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Air expanding into chamber

  1. Sep 28, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider an isolated evacuated chamber. A small adiabatic valve is opened so that the air enters the chamber slowly from the atmosphere. What is the final temperature of the air inside the chamber once it has reached atmospheric pressure? Assume air is an ideal gas with ##\gamma = 1.4##. The air outside the chamber forms a reservoir at P = 1atm and constant temperature 298K throughout.

    2. Relevant equations
    Free expansion, Ideal Gas, Adiabatic process.

    3. The attempt at a solution
    I think I may consider this problem to consist of a two partitioned system with the vacuum on one side and the air on another. As the air enters the chamber, it does no work since there is no external agent inside the chamber hindering the expansion. I think this is then a free expansion of air, which means the temperature difference is given by ##\Delta T = \int_{V_i}^{V_f} \left(\frac{\partial T}{\partial V}\right)_U dV = \frac{P_o}{nR} \int_{V_i}^{V_f} dV##

    My problem is in writing expressions for ##V_i## and ##V_f##. Take the system to be the number of moles, ##n## in the chamber that have expanded adiabatically. I can write ##V_i = nRT_1/P_o##, and ##V_f = nRT_2/P_o## and the expansion of those n moles can be written as ##P_oV_i^{\gamma} = P_oV_f^{\gamma}## for the initial and final states.

    I have tried reexpressing the equations above but I always end up with things like T1=T2.

    Many thanks.
     
  2. jcsd
  3. Sep 29, 2013 #2

    CAF123

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    Does anyone have any hints on how to progress?
    Thanks.
     
  4. Sep 29, 2013 #3
    I haven't worked the problem out completely yet, but I've made some progress, and I'm sure I'm on the right track. The first question to ask yourself is "what is the change in enthalpy of the air (assumed to be an ideal gas) as it passes through the valve from the outside pressure and temperature to the instantaneous pressure within the chamber at time t during the process? I assume you have been studying the first law applied to flow systems in your course.

    Chet
     
  5. Sep 29, 2013 #4

    CAF123

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    The enthalpy of the air before release into chamber is ##h_i = u_i + P_0T_1##, where ##P_0 =## 1 bar and ##T_1 = 298K##.

    At the end of the process where pressure inside chamber = pressure outside = atmospheric pressure, ##h_f = u_f + P_oT_f##.

    The first law in this case is ##Q +W = \Delta E##. where ##\Delta E## is the change in internal energy of the gas, and bulk kinetic energy. I think the LHS of this equation is zero (chamber is thermally insulated and as said in OP, no work done since the flow is a free expansion.)

    So, ##u_f + P_oT_f = -(u_i + P_oT_i)## This eqn has 3 unknowns. I might be able to obtain expressions for the ##u_i## using U=(f/2)nRT, by making some assumptions on f, but this will not use the constant ##\gamma## anywhere in my calculations.

    Many thanks.
     
  6. Sep 29, 2013 #5
    Check your thermo book in the section on the first law applied to flow systems. For the case of an ideal gas flowing through an adiabatic valve, the change in enthalpy (per mole) is zero. This means that the temperature of the air entering the chamber through the valve is the same as it was prior to entering the valve (298). What's happening mechanistically is that the cooling associated with expansion of the gas through the valve is exactly canceled by the frictional heating (viscous heating) of the gas as it passes through the valve (for an ideal gas). Anyway, each parcel of air that enters the chamber through the valve is going to have the same temperature as the air outside, and approximately the same pressure as the air already residing in the chamber at the time that the parcel enters. As more air enters the chamber, the air previously there is going to be compressed (to make room), and the temperature of the air within the chamber will increase with time. Finally, when the pressure within the chamber matches the pressure outside (1 atm), air will stop entering, and the temperature of the air will be higher than outside. See if you can figure out how to take this into account.

    Chet
     
  7. Sep 29, 2013 #6

    CAF123

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    So as the chamber becomes filled by air, the air yet to come inside will do work on the air already inside. The air outside is at a pressure P0 so the amount of work done by a parcel of air entering will be P0 multiplied by the amount that the air parcel compresses the air inside. I am not quite sure how to express this change in volume in terms of known quantities.

    My book didn't say anything about change in enthalpy = 0 for adiabatic flow of an ideal gas. Is this observation key?

    Thank you,
     
  8. Sep 29, 2013 #7
    Not exactly. The pressure of the gas at the entrance to the valve is P0, but, because of the frictional pressure drop through the valve, the pressure of the gas exiting the valve into the chamber is P, where P is the pressure of the gas within the chamber. The pressure and temperature within the chamber after n moles have entered is P and T. The pressure and temperature of the next new parcel of gas leaving the valve and entering the chamber is P and T0. [/quote]
    See section 2.6 of Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics. In this case, the focus of the analysis is the valve, for which the change in enthalpy per mole of gas passing through is zero.
     
