Answer to Question #105410 in Molecular Physics | Thermodynamics for Kule

Question

steam at pressure of 50 psia and a temperature of 281°F has an internal energy of 109.6 BTU/lb. at these conditions the specific volume of the steam is 8.518 ft^3/lb. determine the enthalpy of the steam in BTU/lb

Answer 1

Enthalpy

$$H=U+pV$$

$U=109.6 (BTU/lb); p=50 (psia); V=8.518 (ft^3/lb)$

$$H=109.6+\frac{50\cdot 8.518}{5.40395}=188.4 (BTU/lb)$$