · ft/(gpm) · Q 2

2

Plotting the above equation on the graph of solution for problem 8-34 yields Q  =  3300 gpm or 7.35 cfs.

10