  9. Sep 29, 2013 #8

    CAF123

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    The pressure and temperature of say, an amount N moles of air before flowing through the valve is P0 and T0. When it leaves the valve, thereby entering the chamber it is at a pressure and temperature P and T. In terms of equations, after some manipulation I obtain ##T_0P_0^{\frac{1}{\gamma}-1}=TP^{\frac{1}{\gamma}-1}## since the valve is adiabatic. So I have the temperature of the chamber as a function of the pressure within the chamber. Is this progress?
    Thanks.
     
  10. Sep 29, 2013 #9
    Sorry CAF123. When it leaves the valve, it is at the chamber pressure P, but, since its enthalpy hasn't changed in passing through the valve, its temperature is still T0. You can't apply the equation to adiabatic reversible expansion for the valve because of the extra temperature rise as a result of viscous (frictional) heating. This just cancels out the temperature drop resulting from the adiabatic expansion. One of the keys to this problem is first recognizing that the gas temperature (and enthalpy) of the gas does not change in passing through the valve.

    I have completed my solution to this problem and have a final answer. If you have an answer key, you can check it. That way you can possibly gain more confidence that what I'm saying is correct.

    Incidentally, the remainder of this problem is not very simple either. You need to consider what happens when each little incremental number of moles of gas enters the chamber.

    I really strongly encourage you to look up what happens in flow situations when a gas is flowing through a valve, with a pressure drop occurs across the valve. Check out the section in Wikipedia on the Joule-Thompson Effect entitled "Proof that the Specific Enthalpy Remains Constant."
     
  11. Sep 30, 2013 #10

    CAF123

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    Ok, I'll try again.
    Consider some intermediate step of the process, with some air already inside the chamber. The work done on the gas from other gas outside the chamber is P0V1 (and at temperature of gas outside is a constant T0). The work done by the gas from outside the chamber, compressing the gas inside is PV2 (the gas inside is at a temperature T).
    Net work by gas = PV2-P0V1

    I can then write the first law as uf - ui + PV2 -P0V1 = 0. Rearrange to give uf + RT = ui + RT0, where quantities are per mole.

    I am not really sure how to progress from here. I spoke to my professor today and he said that with the right choice of system, the problem becomes relatively trivial. My choice of system would be the n moles that expand adiabatically. Before the release into the chamber, the gas is at a pressure P0 and T0, Ultimately, it is at a pressure P0 again, but at some higher T. This is the approach I took in the OP, but that didn't lead me anywhere either.

    I have done another problem with the gas initially in a chamber expanding out into the surroundings which, I think, may be solved in a similar manner. Would this be correct?

    Also, I don't have access to the answer key, but I do trust what you're saying :smile:
     
  12. Sep 30, 2013 #11
    I admire your tenacity. It is great that you have been putting such determined thought into this.

    I don't know what your professor meant when he said that, by proper choice of the system, the problem becomes relatively trivial. Maybe I did not discover what his proper choice of system was. However, I'm pretty confident that I got the right answer doing it my way. So let me get you started.

    After n moles have entered the chamber, let Pn and Tn be the pressure and temperature of the gas already in the chamber. Pn and Tn are related to one another and to the chamber volume V by the ideal gas law.

    Now we are going to add Δn moles to the chamber through the inlet valve. The pressure of this gas parcel is P and the temperature of this gas parcel is T0. This parcel is going to mix with the gas already in the chamber (so the temperatures will have a tendency to equilibrate), but both the parcel and the gas already in the container are also going to have to compress a little so that they can both fit into the constant volume chamber (this compression will tend to raise their temperature).

    To accommodate both these mechanisms for a tiny parcel Δn, we can envision a two step process.

    Step 1: The gas parcel enters the chamber, and mixes with the n moles of gas previously present in the chamber at constant pressure P, such that the final temperature of the mixture is at an intermediate temperature value T* (determined by equilibration of the temperatures). This mixing process cannot occur at constant pressure P unless we temporarily allow the volume of the gas within the chamber to rise to a new value V*. At the end to Step 1, the parcel is well mixed with the gas in the chamber, and they form a single entity of n + Δn moles at T*, P, and V*.

    Step 2: In reality, the volume of the chamber is fixed, and cannot increase to V*. So we need to adiabatically compress the n + Δn moles of gas at T*, P, and V* down until the volume is again V. (This is where the γ comes in). Let the final conditions of the n + Δn moles of gas after this adiabatic compression has taken place be given by Pn+Δn, Tn+Δn, and V.

