Thermo problem - obtaining internal energy value
Thermo problem - obtaining internal energy value
(OP)
I don't know if this is the right place to post. I came across this problem for the FE exam and was hoping someone could help me out.
A cylinder and piston arrangement contains saturated water vapor at 110 C. Vapor is compressed in reversible adiabatic process until the pressure is 1.6Mpa. What is the work done by the system.
Sol'n states using the first law of thermo W=deltaU
So I see how they obtained u1=2518.1kJ/kg at 110 C using the steam table
But I don't see how they obtained u2=2950.1kJ/kg.
They say that at state 2, T2 is also 400C
A cylinder and piston arrangement contains saturated water vapor at 110 C. Vapor is compressed in reversible adiabatic process until the pressure is 1.6Mpa. What is the work done by the system.
Sol'n states using the first law of thermo W=deltaU
So I see how they obtained u1=2518.1kJ/kg at 110 C using the steam table
But I don't see how they obtained u2=2950.1kJ/kg.
They say that at state 2, T2 is also 400C





RE: Thermo problem - obtaining internal energy value
If there was heating, the vapor should have been heated up to a superheated state. If you look at the steam table for superheated water and interpolate, you can find that the internal energy should be around 2950.1kJ/kg
RE: Thermo problem - obtaining internal energy value
The entropy of the saturated vapour at 110°C is 7.239 KJ/kg/K.
Water vapour at 400°C and 1.6 MPa also has an entropy of 7.239 KJ/kg/K.
The internal energy taken from the steam tables changes from 2518 kJ/Kg at 110°C, saturated vapour to 2950 kJ/Kg at 400°C and 1.6 MPa.
If the "system" is taken as being the water vapour in the piston, the work done by the system is negative = -(2950 - 2518) = -432 kJ/kg water vapour. The energy to heat the vapour comes from the work supplied to compress the vapour.
Hope that helps.
RE: Thermo problem - obtaining internal energy value
RE: Thermo problem - obtaining internal energy value
RE: Thermo problem - obtaining internal energy value
In this case the steam is confined to a cylinder and does not flow, so a diagram of internal energy vs. entropy would be needed, as no PV work except for that involved in the compression needs to be taken account of.
jnam82: You have to use the S and U data values at 400°C and 16 MPa to do the problem. (My steam tables actually go up to 100 MPa and 1000°C but you can get steam and water properties from the web e.g., http://webbook.nist.gov/chemistry/).