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

1184 kJ/kg

Step-by-step 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* 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 + (5)/(10)*(1213-1154)`

h= 1184 kJ/kg