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Change of pressure, temperature at constant volume?

  1. Oct 2, 2012 #1
    Basically I have a question regarding the heat capacity of a gas and the ideal gas law.

    The heat capacity equation is:

    Q = C(T_f - T_i)

    The ideal gas law equation is:

    PV = nRT.

    The system has n moles of some gas, for example hydrogen in a container with constant volume. The container is heated with Q joules of energy, How do I relate the amount of heat absorbed by the container (adiabatically) being heated to the change in pressure? I assume that for n moles of gas I would need to use the specific heat for hydrogen at constant volume instead so the heat capacity equation becomes,

    Q = nC_v(T_f - T_i)

    I'm also wondering about using the heat conduction equation where:

    P_conductivity = Q/t = (kA(T_f - T_i))/(L)

    for the walls of the container, although only one side is being heated,
    so the container has a cubic shape and the Area is A and the thickness of the wall is L.
    Would the time required to heat up the gas in the container also require the above
    conduction equation and if so what would be the time required for a certain amount of power to change the pressure of the gas at constant volume.

    Basically I'm having trouble relating the ideal gas law to the conductivity equation.


    thanks.
     
    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 3, 2012 #2
    It doesn't make sense to say "How do I relate the amount of heat absorbed by the container (adiabatically) "

    if it is adiabatic it means no heat, by definition.

    As far as the heat conduction is concerned, this is a tricky problem and you have to define it well. Even if you are "heating" one side of the container, the sides of it will get heated as well by conduction. You also have to define your heat source well, and also how the gas is absorbing heat from the container.
     
  4. Oct 3, 2012 #3
    Thanks for catching me on that one, my post is ripe full of errors, I did edit some other aspects before that one but didn't catch it last night, maybe out of being tired.

    Thanks for letting me know its a tricky problem. At first I thought I must have not studied well is school no being able to solve something so simple.

    Anyways lets pretend for simplicity the walls of the cubic container only allow one face to conduct heat to make the problem simpler. Would I need to solve the conduction equation for the type of wall (say iron) and also the gas in the container (hydrogen); equating the two together,

    (iron) Q/t = (kA(T_f - T_i))/(L) = Q/t = (kA(T_f - T_i))/(L) (hydrogen gas).

    Assuming the above model is correct then how would the pressure of the gas change, considering the specific heat at constant volume for hydrogen? And also the amount of power it would take to heat up the gas to a certain temperature?
     
    Last edited: Oct 3, 2012
  5. Oct 3, 2012 #4
    This is a difficult (but not impossible by any means) problem if you need to consider the effects of temperature variations within the cube. The pressure will be, for all practical purposes, constant within the enclosure, but, because of the temperature variations, the density will vary. These spatial variations in density will cause natural convection flow within the enclosure, so you would have to include the fluid dynamics in the modeling calculations. So let's rule out this scenario out until you have more experience.

    You can make sure that the temperature within the enclosure is a constant throughout by running an ideal mixing rod within the container, and assuming that the viscous heating work done by the mixing rod is negligible compared to the power load. So the temperature on the inside wall of the container is the gas temperature, and the temperature at the outside of the heated wall is whatever temperature you impose. Then the rate of heat flow is given by the heat conduction equation, and this is equal to mCv dT/dt. So with constant imposed outside temperature, you have a differential equation to solve for the temperature as a function of time. If you impose a constant power input rate at the wall, then the temperature within the enclosure will increase linearly with time such that mCv dT/dt is equal to the power input. In this case, you would have to raise the imposed outside temperature linearly with time by the same amount to keep the power input constant. So, this is how the heat conduction equation couples with the internal energy equation for the gas in the vessel.
     
  6. Oct 4, 2012 #5

    Philip Wood

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    Gold Member

    Random: Your use of Ti and Tf in the thermal conductivity equation worries me. This is because you've used the same notation in the heat capacity equations. In these equations, the notation is good, with the suffices i and f standing for initial and final. But in the thermal conductivity equation the relevant difference is that between either face of the conductor (e.g. between the inside and outside of the container wall). It isn't a case of initial and final temperatures.
     
