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Adiabatic air compressor

  1. Mar 9, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    A small adiabatic air compressor is used to pump air into a 20-m3 insulated tank. The tank initially contains air at 25°C and 101.33 kPa, exactly the conditions at which air enters the compressor. The pumping process continues until the pressure in the tank reaches 1,000 kPa. If the process is adiabatic and if compression is isentropic, what is the shaft work of the compressor? Assume air to be an ideal gas for which CP = (7/2)R and CV = (5/2)R.


    2. Relevant equations



    3. The attempt at a solution
    I derived the relationship for the enthalpy of an adiabatic as

    dH = V dp

    However, I also know
    ΔH = CpΔT

    When I calculate the two enthalpies, I get different answers for the change in enthalpy, which I know should be equal to the isentropic work. First off, how do I know my equation is only for an adiabatic and isentropic process, and would not work for something that is anisentropic and adiabatic?

    I realize the change in enthalpy is the change in molar enthalpy times the number of moles, but the number of moles is changing throughout the process until it reaches 1000 kPa. The other expression seems to bypass worrying about that. I am wondering which one, if either, is correct and why?

    In other words, how do I calculate ''isentropic enthalpy change'' vs. enthalpy change that is not isentropic?
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2014 #2
    In my judgement, the best way to approach this problem is to focus first on the tank contents at the final set of conditions. We know the volume of air and we know its pressure, but we don't the number of moles n or the final temperature T. However, we do know that all the air in the tank started out at 101.33kPa and that it was compressed isentropically. From this information we can calculate the initial volume that the tank contents occupied. You know that relationship between pressure and volume for an isentropic compression. What do you calculate for the initial volume of the final tank contents? All this volume of air was at 101.33 kPa and 25 C. What is the number of moles of air n in the tank at the final conditions. What was the number of moles of air within the tank at the initial conditions (before running the compressor)? Another thing you know is the relationship between pressure and temperature for an isentropic compression. So, from one form or another of this relationship, what is the final temperature in the tank?

    Let's take stock. You now know the initial and final number of moles of air in the tank, and you also know the initial and final temperatures of the air in the tank. This gives you enough information to calculate the change in internal energy between the initial and final states of the tank. What is the change in internal energy for the tank?

    No mass of air exited the tank between the initial and final states. However, air did enter the tank from the compressor. Using the open system form of the first law, you should be able to calculate integrated enthalpy of all the air that entered the tank, and, more importantly, you should be able to calculate the change in enthalpy for all the gas that left the turbine minus the enthalpy of all the gas that entered. What is that change?

    Chet
     
  4. Mar 9, 2014 #3

    Maylis

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    I am curious, is something wrong with my V dp approach? Also, why would the gas in the tank initially occupy a space less than the volume of the tank, the gas will expand and occupy the volume of its container.

    I calculate the final temperature of the tank to be the same temperature as the stream of air coming in from the compressor, 573.2 K. Is this right?
     
    Last edited: Mar 9, 2014
  5. Mar 10, 2014 #4

    Maylis

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    Here is my second attempt with analysis.

    I disregarded the statement about the volume, because I think it is a typo. I didn't actually have to solve for the internal energy change, as I knew the initial moles in the tank plus what was added was the final number of moles.
     

    Attached Files:

  6. Mar 10, 2014 #5
    I may have misinterpreted the problem, regarding the pressure of the air exiting the compressor. I was going to assume that the compressor is run in such a way that the gas exiting the compressor at any instant is only slightly higher in pressure than the gas within the tank. That is probably not what the problem intended. In the case that you have considered, there is a valve at the entrance to the tank that drops the pressure down from the compressor exit pressure to the tank pressure. I think that that's what they intended.

    Chet
     
  7. Mar 10, 2014 #6

    Maylis

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    Would it be right to say the final temperature of the tank is the same as the temperature of the stream exiting the compressor?
     
  8. Mar 10, 2014 #7
    No. You need to use the flow version of the first law to figure it out. The unknowns are the number of moles that enter and the final temperature in the tank. The temperature is going to be higher than the exit temperature of the compressor. Your equations are the flow version of the first law and the ideal gas law. You know the entering specific enthalpy, and the initial internal energy in the tank.

    Chet
     
  9. Mar 10, 2014 #8

    Maylis

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    Am I to assume that the air is an ideal gas at the final state, at 1000 kPa? That is quite a high pressure, so I am not sure if I should assume it to be ideal.

    I think I am having confusion because there are effectively two compressions going on here. First of all and most obvious, the compressor is compressing the gas that flows into it. Secondly, the gas in the tank is being compressed by the other air that was already compressed. Do they mean that both the compressor and filling up the tank is isentropic?

    The problem satement says the process is adiabatic and compression is isentropic. This makes me confused because it doesn't specify which one is adiabatic and isentropic. Are they saying the processes of the compressor as well as filling up the tank is both adiabatic and isentropic?

