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Pump compresses air into a tank

  1. Jan 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A cylindrical pump of length 25[cm] pumps air in atmoshpheric pressure ant temp 270C into a big tank. the manometric pressure in the tank is 4[atm]. at which distance will the air start entering the tank.

    2. Relevant equations
    In adiabatic process: ##P_1V_1^\gamma=P_2V_2^\gamma##
    γ for air=1.4

    3. The attempt at a solution
    I assume an adiabatic process. i think that manometric pressure is gauge pressure:
    $$P_1V_1^\gamma=P_2V_2^\gamma\rightarrow V_2=\sqrt[\gamma]{\frac{P_1}{P_2}V^{\gamma}}$$
    $$\rightarrow \ln V_2=\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)^{\frac{1}{\gamma}}=\frac{1}{\gamma}\ln\left(\frac{P_1}{P_2}V_1^{\gamma}\right)$$
    $$\ln V_2=\frac{1}{\gamma}\left[\ln\left(P_1V_1^{\gamma}\right)-\ln P_2\right]=\frac{1}{\gamma}\left[\ln P_1+\gamma\ln V_1-\ln P_2\right]$$
    The volume is area A times length x. for our data:
    $$\ln V_2=\ln A+\ln x=\frac{1}{1.4}\left[\ln 0+1.4\ln (A\cdot 25)-\ln 5\right]=\frac{1}{1.4}\left[1.4(\ln A+\ln 25)-\ln 5\right]$$
    $$\ln A+\ln x=\ln A+\ln 25-\ln 5\rightarrow \ln x=\ln\left(\frac{25}{5}\right)=\ln 5\rightarrow x=5[cm]$$
    It should be 7.9[cm] before the end of the cylinder.
    If i don't assume adiabatic process and only use ##PV=nRT## i have 2 unknowns: the distance and the final temperature.
     
  2. jcsd
  3. Jan 4, 2015 #2
    You forgot to divide ln(P2) by 1.4.
    Also, P1 = 1, so you shouldn't have ln(0) in the equation.
    Also, it was not necessary to work this problem in terms of natural logs.

    Chet
     
  4. Jan 4, 2015 #3
    $$\ln A+\ln x=\ln A+\ln 25-\frac{\ln 5}{1.4}\rightarrow x=7.9[cm]$$
    Which other way besides natural logs?
     
  5. Jan 4, 2015 #4
    [tex]\frac{V_2}{V_1}=\left(\frac{P_1}{P_2}\right)^{1/\gamma}[/tex]
    [tex]\frac{x}{25}=0.2^{(1/\gamma)}[/tex]
    Chet
     
  6. Jan 5, 2015 #5
    Thanks Chet
     
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