Liquid water at 180°C and 1002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ?kg?1 and a specific volume of 1.128 cm3? g ?1.
(a) What is its enthalpy?
(b) The water is brought to the vapor state at 300°C and 1500 kPa, where its internal energy is 2784.4 kJ? kg?1 and its specific volume is 169.7 cm3. g?1. Calculate ?U and ?H for the process.
a. By definition, H = U + PV, so H = 762.0 kJ/kg + 1002.7 kPa? 0.001128 m3/kg = 763.1 kJ/kg.
(Note that we converted the specific volume to m3 per kg, and that Pa? m3 = N/m2? m3= N? m = J,
so kPa? m3= kJ).
b. ?U = 2784.4 ? 762 = 2022.4 kJ/kg. The enthalpy of the vapor is
H = 2784.4 kJ/kg + 1500 kPa? 0.1697 m3/kg = 3039.0 kJ/kg. So,
?H = 3039.0 – 763.1 = 2275.9 kJ/kg. ?H is significantly larger than ?U because of the large increase in specific volume during the process.
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