    I am just using the equation here relating T and P for isentropic compression
    http://www.grc.nasa.gov/WWW/k-12/airplane/compexp.html

    That is how I get 573.2 K. That is what I think is the temperature of the stream leaving the compressor and entering the tank. However, if I do the isentropic compression equation for the tank, I get the exact same result, because the stream entering the compressor is at the same conditions as the tank was initially, and the pressure in the tank at the final state is the same as the pressure of the gas leaving the compressor.
     
    Last edited: Mar 10, 2014
  10. Mar 10, 2014 #9

    Maylis

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    Also, I do not know the initial internal energy of the tank, nor do I know the specific enthalpy entering the tank. I could find the change in enthalpy of the gas from the entrance to the compressor to the exit stream, but no absolute values. I know the change in enthalpy of the gas going through the compressor, so all I need is the moles that entered the tank to figure out my shaft work, knowing moles entering multiplied by the change in enthalpy is the shaft work.
     
    Last edited: Mar 10, 2014
  11. Mar 10, 2014 #10
    The problem statement says to assume the air is an ideal gas. The entropy is not constant in the injection of the high pressure air into the tank. The air has to pass through the inlet valve, and this is not a constant entropy process. However, it is a constant enthalpy process.
     
  12. Mar 10, 2014 #11

    Maylis

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    I'm stuck with my mass and energy balance. My unknowns are the initial and final internal energy, and the enthalpy entering the tank, and the number of moles of air entering. I have no idea how to find the enthalpy entering the tank, nor the initial internal energy of the tank.
     
  13. Mar 10, 2014 #12
    You need to express the internal energy relative to a reference state. The form of the first law applicable to the tank is:

    [tex]nh_{in}=(n+n_0)u-n_0u_0[/tex]
    where hin is the enthalpy per unit mass of the gas entering through the valve, n is the number of moles of gas that enters through the valve, n0 is the moles of gas in the tank to begin with, u is the final internal energy per mole of the gas in the tank and u0 is the initial internal energy per mole of the gas in the tank. If we take as the reference state air at 25 C and 1 atm, then

    hin=uin+RTin=5R(573.2-298.2)/2 + 573.2R

    u0=0
    u=5R(T-298.2)/2

    You use the ideal gas law to express n+n0 in terms of the final pressure, the tank volume, and the unknown temperature T, and you already know n0. This will give you what you need to solve for T.

    Chet
     
    Last edited: Mar 10, 2014
  14. Mar 10, 2014 #13

    Maylis

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    I'm not seeing where you are getting the expression for H_in. I would say H_in is 7/2R(573.2-298). I didn't realize that I should set the air as the reference state, that is very good to know.

    I did it using my expression for H_in, and got T=1796.4 K, quite warm..

    Using your expression, I get T=1966.8 K

    http://www.wolframalpha.com/input/?i=solve+%282.406*10^5%2Fx%29*%2820.785%28x-298%29%29-%282.522*10^10%2Fx%29%2B8.58*10^6%3D0
     
    Last edited: Mar 10, 2014
  15. Mar 11, 2014 #14

    Maylis

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    Here is my analysis for my 3rd attempt at this problem. Getting so close to the answer...or maybe it's right?
     

    Attached Files:

  16. Mar 11, 2014 #15
    My first term is the internal energy per unit mass, and my second term is Pvin, where vin is the molar volume of the feed. This is equal to RTin (since Pv=RT for an ideal gas).. Another way of writing this is 7/2R(753.2-298)+298R, where 298R is the value of Pv in the reference state (which is equal to the enthalpy in the reference state, since u in the reference state is zero).

    I haven't run the calculation yet, but I will do so, and see if my result checks with yours.

    Chet
     
  17. Mar 11, 2014 #16
    I wasn't able to wade through the details of your math, but I did the calculation and got T2=686K.

    Chet
     
    Last edited: Mar 11, 2014
  18. Mar 11, 2014 #17

    Maylis

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    I was talking to the graduate student instructor, who was not able to solve the problem either. We ended up looking at the instructor's solution manual, and it seems like the manual suggests that the tank is compressed insentropically as well. It's final temperature is the same as the entering stream.

    Their energy balance includes the shaft work, which makes absolutely no sense to me. The control volume is the tank.

    They say

    d(nU) = n_in*H_in + Ws
     
  19. Mar 11, 2014 #18
    I'm as confused as you are. I'm confident that we did it correctly for the interpretation of the problem that we employed.

    Chet
     
  20. Mar 11, 2014 #19
    This last equation looks like the equation I was going to use in my original interpretation of the problem. The system here would include both the compressor and the tank (so that the shaft work would be included) and the "in"'s would apply to the inlet to the compressor. The equation, which includes shaft work, could not apply to the tank alone.

    Chet
     
  21. Mar 11, 2014 #20

    Maylis

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    Yes. It turns out they choose the control volume to be the tank and compressor. They say the final tank temperature is the same as the stream entering the tank.

    Here is the solution with the correct answer, according to the solution manual.
     

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