kw=gph x delta T / 410 How do you work this out???
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
