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Fluid Question

  • Thread starter Ragnar1995
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  • #1
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Homework Statement



A nozzle at the end of a horizontal pipeline discharges
water into the atmosphere. The rate of discharge is
0.65 m3
/s. The pipeline has a constant diameter of 0.3
m and the downstream end of the nozzle has a
diameter of 0.2 m. Calculate the force exerted by the
water on the nozzle, assuming that there are no
energy losses in the flow.

Homework Equations


Bernoulis Equation
3. The Attempt at a Solution [/B]
I tried plugging in all the date into Bernoulis equation, but cant seem to get the correct answer of 4681N.

Any help would be greatly appreciated thanks.
 

Answers and Replies

  • #2
haruspex
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Homework Equations


Bernoulis Equation
3. The Attempt at a Solution
I tried plugging in all the date into Bernoulis equation, but cant seem to get the correct answer of 4681N.[/B]
It doesn't help to describe what you did. Please post your working.
 
  • #3
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ok firstly i used velocity=flow rate/area to calculate the v1 and v2.
v1=9.20m/s and v2 = 20.69m/s

P1/ρg + V12/2g + z1 = P2/ρg + V22/2g + z2

P1/9810 + 9.2^2/2*9.81 = 1.01x10^5/9810 + 20.69^2/2x9.81

P1=2.73x10^5

Then i subbed into the equation F=PA to give 8567.69 N
 
  • #4
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It doesn't help to describe what you did. Please post your working.
Any idea how to do it?
 
  • #5
haruspex
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That's better, but I strongly encourage you to adopt the habit of working entirely symbolically, only plugging in numbers at the end of the working. There is a raft of reasons for doing that. So the first step should be to create symbols to replace all the numbers you are given. (Reverse engineering your working from the numbers has slowed me down.)
What area did you use at the end to get from pressure to force?

I get an even larger answer than you do. It bothers me that the question says to assume no energy loss. I'm not certain that's realistic, and that creates the danger that there could be two ways of calculating the answer getting different results. I'll continue to work on that.
 
  • #6
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That's better, but I strongly encourage you to adopt the habit of working entirely symbolically, only plugging in numbers at the end of the working. There is a raft of reasons for doing that. So the first step should be to create symbols to replace all the numbers you are given. (Reverse engineering your working from the numbers has slowed me down.)
What area did you use at the end to get from pressure to force?

I get an even larger answer than you do. It bothers me that the question says to assume no energy loss. I'm not certain that's realistic, and that creates the danger that there could be two ways of calculating the answer getting different results. I'll continue to work on that.
Ye i understand what you mean, might start doing it :) I used the area of the nozzle, 0.0314m^2, i also tried the area of the pipeline but it didnt work either.
 
  • #7
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You need to do a momentum balance on the fluid passing through the nozzle. Don't forget that the wall of the nozzle exerts a force on the fluid.

Chet
 
  • #8
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You need to do a momentum balance on the fluid passing through the nozzle. Don't forget that the wall of the nozzle exerts a force on the fluid.

Chet
could you show working please ?
 
  • #9
haruspex
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You need to do a momentum balance on the fluid passing through the nozzle.
I agree, but it seems to me that gives a different (and larger) answer than using Bernoulli. In other words, it is quite wrong for the question to say assume there are no energy losses, since it is provable that there are. Can you confirm my result?
The given answer is clearly much too low. It must exceed atmospheric pressure exerted over the nozzle annulus by some margin.
Nozzle annulus area * Pa = 3966N, so the force on the annulus from the water should be substantially higher.

Using Bernoulli (i.e. conservation of work) I get 6745N + Nozzle annulus area * Pa = 10711N
Using momentum I get 9339N + Nozzle annulus area * Pa = 13306N

Edit:
Thinking about why work cannot be conserved here, I believe it is because we are told that the pipe has constant diameter until it reaches the nozzle, where, presumably, the diameter immediately reduces to 0.2m. As I recall, to use Bernoulli (= work conserved), the diameter should only change gradually.
 
