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