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# Pump power calculation

## Pump power calculation

(OP)
I have the following situation:

Can I use the formula as mentioned below for calculation of the pump power?
We don't have an accurate powerreading on the pumpmotors, but we want to have an idea of the pumppower when changing the flow, pressure and when we do other adjustments in the coolingwater system.

This is the formula:

Ph =(q*rho*g*h)/3.6*10^6
Pshaft = Ph/motorefficiency

unit's are in SI

Cryotechnic

"Math is the ruler of your potential succes...."

### RE: Pump power calculation

Brake Power = motor_efficiency should be pump_efficiency
(read pump's efficiency at its flowrate off the pump curve)
Power Input to Motor = Brake Power/Motor Efficiency

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

### RE: Pump power calculation

The easiest formula to use in metric units is:

l/sec x h(m)x S.G. / 102 x pump% = Kw (pump shaft)

### RE: Pump power calculation

(L/s)(m)(kg/L)(m/s2) = kg.m2/s3 = W

Q → L/s
H → m
ρ → kg/L
g → 9.8 m/s2

If you divide by 1000 and by the system's efficiency as a decimal, you get the equation given by Artisi (expressed in kW)

### RE: Pump power calculation

(OP)
Thanks,

in my formula
Ph =(q*rho*g*h)/3.6*10^6
Pshaft = Ph/motorefficiency

q = m3/h
rho = kg/m3
g = 9,81
h = m

"Math is the ruler of your potential succes...."

### RE: Pump power calculation

No it is not.  If it says so, its a mistake.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

### RE: Pump power calculation

The formula for liquid horsepower (Ph) expressed in kW, is OK.

Pshaft (aka brake horsepower) is wrong. The efficiency should be pump's effy. not motor's.

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