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Nozzles and diffusors(gas dynamics)

  1. Jan 9, 2012 #1
    Hi,

    I was under the impression that nozzles and diffusors are functionally just the opposite of eachother...so their thermodynamics would be accordingly...however, recently we started gas dynamics and the picture doesn't seem to be quite as simple as that.

    Looking at the Mollier diagram for the nozzle and diffusor, referring to pages 36 and 42 respectively in the book:
    http://books.google.com/books?id=-o...ce=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

    it seems that the isentropic(ideal) enthalpy, (represented by subscript 2s) for nozzle is less than the actual irreversible enthalpy (represented by subscript 2) for the nozzle, while they are equal for the diffusor....this further leads to differences in the thermodynamic analysis....why is there such a difference?
     
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  3. Jan 9, 2012 #2

    boneh3ad

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    Think about if you had a wind tunnel with a nozzle and then a diffuser that was its thermodynamic opposite. That would imply that you would need no pressure to run the tunnel. In essence, it is a perpetual motion machine! Clearly this can't physically work.

    The major difference is that while in the nozzle, you can isentropically compress the flow to sonic conditions, then expand it to supersonic conditions, when you get into the diffuser, there is essentially no way to prevent the Mach waves that form when you start re-compressing the flow from coalescing into shocks, which necessarily introduce entropy changes into the diffuser.
     
  4. Jan 11, 2012 #3
    I understand what you're saying about the entropy, but could you explain what implication this has on the enthalpy issue I referred to in the OP?
    Ref:it seems that the isentropic(ideal) enthalpy, (represented by subscript 2s) for nozzle is less than the actual irreversible enthalpy (represented by subscript 2) for the nozzle, while they are equal for the diffusor....this further leads to differences in the thermodynamic analysis....why is there such a difference?

    Also,
    In the book I referred to, they equate the stagnation enthalpies at the beginning and end for the nozzle while they equate the change in kinetic energy for the reverisible and irreverible processes to calculate the final gas velocity...why this difference?
     
  5. Jan 12, 2012 #4
    A thing to remember is that for some forms of fluid flow you need not only to consider the local path (nozzle, diffuser, pipe, whatever) but you also have to include the entry and exit conditions ie the 'receivers' at each end. This is because there may be more than one possible flow regime, dependent upon these conditions. This is particularly true of compressible flow.

    Incidentally for compressible flow from say a large tank or cylinder of gas the entry receiver conditions form one large stagnation point which is obviously at stagnation temperature.

    As regards the enthalpy question, in general work may be done on the gas in compression in the nozzle, resulting in an irreversible temperature rise. This is not recovered/returned to the universe in the cooling expansion in a diffuser, except as heat.
     
  6. Jan 12, 2012 #5

    boneh3ad

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    In the case of the nozzle, the stagnation enthalpy is the the same whether the process is isentropic or not. This is due to conservation of energy. When you factor in entropy change due to losses, you have to lose a bit of stagnation pressure in the process. For a nozzle, you typically specify your own exit pressure, so regardless of entropy changes, the final pressure, $p_2$, is going to be whatever you prescribe. That means that the final two points have to fall on the same isobar, as shown in that figure in your source. That results in an increase in enthalpy.

    For the diffuser, the exit pressure is no longer specified. Instead, they are designed to slow the flow down to a design Mach number, which requires removing the same amount of energy from the flow regardless of whether the process is isentropic. For that reason, the actual process is shown with the same change in kinetic energy as the isentropic process in your source.

    Really what it comes down to is that in each plot, the two cases are achieving the same end-goal as commonly specified by the design of the respective components.
     
  7. Jan 15, 2012 #6
    Thanks, that was very helpful!
     
  8. Jan 17, 2012 #7
  9. Jan 17, 2012 #8
    Hey thanks! I could use this as a reference book.
     
  10. Jan 17, 2012 #9

    boneh3ad

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    This author certainly is pedigreed as far as being a student of Eckert goes, but I flipped through the book and found some errors fairly early on. I only flipped through a chapter or two but I would be wary of using this as a definitive source. It is ambitious and free, but ultimately it isn't peer reviewed and has no one supporting the author with any kind of credentials that are cited. Simple things like the onset of compressibility at Mach 0.3 aren't even correct ( the author cites Mach 0.8).

    It may be an okay source that gets it mostly right, but if you want something definitive, go with Anderson.
     
  11. Jan 31, 2012 #10
    Thanks.....could anyone please tell me why in a convergent-divergent nozzle, after choking has occurred, further lowering of the back pressure causes supersonic flow?

    I'm referring to specifically the fact that after choking, the velocity at the throat is sonic.....and in the divergent part following that, why does it become supersonic, since due to the expansion in cross-sectional area and the continuity equation, I'd expect a decrease in velocity...

    Also, does it directly become supersonic after choking takes place or can we have a duration even after choking that there is subsonic flow in the diverging section?
     
    Last edited: Jan 31, 2012
  12. Jan 31, 2012 #11
    The conditions are more complicated than that.

    Take the following equations
    [tex]\begin{array}{l}
    \frac{{dV}}{V} + \frac{{dA}}{A} + \frac{{d\rho }}{\rho } = 0\quad continuity \\
    VdV + dh = 0\quad firstlaw \\
    dh = \frac{{dP}}{\rho }\quad \sec ondlaw \\
    \end{array}[/tex]

    Combine them and substitute the definition of Mach number, M

    [tex]M = \frac{V}{c}[/tex]

    This yields

    [tex]\frac{{dA}}{{dV}} = \frac{A}{V}\left( {{M^2} - 1} \right)[/tex]

    Where A is Xsection area, V is the velocity, P the pressure , [itex]\rho[/itex] the density and c the local speed of sound.

    This shows that whether the the velocity increases or decreases with changing area depends upon the sign of dA/dV which depends upon the local Mach number. You can tabulate five possibilities. One for M=1 and two each (M<1 and M>1) for an expanding or contracting flow. If you would like to try this first we can compare tables later.
     
  13. Feb 1, 2012 #12
    Oh...I get it...the bottom line is that there are a lot more factors that determine what the exit flow properties would be compared to the incompressible case....so its hard to just look and predict something from intuition.....
     
  14. Feb 1, 2012 #13
    Btw, just 2 very short questions:

    1. Are the entries of flow into a diffusor and nozzle always at M>1 and M<1 respectively?
    2. why can't we have convergent-divergent diffusors also?

    I'm guessing that this is related to: by boneh3ad
    "The major difference is that while in the nozzle, you can isentropically compress the flow to sonic conditions, then expand it to supersonic conditions, when you get into the diffuser, there is essentially no way to prevent the Mach waves that form when you start re-compressing the flow from coalescing into shocks, which necessarily introduce entropy changes into the diffuser."


    But I still don't get the underlined part properly.....
    Thanks!
     
    Last edited: Feb 1, 2012
  15. Feb 3, 2012 #14
    Please clarify!
     
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