Polytropic process where n is not given

In summary, the balloon and ammonia underwent a pressure increase from 0 to 600 kPa. The amount of work done in the process was 117.19 kilojoules.
  • #1
canadiansmith
36
0
Hi I really need help figuring out this problem. A balloon behaves such that the pressure inside is proportional to the diameter squared. It contains 2 Kg of ammonia at 0 degrees celcius and 60% quality. The balloon and ammonia are now heated so that a final pressure of 600 kPa is reached. Considering the ammonia as a control mass, find the amount of work done in the process.
the answer is 117.19 KJ
i was able to find the initial volume from the quality and mass
I got Vinitial = 0.3483
I can't find a way to get a value for n in the polytropic process equation
W = (P2V2-P1V1)/ 1-n
 
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  • #2
canadiansmith said:
Hi I really need help figuring out this problem. A balloon behaves such that the pressure inside is proportional to the diameter squared. It contains 2 Kg of ammonia at 0 degrees celcius and 60% quality. The balloon and ammonia are now heated so that a final pressure of 600 kPa is reached. Considering the ammonia as a control mass, find the amount of work done in the process.
the answer is 117.19 KJ
i was able to find the initial volume from the quality and mass
I got Vinitial = 0.3483
I can't find a way to get a value for n in the polytropic process equation
W = (P2V2-P1V1)/ 1-n


If the pressure is proportional to the square of the diameter, then the diameter is proportional to the square root of the pressure. Now the volume is proportional to what power of the diameter? ... does this help?
 
  • #3
Well not really. The question you asked is exactly the thing I am having problems with. I found that V=(4/3)pi(d/2)^3 but I am not sure if what I am trying to do is to get Volume to the power of something to equal pressure or some random constant.
 
  • #4
canadiansmith said:
Well not really. The question you asked is exactly the thing I am having problems with. I found that V=(4/3)pi(d/2)^3 but I am not sure if what I am trying to do is to get Volume to the power of something to equal pressure or some random constant.

You're almost there. You are trying to find an n for which PVn = constant. So if

V=(4/3)pi(d/2)^3

and

d \propto sqrt(P) → d = d1(P/P1)^0.5

then

V=(4/3)pi(d1(P/P1)^0.5/2)^3

Now collect the P and the V on the same side of the equation, and put it in the form PVn = constant. What is n?
 
  • #5
after simplifying I got (p^1.5)V = 3.9116*10^-5 which gives me 0.3483^n = 9.1052*10^-8
so n would be 15.37?

I got V1 = 0.3483
P1= 429.6
 
  • #6
Thank you so much for your help. I figured out what I was doing wrong. I got n = -2/3 now. Which gave me my relation. Thanks again
 
  • #7
Good luck!

BBB
 
Last edited:

Related to Polytropic process where n is not given

1. What is a polytropic process?

A polytropic process is a thermodynamic process in which the pressure and volume of a gas are related by the equation P^n = constant, where n is a constant.

2. How is a polytropic process different when n is not given?

When n is not given, the polytropic process is considered to be a special case known as an isothermal process, in which the temperature remains constant. In this case, the equation becomes P x V = constant.

3. What factors determine the value of n in a polytropic process?

The value of n in a polytropic process is determined by the type of gas, the conditions of the process (such as temperature and pressure), and the type of energy transfer (e.g. heat or work).

4. Can n have a negative value in a polytropic process?

Yes, n can have a negative value in a polytropic process. This indicates that the process is an expansion process (work is done by the gas) rather than a compression process (work is done on the gas).

5. What is the importance of the polytropic process in thermodynamics?

The polytropic process is important in thermodynamics as it allows us to analyze and understand the behavior of gases and their energy transfers in various processes, such as in heat engines and refrigerators. It also helps in predicting the work, heat, and temperature changes in these processes.

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