Thermodynamics Question

[Woodchuck2000]

2kg/s of air enters a reversible adiabatic compressor at a pressure of 2 bar and a temperature of 27 degrees centigrade. It is compressed to 6 bar. What are the exit temperature and the required power input?

I'm getting to the 'Eat my own head' point trying to revise thermo, any explanations of the whats and the whys greatly appreciated.

 

[StageLeft]

Bout 120 F and you need around 250 W power supply AKA 350 horsepower with NOS.

 

[Woodchuck2000]

Originally posted by: Skoorb
Bout 120 F and you need around 250 W power supply AKA 350 horsepower with NOS.

I was thinking more along the lines of modding the compressor with a window and a few cold cathodes to increase its efficiency but I get the feeling my lecturer would kill me 🙁

 

[Shooters]

My thermo is kinda rusty (took it about 3 years ago), but if the process is reversible and adiabatic, then couldn't it be considered to be isentropic? If that's the case, then you can just use the isentropic relation for pressure and temperature to solve for T2:

T2/T1 = (P2/P1)^((k - 1)/k)

where k is the ratio of specific heats (Cp/Cv)....1.4 for air if I remember correctly.

 

[Woodchuck2000]

Originally posted by: Shooters
My thermo is kinda rusty (took it about 3 years ago), but if the process is reversible and adiabatic, then couldn't it be considered to be isentropic? If that's the case, then you can just use the isentropic relation for pressure and temperature to solve for T2:

T2/T1 = (P2/P1)^((k - 1)/k)

where k is the ratio of specific heats (Cp/Cv)....1.4 for air if I remember correctly.

Ah, thanks a lot. I've never seen that formula before but it allows me to get the answer. I'll try and find the derivation for it.

Cheers etc...

 

[Triumph]

The derivation comes from the 2nd law of thermo, I believe. I don't have my book anywhere near, but since it is reversible, then the entropy change is zero, s2-s1 = 0 = some stuff with T and P and K. Lemme try to remember...

Tds =

Then you use a natural logarithm trick to get it into the form that Shooters posted. Ok, I KNOW that my equations aren't right, but I'm trying to guide you in the right direction. The trick to thermo problems is to realize what is constant throughout a process, and you can then find an equation where state 2 = state 1, so they'll cancel and everything gets easier.

What I always always did was to write out the 1st and 2nd laws, COMPLETELY, every time, even though it was tedious. Then cross out everything that you know will cancel between the states (for your problem, entropy and heat will cancel). Now with everything in front of you, simplified, it's easier to see what you need to do next.

 

[Evadman]

In this house we obey the laws of Thermodynamics!!!

Sorry, had to be said.

 

[silverpig]

Ugh, I found thermo to be so dry it wasn't even funny. 50% pass thank you very much. I didn't go half the time.

 

[Triumph]

Ok now I got it! My teacher always said, "Don't forget your Tds equations!"

Tds = dU + PdV
and
Tds = dH - VdP

For an ideal gas (air is ideal enough):
dU = Cv(dT)
dH = Cp(dV)
pV = RT
k=Cv/Cp

now the trick to deriving it into Shooter's form is to remember your natural logarithm rules....