The pump is driven by a 75 kw motor at 1475 rpm, the impeller dia is 17-1/2". the suction manifold is 8' and the discharge manifold is 6". We are circulating a tank with a SUCTION head of 16 ft of sea water and the pump pulls FLA 133 Amps with the discharge butterfly valve 40% open and the discharge pressure is 100psi. Volts 400v - 50 Hz. What is correct hp to operate pump this size. Gpm on data sheet is 900 gpm.
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Overload pump motor
- Thread starter chief
- Start date
Suction manifold 8 "?
What's the circulation flowrate? Qc
What's the pump efficiency supposed to be at the circulation flowrate? Should be as shown on the power curve. Eff_at_Qc =
"Gpm on data sheet is 900." Is that the BEP flowrate?
Assuming density of sea water as 63.65 lbs/ft3, HP input to the pump shaft at this condition should be approx equal to
HP ~= 63.65 x (226-16) x 449 x flow_USgpm / 550 / Eff_at_Qc
Going the Big Inch!![[worm]](/data/assets/smilies/worm.gif "[worm] [worm]")
What's the circulation flowrate? Qc
What's the pump efficiency supposed to be at the circulation flowrate? Should be as shown on the power curve. Eff_at_Qc =
"Gpm on data sheet is 900." Is that the BEP flowrate?
Assuming density of sea water as 63.65 lbs/ft3, HP input to the pump shaft at this condition should be approx equal to
HP ~= 63.65 x (226-16) x 449 x flow_USgpm / 550 / Eff_at_Qc
Going the Big Inch!
I am assuming that the flow given is Imperial and not US gallons - and the "suction" head is positive ie, above the pump inlet.
The motor power is in kW, therefore use the following to calculate the water power - this does not account for the pump effiency.
kW = Q (l/sec) x H (metres) x SG / 102
SG sea water = assumed 1.02
Discharge = 100PSI x 2.31 / 1.02 / 3.28 = 69 metres
Inlet = 16ft /3.28 x 1.02 = 5 metres
flow = 900 IGPM / 13.2 = 68 l/sec
Therefore:
68 x 64 x 1.02 / 102 = 43.5 kW. This does not account for the pump efficiency which will increase the absorbed power.
You need to meassure the flow rate accurately - not rely on the data sheet.
However, if the pump is only 60% efficient at the flow and head at which it is operating you will use all the available motor ie, water power 43.3 / pump eff. 0.6 = 73 kW
Naresuan University
Phitsanulok
Thailand
The motor power is in kW, therefore use the following to calculate the water power - this does not account for the pump effiency.
kW = Q (l/sec) x H (metres) x SG / 102
SG sea water = assumed 1.02
Discharge = 100PSI x 2.31 / 1.02 / 3.28 = 69 metres
Inlet = 16ft /3.28 x 1.02 = 5 metres
flow = 900 IGPM / 13.2 = 68 l/sec
Therefore:
68 x 64 x 1.02 / 102 = 43.5 kW. This does not account for the pump efficiency which will increase the absorbed power.
You need to meassure the flow rate accurately - not rely on the data sheet.
However, if the pump is only 60% efficient at the flow and head at which it is operating you will use all the available motor ie, water power 43.3 / pump eff. 0.6 = 73 kW
Naresuan University
Phitsanulok
Thailand
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