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To find final temperature when an evacuated bottle is opened

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    The internal energy of air is given, at ordinary temperatures, by

    ## u= u_o + 0.718t ##
    where ##u## is in kJ/kg, ##u_o## is any arbitrary value of ##u## at 0oC, kJ/kg, and t is temperature in oC.

    Also for air,
    ##pv= 0.287(t+273)##

    where p is in kPa and ##v## is in m3/kg.
    (a) An evacuated bottle is fitted with a valve through which air from the atmosphere, at 760mm of Hg and 25oC, is allowed to flow slowly to fill the bottle. If no heat is transferred to or from the air in the bottle, what will it's temperature be when the pressure in the bottle reaches 760mm of Hg ?

    Ans 144.2oC

    (b) If the bottle initially contains 0.03 m3 of air at 400mm Hg and 25oC, what will the temperature be when the pressure in the bottle reaches 760mm Hg ?

    Ans 71.6oC

    2. Relevant equations

    ## u= u_o + 0.718t ##

    ##pv= 0.287(t+273)##

    ##PV= mRT ##

    ##Q-W= \Delta u ##

    Q- heat transfer
    W- work transfer
    ##\Delta u ## = Internal energy change.

    3. The attempt at a solution

    (a)

    ##P_1 = 13600*9.81*0.760 = 101396.16 Pa##
    ##T_1## = 25oC

    ##u_1 = u_o +0.718(25) = u_o +17.95 kJ/kg ##

    To find ##T_2##

    ##Q-W= \Delta u ##

    Q= 0 as no heat is transferred , W=0 no work is done in free expansion ??

    So ##\Delta u## has to be zero and T1 =T2 ??

    I don't know how they got an increase in temperature.

    (b)
    ##V_1 = 0.03 m^3 ##
    ##P_1 = 13600*9.81*0.400 = 53366.4 Pa##
    ##T_1 = 298K ##
    ##P_2= 13600*9.81*0.760 = 101396.16 Pa##

    To find ##T_2 ##

    Since volume of the bottle is constant ##V_1= V_2## ??

    Initial mass = ##m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936 ## kg

    Final mass = ##m_2 =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2} ## kg ...is this useful ???



    Can i use ## \frac{P_1 V_1} {T_1} = \frac{ P_2 V_2} {T_2} ## ??

    I get T_2 = 566.2 K ......clearly wrong , how to continue ?














     
  2. jcsd
  3. Nov 12, 2016 #2
    This is an example of a problem in which you are not dealing with a closed system, because mass is entering the system. So you can't use the closed-system version of the first law of thermodynamics to solve it. Are you familiar with the open-system (control volume) version of the first law of thermodynamics? If so, please write the equation.
     
  4. Nov 12, 2016 #3

    I like Serena

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    Homework Helper

    Hi Monsterboy! :oldsmile:

    It's not free expansion, which is happens when the fluid is already inside the bottle.
    Instead the outer atmosphere does flow work on the fluid as it moves inside the bottle.
    On one unit of mass, the flow work is the pressure outside ##P_0## times the volume displaced, which is ##v##.
    So the final internal energy is ##u_{final} = u_{initial} + P_0 v##.
     
    Last edited: Nov 16, 2016
  5. Nov 12, 2016 #4
    You mean ## \dot m_1 (h_1 + \frac{V_1^2} {2} +Z_1 g) + \frac{dQ} {dT} = \dot m_2 ( h_2 + \frac{V_2^2} {2} +Z_2 g) + \frac {dW} {dT} ## ??

    (a)
    As heat transfer is absent

    ## \frac {dQ} {dT} = 0##

    ## h_1 =h_2 ## ??
     
    Last edited: Nov 13, 2016
  6. Nov 12, 2016 #5
    No , in problem (a) the bottle is initially evacuated and air is slowly allowed to flow inside.
     
    Last edited: Nov 12, 2016
  7. Nov 12, 2016 #6
    No. This is not a steady state problem. You omitted a key term from this equation.
    No. There is no stream leaving the bottle, so ##m_2=0##
     
  8. Nov 13, 2016 #7
    I am not familiar with the unsteady state equation , is the term i omitted related to pressure energy ? like ##\frac{1}{2} \rho v^2 ## or ##\frac{P}{\rho}## something?
     
  9. Nov 13, 2016 #8

    I like Serena

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    Homework Helper

    Indeed. It's not free expansion. Instead...

    The first law states that energy is conserved.
    That is:
    $$U_{initial} + E_{in} = U_{final} + E_{out}$$
    where
    $$E_{in} = m_{in}(h_{in} + \frac{\mathscr V_{in}^2}{2} + Z_{in} g) + Q$$
    and
    $$E_{out} = m_{out}(h_{out} + \frac{\mathscr V_{out}^2}{2} + Z_{out} g) + W_{shaft}$$
     
    Last edited: Nov 15, 2016
  10. Nov 13, 2016 #9

    Andrew Mason

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    Think of the process in a) in two parts starting with two bottles, one full at P = 760 mm Hg; T = 298K; V = V0. and the other empty. The bottles are connected and air is allowed to flow between them. The first part is a free expansion which results in both bottles at P = 380 mm Hg and T = 298K; V = 2V0. Then you adiabatically compress the air into one of the bottles. Use the adiabatic relation to determine the temperature when the pressure reaches 760.

