Thermodynamics compression problem

In summary, the conversation discusses the process of a compressor taking air at 300K and 1 atm and delivering compressed air at 2 atm while using 200W of power. The question asks for the rate of air delivery and the final temperature of the air if the process is adiabatic and reversible. Using the ideal gas law, the final temperature is calculated to be 366K. The conversation also considers the work done and the expression for it is derived as W = nR(T1-T2)/(1-K), which is negative due to T2 being greater than T1. The conversation concludes with a discussion about the potential error in the derived expression and the correct factor being [1-K] instead of [1-(
  • #1
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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  • #2
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.
 
  • #3
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!
 

FAQ: Thermodynamics compression problem

What is thermodynamics compression problem?

The thermodynamics compression problem is a common issue encountered in thermodynamics, where a substance undergoes compression and its temperature increases without any external heat input. This phenomenon is often seen in gases and is a result of the conversion of work into heat.

How is the thermodynamics compression problem solved?

The solution to the thermodynamics compression problem involves using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. By understanding the energy balance of the system, the temperature increase during compression can be explained.

What factors affect the thermodynamics compression problem?

The main factors that affect the thermodynamics compression problem are the initial temperature and volume of the substance, the type of gas being compressed, and the amount of work being done on the gas. These factors ultimately determine the amount of heat produced during compression, and thus, the temperature increase.

How does the thermodynamics compression problem relate to real-life situations?

The thermodynamics compression problem can be observed in various real-life situations, such as in the compression of air in a bicycle pump or the compression of refrigerant in an air conditioning system. It also plays a crucial role in the functioning of internal combustion engines, where the compression of fuel-air mixture leads to an increase in temperature and pressure, resulting in ignition and power generation.

What are some practical applications of understanding the thermodynamics compression problem?

Understanding the thermodynamics compression problem is essential in various fields, such as engineering, physics, and chemistry. It allows for the design and optimization of various systems, such as engines and refrigeration systems, and also helps in the development of new technologies. Additionally, it helps in the study of thermodynamic processes and the behavior of gases under different conditions.

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