Question

At standard atmospheric pressure, saturated steam has a specific volume of 26.80 ft3 /lb. If the enthalpy of that same vapor is 1150.5 Btu/lb, calculate the internal energy of the steam in Btu/lb.

Answer 1

$26.80 \frac{ft^3}{lb}=1.673 \frac{m^3}{kg}$

$1150.5 \frac{Btu}{lb}=2676.063 \frac{kJ}{kg}$

So,

$H=U+pV\to \frac{H}{m}=\frac{U}{m}+p\frac{V}{m}\to h=u+pv$

$u=h-pv=2676.063\cdot10^3-101325\cdot 1.673=2506.546 \frac{kJ}{kg}$

$2506.546 \frac{kJ}{kg}=1077.62$ $\frac{Btu}{lb}$

$u=1077.62$ $\frac{Btu}{lb}$ Answer