1. Oct 25, 2016

Incand

1. The problem statement, all variables and given/known data
A turbine is driven by compressed air. The air enter the turbine with temperature $T_1$ and pressure $P_1$. When the air leaves the turbine it's pressure is lowered to $P_2$. Calculate the work done for one mol air.

The expansion of the air can be seen as reversible and adiabatic. The contribution from the airflow as the air enters/leave the turbine can be ignored. The air can be treated as a two atomic ideal gas.

2. Relevant equations
$Tp^{\frac{1}{\gamma}-1} =$ constant
First law of thermodynamics (We consider the work done by gas, not the work done on the gas)
$\Delta U = -W+Q$
Internal energy for ideal gas (one mole)
$U=C_VRT=\frac{f}{2}RT$

$C_P = \left(\frac{\partial H}{\partial T}\right)_P$
3. The attempt at a solution
I'm having trouble seeing why my solution is incorrect:
It follows that $T_2 = T_1 \left( \frac{P_2}{P_1}\right)^{1-\frac{1}{\gamma}}$
Since adiabatic process $Q=0$ which gives
$W = U_2-U_1 = \frac{5}{2}RT_1\left[1-\left( \frac{P_2}{P_1}\right)^{1-\frac{1}{\gamma}}\right]$

Correct solution:
$H_1 = W+H_2$ gives that
$W = H_1-H_2 = C_P(T_1-T_2) = \frac{7}{2}RT_1\left[ 1-\left(\frac{P_2}{P_1}\right)^{1-\frac{1}{\gamma}} \right]$

2. Oct 25, 2016

Staff: Mentor

Are you familiar with the version of the first law of thermodynamics applicable to an open system (aka control volume) like a turbine? Are you familiar with the difference between the total work, and the so-called "shaft work?" Which of these are you trying to determine in this problem?

3. Oct 25, 2016

Incand

I don't think so. I looked up the terms and it doesn't seem to be something that we covered in our course. What I'm trying to determine is the work the turbine outputs which should be the second solution. So what you're hinting at is that this version of the first law doesn't apply for open systems where matter may enter/leave the system.

I guess what I'm missing is the pressure-volume work needed to make room for the gas. The key being that the air enter and leave the giving us a similar system to a throttle except with work coming out.
So the the first "solution" would be the difference in internal energy between two closed system with those parameters but it doesn't account for the "external forces" from the process itself.

When would I need to use Gibbs free energy instead?

Edit: Okey so after reading up on this a bit more it seems to me that for both enthalpy and Gibbs free energy what happens is that we use the environment to do part of the work. Either as pressure-volume work or as heat extracted from the environment.

Last edited: Oct 25, 2016
4. Oct 25, 2016

Staff: Mentor

Yes. Very perceptive.
Yes. Very perceptive. Here is a reference that you may not have seen: http://www.learnengineering.org/2013/03/frist-law-of-thermodynamics-open-system.html

I'm surprised that you were assigned this problem without having been first assigned reading on (or being taught about) the open system version of the first law.

Why do you feel that the Gibbs free energy would be appropriate to use on this kind of problem?

5. Oct 25, 2016

Incand

Thanks! That's good to know.

We never learned that version of the law but we have derived the behaviour of specific systems like the equation of a throttle $H_i +Q_\text{in} + mv_i^2/2 = H_f + W_\text{out} +mv_f^2/2$. Just sometimes you don't remember that method and start doing something else and wonder why that doesn't work as in this case.

Well not for this problem but they seem quite similar in that you add on another term to account for something extra. First we have internal energy and then we add the pressure volume work to get enthalpy and lastly we can add the temperature-entropy part to get Gibbs free energy.

6. Oct 25, 2016

TSny

7. Oct 25, 2016

Staff: Mentor

Very perceptive again. Gibbs free energy is very useful for open systems operating at constant temperature.