Question

What is the quality and internal energy of a tank of water at 250°C with 0.02 m3 /kg.

Answer 1

Temperature of water = 250°C (saturated steam)

Specific volume v = 0.02 m3/kg

At 250 °C

Specific Volume of Saturated Steam vg = 0.0500866 m3/kg

Specific Volume of Saturated Water vf = 0.00125174 m3/kg

Specific volume balance

v = vf + x(vg - vf)

0.02 = 0.00125174 + x(0.0500866 - 0.00125174)

0.02 - 0.00125174 = x(0.04883486)

0.01874826 = x(0.04883486)

Steam quality x = 0.3839

specific internal energy of water Uf = 1080.7099 kJ / kg

specific internal energy of steam Ug = 2601.8709 kJ / kg

Internal energy balance

U = Uf + x(Ug - Uf)

= 1080.7099 + 0.3839*(2601.8709 - 1080.7099)

= 1664.6836 kJ/kg

All data have been taken from steam table