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  • #10
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could you show working please ?
[tex]P_1A_1-P_2A_2 -F=650(V_2-V_1)[/tex]

Chet
 
  • #11
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[tex]P_1A_1-P_2A_2 -F=650(V_2-V_1)[/tex]

Chet
That gives the correct answer, thanks alot ! il need to go over momentum balance to understand it now :)
 
  • #12
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I agree, but it seems to me that gives a different (and larger) answer than using Bernoulli. In other words, it is quite wrong for the question to say assume there are no energy losses, since it is provable that there are. Can you confirm my result?
The given answer is clearly much too low. It must exceed atmospheric pressure exerted over the nozzle annulus by some margin.
Nozzle annulus area * Pa = 3966N, so the force on the annulus from the water should be substantially higher.

Using Bernoulli (i.e. conservation of work) I get 6745N + Nozzle annulus area * Pa = 10711N
Using momentum I get 9339N + Nozzle annulus area * Pa = 13306N

Edit:
Thinking about why work cannot be conserved here, I believe it is because we are told that the pipe has constant diameter until it reaches the nozzle, where, presumably, the diameter immediately reduces to 0.2m. As I recall, to use Bernoulli (= work conserved), the diameter should only change gradually.
Turns out i need to use momentum balance to solve it thanks to Chestermiller :) Thanks for your help !
 
  • #13
haruspex
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Turns out i need to use momentum balance to solve it thanks to Chestermiller :) Thanks for your help !
OK, but three things...
- do you understand where he got that equation from?
- what answer do you get?
- do you understand that this means the advice "assuming that there are no energy losses in the flow" is incorrect?
 
  • #14
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[tex]P_1A_1-P_2A_2 -F=650(V_2-V_1)[/tex]

Chet
Why is P2A2 and Force negative?
 
  • #15
haruspex
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Why is P2A2 and Force negative?
Draw a free body diagram on the water in the pipe. Consider:
- the force at the inlet
- the force on the water exerted by the nozzle around the outlet
- the back force from the water exiting the nozzle
Do the same for the water leaving the nozzle:
- the force exerted on it by the water in the pipe
- the force exerted on it by air pressure
- the net force required to produce its rate of change of momentum
 
  • #16
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OK, but three things...
- do you understand where he got that equation from?
- what answer do you get?
- do you understand that this means the advice "assuming that there are no energy losses in the flow" is incorrect?
Im after watching a video on it, but i dont understand why P2A2 and the force is negative, the right hand side i understand.
 
  • #17
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Im after watching a video on it, but i dont understand why P2A2 and the force is negative, the right hand side i understand.
P1A1 is the force exerted by the fluid that is about to enter the control volume (the nozzle) on the fluid within the control volume. It is in the forward direction of fluid motion, and is thus positive. P2A2 is the force exerted by the fluid that has just exited the control volume (at the exit of the nozzle) on the fluid within the control volume. It is in the backward direction relative to the fluid motion, and is thus negative.

Chet
 
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  • #18
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- do you understand that this means the advice "assuming that there are no energy losses in the flow" is incorrect?
This is not correct. The assumption that there are no energy losses in the flow is the thing that gives the student permission to use the Bernoulli equation to solve the problem, as Ragnar1995 successfully did. Assuming that Bernoulli equation applies is tantamount to assuming conservation of energy.

The force F exerted by the converging walls of the nozzle on the fluid flowing through the nozzle is a pressure force, and, as such, acts perpendicular to the wall. For the present energy-conserving inviscid fluid flow, the fluid velocity at the wall is tangent to the wall. The force F therefore does no work on the fluid passing through the nozzle, because the local pressure (force per unit area) acts perpendicular to the fluid velocity (and local displacement).

Chet
 
  • #19
haruspex
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The assumption that there are no energy losses in the flow is the thing that gives the student permission to use the Bernoulli equation to solve the problem, as Ragnar1995 successfully did. Assuming that Bernoulli equation applies is tantamount to assuming conservation of energy.
Yes, I understand that, but using the momentum balance is independent of that assumption and seems to yield a different answer.

