1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Internal energy when filling an evacuated tank

  1. Sep 4, 2013 #1
    This is the example 2.12 from "Introduction to Chemical Engineering Thermodynamics", 6th. ed, by Smith, Van Ness & Abbott:

    "An evacuated tank is filled with gas from a constant-pressure line. What is the relation between the enthalpy of the gas in the entrance line and the internal energy of the gas in the tank? Neglect heat transfer between the gas and the tank."

    The solution is that the internal energy of the gas in the tank is equal to the enthalpy of the gas in the entrance line.

    Although I could solve the exercise, I can't say I've really understood/accepted the result.

    Two objections I have to it:

    a) the gas filling the evacuated tank is not doing any work, so it seems to me that the product PV (H = U + PV) is of no interest here.

    b) I can imagine that the entrance line is connected to a huge tank that supplies gas at the same pressure available at the line and that the filling of the evacuated tank will have a neglegible effect over that huge tank. When the flow ceases, the gas at both reservoir will have the same properties, equal to the original ones of the huge tank. As a consequence, the internal energy of the gas in the tank will be equal to the internal energy of the gas in the entrance line, and not equal to its enthalpy.

    Can anyone please clarify which concept I have not understood correctly?
     
  2. jcsd
  3. Sep 5, 2013 #2
    Point a) the gas has work done on it so it enters into the tank. ( from the gas in the line behind it )
     
  4. Sep 5, 2013 #3
    Since the tank is an evacuated one, the gas entering the tank is submited to a free expansion, and during free expansion, no work is done by the gas, is it?
     
  5. Sep 6, 2013 #4
    In a free expansion there is no back pressure - the volume increases and the pressure decreases, and temperature stays the same.

    With filling a tank, once some gas has entered the tank, any more gas entering into the tank has to do some work to compress that gas already in the tank.
     
  6. Sep 6, 2013 #5
    I agree with you: "once some gas has entered the tank, any more gas entering into the tank has to do some work to compress that gas already in the tank." and this leads me to rephrase the objection (a) written above:

    a) the gas filling the evacuated tank has to do some work to compress that gas already in the tank. This work is dependent on the pressure of that gas already in the tank. However, when deriving the result that the internal energy of the gas in the tank is equal to the enthalpy of the gas in the entrance line, it is assumed that the work PV is constant, with P equal to the pressure in the entrance line.
     
  7. Sep 6, 2013 #6
    Hi Thomas
    I have a little more time to give a more comprehensive answer.

    Actually there are 2 ways to approach this problem: 1. by considering a control volume around the evacuated tank, and 2. considering the gas in the pipe as a system.

    Since you mention work = PV = constant, that is the system approach.

    So, a system boundary is put around all the gas that will enter the evacuated tank, and this will be a moving system boundary.
    The gas behind the tank does work on this system, and by moving the boundaries of the system, does work of
    W = -P1 V1 ( work being done on the system )

    This is the same as if the gas behind the system is acting as a as piston with force F a distance x.
    ie W = - F/A xA = Fx.
    Perhaps if you consider that a force on a mass can move the mass a distance x at constant velocity if there is an opposing force of equal magnitude, OR accelerate the mass if there is no opposing force. The gas in a system in the line has opposing force so moves at constant velocity. The system gas entering the evacuated tank is unopposed so there should be some change from state 1 in the line to state 2 in the tank.

    ( PS. do not take that too too far, since gases are not solid objects. )

    From the first law: 1Q2 = 2U1 + 1W2
    where
    1Q2 is the heat added to the sytem from state 1 to 2 = 0
    2U1 is the change in internal energy
    1W2 is the work done on the system from state 1 to 2

    therefor
    0 = U2 - U1 - P1V1

    0 = U2 - ( U1 -P1V1 )

    0 = U2 - H1

    or U2 = H1

    Is that how your book derived the concept?

    PS. It is usually common to use specific internal energiy and enthalpy, but my latex isn't working.
     
    Last edited: Sep 6, 2013
  8. Sep 10, 2013 #7
    256bits,

    Thank you very much for your time!

    Your hints on
    "considering a control volume around the evacuated tank and considering the gas in the pipe as a system" and
    "if you consider that a force on a mass can move the mass a distance x at constant velocity if there is an opposing force of equal magnitude, OR accelerate the mass if there is no opposing force"
    really opened my eyes to how to consider the problem.

    I may suggest for another person facing this problem to read also the section 6.5 of Fundamentals of Thermodynamics 6th ed, by Gordon J. Van Wylen, Richard E. Sonntag, Claus Borgnakke, where it's tackled the same two ways you have described.

    Best regards,
    Thomas
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Internal energy when filling an evacuated tank
  1. Internal Energy (Replies: 1)

Loading...