How many calories are released when 1 ton of saturated steam at 0.1 MPa and 120 degrees Celsius is cooled to condensate at 40 degrees Celsius?
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How many calories are released when 1 ton of saturated steam at 0.1 MPa and 120 degrees Celsius is cooled to condensate at 40 degrees Celsius?
It is approximately 132,820 KJ or 31,730 kcal.
1000X2202.6+1000X4.211X(120-40)=2539480kJ≈607531kCal
Algorithm using enthalpy values: The enthalpy of steam at 0.1 MPaA and 120°C is 2716.680 kJ/kg, while the enthalpy of water at 0.1 MPaA and 40°C is 167.623 kJ/kg. Therefore, (2716.608 – 167.623) * 1000 / 4.1868 = 608814.6 kcal. Source for steam enthalpy values: http://steam.dreamworld.ltd
The specific heat of water vapor is different from that of water, so it cannot be simply subtracted
Algorithm using enthalpy values: The enthalpy of steam at 0.1 MPaA and 120°C is 2716.680 kJ/kg, while the enthalpy of water at 0.1 MPaA and 40°C is 167.623 kJ/kg. Therefore, (2716.608 – 167.623) * 1000 / 4.1868 = 608814.6 kcal. The source for the steam enthalpy values is http://steam.dreamworld.ltd; this is correct
Sensible heat indeed has little impact on the calculation results. My question is: if the temperature is above 100 degrees and it’s water vapor, isn’t the specific heat capacity of water vapor 1.8 kJ/kg°C? The specific heat capacity of condensed water is 4.2 kJ/kg°C. Shouldn’t the specific heat capacity of water vapor be used for temperatures between 120 degrees and 100 degrees? Maybe I’m misunderstanding something.