# How to solve state parameters using these givens for an ideal gas?

In a thermodynamic turbine using air as an ideal gas, given that you have a known inlet temperature value $T_i$, a known exit pressure value $P_e$, a known inlet and exit velocity $V_i$ and $V_e$, a known value for the actual work of the turbine $w_{actual}$, and an isentropic efficiency $\eta$, how can you find the other state parameters?

I know the efficiency of the turbine is

$$\eta = \frac{w_{actual}}{w_{isentropic}} = \frac{h_i - h_e}{h_i - h_{e)s}}$$

Through getting $w_{isentropic} = \frac{w_{actual}}{\eta}$, we'll know that value. My approach was then to find $h_i$.

$$h_i = w_{actual} + h_e = w_{isentropic} + h_{e)s}$$ $$w_{actual} - w_{isentropic} = h_e - h_{e)s}$$

And then I'm stuck there. I tried a different approach.

$$s_e - s_i = C_{p0}ln\frac{T_e}{T_i} - Rln\frac{P_e}{P_i}$$

$C_{p0}$ and $R$ are known here because they are constants. If I assume an isentropic process, then we'll have $s_e - s_i = 0$. But our unknowns are still $T_e$ and $P_i$. I can use the ideal gas equation:

$$\frac{T_e}{T_i} = (\frac{P_e}{P_i})^\frac{k-1}{k}$$

And $k$ is known here because it's air and I assumed it's isentropic, and it's a constant. So with two equations and two unknowns, I'll find $P_i$ and $T_e$ assuming the process is isentropic. So then, from here, how do you find $h_{e)s}$? I think I should be able to find it, but with what relationship?

I see that once I find $h_{e)s}$ I'll be able to solve for $h_i$, and then $h_e$. Once I have $h_e$, what relationship can I use with my known $T_i$ to derive the other state parameters?

The two key things here are the $h_i$ and $h_e$, I think, because I already have $P_e$ and $T_i$ given to me. And if I have those four, I'll be able to find all the other state parameters ill both the inlet and exit states. But what relationship can I use to find them?

My thermodynamic tables don't have any values for air, just steam and other elements and refrigerants. I do have a table for air, but only for a 1-bar pressure, and I suspect I can't assume that pressure in this question. I suspect that since it's an ideal gas, I won't need the tables to solve for it.

• If you have correctly found $T_e$, then the state equation should give you the exit enthalpy. Since at the exit you have temperature and pressure, you can get any other property, case closed. Similarly, the isentropic exit entropy is the inlet entropy and you know that from (P,T) at inlet. Problem is, you haven't used velocity. I can't help you there because you need to look in your own textbook for the definition of turbine efficiency. Velocity is often neglected, but sometimes efficiency applies to a "total" energy metric, $h+1/2 V^2$. – Alan Rominger Mar 20 '13 at 14:27
• The only thing it says in there about turbine efficiency is already written above: $\eta = \frac{w_{actual}}{w_{isentropic}} = \frac{h_i - h_e}{h_i - h_{e)s}}$. That's my problem, because I only solved for the isentropic $T_e$ but I know that my process isn't really isentropic, and I'm having a hard time finding the real value. Isn't that metric $h+1/2V^2$ applicable only to a reversible, isentropic case? Or am I wrong? – markovchain Mar 20 '13 at 14:40
• I see the issue with the exit (P,T) now, sorry I didn't read carefully enough. Still, if you're looking for the isentropic enthalpy, why not just en.wikipedia.org/wiki/Ideal_gas, $h = \hat{c}_V nRT + \nu P$? The isentropic $P_e$ is still $P_e$. The 1/2 V^2 term is using velocity. I mean this energy metric in the sense of $E=h+1/2 V^2 = u + \nu P + 1/2 V^2$. Any basic thermo class would probably neglect the velocity contribution to energy, but I don't know what level you're at, and I don't know why you have inlet and exit velocity in the problem statement. – Alan Rominger Mar 20 '13 at 15:54
• But then I would need to know $E$, $v$ and/or $u$, since I don't know these values yet. Honestly, this is a basic thermo class, and I'm not sure if the inlet and exit velocities are useful or just a red herring, but my teachers feel like giving us a hard time. – markovchain Mar 21 '13 at 0:47