Evaluate total heat of steam:
10 kg of wet saturated steam at 15 bar pressure is superheated to the temperature of 290°C at constant pressure. Find heat needed and the total heat of steam. The dryness fraction of steam is 0.85.
Sol.: From steam table, we get following data:
Absolute pressure (P)
bar

Saturation
temperature (t) ºC

Specific enthalpy kJ/kg

Water (h_{f})

Latent heat (h_{fg})

15

198.3

844.6

1947

Total heat of 10 kg wet saturated steam
=10 × 2499.55 = 24995.5 kJ .......ANS
Total heat of 1 kg superheated steam can be given by Hsup = h_{f} + h_{f}_{g} + Cp(t_{sup}  t_{s}) kJ
= 844.6 + 1947 + 2.1 × (290  198.3) kJ
= 2984.17 kJ .......ANS
Total heat of 10 kg superheated steam =10 × 2984.17 = 29841.7 kJ .......ANS
= h_{f} + h_{f}_{g} + C_{p}(t_{sup}  t_{s})  (h_{f} + x_{h}_{fg})
= h_{f}_{g} + C_{p}(t_{sup}  t_{s})  x_{h}_{fg}
=1947 + 2.1 x (290  198.3)  0.85 x 1947 kJ = 484.62 kJ
Heat which is required to convert 10 kg wet saturated steam into 10 kg superheated steam
=10 × 484.62 = 4846.2 kJ .......ANS
Total heat of 1 kg wet saturated steam can be given by H_{wet} = h_{f } + x_{h}_{f}_{g} kJ
= 844.6 + 0.85 × 1947 kJ = 2499.55 kJ
Heat which is required to convert 1 kg wet saturated steam into 1 kg superheated steam
= H_{sup }  H_{wet},
here Hsup = enthalpy of 1 kg superheated steam = h_{f} + h_{f}_{g} + Cp(t_{sup}  t_{s}) kJ
Hwet = enthalpy of 1 kg wet saturated steam = h_{f} + x_{h}_{f}_{g} kJ
Heat which is required to convert 1 kg wet saturated steam into 1 kg superheated steam