    After doing the analysis for this two step process, we will take the limit as Δn becomes small, and end up with a differential equation for the derivative of the pressure P with respect to n. This equation will be readily integrated from P = 0 to P = P0 to find the final number of moles n. From this integrated result, we can determine the final temperature using the ideal gas law.

    Why don't you start out by analyzing Step 1, and determining T* and V*? Let us know what you get.

    Chet
     
  13. Sep 30, 2013 #12

    D H

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    Rhetorical question: How does this differ from the standard free expansion setup, a pair of vessels connected by a valve, isolated from the external environment, one vessel initially containing a gas, and the other evacuated? In this standard setup, that the gas is isolated from the environment means there is no work, no heat transfer, and thus no change in the energy of the gas. Therefore there is no change in temperature (assuming an ideal gas).

    Which is what you should get. Only you did it the hard way.
     
  14. Sep 30, 2013 #13

    CAF123

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    So the final volume may be expressed as V* = (n+Δn)RT*/P. The gas parcel enters the chamber at a pressure P, so it does work in expanding the gas from Vn to V*. W for a constant pressure process is W=P(V*-Vn) = P(V*-nRTn/Pn). By the first law, this is equal to the change in internal energy of the gas.
    T*-Tn is the change in T when Δn moles have been added.
     
  15. Sep 30, 2013 #14

    CAF123

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    When I was discussing this today in class, this is what I and some of my classmates were getting confused about. Are the differences being that the gas entering the chamber comes from a reservoir and there is no boundary to the partition where the air resides initially. I think that the very first parcel of air that moves into the chamber will have to do zero work (expansion against vacuum) but as the number of moles increases in the chamber, since V is fixed, T must increase.

    I am not sure what you mean here.
     
  16. Sep 30, 2013 #15

    D H

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    What I mean is that the temperature doesn't change. There are a number of easy ways to arrive at that conclusion. You picked a hard way to arrive at that same conclusion.
     
  17. Sep 30, 2013 #16
    Well, for one thing, in this situation, the total volume of the two "containers" housing the gas decreases. Initially, you have the chamber of volume V which is under vacuum, plus you have the volume of the gas outside the chamber that will eventually enter the chamber. Finally, you just have the volume V of the chamber. None of the gas that entered is outside anymore. [/quote]
     
  18. Sep 30, 2013 #17
    In Step 1, if the n moles at temperature Tn are mixed with the Δn moles at temperature T0 (at constant pressure Pn), the temperature T* will be
    [tex]T^*=\frac{nT_n+(Δn)T_0}{n+Δn}[/tex]
    This can also be substituted into your equation for V* to obtain:
    V*=(nT_n+(Δn)T0)R/Pn

    Now, you are up to Step 2. You are adiabatically compressing air from a volume of V* at pressure Pn to a volume V at pressure Pn+Δn. In terms of γ, what is the relationship between V*, Pn, V, and Pn+Δn?
     
  19. Sep 30, 2013 #18
    Good news CAF123. I finally figured out the system your professor was alluding to. And it's very easy to do the problem by working with this system (even though you get the exact same answer as I did with my more detailed and complicated approach). His system consists of the total n moles of air that eventually enter the chamber plus the chamber itself. If n moles of air enter the chamber, the starting volume of these n moles of air is nRT0/P0. The idea is to solve for n. The surrounding air does nRT0 amount of work on the system when it drives the n moles into the chamber. From the first law, this must be matched by the increase in internal energy of the combined system: nCv(Tfinal-T0). Another constraint is that the ideal gas law must be satisfied in the chamber under the final conditions: nRTfinal=P0V. You eliminate Tfinal from these equations, and solve for n in terms of P0,V, T0, and Cv. In the equation you obtain for n, you will recognize the presence of γ. You then combine this equation with the ideal gas law at the final conditions to obtain Tfinal in terms of T0.
     
  20. Sep 30, 2013 #19
    After further consideration, the solution is even simpler than that. From the first law,

    Cv(Tfinal-T0)=RT0.

    Just solve for Tfinal.
     
  21. Oct 1, 2013 #20

    CAF123

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    Is this not the system that I chose initially in the OP? My system in the OP was taken as the n moles that expand adiabatically into the chamber.
    I am still trying to work out how this was obtained. Since W=P0∫dV => W = P0(Vf-Vi). If Vfis the total vol occupied by the air at the end and Vi initially (=nRT0/P0), wouldn't work=0? Imagining this, I see that work=0 makes no sense, but I don't see it right now from the definition of work. Also if U is proportional to CV, then does this not assume dV=0, i.e work=0?

    The final answer I got was T0γ. Your method is nice and I would go back to trying it later.
     
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