Thermo problem - obtaining internal energy value 1

[jnam82]

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

 

[dtse86]

Your question is a little confusing. If the only process done on the system is the adiabatic compression, then the temperature of the system should be lower than temperature unless there was heating to the system as well to get it up to 400 degrees C.

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

 

[Michael2006]

Since the process is adiabatic there is no transfer of heat from or to the vapour, and since the compression process is also reversible the entropy change is zero, since dS = dq/T in a reversible process and dq, the heat transferred, is zero.

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.

 

[jnam82]

Are you using the superheated table to obtain u2=2950kJ/kg? Because it only goes up to 1MPa. Or does superheated steam change very little after that pressure? If that's the case then entropy at 400C(7.4651kJ/Kg-k) does not equal the original entropy of 7.239Kj/KgK). I'm confused...

 

[CH5OH]

wouldn't it be easier to plot on molier diagram?

 

[Michael2006]

A Mollier (H-S) diagram would be useful if the vapour compression took place in a flowing system, i.e. with a continuous stream of saturated vapour being compressed in a machine and leaving at 1.6 MPa, since the enthalpy takes account of the PV work done by / on the flowing medium.

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/).