Pump head

  • Thread starter balotpinoy
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  • #1
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a pump capacity of 10 m^3/hr at 30 meters head. what does the 30 meters head mean?

thank you..
 

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  • #2
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The pump increases the pressure downstream, relative to the pressure upstream. The 30 meters head is the difference in pressure 'across' the pump (discharge pressure minus suction pressure). Pressure (force per area) can be expressed in height (meters) by considering the pressure at the base of a column of the pumped fluid 30 meters tall. To change the units of 'head' to units of 'pressure' you need to know the weight density of the fluid.
 
  • #3
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so can i use it in Bernoulli's equation as work given to the fluid?

- head x density x gravitational constant

is that right?
 
  • #4
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Right. [tex]V^2/2g[/tex] [tex] z [/tex] [tex] P/\rho g[/tex] all have units of length.

And if you multiply each by [tex]g \rho[/tex] you get units of pressure
 
Last edited:
  • #5
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no more deductions for pump efficiency?
 
  • #6
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kindly help me with this one..

i want to compute the actual volumetric flowrate in line 1 & 2. discharge for both lines are in different elevations. given data for me is these;

mono pump capacity= 10 m3/hr at 30 meters
line 1 from tee to discharge is a 2-inch pipe
line 1 discharge elevation= 12 meters
line 2 from tank bottom to discharge is 3-inch pipe
line 2 discahrge elevation=11.5 meters
fittings/valves are present along the line, pls bear with my drawing..

i've been trying to figure out how im gonna use the available pump data in my bernoulli's equation.

and another thing is the flowrate. can i directily solve for velocity of fluid at the pump discharge from the given flowrate..

thank you in advance..
 

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  • #7
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You need more than one data point for the pump. Typically you have a 'pump curve' which shows the head at various flowrates. As the flowrate is decreased towards zero, the head will rise to the 'shutoff head', and as flow is increased the head will drop off. Then you can plot the pump curve along with the system resistance curve (basically bernoulli solved for head loss). Where the two curves cross, that's the operating point.

Above describes the situation for a centrifugal pump. If you have a positive displacement pump, the flowrate is constant; all you need is bernoulli. But I don't think you have a PD pump, because then the 'at 30 meters head' doesn't make sense.
 
  • #8
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its a PD pump and that's the only data available for that pump since its old already.

so far here is what i did with my problem;

to solve for Q in lines 1 & 2, i first computed for Pressure at point B (pls refer to attached drawing) using bernoulli's equation in pipe A-B using 30 meters head as pump work and obtained my velocity at point B using the flowrate/area of pipe.

is it correct? and if so, how can i proceed to solve for flowrates in both lines?

thank you again..
 

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  • #9
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If it is a PD pump then you know the flowrate, right? The 30 meters head is misdirection - a red herring. All you need to figure out is the split between Line 1 and Line 2. What can you say about the pressure drops BD and BC? How do those pressure drops vary with flowrate?
 

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