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Throttling process

  1. May 16, 2015 #1
    hi all!
    I am learning basic thermodynamics.
    I already know that expansion process of fluid through a throttling valve is irreversible process.
    what i wonder, why this process is irreversible.
    as far as i know, if delta S>0 <-- irreversible process.

    i want to know why Entropy of expansion process of fluid through a throttling valve is greater than zero.
     
  2. jcsd
  3. May 16, 2015 #2
    This is a really good question that has puzzled many people. Ordinarily, if you had an adiabatic reversible expansion, the entropy would be constant. However, in a throttling process, there is viscous dissipation in the valve, and this just about cancels the temperature decrease from expansion. So the net effect is an entropy increase.

    Chet
     
  4. May 16, 2015 #3
    thank you for your helpful answer.
    i think that you are talking about ideal gas., right ?? if so, I would like to know about real gas .
    and... I should be appreciated if you would answer the question as formula or symbol.
     
  5. May 16, 2015 #4
    No. I am talking about real gases and ideal gases. An ideal gas is just a real gas in the limit of low pressure and high temperature.
    I don't quite know what you mean here. In an adiabatic throttling process, the enthalpy change is zero. You go from high pressure on one side of the valve to low pressure on the other side, while the temperature change is nearly zero, or slightly positive. So you are interested in the effect of pressure on entropy at constant enthalpy. Because of the viscous heat generation in the throttling operation, the temperature change is small, and so the decrease in pressure (increase in volume) will result in an increase in entropy.

    Chet
     
  6. May 17, 2015 #5
    sorry,
    i understand what you say.

    what i want to know more proves this process.(entropy > 0 ). if it possible, please show me the solving process.
     
  7. May 17, 2015 #6
    Do you know the general relationship between dH, dS, and dP?
     
  8. May 17, 2015 #7
    yes, i know the general relationship... dh, ds, dp, du, residual properties, and maxwell equation etc.
     
  9. May 17, 2015 #8
    OK. So, what's the equation for dH in terms of dS and dP?

    Chet
     
  10. May 17, 2015 #9
    dH= TdS + VdP
     
  11. May 17, 2015 #10
    Now, in a throttling process, you are aware that the enthalpy in the initial state is equal to the enthalpy in the final state, correct? If H is constant, what is dH equal to in the above equation. What does that give for dS? If dP is negative, what does that tell you about dS?

    Chet
     
  12. May 17, 2015 #11
    um... dh is zero ( dh = 0 ) so, TdS= -VdP <--> dS=-VdP/T and dP is negative. however, as fas as i know, V and T are both variables.
    in the case of (dH= TdS + VdP), i don't understand that

    1. T and V are both constant.

    2. dV is positive and generally dT is negative ( exceptionally helium has a positive). then, indeed dS is a positive ??
     
  13. May 17, 2015 #12
    Your equation says that in going between these two differentially separated equilibrium states at constant H, the temperature T and the volume V are to be taken as nearly constant. But this requires you to know how to handle math, not the physics. As things stand now, you don't know how to handle the math. To learn how to handle the math correctly, you need to study the math. If you still have problems after this, I suggest you pose your questions to the Calculus and Beyond Forum.

    Chet
     
  14. Jun 25, 2015 #13
    if we keep V and T const. , and observe that dP becomes +ve because the pressure reduces from high to low . therefore the chng. in pr. is -ve, thus -ve x -ve becomes +ve and dS becomes greater than 0.
    I hope I am right.
     
  15. Jun 28, 2015 #14
    Does this make any sense to anyone else? It doesn't to me.

    Chet
     
  16. Dec 12, 2016 #15
    i know this is late but it is for anyone coming here looking for an answer:

    Start using Gibbs's Equation: Tds=du+Pdv
    In a throttling process, the enthalpy is constant, so du=0. Rearranging for ds gives us:
    ds=(P/T) dv
    Since in a throttling process we HAVE to expand the fluid, the change in volume term (dv) has to be positive, meaning that the change in entropy has to be positive as well.
    The two terms P and T also are positive constants( T is in Kelvin for Gibb's equation, so no value of T can be negative), meaning they do not affect the sign of ds.
     
  17. Dec 12, 2016 #16
    This is not done correctly and, moreover, it doesn't address the underlying question posed by the OP regarding the physical mechanism responsible for the increase in entropy (i.e., not in terms of mathematics).

    In your starting equation, u is the internal energy, not the enthalpy. In terms of enthalpy, the correct equation should read: Tds=dh-vdP. At constant enthalpy, Tds=-vdP. The specific volume is positive and dP is negative in throttling, so Tds>0. This is the mathematical development, but not the mechanistic explanation.

    The mechanistic explanation for the increase in entropy is that the throttling process constitutes an irreversible adiabatic change (involving a decrease in pressure). Such a process causes viscous dissipation within the fluid, which generates entropy. There is no heat transfer in an adiabatic system to allow heat and entropy to leave the system, so the generated entropy remains in the system, which results in an increase in the system entropy.

    This thread is more than a year old, and the original question has been adequately addressed previously. Therefore, I am hereby now closing this thread.

    Chet
     
    Last edited: Dec 12, 2016
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