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Homework Help: Bernoulli's Equation- on a jet nozzle

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A jet of liquid of density ρ surrounded by air flows vertically downwards from a nozzle
    of area Ao. The stagnation pressure of the jet at exit from the nozzle is po and the
    surrounding air is at pressure B. The acceleration due to gravity is g.

    If the jet flow is one-dimensional, incompressible and frictionless, and the air density
    is regarded as negligible,

    (i) Derive an equation for the jet speed at exit from the nozzle

    (ii) Derive an equation for the jet speed a vertical distance z below the nozzle
    assuming the surrounding air pressure is unchanged

    (iii) Derive an equation for the cross-sectional area A of the jet a vertical distance z
    below the nozzle

    2. Relevant equations

    3. The attempt at a solution

    (i) jet speed from exit

    continuity equation Q(dot)= A1V1

    V1=Q/A1 ?


    Po= B+0.5*ρ*v^2+ ρgz

    (iii) help is needed!

    are the parts for (i) and (ii) correct guidance to put me in the right direction would be very much appreciated!
  2. jcsd
  3. Aug 22, 2012 #2
    i)The jet speed at the nozzle is computed from the Bernoulli equation. The pressure at the nozzle exit is B.

    ii) This part is similar to a projectile being shot vertically downward with an initial velocity computed in part i.

    iii) Apply continuity (conservation of mass) equation to determine new area.
  4. Aug 22, 2012 #3
    i) would the equation be....

  5. Aug 23, 2012 #4

    The above does not have an equals mark so how can it be an equation? Stagnation pressure has two components here.
  6. Aug 23, 2012 #5
    P +1/2*ρ*v^2+ρgz = B

  7. Aug 23, 2012 #6
    If you seek the velocity for question i, the velocity should be the only term on one side of the equation. Secondly, the problem states the stagnation pressure at the nozzle exit is P0. Write an equation for the velocity in terms of stagnation pressure and external pressure B. At the nozzle exit, the initial altitude has nothing to do with it.
  8. Aug 23, 2012 #7
    stagnation Po = B+1/2* ρ*v^2

    v= sqrt (B+1/2*ρ)
  9. Aug 23, 2012 #8
    You did not solve the equation correctly.

    P0 = B + rho*V^2/2

    Solve it for V.
  10. Aug 23, 2012 #9
    v= sqrt (2(B-Po)/rho) ??
  11. Aug 23, 2012 #10
    You have a mistake in your solution.
  12. Aug 23, 2012 #11
    v= sqrt (2(Po-B)/rho) ?
  13. Aug 23, 2012 #12
    Ok, that is correct for part i.
  14. Aug 23, 2012 #13
    ii) P0 = B + rho*V^2/2 + rho*g*z'

    v= sqrt (2(Po-B-rho*g*z')/rho)
  15. Aug 23, 2012 #14
    ii) P0 = B + rho*V^2/2 + rho*g*z'

    v= sqrt (2(Po-B-rho*g*z')/rho)

    Whenever you derive an equation you should check it to see if it makes sense with regard to reality. In your equation above, if z' is a positive number does the equation make sense with what you perceive to be reality?
  16. Aug 23, 2012 #15
    P0 = B + rho*V^2/2 - rho*g*z'

    v= sqrt (2(Po-B +rho*g*z')/rho)
  17. Aug 23, 2012 #16
    That looks good. Now work on part iii.
  18. Aug 23, 2012 #17
    From earlier you said apply continuity equation

    Q(dot)=AV A=Q/V

    but now I am unsure what to do with this?
  19. Aug 23, 2012 #18
    If you pick a point right at the nozzle exit, you have a specific velocity that you already computed based on the stagnation pressure. At a distance z below the nozzle exit, you computed another velocity, and after you corrected the sign error I pointed out you can conclude the stream is going faster. But what can you say about the mass flow considering conservation of mass?
  20. Aug 23, 2012 #19
    that it will always remain constant?
  21. Aug 23, 2012 #20
    Yes, because the problem is steady state. The amount of mass that leaves the nozzle does not change with time. So write an equation for the mass flow at the nozzle and one for the mass flow at a point z below the nozzle exit. Equate them and solve part iii.
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