# Homework Help: Nozzle Conservation of Momentum

1. Jan 24, 2017

### joshmccraney

1. The problem statement, all variables and given/known data
I am happy to re-write the question, but I'm just leisurely working through a list of problems I found online, so here is the link (with picture).
http://web.mit.edu/2.25/www/5_pdf/5_01.pdf

2. Relevant equations
Cons of Momentum
Cons of Mass

3. The attempt at a solution
a) I would use Bernoulli's equation since flow is inviscid and incompressible and conservation of mass. Steady flow implies Bernoulli's equation is $P_1+1/2 \rho V_1 = P_A+1/2 \rho V_2$ and Conservation of mass is $R_1^2 V_1 = R_2^2 V_2$. Thus we may solve for and know $P_1$.

b) Take a CV to be the fluid in the nozzle, starting somewhere at $R_1$ and ending at $R_2$. Since we only consider this force during steady flow, conservation of momentum is $$\iint \rho \vec{V} (\vec{V_{rel}}\cdot \vec{n})\,dA = \sum \vec{F} \implies\\ -\rho V_1^2 \pi R_1^2+\rho V_2^2 \pi R_2^2 = -P_1 \pi R_1^2+P_a \pi R_2^2 +\vec{F_{bolt}}$$

Does this look correct? If so, what would change if we didn't have the bolt, but rather the pipe was manufactured with the nozzle directly apart of itself? The tensile force $\vec{F_{bolt}}$ wouldn't be in the equation, but surely the force would not simply be gone, right?

Thanks!

2. Jan 24, 2017

### Nidum

The force on the nozzle is the same regardless of how the nozzle is fixed .

The forces acting in the fixing bolts or in a solid connection are not relevant in the flow analysis .

3. Jan 24, 2017

### joshmccraney

Are you saying my solution is wrong? Did I miss something?

4. Jan 24, 2017

### Staff: Mentor

This equation does not look correct to me. Check the signs of the terms. Also, in this equation, please specify whether F is supposed to be the force in the positive x direction that the nozzle is exerting on the fluid, or the force in the positive x direction that the fluid is exerting on the nozzle.

5. Jan 24, 2017

### joshmccraney

What about it doesn't look right to you?
Sorry, $F$ is suppose to be the force the nozzle is exerting on the fluid. I did flip the signs for pressure, so the balance is
$$-\rho V_1^2 \pi R_1^2+\rho V_2^2 \pi R_2^2 = P_1 \pi R_1^2-P_a \pi R_2^2 -{F_{bolt}}$$

6. Jan 24, 2017

### Staff: Mentor

Much better.

7. Jan 24, 2017

### joshmccraney

Thanks! But I'm still curious how to deal with the bolts, or rather my $F_{bolt}$. For instance, if the problem sketch did not have the bolts, but instead showed a continuous nozzle, how would my analysis be wrong (it must be, right?)

8. Jan 24, 2017

### Staff: Mentor

If it were continuous, this would give the tensile force in the straight section of the pipe upstream of the nozzle.

Incidentally, you should be solving this using gauge pressure in the equation. Otherwise, when you want to find the net force on the nozzle, you will have to include the air pressure on the outside surface of the nozzle in the force balance equation. F from the momentum balance on the fluid is not the same as the force exerted by the bolts.

9. Jan 24, 2017

### joshmccraney

So the air pressure on the outside would be $-\iint P\, dS$ which is $P_a$ times the surface area of the pipe, right? If $F$ is not the force exerted by the bolts, then what is it exactly? You said it is the tensile force; how is this not the force the bolts would offset?

10. Jan 24, 2017

### Staff: Mentor

F is the force that the fluid exerts on the nozzle and also the force that the nozzle exerts on the fluid. But, this is not the total force on the nozzle. And it is not the force that the bolts exert on the nozzle. Draw a free body diagram of the nozzle. If you solve this problem using gauge pressure instead of absolute pressure, you will get a different result. Try it and see. What do you get for F in terms of $V_1$, $\rho$, $P_a$, $A_1$, and $A_2$ if you use absolute pressure? What do you get if you use gauge pressure?

11. Jan 25, 2017

### joshmccraney

$$F = (\rho V_1^2+P_1)A_1-(\rho V_2^2+P_a)A_2-P_a\cdot A_{nozzle}\\ F = (\rho V_1^2+P_1-P_a)A_1-\rho V_2^2A_2$$
for absolute and gauge pressure respectively. I don't think I did this correctly though, since we could now solve for the nozzle area; what do you think?

12. Jan 25, 2017

### Staff: Mentor

Hi Josh,

When I asked "for F in terms of $V_1$, $\rho$, $P_a$, $A_1$, and $A_2$," I didn't expect to see $P_1$ and $V_2$ in your final equation. I expected you to eliminate these parameters from the final result. Can you please go back and do this? Thanks.