I took the nozzle to consist of an annulus of area Ap-An. This may be where I went wrong. Perhaps we are to assume the nozzle tapers gradually to that exit diameter. I'll return to that later.
I independently came up with your equation in post #10. This does not assume work is conserved.
We have a pressure P in the pipe and an area Ap. The water exits the nozzle through an aperture area An at mass rate r. The linear velocity is vp in the pipe, vn through the nozzle. Let the force to be calculated be F.

The increase in linear velocity ##\Delta v = v_n- v_p = r\left(\frac 1{A_n} - \frac 1{A_p}\right)##.
The net thrust on the exiting water is ##F_w = r \rho \Delta v = r^2\rho\left(\frac 1{A_n} - \frac 1{A_p}\right)##.
Since there is a backpressure ##P_a A_n## from the external atmosphere, the force the water in the pipe exerts on the exiting water is ##P_a A_n+F_w = P A_n##.
The force the water exerts on the annulus ##F = P (A_p - A_n) = \left(P_a +\frac{F_w}{A_n}\right)(A_p - A_n) = P_a (A_p - A_n) + r^2\rho \frac{(A_p - A_n)^2}{A_p{A_n}^2}##.
Plugging in the numbers, this gives me 13306N.

Using Bernoulli instead I get ##F = P_a (A_p - A_n) + r^2\rho \frac{(A_p - A_n)^2(A_p+A_n)}2{{A_p}^2{A_n}^2}##.
Leaving aside the ##P_a (A_p - A_n) ## term they have in common, the ratio is ##\frac{A_p+A_n}{2A_p}##.
Plugging in numbers, I get 10711N.

The question arises, why cannot work be conserved here? As I posted, I believe Bernoulli only applies where the diameter changes are reasonably gradual.
So let's consider a gradual taper. Now our/your post #10 equation breaks down because the pressure varies along the taper.

Either way, I do not see how the answer can be a mere 4681N. If I omit the atmospheric pressure, I get 6745N for Bernoulli.
 
  • #20
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Hi haruspex,

I can't say that I follow every detail of your previous post, but I'm very confident that I can alleviate your doubts.

You alluded to two situations, one in which there is a sudden jag in the nozzle diameter, and the other where there is a gradual taper. I can tell you that, for this flow of an inviscid fluid, it doesn't matter whether there is a sudden jag or a gradual taper. The answer to the problem will come out the same either way, as long as we are looking for the overall effects.

The flaw in the analysis of the annular jag in the pipe is the assumption that the pressure distribution is uniform over the face of the annulus. This will not be the case, because there will be a substantial inward radial component of flow approaching the annulus (as the flow has to converge to a smaller cross sectional area), and the axial component of flow will be decreasing substantially in the region of the fluid closely approaching the annular region (the outer flow streamline has to make a right angle turn). So, the radial pressure variations will be substantial over the face of the annulus. The analysis we have done so far gives us the total force F on the annulus, and, if we divide that by the cross sectional area of the annulus, we can get the average pressure. But, in the case of the annulus, the only way of getting the detailed radial pressure variations on the annulus it to solve the inviscid flow equations (PDEs).

For the case of a very gradual taper of the nozzle cross section, the situation is much simpler, because we can treat the flow as quasi 1D, so that the radial variations of pressure and velocity can be neglected. This would allow us to determine the pressure distribution on the wall, and then integrate this pressure distribution to get the force F. The differential momentum balance equation for this case will automatically deliver both the overall Bernoulli equation that we got previously by applying the Bernoulli equation across the entire nozzle, and also the overall momentum balance result that we got. If you are interested, I can provide this analysis for the tapered case (which is pretty straightforward). Any interest?

Chet
 
  • #21
haruspex
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Hi haruspex,

I can't say that I follow every detail of your previous post, but I'm very confident that I can alleviate your doubts.

