# Thermodynamics compression problem

1. Oct 18, 2007

### T-7

1. The problem statement, all variables and given/known data

"A compressor takes air at 300K, 1 atm. It delivers compressed air at 2 atm, using 200W of power.

If the process is adiabatic and reversible, what is the rate that air is delivered, and what is the final temperature of the air? [k of air = 1.40]"

3. The attempt at a solution

Hi. I'd like to bounce off someone these expressions I derive, to see if they're sensible.

Since it's adiabatic, I infer that

$$PV^{K} = const$$

Using the ideal gas law, and rearranging a bit, I make it that

$$T_{2} = \frac{P_{1}^{1/K-1}}{P_{2}^{1/K-1}}T_{1}$$

I think that's sensible (?). It comes out as 366K.

For the last bit, I ended up deriving the expression

$$W = nR\frac{T_{1}-T_{2}}{1-1/K}$$

which to me just looks too simple (it comes out as -1920J)

I started with the usual

$$W = \int^{V_{1}}_{V_{2}}p dV$$

replacing dV with

$$-nRT\frac{dP}{P^{2}}$$

and the T with

$$bP^{1-1/K}$$

(b is a constant)

to obtain, in the end, the integral

$$W = -nRb \int^{P_{2}}_{P_{1}} P^{-1/K} dP$$

On integrating, and then substituting for the b's in terms of T1 and P1 or T2 and P2, I found they disappeared, leaving me with

$$W = nR\frac{T_{1}-T_{2}}{1-1/K}$$

But I'm not sure that's right (?). How does it look?

Last edited: Oct 18, 2007
2. Oct 19, 2007

### dynamicsolo

I concur.

$$W = nR\frac{T_{1}-T_{2}}{1 - K}$$ .

So I find dn/dt = [ (dW/Dt)(1 - K) ] / [ R · (T1 - T2) ] .

I'm pretty sure the factor can't be [ 1 - (1/K) ]: since K > 1, this would be positive when it should be negative.

3. Oct 21, 2007