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Nozzle problem between subsonic and supersonic speed & nozzle area

  1. Mar 21, 2015 #1
    Hi all,

    I have really a confusion between subsonic and supersonic speed evolution when the area decrease or increase

    I have a problem to understand why in supersonic regime the velocity evolution in a nozzle is adverse of the subsonic regime?

    for example in subsonic regime, when the area decrease so the velocity increase (because the static pressure increase) this analyse is logic for me,

    In supersonic the density is not constant (so we have ruA=constant), is the density decrease when the area decrease? if yes can you give me an explanation please (because for me if the volume or the area decrease the density of my fluid must increase!?)

    Another question why Mach number equal to 1 in the nozzle (when the area is minimum ?
     
  2. jcsd
  3. Mar 21, 2015 #2

    boneh3ad

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    Alright so start from your conservation of mass equation. Since mass flow is constant, ##d(\rho u A) = 0##. This can be rewritten as
    [tex]\dfrac{du}{u} + \dfrac{d\rho}{\rho} = -\dfrac{dA}{A}.[/tex]
    Similarly, using conservation of momentum,
    [tex]\dfrac{d\rho}{\rho} = -M^2\dfrac{du}{u}.[/tex]
    So combining those,
    [tex]\dfrac{du}{u} = -\dfrac{1}{1 - M^2}\dfrac{dA}{A}.[/tex]
    So when ##M^2<1##,
    [tex]\dfrac{du}{dA} < 0,[/tex]
    and velocity decreases as area increases. For ##M^2 > 1##,
    [tex]\dfrac{du}{dA} > 0,[/tex]
    and velocity increases as area increases.

    Regarding ##M=1## where the area is at a minimum, this is the throat of the nozzle, not the whole nozzle. That was probably just a typo in your post, but just in case I figured I would point it out. At any rate, if you go back to the area-velocity relationship,
    [tex]\dfrac{dA}{A} = (M^2 - 1)\dfrac{du}{u},[/tex]
    it is clear that ##M=1## corresponds to ##dA=0##, which is the throat.

    You can find a lot of this any book on gas dynamics. For example of one cheap and very good options, check out https://www.amazon.com/Elements-Gasdynamics-H-W-Liepmann/dp/0486419630 by Liepmann and Roshko.
     
    Last edited by a moderator: May 7, 2017
  4. Mar 21, 2015 #3

    jack action

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    First, we need to mention that the mass flow rate chokes at M = 1.

    Why does the flow chokes at M = 1?

    Imagine you have a constant air source (pressure & temperature) and a tube connected to it with a closed valve. The tube is closed at the other end. When you open the valve, a pressure wave will be sent through the tube, traveling at the speed of sound. You can look at that pressure wave as the way molecules communicate with each other. That pressure wave says: «We want to go in, get ready». When the pressure wave will reach the tube end, the local molecules will send back the pressure wave, a way to say: «There is no space here, don't come in.» When the pressure wave will be back at the tube entrance, it will tend to slow down the fluid entry and the higher the pressure inside the tube it will be, the slower will be the fluid flow until it stabilizes (to zero).

    Now imagine the same set up, but with tube exiting to the atmosphere. Now when the pressure wave reach the exit, the local molecules have a different message to send back: «We have lots of free space, keep them coming!» So instead of high pressure wave, they send back a low pressure wave (think vacuum). When this low pressure wave arrives at the tube entrance, it favors a faster flow. It does so until the flow stabilizes.

    When you have a convergent or divergent nozzle, instead of reflecting an intense pressure wave at the end, it reflects a series of weaker pressure waves, i.e. every time the cross-sectional area changes.

    What happens when the velocity of the fluid reaches M = 1? Well, the pressure front of the flow actually follows the pressure wave it sent. So they both arrive at the tube exit at the same time. So, in the case where the exit end is opened, the local molecules send a low pressure wave back to say: «We have lots of free space, keep them coming!» but the full pressure is already at the exit, so whatever the flow you have, it won't increase more.

    If you try to decrease the area, as the speed attempt to increase, a high pressure wave will be returned to set the Mach number to 1. This doesn't necessarily mean that the fluid velocity is not increase, it can also mean that its temperature is increase as M = V/√(kRT).

