1.

Steam at 140 bar has an enthalpy of 3001.9 kJ/kg, find the temperature, the specific volume and the internal energy.

Answer»

Pressure of steam, p = 140 bar 

Enthalpy of steam, h = 3001.9 kJ/kg 

(i) Temperature : 

At 140 bar, hg = 2642.4 kJ, which is less than the actual enthalpy of 3001.9 kJ/kg, and hence the steam is superheated. 

From superheat tables at 140 bar, h = 3001.9 kJ/kg at a temperature of 400°C

(ii) The specific volume, 

v = 0.01722 m3/kg. 

∴ The internal energy (specific),

u = h – pv = 3001.9 – \(\cfrac{140\times10^5\times0.01722}{10^3}\) 

= 2760.82 kJ/kg.



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