Volume of wet steam, v = 0.15 m3
Pressure of wet steam, p = 4 bar
Dryness fraction, x = 0.8
At 4 bar. From steam tables,
vg = 0.462 m3/kg, hf = 604.7 kJ/kg, hfg = 2133 kJ/kg
∴ Density = \(\cfrac{1}{xv_g}\) = \(\cfrac{1}{0.8\times0.462}\) = 2.7056 kg/m3
Mass of 0.15 m3 of steam
= 0.15 × 2.7056 = 0.4058 kg.
Total heat of 1 m3 of steam which has a mass of 2.7056 kg
= 2.7056 h (where h is the total heat of 1 kg of steam)
= 2.7056 (hf + xhfg)
= 2.7056(604.7 + 0.8 × 2133)
= 6252.9 kJ.
