Thermodynamics compression problem

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SUMMARY

The forum discussion centers on a thermodynamics problem involving an adiabatic and reversible compressor that takes air at 300K and 1 atm, delivering it at 2 atm with a power input of 200W. The derived final temperature of the air is 366K, calculated using the formula T2 = (P11/K-1 / P21/K-1)T1. The work done by the compressor is expressed as W = nR(T1 - T2) / (1 - 1/K), leading to a negative value of -1920J, indicating that the calculations align with the principles of thermodynamics.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law
  • Knowledge of the specific heat ratio (k) for air, specifically k = 1.40
  • Basic calculus for integrating thermodynamic equations
NEXT STEPS
  • Study the derivation of the adiabatic process equations in thermodynamics
  • Learn about the implications of the specific heat ratio (k) on gas behavior
  • Explore the concept of work done in thermodynamic systems, particularly for compressors
  • Investigate the relationship between pressure, volume, and temperature in ideal gases
USEFUL FOR

This discussion is beneficial for students and professionals in mechanical engineering, particularly those focusing on thermodynamics, as well as anyone involved in the design and analysis of compressors and similar systems.

T-7
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Homework Statement



"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]"

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

[tex]PV^{K} = const[/tex]

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

[tex]T_{2} = \frac{P_{1}^{1/K-1}}{P_{2}^{1/K-1}}T_{1}[/tex]

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

For the last bit, I ended up deriving the expression

[tex]W = nR\frac{T_{1}-T_{2}}{1-1/K}[/tex]

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

I started with the usual

[tex]W = \int^{V_{1}}_{V_{2}}p dV[/tex]

replacing dV with

[tex]-nRT\frac{dP}{P^{2}}[/tex]

and the T with

[tex]bP^{1-1/K}[/tex]

(b is a constant)

to obtain, in the end, the integral

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

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

[tex]W = nR\frac{T_{1}-T_{2}}{1-1/K}[/tex]

But I'm not sure that's right (?). How does it look?
 
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T-7 said:
Using the ideal gas law, and rearranging a bit, I make it that

[tex]T_{2} = \frac{P_{1}^{1/K-1}}{P_{2}^{1/K-1}}T_{1}[/tex]

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

I concur.

[tex]W = nR\frac{T_{1}-T_{2}}{1-1/K}[/tex]

As I say in your other thread, I get

[tex]W = nR\frac{T_{1}-T_{2}}{1 - K}[/tex] .


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.
 
dynamicsolo said:
I concur.



As I say in your other thread, I get

[tex]W = nR\frac{T_{1}-T_{2}}{1 - K}[/tex] .


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.

Thanks for your reply!

My W does come out negative, as T2 > T1. But it is a somewhat larger value of W than yours, of course. At the moment, I can't see where I've gone wrong. I can derive your result, and mine. I'll keep trying, I guess!
 

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