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Cylinder with piston, stationary state

  1. Aug 25, 2014 #1
    1. The problem statement, all variables and given/known data

    We have a horizontal cylinder with an ideal piston, both are made from a heat nonconductive material. There is 10 liters of steam at 3 bars and 200 degrees Celsius in the cylinder. The cylinder is surrounded with atmosfere with pressure of 1 bar and temperature at 20 degrees Celsius.

    At first we hold the piston, then we release it and wait for a standstill. What is the final temperature of steam at that state? How did the enthropy change?

    In the second part, we put the system in a stationary electric field E. What is the final temperature of steam in the stationary state of piston?

    [tex]E = 5\times 10^{9} \frac{V}{m}[/tex]
    [tex]a = 4 \times 10^{-13} \frac{Asm^2 K}{V}[/tex]
    so we calculate
    [tex]c_p = 1850\frac{J}{kg K}[/tex]
    [tex]M = 18\frac{kg}{kmol}[/tex]
    [tex]p = 3\text{ bar}; V = 10 l;T = 200°C;[/tex]
    [tex]p_0 = 1\text{ bar}; T_0 = 20°C;[/tex]

    2. Relevant equations

    [tex]p_e = \frac{a E}{T}[/tex]

    3. The attempt at a solution
    I solved the first part, not sure if it is correct. Could there be a phase change of steam? I don't know how to calculate that.

    [tex]dU = dQ + dW[/tex] and [tex]dQ = 0[/tex]

    so I integrated

    [tex] c_v dT = -\frac{R T}{M V}dV[/tex]

    from [itex]p, V, T[/itex] to [itex]p_0, V_1, T_1[/itex].

    The solution is
    [tex]T_1 = T \left ( \frac{p_0}{p}\right )^{\frac{\beta}{\beta+1}} \approx 360 K[/tex]
    where
    [tex]\beta = \frac{1}{\frac{M}{R}c_p - 1}[/tex].

    I calculated enthropy using
    [tex]\delta S = m c_p \ln \left(\frac{T_1}{T}\right) - \frac{m R}{M} \ln\left( \frac{p_0}{p} \right) \approx -0.7 \frac{J}{K}[/tex]

    For the second part, when we turn on the electric field [itex]E[/itex], I get a term in [itex]dU = dW = -pdV - E dp_e[/itex], so I need to integrate

    [tex]\frac{mR}{MV}dV = \left( \frac{a E^2}{T^3} - \frac{m c_v}{T}\right) dT[/tex]

    but this makes little sense as [itex]T_1[/itex] cannot be calculated easily from this?
     
    Last edited: Aug 25, 2014
  2. jcsd
  3. Aug 28, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Aug 29, 2014 #3

    rude man

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    You're implicitly assuming a quasi-static adiabatic process; unfortunately this is not such a process.

    Also, your entropy calculaton is incorrect. Since dQ = 0 what is delta entropy?

    There is no phase change of the steam since work BY the system is obviously positive so internal energy change is negative whereas it takes positive work ON the system to compress steam into water adiabatically.

    Chet, TSny et al - help!
    Would a massive piston approximate a quasi-static process? Can the problem not be solved unless a quasi-static process is assumed?
     
  5. Aug 29, 2014 #4
    But there's nothing that could exchange heat with steam, it says that the piston and the interior doesn't conduct heat. Weird.
     
  6. Aug 29, 2014 #5

    rude man

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    I've asked for help on this. Stand by!
     
  7. Aug 29, 2014 #6
    Hi Dave84. It looks like you approached the problem basically correctly, except for assuming that the process is reversible. This is not the case if the gas is allowed to expand against the constant confining force per unit area of 1 bar. In this case, the work should be (see my PF blog before it disappears on 9/2):

    ##W = P_{ext}(V_2-V_1)=p_0(V_2-V_1)##

    where V2 is the final volume, V1 = 10 l, and p0=1 bar. If you substitute this into the first law energy balance, you get:

    [tex]nC_v(T_2-T_1)=-p_0(V_2-V_1)=p_0(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1})=-nR(T_2\frac{p_0}{p_2}-T_1\frac{p_0}{p_1})[/tex]

    where p1 = 3 bars, T1=273K, and Cp is the molar heat capacity. In addition, the final pressure will be p2=p0=1 bar. So,

    [tex]C_v(T_2-T_1)=-R(T_2-T_1\frac{p_0}{p_1})[/tex]

    So, from this equation, you can calculate the final temperature.

    You used the correct equation to calculate the entropy change, but, for the reversible case you ran, the change in entropy should have come out to zero. For the irreversible case we are looking at now, the entropy change should come out positive (by Clausius's inequality). See what you get.

    Chet
     
  8. Aug 29, 2014 #7

    rude man

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    @Chet, thanks for reminding that just because dQ = 0 that doesn't mean delta S = 0 if the process is irreversible!

    However, I fail to see why W = p0(V2 - V1). Doesn't the mass of the piston enter into this? If the piston has finite mass then the pressure p on it from the inside will certainly not be p0 all the time. Will it not start at 3 at., accelerate the piston, which, in the absence of friction, will "eventually" oscillate about a value V2 depending on the mass of the piston? Without friction the piston would retain kinetic energy indefinitely. So some of the work done by the gas goes into K.E. of the piston. During an oscillating cycle, whenever p = p0, the K.E. is in fact at its maximum. Make any sense?
     
  9. Aug 29, 2014 #8
    This is a very good question. Actually, even without friction, the piston will eventually come to rest as a result of viscous dissipation in the gas. The oscillation of the piston will be damped by the viscous dissipation. When the piston finally does come to rest, the system will attain its final thermodynamic equilibrium state, and the energy change associated with the piston will be zero. So the equation W = p0(V2 - V1) will give the total amount of work done by the gas on the surroundings when the system reaches its final thermodynamic equilibrium state.

    Chet
     
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