Propane will be used to heat 5500 ml of water to run a turbine. The volume of th
ID: 3162614 • Letter: P
Question
Propane will be used to heat 5500 ml of water to run a turbine. The volume of the boiler is 6123 ml, how much propane will be used to heat the water from ambient conditions (room temperature and pressure) to the specified 758 kPa (gage) pressure? Also determine the state of the water (saturated mixture or superheated steam) as well as temperature, specific volume, internal energy, and entropy. You can neglect the effects of air in the system when it is started. Final Temperature of steam is 168.132 C.
Explanation / Answer
at 758 kPa G pressure, saturated water temperature is 173.402 oC
As given T= 168.132 oC => It is subcooled liquid
Heat needed Q = mw*Cpw*(168.132 -25) = 5500*10^-3*4184*(168.132 -25) =3293.75 kJ
Amount of propane needed = Heat needed / heat content of propane = 3293.75*10^3/(50.35 *10^6 J/kg) =0.0654kg
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