    Last edited: Oct 4, 2012
  7. Oct 4, 2012 #6
    I know I wanted to keep it simple. I'll just assume that the cube is small enough so that the temperature variations from the top to the bottom are maybe within 10 degrees Kelvin of each other and that should be reasonable. I'm pretty sure this problem could get the royal treatment but so far I have this instead.

    The temperature of the inside wall is the lower temp and the number of Joules of energy transferred by an adiabatic flame are also known so I rearranged the conductivity equation as follows,

    [itex]\frac{P_cL}{Ak}[/itex] = [itex]\frac{QL}{tAk}[/itex] - THigh = -TLow

    and then rearranging the ideal gas law in terms of P knowing TLow,n,R,V,

    P = [itex]\frac{ nRT_Low }{V}[/itex].

    I'm just not sure if the specific heat capacity is important somewhere. My guess is that it would take time to heat up a gas of n moles and the specific heat would go something like this knowing Q and Tfinal = TLow,
    Q = cn( Tfinal - Tinitial)
    with Q being the same as the one above? Not sure since the conductivity factor k only allows a certain amount of energy into the container which is less than the total amount Q?

    Thats what I was thinking although wouldn't an approximation like mine work though. Otherwise could you give more detail in the calculation of it. thanks.

    I think I fixed it in the above post. Feel free to mention any errors to me if its wrong though.

    thanks guys for the replies so far.
     
  8. Oct 5, 2012 #7
    Let's be more specific. If all the sides of the cube are insulated except for the base, and the thickness of the container at the base is h, and the gas within the cube is ideally well mixed, and the temperature at the bottom of the base is Toutside, and the instantaneous temperature within the cube is T, then the time-dependent variation of T will be given by:

    mC (dT/dt) = (kA/h) (Toutside - T)

    where m is the mass of gas within the container, C is the heat capacity of the gas at constant volume, k is the thermal conductivity of the base, and A is the area of the base. The initial condition is T = Tinitial at t = 0.

    If the gas in the cube is not ideally well mixed, then there will be temperature variations within the container that must be taken into account by solving the transient differential heat balance equation. This includes the effects of differential heat conduction within the gas, based on the thermal conductivity of the gas. Even in the absence of gravity, where natural convection effects would no longer be present, you would still have to take into account the limited redistribution of mass within the cube as a result of gas expansion coupled with the non-uniform temperature distribution. To do this, you would have to use a modified version of the transient heat conduction equation to account for the modified mass distribution. Also, in this case, because of the much lower thermal conductivity of the gas compared to the container base (which is presumably metal), the main resistance to heat transfer would reside in the gas, rather than the wall. The temperature in the cube would not only be a function of time, but it would also be a function of position, which you would necessarily have to solve for. As I said, this is a doable problem, but you will need to take a course in transport phenomena to be able to set up the equations and solve it. It is also possible to do this with less fundamental experience by using a packaged computational fluid dynamics code, but there is a significant learning curve associated with using such computer codes too.
     
  9. Oct 5, 2012 #8
    I prefer the non numerical method so I'll just use what you wrote above, But then what temperature do I relate to the pressure P = nRT/V, where T is the variable in your equation.
    I guess the above must be solved as a first order differential equation so it will be Eulers constant raised to the formula on the right hand side (by inspection, not exactly). But I still have trouble seeing how that solves for T inside the container since the variable is on the other side of the equation (on the right) and you're solving for dT/dt on the left side.

    Thanks for the info, I'll just try to approximate with a simple differential equation as much as possible.
     
  10. Oct 5, 2012 #9
    I already said in my previous response that T is the gas temperature in the cube. Need I say more?

    I'm not going to solve this simple differential equation for you. I leave that up to you.

    Note that, because of the assumed ideal mixing within the cube, this solution will correspond to the limiting case of the fastest you can possibly heat up the gas. In actual situations where the gas in the cube is not ideally mixed, the actual heating time will probably be much longer.
     
  11. Oct 5, 2012 #10
    I rearranged the formula you gave me accordingly as follows,

    mC_v(dT/(T_o - T)) = (kA)/h dt

    and I integrated both sides and got,

    mC_v LN(T_o - T) = ((kA)/h)t +C

    I hope I'm right. There doesn't seem to exist a simple rule for dealing with LN(T_o - T). I'm trying to move the T outside somehow in order to write the equation in terms of T. Please help.


