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CLE1 - Prob 5

The document describes a problem calculating the power of a pump that moves oil through a pipe system. The oil is pumped from an open tank 20 meters above a pressurized tank through 122 meters of steel pipe with two elbows and a partially open valve. Given information about the oil properties, pipe dimensions, flow rate and system layout, the summary calculates the head loss and friction factor to determine the pump power required is 2.80785 kW.

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1K views2 pages

CLE1 - Prob 5

The document describes a problem calculating the power of a pump that moves oil through a pipe system. The oil is pumped from an open tank 20 meters above a pressurized tank through 122 meters of steel pipe with two elbows and a partially open valve. Given information about the oil properties, pipe dimensions, flow rate and system layout, the summary calculates the head loss and friction factor to determine the pump power required is 2.80785 kW.

Uploaded by

Drake Austria
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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PROBLEM 5:

Oil having a density of 833 kg/m 3 and a viscosity of 3.3 x 10 -3Pa.s is pumped from and open tank to a pressurized tank
held at 345 kPa gage. The oil is pumped from an inlet the side of the open tank through a line of commercial steel pipe
having an inside diameter of 0.07792 m at the rate of 3.494 x 10 -3 m3/s. The length of the straight pipe is 122 m, and
the pipe contains two elbows (90) and a globe valve half open. The level of the liquid in the open tank is 20 m above
the liquid level in the pressurized tank. The pump efficiency is 65% Calculate the kW power of the pump.
Given:
Lpipe = 122 m
Di = 0.07792 m [commercial steel type]
Q = 3.494 x 10-3 m3/s

( DL )

2 - 90 Standard Elbow Fittings

( DL )

Globe Valve half open

= 30

= 475

z = 20 m
65% pump efficiency
=
=

3.3 x 10 Pa s
833

kg
m3

P1 = 101.325 kPa
P2 = 101.325 + 345 kPa = 446.325 kPa
Find: Power (kW)
Solution:

N =

Dv
=

( 0.007792 m ) 0.732715082

m
kg
833 3
s
m

)(

3.3 x 10 Pa s
N , Pl> 4000,

B=

{ [(

A= 2.457 ln

L:

7
N

+ 0.27

16

) (
=

Turbulent Flow Regime, = 1

37530
14411.69747

16

) =4473164.191

16

1
0.9

37530
N

) =14411.69747

( D )

]} { [ (
= 2.457 ln

16

1
7
14411.697473

0.9

+0.27 ( 0.0006 )

Pipe = 122 m
90 Standard Elbow Fittings =

Globe valve half open =

= 161.3496 m

( DL )=( 30 )( 0.007792 m )=2.3376 m

( DL )=( 475 ) ( 0.007792m )=37.012 m

]}

=3.067762401 x 1019

m
=2070.708419
( DL )= 161.3496
0.007792 m

[( )

8
f F =2
N

12

1
12

( A+ B )

3
2

] [

8
=2
14411.69747

12

) + (3.067762401 x 10 + 4473164.191 )
19

3
2

1
12

=7.331237359 x 103

m
m 2
0.732715082
s2
s
L v2
J
=( 122 m)
+ 4 ( 7.331237359 x 103 ) ( 2070.7084193 )
=626.594
D 2 gc
kg
kg m
kg m
1
2 1
(1 )
2
Ns
Ns 2

W s = z

g
+4 f F
gc

626.594

J
1
x
=2.80785 kW
kg 0.65

(
(

9.81

)
)

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