    Why does the flow becomes supersonic in a divergent nozzle when M = 1 at the entrance?

    When there is an increase in area, the mass flow rate (ρVA) must stay constant. Normally the velocity decreases to compensate. But if you do this when the flow at the entrance is at sonic speed, it will mean that the pressure wave will begin to go faster than the pressure front and a low pressure wave will be reflected to aid increasing the flow. But the flow cannot increase since it is choked. So V can't go down. If it doesn't, then the density must go down. Density is equal to PR/T, so either decrease the pressure or increase the temperature. If you increase the temperature, you also increase the speed of sound [√(kRT)], which in turn reduces the Mach number if you keep the velocity constant; you have the same problem as before, so this solution is out also. All nature can do is lower the pressure. But when you lower the pressure, you also lower the temperature (isentropic process). If you lower the temperature, the speed of sound decreases, so the Mach number increases (even if the speed doesn't change). Another effect of having a decrease in pressure and temperature is that all that internal energy has to go somewhere and will be transformed into kinetic energy, i.e. the velocity will also increase (Bernoulli principle).

    When you do all the calculations, you find out that overall, the Mach number will increase in a divergent nozzle when M = 1 at the entrance. So now the pressure front goes faster than the pressure wave it sends. These pressure waves will start to accumulate one over the other until it creates a shock. This shock is a local increase of pressure. If the pressure increases, the density increases, as well as the temperature, thus the speed of sound increases as well. So the system can take it anymore and reverts back to subsonic flow. The area change needed to create that shock depends on the energy contained in the flow at the entrance.

    More info:
    Shock wave
    Isentropic nozzle flow
    de Laval nozzle

    Flowpatterns.gif
     
  5. Mar 22, 2015 #4
    Thanks a lot for your answer,

    So mathematically is correct, but physically how we can explain this (naturally when the area decrease so the density increase "because the friction between moving fluid particles becomes more important" so how we can explain the decrease in velocity "physically", if there is an explanation?)

    You are right I talked about the throat of nozzle (physically what means dA=0??

    Thank youin advance


     
  6. Mar 22, 2015 #5

    jack action

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    The density is only a matter of mass and volume by definition (ρ = m/V).

    The decrease in velocity is explained by the Bernoulli principle, which basically stipulate the conservation of energy. When the density increases, the pressure also increases. That requires energy. That energy is taken from the kinetic energy of the fluid, which means that it will decelerate to a lower velocity.

    If there is enough friction to heat up the fluid, its temperature will increase, which means that the pressure and density increase as well. Of course, that energy to create this friction will come from somewhere and, again, its the kinetic energy that will suffer the consequence.

    But you don't necessarily need friction to accomplish the energy transfer in a decreasing area.

    dA means the change in area. It is equal to Af - Ai (the final area minus the initial area) for an infinitesimal change. If it is equal to zero, then the area keeps the same size. At the throat, since the area change goes from decreasing just before to increasing just after (dA goes from a negative value to a positive value); thus it must be equal to zero.
     
  7. Mar 22, 2015 #6

    boneh3ad

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    ##dA = 0## means the location where the change in the area (##dA##) is zero. In a converging-diverging nozzle, this is called the throat. It's the region where the nozzle changes from the converging (##dA < 0##) to the diverging (##dA > 0##) section, so at that location, ##dA = 0##.

    Well the mathematics and the physics are connected. The mathematical relationship was derived from the physical requirement that mass and momentum are conserved. In essence, then, the situation is physically explained by the fact that these two quantities must be conserved. You don't need any other values like density or temperature to understand the situation.

    If the expansion is isentropic (and it can be reasonably considered to be in a nozzle), then you can use that fact and some thermodynamics to show that when the flow is expanding in a supersonic region, temperature, density, and pressure all decrease. You can think about that being a result of the fact that total enthalpy is conserved (##h_0 = c_pT +u^2/2## and ##dh = 0##, from the first law of thermodynamics), so as ##u## increases, ##T## decreases. You can then derive the evolution of the rest of the thermodynamic quantities from these first principles. It's just the typical conservation laws for a fluid: mass, momentum, and energy.

    Bernoulli's principle is not valid for compressible flows. There is nothing quite so convenient as the Bernoulli equation for compressible flows, though it's still pretty easy to calculate all of the pertinent information provided the flow is isentropic (or other similarly simple relationships).
     
