Basic Thermodynamics Sample Problem 2

Question

A rigid vessel contains 2 kg of steam at 5.25 MPa and 300°C. Determine its enthalpy in kJ using steam tables only.

Answer 1

A Table Of Properties Of Superheated Water Vapor gives specific enthalpy readings (at 4.0 MPa) of 

°C------------------kJ/kg

280---------------2901.8

300------------------h_4.0

320---------------3015.4

Then (320 - 300)/(320 - 280) = (3015.4 - h_4.0)/(3015.4 - 2901.8) or h_4.0 = 2958.6 kJ/kg.

For 6.0 MPa, the same table gives specific enthalpy readings at (6.0 MPa) of 

°C------------------kJ/kg

280---------------2804.2

300------------------h_6.0

320---------------2952.6

Then (320 - 300)/(320 - 280) = (2952.6 - h_6.0)/(2952.6 - 2804.2) or h_6.0 = 2878.4 kJ/kg.

Finally, interpolation of 

MPa--------------kJ/kg

4.0---------------2958.6

5.25----------------h_5.25

6.0---------------2878.4

gives (6 - 5.25)/(6 - 4) = (2878.4 - h_5.25)/(2878.4 - 2958.6) or h_5.25 = 2908.475 kJ/kg,

which goes to 5816.95 kJ for 2 kg of steam at 300°C and 5.25 MPa.

Note that an Online Enthalpy Calculator gives specific enthalpy as 2912.07 kJ/kg

(or 5824.14 kJ for 2 kg of steam) at 300°C and 5.25 MPa(Gauge).