Chet

13. Jan 25, 2017

### joshmccraney

$$F = \left(\rho V_1^2+ P_A+1/2 \rho \frac{A_1 V_1}{A_2}-1/2 \rho V_1\right)A_1-\left(\rho \left( \frac{A_1 V_1}{A_2} \right)^2+P_a\right)A_2-P_a\cdot A_{nozzle}\\ =\rho V_1^2A_1+ P_AA_1+1/2 \rho \frac{A_1^2 V_1}{A_2}-1/2 \rho V_1A_1-\rho \frac{A_1 V_1^2}{A_2}+P_a A_2-P_a\cdot A_{nozzle}\\ F = \left(\rho V_1^2+ P_A+1/2 \rho \frac{A_1V_1}{A_2}-1/2 \rho V_1-P_a\right)A_1-\rho \left( \frac{A_1 V_1}{A_2} \right)^2A_2\\ =\rho V_1^2A_1+ P_AA_1+1/2 \rho \frac{A^2_1V_1}{A_2}-1/2 \rho V_1A_1-P_a A_1-\rho \frac{A_1^2 V_1^2}{A_2}$$
How does this look?

14. Jan 25, 2017

### Staff: Mentor

It looks like there are some algebra errors. I get:
$$F=P_a(A_1-A_2)+\frac{1}{2}\rho V_1^2A_1\left(\frac{A_1}{A_2}-1\right)^2$$

15. Jan 28, 2017

### joshmccraney

Oops, dang algebra!
So if $F$ is not the tensile force exerted on the bolts then what is?

16. Jan 28, 2017

### Staff: Mentor

The increase in tensile force at the bolts is $\frac{1}{2}\rho V_1^2A_1\left(\frac{A_1}{A_2}-1\right)^2$. This is also what you had gotten if you had used gauge pressures instead of absolute pressures.

Can you provide an explanation of why the air pressure term drops out?

17. Jan 28, 2017

### joshmccraney

I would guess because there is also pressure on the outside of the nozzle that is acting normal to the nozzle, and this pressure cancels the pressure gradient inside the nozzle?

18. Jan 28, 2017

### Staff: Mentor

Yes. If you integrate the atmospheric pressure vetorially over the outside surface of the nozzle, it exactly cancels the atmospheric pressure term from the momentum balance.

19. Jan 28, 2017

### joshmccraney

So without integrating explicitly how would I know this?

20. Jan 28, 2017

### Staff: Mentor

It's not hard to integrate formally for any arbitrary shape r = r(z). But you find that the outside force is just equal to the pressure times the difference in projected area. Try it and see.

21. Jan 28, 2017

### joshmccraney

Wait, I see it now! So all the vertical components of the pressure will cancel out by symmetry, leaving only the horizontal components. Adding up those horizontal pressure components will be $A_2-A_1$, the disk. Does this always happen with pressure (for instance, if we were taking the hydrostatic pressure on, say, a 2-D board that goes from $(0,0)\to(1,1)$ and another that goes from $(0,0)\to(0,1)\to(1,1)$, would we have the same horizontal hydrostatic pressure?

I'm curious for your integral though!

22. Jan 28, 2017

### Staff: Mentor

Yes.

23. Jan 29, 2017

### Staff: Mentor

Differential area on surface of nozzle = $rd\theta \sqrt{(dr)^2+(dz)^2}$

Unit tangent along surface of nozzle = $\vec{i}_t=\frac{\vec{i}_rdr+\vec{i}_zdz}{\sqrt{(dr)^2+(dz)^2}}$

Inward directed unit normal to nozzle surface = $\vec{i}_n=\frac{-\vec{i}_rdz+\vec{i}_zdr}{\sqrt{(dr)^2+(dz)^2}}$

Pressure force acting on differential area on differential area on surface of nozzle =
$(rd\theta \sqrt{(dr)^2+(dz)^2})(\frac{-\vec{i}_rdz+\vec{i}_zdr}{\sqrt{(dr)^2+(dz)^2}}p_a)=rd\theta(-\vec{i}_rdz+\vec{i}_zdr)p_a$

If we integrate this over the surface of the nozzle, we get $(A_2-A_1)p_a\vec{i}_z$

24. Jan 29, 2017

### Staff: Mentor

Differential area on surface of nozzle = $rd\theta \sqrt{(dr)^2+(dz)^2}$

Unit tangent along surface of nozzle = $\vec{i}_t=\frac{\vec{i}_rdr+\vec{i}_zdz}{\sqrt{(dr)^2+(dz)^2}}$

Inward directed unit normal to nozzle surface = $\vec{i}_n=\frac{-\vec{i}_rdz+\vec{i}_zdr}{\sqrt{(dr)^2+(dz)^2}}$

Pressure force acting on differential area on surface of nozzle =
$(rd\theta \sqrt{(dr)^2+(dz)^2})(\frac{-\vec{i}_rdz+\vec{i}_zdr}{\sqrt{(dr)^2+(dz)^2}}p_a)=rd\theta(-\vec{i}_rdz+\vec{i}_zdr)p_a$

If we integrate this over the surface of the nozzle, we get $(A_2-A_1)p_a\vec{i}_z$

Last edited: Jan 29, 2017
25. Jan 29, 2017

### joshmccraney

Thanks a ton!!!!!