  8. Mar 22, 2015 #7
    Thane you for your response boneh3ad and jacK,

    Know, I understand beter, I agree Witherspoon you concerning thé Bernoulli principle is used just in incompressible flow in my case I Will use thé relationship between pressure ratio and Mach number thanks à lot

    I have another Question concening thé momentum equation in thé nozzle when the area is thé same we write p1A+r u2/2 =thé same eq but in position 2

    But when the area changes WE Have p1A1+r u1 (square)/2 +1/2 (p1+p2)*(A2-A1)=.... but I does not understand the third term 1/2 (p1+p2)*(A2-A1) ?
    The first term represents thé pressure at the inlet position 1 then
    thé second term represents thé pressure Applied on area 2 outlet but thé third I can not explain it. Do you have an idea





     
  9. Mar 22, 2015 #8

    boneh3ad

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    Did you derive this on your own from first principles? If you start form the integral form of the momentum balance in a control volume, you can arrive at the equation
    [tex]\dfrac{\partial}{\partial t}(\rho u A) + \dfrac{\partial}{\partial x}(\rho u^2A) + \dfrac{\partial}{\partial x}(pA) = p\dfrac{dA}{dx},[/tex]
    where ##x## is just the coordinate in the direction of your duct. If you assume steady flow (##\partial/\partial t = 0##), then you get
    [tex]\boxed{d(\rho u^2 A + pA) = p\;dA.}[/tex]
    For a duct of constant area, ##dA = 0## and you get ##d(\rho u^2 A + pA) = 0##, or
    [tex]\rho_1 u_1^2 A + p_1A = \rho_2 u_2^2 A + p_2A,[/tex]
    which is your equation. If the area is not constant, then the ##dA## term in the boxed equation does not fall out and you have to account for it. The left side is all in a differential, so it pretty easily can be turned into two terms representing the inlet and outlet conditions for the control volume. The problem for the right side of the equation is that, while the ##dA## term can easily be changed to a difference, choosing which ##p## to use for the approximation is ambiguous. It appears that whatever source you are using has accounted for this problem byapproximating ##p## as an average over the two locations in question, which is an approximation, but a decent one. Essentially, in order to relate that term to the differential terms, it is replacing ##p## at one location by ##(p_1+p_2)/2## so that the boxed equation becomes
    [tex](\rho_2 u_2^2 A_2 + p_2A_2) - (\rho_1 u_1^2 A_1 + p_1A_1) = \dfrac{p_1 + p_2}{2}(A_2-A_1),[/tex]
    or
    [tex]\rho_2 u_2^2 A_2 + p_2A_2 = \dfrac{1}{2}(p_1 + p_2)(A_2-A_1)+ \rho_1 u_1^2 A_1 + p_1A_1 .[/tex]

    More interestingly, if you start with ##d(\rho u^2 A + pA) = p\;dA,## you can expand it to get
    [tex]u\;d(\rho u A) + \rho u A\;du + A\; dp = 0.[/tex]
    Since ##d(\rho u A) = 0## is just the continuity equation, this simplifies to
    [tex]\rho u\;du = -dp.[/tex]
    This is simply Euler's equation.

    If you aren't familiar with how to get the boxed equation, I can go over that as well, but it is a longer derivation so it would be helpful if you were already familiar with a few things like control volume analysis.
     
    Last edited: Mar 22, 2015
  10. Mar 22, 2015 #9
    Its verry clear,

    So this force represents the force excerted o the throat of the nozzle? right? since the first term is the force at the inlet and the 2sd is the force at the outlet ?



