Question

A rigid vessel of volume 0.86 m3 contains 1 kg of steam at a pressure of 2 bar. Evaluate the specific volume, temperature, dryness fraction, internal energy, enthalpy, and entropy of steam.

Answer 1

Specific volume=$\frac{volume}{mass}=\frac{0.86}{1}=0.86m^3/kg$

Therefore, specicifc volume $0.885m^3$ at 2 bar temperature

This is wet steam therefore saturation temperature is$120\degree$

dryness fraction=$v=v_f+x(v_g-v_f)$

Therefore $x=\frac{v-v_f}{v_g-v_f}=\frac{0.86-0.001061}{0.885-0.001061}=0.97172$

internal energy$(u)=h-pv=2644-200\times0.86=2472kj/kg$

Enthalpy $h=h_f+xh_fg=504.7+0.97172\times2201.6=2644KJ/Kg$

Entropy $s=s_f+xs_fg=1.5301+0.97172\times5.5967=6.9685KJ/Kg-k$