Question

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.

Answer 1

Answer:

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

$U= U_{f}= 987.6 kJ/Kg$

$v_{f}=v=1.210\times 10^{-3}m^{^{3}}/kg$

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)

$h= 1154.5 + \frac{5}{10}\times(1213-1154)$

h= 1184 kJ/kg