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kw=gph x delta T / 410


How do you work this out???

2007-11-07 10:20:09 · 1 answers · asked by MillK J 2 in Science & Mathematics Engineering

1 answers

The Boiler's Horsepower = (pounds per hour of water entering the boiler)*(h1 - h2)/33,475

Where h2 = enthalpy of the water entering the boiler
and h1 = enthalpy of the steam leaving the boiler.

Then 1 horsepower = 0.7457 KW, so multiply the Horsepower answer by 0.7457 to get the KW rating of the
boiler.

To get your other equation you have to assume that the boiler doesn't form any steam.
The basic equation is Q = M*cp*(T2-T1)
where:
Q = BTU/HR
cp=specific heat of water = 1.0
M=weight of water entering the boiler, Pounds per hour
T2=Temp. of water leaving
T1=Temp of water entering.
1000 BTU per hour = 0.2928104 KW
1 gallon = 62.4/7.481 = 8.341 pounds

Q = GPH*8.341*1.0*(T2-T1) = BTU/HR
KW = 0.29291*BTU/HR/1000

KW = 8.341*.29281*GPH*(T2-T1)/1000
KW = .00244*GPH*(T2-T1)
KW = GPH*(T2-T1)/410

2007-11-07 10:48:39 · answer #1 · answered by gatorbait 7 · 0 0

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