The liquid is to be pumped from the underground storage vessel which is vented to atmosphere (assume 1 bar pressure) to a pressurised container supported some distance above ground level. The pump is...

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The liquid is to be pumped from the underground storage vessel which is vented to atmosphere (assume 1 bar pressure) to a pressurised container supported some distance above ground level. The pump is sited at ground level and must be capable of delivering 0.01 m3 r1 with a maximum velocity of 1.8 m s-1.
(a) Calculate the theoretical diameter of the pipe; the nominal pipe diameter chosen from the table; and the actual average velocity in the nominal pipe.


Answered Same DayDec 29, 2021

Answer To: The liquid is to be pumped from the underground storage vessel which is vented to atmosphere (assume...

Robert answered on Dec 29 2021
106 Votes
Discharge from pump Q = 0.01 m3s-1
Maximum velocity Vmax = 1.8 ms-1
Let us suppose that the dia
meter is d
Area of cross section A = πd2/4
So, we can write AVmax = Q => Vmaxπd
2/4 = Q
Solving this, we get d = 0.0841 m = 84.1 mm
To be in safe side, we provide a nominal diameter of 90 mm
Area of the nominal pipe An = π×0.09
2/4 = 0.0064 m2
Actual velocity V = Q/An = 0.01/0.0064 = 1.5719 ms-1
b) Reynold’s number
960 1.5719 0.09
1676.7
0.081
Vd
R


 
  
Total length L=(3+2+5+15+6+1)m = 32 m
Head loss due to friction
2 21.570.03 32
1.3433
2 2 0.09 9.8
19
1
Lf
fLV
h m m
dg
 
  
 
We have assumed the factor f= 0.03
c) Minor losses using the equivalent length method
Standard 900 bend =
2 2
1
1.571937 37 0.03
0.1398
2 2 9.81
L
fV
h m m
g
 
  


Entry to pipe =
2 2
2
1.571915 15 0.03
0.0567
2 2 9.81
L
fV
h m m
g
 
  


Exit from pipe =
2 2
3
1.571950 50 0.03
0.1889
2 2...
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