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cryotechnic (Chemical)
21 May 08 14:06
I have the following situation:
3 coolingwaterpumps, 3 different coolingwater users with gascoolers, coolingwaterpressure reading, coolingwaterflowreading.
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
Thank you in advance,
Cryotechnic "Math is the ruler of your potential succes...."
BigInch (Petroleum)
21 May 08 14:28
Brake Power = motor_efficiency should be pump_efficiency
(read pump's efficiency at its flowrate off the pump curve) 字串6
Power Input to Motor = Brake Power/Motor Efficiency
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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain
Artisi (Mechanical)
21 May 08 20:47
The easiest formula to use in metric units is:
l/sec x h(m)x S.G. / 102 x pump% = Kw (pump shaft)
25362 (Chemical)
22 May 08 11:50
(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)
cryotechnic (Chemical)
22 May 08 14:14
Thanks, 字串5
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...."
BigInch (Petroleum)
22 May 08 18:14
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
25362 (Chemical)
23 May 08 0:40
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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