    AM
     
  11. Nov 13, 2016 #10
    In your control volume first law equation, you omitted the rate of accumulation of internal energy within the bottle. The open-system version of the first law for part (a) reduces to:
    $$h_{in}\frac{d(m_{in})}{dt}=\frac{d(mu)}{dt}$$where m is the mass of air in the bottle at time t, u is the internal energy per unit mass in the bottle at time t, ##h_{in}## is the enthalpy per unit mass of the air entering the bottle, and ##d(m_{in})/dt## represents the rate at which mass of air is flowing into the bottle. For part (a), if we integrate this equation with respect to time, we get:
    $$m_{in}h_{in}=mu$$where ##m_{in}## is the cumulative mass of air that has flowed into the bottle up to time t and m is the mass of air in the bottle at time t. How are ##m_{in}## and m related? What does that leave you with?
     
  12. Nov 13, 2016 #11
    In my judgment, this is not correct. Certainly the adiabatic compression will require work to be done, and that work should not be considered reversible (as required by the adiabatic relation). The amount of work that the surrounding air does in forcing air into the bottle is at constant pressure, and, in passing through the valve, the air undergoes an irreversible Joule-Thompson pressure reduction. It would be interesting to see what this approximation provides in comparison with the correct control volume (open system) application of the first law of thermodynamics.

    This student is in the process of learning about the open-system (control volume) version of the first law of thermodynamics, and he should be applying that specific approach to this problem.
     
  13. Nov 13, 2016 #12
    ##m_{in} h_{in} =mu ##

    ## m_{in}[ u_{in} +p_1v_1] = mu ##

    Since ## P_1v_1 = RT_1 ##
    ## m_{in}[u_o +718(25) + 287(298) ] = mu ##

    ##m_{in}[ u_o +103476] =mu = m[u_o + 718(25)] ##

    am i going the right way ?
     
  14. Nov 13, 2016 #13
    Almost. Since the amount of air that enters is equal to the amount of air in the bottle, ##m=m_{in}##. I guess you are assuming that the m's are in grams, rather than kg, correct? Finally, the term in brackets on the right hand side should have a T instead of a 25, where T is the final temperature in C.
     
  15. Nov 14, 2016 #14
    Thanks , got the first problem

    ## T = \frac{103476} {718} ## = 144.116oC
     
  16. Nov 14, 2016 #15
    (b)

    Initial mass = ##m_1 = \frac{P_1 V_1} { RT_1} = \frac{53366.4*0.03} {287*298} = 0.01871936 ## kg

    Final mass = ##m_{final} =\frac {P_2 V_2} {RT_2} = \frac{101396.16*0.03} {287*T_2} = \frac {10.5890} {T_2} ##

    $$ m_{in}h_{in} + m_1u_1 = m_{final}u_{final}\tag{1} $$ correct ?

    ## m_{in}[u_{in} + 287(298)] + 0.01871936[ u_o + 0.718(25)] = \frac {10.5890} {T_2}(u_o +0.718T_2) \tag{2}## correct ?
     
    Last edited by a moderator: Nov 14, 2016
  17. Nov 14, 2016 #16
    Yes.
    Yes, but a couple of corrections. In the first term on the left hand side, replace ##u_{in}## with ##u_o##. For the term on the rhs, ##T_2## from the ideal gas law should be replaced by ##(T_2+273)##. From a mass balance, how are ## m_{in}##, ##m_1##, and ##m_{final}## reltated?
     
  18. Nov 14, 2016 #17
    ##m_{in}=m_{final}-m_1 ## ??

    ##( \frac{10.589}{T_2 + 273} - 0.01871936)(u_o +85526) + 0.01871936[u_o + 17.95] = \frac{10.589}{T_2+273} (u_o +0.718T_2) ## ??

    ##u_0## terms cancel out ,I am getting ##T_2 ## =291.4oC , something is wrong.
     
  19. Nov 14, 2016 #18
    Shouldn't ##h_{in}## be given by $$h_{in}=u_o+(0.718)(25)+(0.287)(298)$$?
     
  20. Nov 15, 2016 #19
    Yea , i got it ,thanks for the help.

    I want to find the work done by the gas in both (a) and (b) , "I like Serena" said it is not zero in the case (a) as well ,i don't how, i thought when a gas freely expands into an evacuated chamber zero work is done.

    In (b) work is done against the pressure of 400mm Hg ? so is it ##(m_2- m_1)(p_1v_1) ## ?

    where ##p_1## and ##v_1 ## are initial values inside the bottle.

    In the first case ##m_1 =0 ## so it is ##m_{in}(0) =0 ? ##
     
    Last edited: Nov 15, 2016
  21. Nov 15, 2016 #20
    We need to be more precise here. You are aware that this problem is an exercise in applying the open-system (control volume) of the first law, correct? You are aware that, in the open-system version, the work is subdivided into two distinct parts (a) the work done to push fluid into and out of the control volume and (b) all other work (typically called shaft work), correct? You are aware that the former is included in the enthalpy of the entering and exit streams, correct?

    Are you trying to also interpret this problem in terms of the ordinary closed system version of the first law (which is mathematically equivalent to the open system version)? If so, you need to precisely define the constant mass included in your system in order to calculate the work done on the system. The work changes when you change what you call the system.
     
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