# Homework Help: Thermodynamics- Calculating work done by turbine.

1. Oct 7, 2012

### supahtofu

Hello, everyone I would very much appreciate some help on this Intro to Thermo 2 problem if someone could give advice even a little would be helpful.

1. Air (for this example, an ideal gas) is being used to produce work in a turbine and then is expanded to atmospheric pressure (101.2 kPa) through a Joule-Thomson expansion then released to the environment. The turbine is producing work and takes in 100g/min of air at
1 MPa at 500 C. Air exits the turbine at 150C. Cv=20J/mol*K Mw=29g/mol
a.) The turbine loses 2kJ/min of heat, determine the work produced by the turbine.
b.) Find T2 of the joule-thomson expansion.

2. Relevant equations

energy balance: d/dt(U+M(V^2/2+gh)=Mk(mass flow) (H+v^2/2+gh)+Q+W

deltaH=Cp(deltaT)

Cp=Cv+R

PV=nrT (possibly relevant?)

Here is my attempt at a solution:

a.) So in the energy balance you can get rid of irrelevant terms and simplify down to 0=Mk(H)+Q+w. Am I correct in making the left side 0, since it is most likely steady state? Nothing is changing with respect to time and I don't believe there is internal energy change.

now I solved for H using deltaH=Cp(deltaT) finding cp with cv and R. Then I solved for W and found the answer in kW. I believe this is right but I don't know if my assumption of the left side being 0 is wrong, also I feel weird not using that 1 MPa for anything.

b.) In a Joule-Thomson expansion Hin=Hout so the energy balance becomes Mk(deltaH) M1=M2 so its just deltaH thewn I used delta H equation to get 0=Cp*deltaT and solved for T2, I feel confident but I didnt even use Pin or Pout, am I missing something here???

Last edited by a moderator: Mar 4, 2017
2. Oct 8, 2012

### 256bits

a) seems correct. Within the turbine there is considered stady flow, steady state so your assumption is correct.

b) Yes H1=H2 for Joule Thompson in many cases.
If though you solve for 0 = Cp delta T , do you not get T2=T1, so I must be missing something here also on how to solve this.