Bernoulli's Equation- on a jet nozzle

  • Thread starter andyb1990
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  • #1
andyb1990
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Homework Statement



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

Homework Equations





The Attempt at a Solution



(i) jet speed from exit

continuity equation Q(dot)= A1V1

V1=Q/A1 ?

(ii)

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!
 

Answers and Replies

  • #2
LawrenceC
1,198
5
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.
 
  • #3
andyb1990
34
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i) would the equation be....

B+1/2*ρ*v^2+ρgz'
 
  • #4
LawrenceC
1,198
5
B+1/2*ρ*v^2+ρgz'

The above does not have an equals mark so how can it be an equation? Stagnation pressure has two components here.
 
  • #5
andyb1990
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P +1/2*ρ*v^2+ρgz = B

??
 
  • #6
LawrenceC
1,198
5
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.
 
  • #7
andyb1990
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stagnation Po = B+1/2* ρ*v^2

v= sqrt (B+1/2*ρ)
 
  • #8
LawrenceC
1,198
5
You did not solve the equation correctly.

P0 = B + rho*V^2/2

Solve it for V.
 
  • #9
andyb1990
34
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v= sqrt (2(B-Po)/rho) ??
 
  • #10
LawrenceC
1,198
5
You have a mistake in your solution.
 
  • #11
andyb1990
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v= sqrt (2(Po-B)/rho) ?
 
  • #12
LawrenceC
1,198
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Ok, that is correct for part i.
 
  • #13
andyb1990
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ii) P0 = B + rho*V^2/2 + rho*g*z'

v= sqrt (2(Po-B-rho*g*z')/rho)
 
  • #14
LawrenceC
1,198
5
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?
 
  • #15
andyb1990
34
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P0 = B + rho*V^2/2 - rho*g*z'

v= sqrt (2(Po-B +rho*g*z')/rho)
 
  • #16
LawrenceC
1,198
5
That looks good. Now work on part iii.
 
  • #17
andyb1990
34
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From earlier you said apply continuity equation

Q(dot)=AV A=Q/V

but now I am unsure what to do with this?
 
  • #18
LawrenceC
1,198
5
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?
 
  • #19
andyb1990
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that it will always remain constant?
 
  • #20
LawrenceC
1,198
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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.
 
  • #21
andyb1990
34
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M= rho*A*V = rho*A*V - rho*g*z' ?
 
  • #22
LawrenceC
1,198
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Where do you get this stuff? Your units don't match. How can you subtract rho*g*z from rho*A*V?
 
  • #23
andyb1990
34
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M(dot)= A1 sqrt (2(Po-B)/rho) = A2 * sqrt (2(Po-B-rho*g*z')/rho)
 
  • #24
LawrenceC
1,198
5
1. Problem designates exit area of nozzle as A0.
2. You have made the same sign error as before.
3. Problem asks for area so solve the thing for the area.
4. You define M above as rho*A*V. What is Mdot? What is A1? What is A2? Symbols should be defined.
 
  • #25
andyb1990
34
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M(dot) is the mass flow rate

M(dot)= A0 sqrt (2(Po-B)/rho) = A2 * sqrt (2(Po-B+rho*g*z')/rho)

where A0 is exit area at nozzle
A2 is a vertical distance z below nozzle

i have also subbed in v from part i and part ii
 
  • #26
LawrenceC
1,198
5
OK, so solve for the variable A2 and you'll have the result for part iii. Should A2 be greater or less than A0?
 
  • #27
andyb1990
34
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A2= A0 sqrt (rho*g*z)

A2 will be greater
 
  • #28
LawrenceC
1,198
5
You seem to have a problem with basic algebraic manipulations. You have the equation

A0 sqrt (2(Po-B)/rho) = A2 * sqrt (2(Po-B+rho*g*z')/rho)

but when you solve it for A2 you get

A2= A0 sqrt (rho*g*z)

which is completely wrong. The units don't match.

Moreover, if the fluid is going faster how can the area be larger. You are violating the law of conservation of mass.

If I were you I'd brush up on algebra. From the subject matter, I assume you are an engineering student. You are going to have a great deal of trouble with coursework if you are not thoroughly proficient in algebra.
 
  • #29
andyb1990
34
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to get A2 dived the right hand side by sqrt...... then by cancelling out wouldn't that leave sqrt rho*g*z??
 
  • #30
LawrenceC
1,198
5
Nothing cancels out other than rho in the denominators. It cannot be what you said because the units no longer balance.
 
  • #31
andyb1990
34
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Would you subtract the square root from the right hand side to get A2 on its own, then group like terms?
 
  • #32
LawrenceC
1,198
5
Subtracting the radical does not get A2 alone. You have missed something in your algebra studies.
 
  • #33
andyb1990
34
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Can you help me to what the equation would work out to?
 
  • #34
LawrenceC
1,198
5

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