You alluded to two situations, one in which there is a sudden jag in the nozzle diameter, and the other where there is a gradual taper. I can tell you that, for this flow of an inviscid fluid, it doesn't matter whether there is a sudden jag or a gradual taper. The answer to the problem will come out the same either way, as long as we are looking for the overall effects.

The flaw in the analysis of the annular jag in the pipe is the assumption that the pressure distribution is uniform over the face of the annulus. This will not be the case, because there will be a substantial inward radial component of flow approaching the annulus (as the flow has to converge to a smaller cross sectional area), and the axial component of flow will be decreasing substantially in the region of the fluid closely approaching the annular region (the outer flow streamline has to make a right angle turn). So, the radial pressure variations will be substantial over the face of the annulus. The analysis we have done so far gives us the total force F on the annulus, and, if we divide that by the cross sectional area of the annulus, we can get the average pressure. But, in the case of the annulus, the only way of getting the detailed radial pressure variations on the annulus it to solve the inviscid flow equations (PDEs).

For the case of a very gradual taper of the nozzle cross section, the situation is much simpler, because we can treat the flow as quasi 1D, so that the radial variations of pressure and velocity can be neglected. This would allow us to determine the pressure distribution on the wall, and then integrate this pressure distribution to get the force F. The differential momentum balance equation for this case will automatically deliver both the overall Bernoulli equation that we got previously by applying the Bernoulli equation across the entire nozzle, and also the overall momentum balance result that we got. If you are interested, I can provide this analysis for the tapered case (which is pretty straightforward). Any interest?

Chet
Ok, I understand - thanks.
But... What about the numerical answer to this question?
 
  • #22
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Ok, I understand - thanks.
But... What about the numerical answer to this question?
The numerical answer to this question is the result that Ragnar1995 got using the equation that I recommended, and which is in agreement with the book answer. To get an understanding of why this is correct, here is a brief analysis of the problem solution for the case of a gradually varying nozzle cross sectional area.

If we do a momentum balance on the fluid located between axial cross sections at x and x + Δx in the nozzle, we get:

##P(x)A(x)-P(x+Δx)A(x+Δx)+P\left(x+\frac{Δx}{2}\right)[A(x+Δx)-A(x)]=##
##ρA(x)v(x)[v(x+Δx)-v(x)]##

In this equation, the term ##P\left(x+\frac{Δx}{2}\right)[A(x+Δx)-A(x)]## represents the force that the converging (or diverging) section of wall exerts on the fluid in the x direction.

If we divide the above equation by Δx, and take the limit as Δx approaches zero, we obtain:
[tex]-\frac{d(PA)}{dx}+P\frac{dA}{dx}=ρAv\frac{dv}{dx}[/tex]
Here again, the term ##P\frac{dA}{dx}## represents the force per unit length that the wall of the nozzle exerts on the fluid in the x direction.

If we now differentiate the first term according to the product rule, we obtain:

[tex]-P\frac{dA}{dx}-A\frac{dP}{dx}+P\frac{dA}{dx}=-A\frac{dP}{dx}=ρAv\frac{dv}{dx}[/tex]
Note that one of the terms from the product differentiation has cancelled with the force exerted by the wall. If we now divide both sides of this equation by A, we obtain:
[tex]-\frac{dP}{dx}=\frac{d}{dx}\left(\frac{ρv^2}{2}\right)[/tex]
This is the Bernoulli equation. Note that it has been derived directly from the differential momentum balance on the fluid, including the force exerted by the wall. So we can solve this equation for the inlet pressure, and then use that together with the integrated momentum balance to get the force exerted by the wall on the fluid:
[tex]P_1A_1-P_2A_2+\int{P\frac{dA}{dx}dx}=ρAv(v_2-v_1)[/tex]
Note that the integral is the force F that the nozzle wall exerts on the fluid (and the fluid exerts on the nozzle).

I hope this helps.

Chet
 

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