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Why do we use Cp while finding the work of the compressor?

  1. Oct 29, 2015 #1
    why do we use Cp value while finding work of the compressor or the expander i.e
    W=CpdT=H2-H1 H= enthalpy
    when we know pressure inside the expander or compressor drops
    it basically is isentropic process why not we use isentropic work eqn
  2. jcsd
  3. Oct 29, 2015 #2
    Hi Ali, welcome to PF!

    First of all, assuming ideal gas behavior. Heat capacity at constant pressure, Cp is a characteristic property of the working fluid, so, for ideal gases
    [tex]\Delta H = \int_{T_1}^{T_2} C_p \ dT[/tex]
    So, for any ideal gas undergoing any kind of process (isentropic, isobaric, etc.) ΔH will always be given by the above equation.

    Regarding the isentropic work equation you mention, I assume you're talking about this one
    [tex]W = \frac{\gamma \ R}{\gamma - 1} (T_2 - T_1)[/tex]
    Again for ideal gases, it can be shown that [itex]\frac{\gamma \ R}{\gamma - 1} = C_p[/itex], so the two equations you mention are actually the same.
    Last edited: Oct 29, 2015
  4. Oct 29, 2015 #3
    Are you familiar with the open-system version of the First Law of Thermodynamics. If so, what is the equation for the case of steady-state operation?

  5. Oct 30, 2015 #4
    Its Sum of all energies coming inside a control volume is equal to sum of all energies going out of the control volume (if energy generation per unit volume is zero)
    i.e Q-W=DeltaH+deltaKE+deltaPE
  6. Oct 30, 2015 #5
    How about this ?

    Attached Files:

  7. Oct 30, 2015 #6
    OK. Then you know that, if the changes in KE and PE are negligible and the device is adiabatic, then ##\dot{W}=-\dot{m}\Delta h##.

  8. Oct 30, 2015 #7
    Thanks A lot Sir
  9. Oct 30, 2015 #8
    Ok would this equation mean that, the work done by the compressor is partly used in rising the internal energy of the fluid and partly giving rise to the pressure
    then can we express it as
    dW=dU+Vdp (In case of liquids the volume remains constant and only pressure changes)
    Is this correct or not sir ?
  10. Oct 30, 2015 #9
    No. It means that part of the shaft work goes into raising the internal energy of the gas, and the rest goes into drawing the gas in at a low pressure and pushing it out at a higher pressure. The equation dW=dU+Vdp is not correct. Both the specific volume and the pressure of the gas change, so dH = dU+d(pV).

    I don't like the idea of using the differential of the shaft work at any time, because the shaft work is generally path dependent. Of course, since H is a state function, differentials can be used when they refer to differences between closely neighboring thermodynamic equilibrium states (i.e., changes along a reversible path).

  11. Oct 30, 2015 #10
    i was referring to Liquids but yes my mistake we Use the term "Pump" and not "Compressor" for liquids, if so my equation seems valid maybe ;)
  12. Oct 30, 2015 #11
    I still don't like the idea of the dW. So I won't comment on the correctness of the equation, because I feel it is incorrect to work with dW.

    The equation I wrote says that the rate at which enthalpy exits the system minus the rate at which enthalpy enters the system is equal to the rate at which shaft work is done on the material passing through the system. This considers the total shaft work and not the differential shaft work (whatever that means), and the total overall change in enthalpy between entrance and exit.

  13. Oct 30, 2015 #12
    Well, the equation you attached is for closed systems. However, we can derive the same equation you attached for open systems, starting from the equation I posted. Here we go
    [tex]W = \frac{\gamma R}{\gamma -1} (T_2 - T_1)[/tex]
    We know [itex]R(T_2-T_1)= P_2 V_2 - P_1 V_1[/itex], so
    [tex]W = \frac{\gamma}{\gamma -1} (P_2 V_2 - P_1 V_1)[/tex]
    Factoring [itex]P_1 V_1[/itex]
    [tex]W = \frac{\gamma P_1 V_1}{\gamma -1} \left(\frac{P_2 V_2}{P_1 V_1} - 1 \right)[/tex]
    We know that [itex]\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}[/itex], so we have
    [tex]W = \frac{\gamma P_1 V_1}{\gamma -1} \left( \left( \frac{V_1}{V_2} \right)^{\gamma} \frac{V_2}{V_1} - 1 \right)[/tex]
    [tex]W = \frac{\gamma P_1 V_1}{\gamma -1} (V_1^{\gamma - 1} V_2^{1 - \gamma} - 1)[/tex]
    Factoring [itex]V_1^{\gamma - 1}[/itex]
    [tex]W = \gamma P_1 V_1 V_1^{\gamma - 1} \left( \frac{V_2^{1 - \gamma} - V_1^{1- \gamma}}{\gamma - 1} \right)[/tex]
    Finally, we have
    [tex]W = \gamma P_1 V_1^{\gamma} \left( \frac{V_2^{1 - \gamma} - V_1^{1- \gamma}}{\gamma - 1} \right)[/tex]
    The difference between this equation and the one you posted is the extra [itex]\gamma[/itex] to account for the difference between open and closed systems. This is more evident when using the equation in the form I posted it, given that
    [tex]C_V = \frac{R}{\gamma - 1}[/tex]
    [tex]C_P = \frac{\gamma R}{\gamma - 1}[/tex]
    The other difference is that the equation I posted has [itex]\gamma - 1[/itex] and yours has [itex]1 - \gamma[/itex], but that's just because your equation is consistent with the Q - W convention, whereas the one I posted follows the Q + W convention.
  14. Oct 31, 2015 #13
    Thanks Chester miller Sir and MexCheme sir!
    Well, thermo is so confusing, the more i read thermo the more i get stuck. plus this language problem. and its my favorite area so i can't leave it like this
    Thanks once again both of you
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