Given that the internal energy of water at 28 bar pressure is 988 kj kg–1 and that the specific volume of water at this pressure is 0.121 × 10–2 m3 kg–1, calculate the specific enthalpy of water at 55 bar pressure.

1184 kJ/kg

Explanation:

Given:

water pressure P= 28 bar

internal energy U= 988 kJ/kg

specific volume of water v= 0.121×10^-2 m^3/kg

Now from steam table at 28 bar pressure we can write

therefore at saturated liquid we have specific enthalpy at 55 bar pressure.

that the specific enthalpy h =  h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)

h= 1184 kJ/kg

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