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How does a nozzle work (in a steam turbine)?

  1. Feb 21, 2008 #1

    For sometime now i've been into electronics and i have forgotten most of my meager knowledge in thermodynamics. .

    Could you explain how a convergent-divergent nozzle works ?

    Steam at high temperature and high pressure but relatively low speed enters into the nozzle. Then why should the enthalpy decrease ? and how do i get high velocity but low pressure steam at the other end of the nozzle ?

  2. jcsd
  3. Feb 21, 2008 #2


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    Because the other end of the nozzle is connected to a low pressure environment?
  4. Feb 21, 2008 #3


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    A nozzle does no work. The rule to remember is that the total temperature through the nozzle remains constant. So, what does the pressure distribution look like through the nozzle, especially when approaching the throat? That should help with what happens to the static enthalpy issue. Usually in these applications, total enthalpy is looked at, not static.

    The pressure drops to a critical value at the throat at which point the nozzle chokes at the throat. Then, in a C-D nozzle, the pressure there can still drop because the flow is accelerating out the back in the diverging section (supersonic flow acts exactly opposite to sub sonic in this regard).
  5. Feb 21, 2008 #4
    Isn't this Bernouilli-principle-type fluid dynamics stuff rather than thermodynamics? Not sure myself, just asking.
  6. Feb 21, 2008 #5
    thanks. temperature remains constant . ok. once the steam enters a c-d nozzle , the velocity of flow increases as the mass flow rate remains constant. .why is the pressure decreasing ?

    Could you elaborate more on enthalpy , particularly how it decreases while reaching the throat ?

    To make things easier for me, i think we can assume that sound speed has not been reached yet.

    thanks gain
  7. Feb 21, 2008 #6


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    When the flow flows throw the nozzle, what force causes it to accelerate- If the pressure drops because the flow is accelerating, which force accelerates the flow?
  8. Feb 21, 2008 #7


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    Depends on the nozzle. If the pressure drops are large, then you have to deal with temperature change and thermodynamic properties of the fluid. If it's just a low pressure system like an air conditioner vent, yes.
  9. Feb 21, 2008 #8
    Oh, of course, I never thought of that. Wow, that's nuts. No wonder the guys who try to model the atmospheres of gas giants never get anywhere if they have to deal with storm pressure changes causing temperature changes, chemical reactions, enormous magnetic fields, and GR effects at the same time.
  10. Feb 21, 2008 #9
    In a convergent nozzle the fluid can reach only up to sonic velocity (which would be the speed of sound in that fluid, not in air, right?) at the narrowest point. Higher pressures won't lead to higher velocities. But since the fluid can also heat up, the temperature changes would result in a higher sonic velocity which would allow a higher speeds.
    It is said that adding a divergent nozzle downstream to the narrowest part would lead to even higher velocities, if the fluid became sonic in the narrowest point. This is due to properties of super sonic fluids. But does anyone know why supersonic fluids act opposite to subsonic fluids, accelerating with expanding volume, not according to Bernoulli's principle?
    Furthermore, why then to add a divergent nozzle, isn't the the open environment a (very) fast divergent nozzle, in fact?

    Also, I do not fully understand what influences the outcome of PV=nRT. What processes results in heat loss or conservation, how do I ensure an isothermic process, etc. This is a general question, can anyone help?
  11. Feb 21, 2008 #10
    i think a divergent part is added so that steam expands along the axis of the nozzle and hence velocity further increases.

    But in the convergert part , as steam approaches the throat why does the pressure drop( and the enthalpy drops , why ? ) ? logic tells me that the pressure should increase .
  12. Feb 22, 2008 #11


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    The fluid speeds up and pressure drops to maintain conservation of mass, or the continuity equation. If you look at both continuity and Newton's second law for a particle along a streamline, you can eventually come up with the following two relationships:

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

    [tex]\frac{d \rho}{\rho} = \frac{dA}{A}\left[\frac{M^2}{1-M^2}\right][/tex]

    [tex]V[/tex] = velocity
    [tex]\rho[/tex] = density
    [tex]A[/tex] = flow area
    [tex]M[/tex] = Mach number

    So, if you notice that for subsonic flows (M<1), the area and the density changes follow each other and the velocity and area go opposite to each other. The exact opposite is true for supersonic flows (M>1). Just plug in some numbers to get a feel for it.

    In terms of the enthalpy, you have to look at the constraint that for isentropic flow in a nozzle where no work is done on the fluid, the stagnation enthalpy is considered constant across the entire nozzle. Stagnation enthalpy is different from static enthalpy. Stagnation enthalpy is defined as:

    [tex]h_o = h + \frac{V^2}{2}[/tex]

    So for conservation of energy to hold true, then you can say

    [tex]h_x - h_y = \frac{1}{2}(v_x - v_y)[/tex]

    [tex]h_o[/tex] = stagnation or flow enthalpy
    [tex]h_x[/tex] and [tex]h_y[/tex] = entalpy and locations x and y along the nozzle
    [tex]v_x[/tex] and [tex]v_y[/tex] = velocity at x and y locations

    So you can say that the flow is trading its thermal content, i.e. enthalpy, for its kinetic energy, velocity.
    Last edited: Feb 22, 2008
  13. Feb 22, 2008 #12
    ok . thanks
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