     
  11. Mar 22, 2015 #10

    boneh3ad

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    No, this equation applies regardless of of whether any throat even exists. You could have a duct with crazy wavy walls or a duct that is simply diverging or converging (as opposed to both) and it would still apply. However, with the approximation of ##p## as its simple average, it's really mostly valid for a duct without any craziness such that the pressure varies linearly and the average would then be an exact answer (or very nearly so). Think of it this way, go back to that boxed equation,
    [tex]d(\rho u^2 A + pA) = p\;dA.[/tex]
    One way to solve this is to integrate over your direction of chance, which we will call ##x##, so from position 1 to position 2,
    [tex]\int\limits_1^2 d(\rho u^2 A + pA) = \int\limits_1^2 p\;dA.[/tex]
    [tex](\rho u^2 A + pA)|_1^2 = \int\limits_1^2 p\;dA.[/tex]
    [tex](\rho_2 u_2^2 A_2 + p_2A_2) - (\rho_1 u_1^2 A_1 + p_1A_1) = \int\limits_1^2 p\;dA.[/tex]
    That takes care of the left side. For the right side, Recall that there are a number of ways to approximate an integral. In this case, use the trapezoid rule so that
    [tex]\int\limits_1^2 p\;dA \approx \dfrac{p_1 + p_2}{2}(A_2-A_1).[/tex]
    Subbing this back into the previous equation gets back to your original equation,
    [tex]\rho_2 u_2^2 A_2 + p_2A_2 = \rho_1 u_1^2 A_1 + p_1A_1 + \dfrac{p_1 + p_2}{2}(A_2-A_1).[/tex]
    Since the trapezoid rule is just an approximation, so, too, is this equation. It is more accurate in a hypothetical situation where pressure is a linear function of the area. Unfortunately, that just isn't true in nature. Pressure typically has a power law relationship with area, so really this equation is only a decent approximation over very small area changes.

    As far as what it represents, it is harder to really explain physically. It is basically a term that represents the change in the pressure force as a result of the change in area over the region of interest.

    Also, I had to go back and change a typo in my last post. I mistakenly had ##(A_2 - A_1)## in the denominator.
     
  12. Mar 22, 2015 #11
    Thank you a lot!

    So if I undestood, this term is applicable in incompressible and compressible flow? or juste compressible *


     
  13. Mar 22, 2015 #12

    boneh3ad

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    Nothing to this point has required that the flow be incompressible, so all of this is valid for any inviscid flow.
     
  14. Mar 25, 2015 #13
    Hi,

    In the some Nozzle prob we did not take into account the pressure integral (pdA)? why? in some case we take this term and in others no?? there is an explanation please

    Thanks

     
  15. Mar 25, 2015 #14

    boneh3ad

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    I am not entirely sure what you are asking.
     
  16. Mar 25, 2015 #15
    Its OK I found the response :) :)
    Thank you very much
     
  17. Apr 1, 2015 #16
    Hi Boneh3ad,

    I have a question about the stagnation properties (stagnation pressure, temperature, density, etc.), normaly we have the stagnation state if the velocity equal to zero, but why we have the stagnation point in the nose of airfoil? why he velocity in the nose of airfoil equal zero (always), and we can consider that the boundary layer as a stagnation state ? since the velocity at the vecinity of the solid boundaries equal to zero??

    Then concerning the duct, why we have the stagnation points in the duct while the flow has some velocity?

    Thank you very much for your help


     
  18. Apr 1, 2015 #17
    I asked for the stagnation properties? what is the definition of stagnation properties? and why at the nose of an airfoil the velocity is zero?

    I can't understund the purpose of stagnation point/properties

     
  19. Apr 1, 2015 #18

    jack action

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    Because at this point, the airfoil is perfectly vertical (slope = ∞ with respect to flow) such that the flow has to change direction by 90° (up, down, left or right). The longitudinal velocity is then 0; the flow literally stops before turning.
     
  20. Apr 1, 2015 #19
    Thanks Jack for your answer,

    Ok, but why we have the stagnation points in a Nozzle (duct) (stagnation point in the converging part and stagnation point in tne diverging part )??

     
  21. Apr 1, 2015 #20

    boneh3ad

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    Jack already mentioned the reason for a stagnation point at the front of an object. I don't know why you have the impression that there would be a stagnation point in a open duct, though. That wouldn't happen.

    The boundary layer causes stagnation in the sense that it causes the velocity to drop to zero, but not in the sense that we usually discuss it in most fluid applications. Boundary layers are not isentropic, so the pressure at the wall where you have zero velocity will not be the stagnation pressure. In fact, it is very nearly identical to the pressure at the edge of the boundary layer. The flow is slowing down, but te boundary layer is dominated by viscosity and is highly dissipating, so it will never reach what is typically called stagnation pressure.
     
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