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Thermodynamics of a pump

  1. Sep 27, 2009 #1
    Can anyone please help me with this problem?

    Consider the act of evacuating a chamber with a vacuum pump. Vacuum pumping rates are usually given in volumetric pumping speeds, Sp, whose units are typically volume/unit time, e.g., 1 m3/min. In other words, the number of molecules in 1 m3 at the chamber pressure and temperature are removed in one minute. A 0.1 m3 chamber is initially at atmospheric pressure (1 bar) and 298 K. The gas may be considered to be an ideal diatomic gas. Calculate and plot the temperature and pressure in the chamber as a function of time with a pump rated at Sp= 1 m3/min if the pump down process may be considered to be

    (a) adiabatic
    (b) isothermal.

    I've done everything I can think of with energy and entropy balances. I think the problem is the temperature and pressure of the influx stream are changing as the chamber is evacuated so I don't know how to relate them to time. I need an equation for Temperature as a function of time and nothing else and the rest is easy.

    I think dn/dt is the Sp but in units of mol/min with Volume held constant. Now I just need to know T without relying on P (or vice versa)

    Thanks.
     
  2. jcsd
  3. Sep 27, 2009 #2

    Mapes

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    Have you applied the ideal gas law yet? It's not clear from your post.
     
  4. Sep 27, 2009 #3
    Yes, I took the time derivative of [tex]PV=nRT[/tex] holding V constant.

    [tex]V \left(\frac{dP}{dt}\right)=R\left(T \frac{dn}{dt}+n \frac{dT}{dt}\right)[/tex]

    dU=nCvdT

    [tex]\frac{dU}{dt}=Cv\left(n\frac{dT}{dt}+T\frac{dn}{dt}\right)[/tex]

    from energy balance, dU=Hdn integrating gets U=Hn which equals U=(U+PV)n

    Nothing else seems to work. Intuitively I would think the pressure would decrease non-linearly but I see no way of introducing that into the equations.
     
  5. Sep 27, 2009 #4

    Mapes

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    OK, let's apply the energy differential equation to the adiabatic case. Take the system to be the amount of gas that doesn't leave the box. What can you say about [itex]dU[/itex]?
     
  6. Sep 27, 2009 #5
    You get

    [tex]Cv\frac{V}{R}\frac{dP}{dt}=\frac{dU}{dt}[/tex]

    So we can relate U with P but you still don't have an expression for P so only one equation with two unknowns. You can't replace P with an expression for T because T is also unknown.

    Here's what I did.

    [tex]dU=d(Hn)[/tex] but both H and n are not constant so

    [tex]\frac{dU}{dt}=H\frac{dn}{dt}+n\frac{dH}{dt}[/tex]

    [tex]\frac{dU}{dt}=\left(U+PV\right)\frac{dn}{dt}+n\left(\frac{dU}{dt}+\frac{dP}{dt}V\right)[/tex]

    Through some algebra:

    [tex]\frac{dU}{dt}(1-n)-U\frac{dn}{dt}=PV\frac{dn}{dt}+\frac{dP}{dt}nV[/tex]

    V and [tex]\frac{dn}{dt}[/tex] are given. I think the [tex]n[/tex] in this case is the total number of moles you start out with so that is known from [tex]PV=nRT[/tex]. Now I have two first order ODE that I can't solve.
     
  7. Sep 27, 2009 #6

    Mapes

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    I was thinking more like, the rate of work that that amount of gas does is [itex]P(dV/dt)[/itex], which is [itex]PF[/itex] where [itex]F[/itex] is the volumetric flow rate.

    It was easier for me to start with the isothermal case. What's [itex]dn/dt[/itex], where [itex]n=PV/RT[/itex]? What's [itex]dn/dt[/itex] in terms of the volumetric flow rate? Set these equal and integrate to solve for pressure.
     
  8. Sep 27, 2009 #7
    But this is what I don't understand. The volumetric flow rate is not constant because the tank is draining so presumably the amount of molecules flowing out of the tank is changing with time. So I don't know what the volumetric flow rate is. I understood the Sp in the question to mean, it takes 1 minute to evacuate a 1 m3 chamber so it would take a tenth of that time to evacuate a 0.1 m3 chamber.
     
  9. Sep 27, 2009 #8

    Mapes

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    The volumetric flow rate is constant, but the molar flow rate is not constant. A better way to think of it is that 17% of the chamber's contents are removed every second. Employing the continuum assumption, the chamber is never completely emptied.
     
  10. Sep 28, 2009 #9
    I still don't quite follow what you're suggesting. The volume does not change because we took the system to be the inside of the chamber which is rigid. Using [tex]dV/dt[/tex] would only work if the chamber was not rigid. But then if the volumetric flow rate is not the molar flow rate than it has no place in any equation and that wouldn't make sense.

    [tex]n=\frac{PV}{RT}[/tex]

    [tex]\frac{dn}{dt}=\frac{1}{RT}\left(V\frac{dP}{dt}+P\frac{dV}{dt}\right)[/tex]

    And the rate of work can't be [tex]P\dot{V}[/tex] because pressure is not constant.
     
  11. Sep 29, 2009 #10

    Mapes

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    OK, let's back up a second. For the isothermal case, I'm suggesting you consider the system to be the chamber, determine [itex]dn/dt[/itex] by the ideal gas law (which you did in your last post, [itex](V/RT)(dP/dt)[/itex]), and equate that to [itex]dn/dt[/itex] determined by the volumetric flow rate [itex]F=dV_0/dt[/itex]. You can use the ideal gas law to get this molar flow rate because the gas removed has a constant volume and follows the ideal gas law. Set these two molar flow rates equal and solve for [itex]P[/itex] as a function of time.

    For the adiabatic case, I'm suggesting you take the system to be the amount of gas that stays in the chamber as it pushes out an infinitesimal amount of gas during evacuation. This system is adiabatic and does work on the exiting gas, so its energy decreases, and the rate of energy decrease is [itex]P\,dV_0/dt[/itex]. Stick in the relationship between specific heat and energy and continue from there.
     
  12. Sep 29, 2009 #11
    ok, I got it.

    I asked the TA and he said the key relationship that I was missing is this:

    [tex]\frac{dn}{dt}=n\frac{Sp}{V}[/tex] The units work out.

    So I worked it out from there. What you can do then is say [tex]U=H_{out}n_{out}[/tex] since [tex]H_{in}n_{in}[/tex] is zero because there is no mass flowing into chamber.

    Then you can say dU is nCvdT so dU/dt is same as above post. Equate [tex]nCvdT[/tex] with [tex]H_{out}n_{out}[/tex]n (you can replace H with U+RT) and you take it from there. I worked it out and basically you get a P=noRTo/V times an exponential e.
     
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