    Nevermind: I think its:

    T_o - T = e^((((kA)/h)t +C)/mC_v) ->
    T = T_o - e^((((kA)/h)t +C)/mC_v)

    Let me know if I'm correct or completely wrong, thanks for the help so far. I also wondered what the solution might be if the time to heat up the gas would take long. Would it be possible to do it without numerical simulation or is that really a CFD problem?
     
    Last edited: Oct 5, 2012
  12. Oct 5, 2012 #11
    In obtaining the following equation, you made a sign error in the integration. Take the derivative of both sides with respect to t to check why.

    mC_v LN(T_o - T) = ((kA)/h)t +C

    The constant of integration C should next be obtained by applying the initial condition. This will give you another ln term, which you should then combine with the ln on the left hand side before inverting the ln.
     
  13. Oct 5, 2012 #12
    Starting from scratch I'll do this again but the result is a little funny at the end since the temperature inside the container seems to cancel itself????

    anyways,

    mC(dT/dt) = ((kA(T_o-T_i))/h)

    rearranging the above,

    mC(dT/(T_o-T_i)) = (kA/h)dt

    integrating both sides,

    -m C LN(T_outside - T_inside) = (kAt)/h + C

    Setting t = 0 and T_outside = T_initial = 0 (essentially)

    the integrating factor becomes,

    C = -mC LN(T_initial - T_inside) , this might have been the mistake btw.

    next I plug back the integrating factor into the 3rd equation from the top,

    -mCLN(T_outside - T_inside) + mCLN(T_initial - T_inside)= (kAt)/h

    rearranging the specific heat terms and raising both sides to e,

    -(T_outside - T_inside) + (T_initial - T_inside) = e^((kAt)/(hmC))

    and T_inside cancels for some reason,

    -T_outside + T_initial = e^((kAt)/(hmC))

    what am I doing wrong?
     
  14. Oct 5, 2012 #13
    You need to review Algebra.

    ln(A) - ln(B) = ln(A/B)
     
  15. Oct 6, 2012 #14
    Facepalm! I even looked at that ten minutes before making that mistake, so trying again lets hope this is correct.

    mC(dT/dt) = ((kA(T_o-T_i))/h)

    rearranging the above,

    mC(dT/(T_o-T_i)) = (kA/h)dt

    integrating both sides,

    -m C LN(T_outside - T_inside) = (kAt)/h + C

    Setting t = 0 and T_outside = T_initial = 0 (essentially)

    the integrating factor becomes,

    C = -mC LN(T_initial - T_inside) , this might have been the mistake btw.

    next I plug back the integrating factor into the 3rd equation from the top,

    -mCLN(T_outside - T_inside) + mCLN(T_initial - T_inside)= (kAt)/h

    dividing both sides by mC and setting (kAt)/(mCh) -> X

    -LN(T_outside - T_inside) + LN(T_initial - T_inside) = X

    using LN(A/B) = LNA - LNB and raising both sides to e,

    ((T_initial - T_inside)/(T_outside - T_inside)) = e^X

    rearranging terms,

    (T_initial - T_inside) = (T_outside - T_inside)e^X

    again...

    (T_initial - T_inside) = (T_outside)e^X - (T_inside)e^X

    and again...

    ( -T_inside) + (T_inside)e^X = (T_outside)e^X - T_initial

    and finally...

    (T_inside)(e^X - 1) = ((T_outside)e^X - T_initial)

    which yields...

    (T_inside) = ((T_outside)e^X - T_initial)/(e^X - 1) where X = (kAt)/(mCh)

    please let me know where I goofed up this time, somethings not right.
     
  16. Oct 6, 2012 #15
    You did goof up again. I'm not going to go through your "arithmetic" to find the error, but the correct answer should be:

    T = Tinitial exp (-X) + Toutside (1 - exp (-X))
     
  17. Oct 7, 2012 #16
    Well thanks Chester, I'll try looking through it to make sense of my post.


    I checked every possible way now and I have no idea where I went wrong. Could you share your derivation?
     
    Last edited: Oct 7, 2012
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