# Homework Help: Forces & Hydrodynamics

1. Feb 19, 2013

### danielakkerma

(Hello everyone!)
1. The problem statement, all variables and given/known data
Water is gathered in a column of height h, above above a distributing pipe.
The pipe then narrows from cross sectional area S1 to S2(S1>S2), after which, the fluid exits(as shown in the attached figure)
Find the total horizontal force(in the direction of the fluid's flow) acting on the pipe.
(s1 = 3.0 cm^2, s2 = 1 cm^2, h = 4.6 m).
2. Relevant equations
Bernoulli's law(dynamic pressure):
1. P = rho*v^2/2
The continuity equation 2. v1*s1=v2*s2=const.
3. The attempt at a solution
I started by defining the force:
Over a certain, small, pressure drop, the (minute) force generated is:
$$dF = dP\cdot S(x)$$
Where S(x) is cross-sectional area over the pressure gradient).
I assume that, before the pipe starts to constrict, as the fluid is taken to be ideal(inviscid), there are no force contributions there.
Therefore, according to Bernoulli, a small chance in pressure would be directly related to an alteration of the velocity, by differentiating 1.:
$$dP = \rho v dv$$
And from continuity conditions:
$$S_1 v_1 = S(x)\cdot v(x) = const \\ v(x) = \frac{S_1 v_1}{S(x)} \\ dv = \frac{-S_1 v_1}{{S(x)}^2}dS \\$$
Plugging both into dF:
$$dF = dP\cdot S(x) = \rho vdv S(x) = -\rho\frac{S_1 v_1}{{S(x)}^2} \frac{S_1 v_1}{S(x)}S(x)dS = -\frac{\rho{S_1}^2{v_1}^2}{{S(x)}^2}dS$$
Integrating, to obtain F, S, going from S1, to S2:
$$F = \rho {S_1}^2{v_1}^2 \int_{S_1}^{S_2}\frac{-dS}{{S^2}} = \rho {S_1}^2{v_1}^2 \cdot (\frac{1}{S_2}-\frac{1}{S_1})$$
Now, since v1 is the velocity the fluid gains from its column, i.e., due to gravity, Bernoulli gives us:
$$v_1 = \sqrt{2gh}$$
Inserting all the given data gives me:
(rho is the density of water: 1 g/cm^3)
$$F \approx 55 N$$
Yet the correct answer is, F = 6 N!
What have I done wrong?
Any help would be greatly appreciated,
And I am as always thankful for your time and attention,
Daniel

#### Attached Files:

• ###### dia2.jpg
File size:
56 KB
Views:
141
Last edited: Feb 19, 2013
2. Feb 19, 2013

### SteamKing

Staff Emeritus
If you show your work (with numbers plugged into formulas) you might get some help in tracking down your error.

3. Feb 19, 2013

### danielakkerma

(You can see the process unfold before(that is, the derivation));
I applied the given data as follows:
$$F= \rho {v_1}^2{S_1}^2(\frac{1}{S_2}-\frac{1}{S_1})$$
I inserted:
$$\rho = \frac{1g}{{cm}^3} \\ S_1 = 3.0 {cm}^2 \\ S_2 = 1 {cm}^2 \\ v_1 = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 4.6 \frac{m^2}{s^2}} \\$$
Putting all that together:
$$F = \frac{1g}{{{cm}^3}}2 \cdot 10 \frac{m}{s^2} \cdot 460 {cm} \cdot 9 {{cm}^4}(\frac{2}{3}\frac{1}{{{cm}^2}})=55,200 \frac{g \cdot m}{s^2} = 55.2 N$$
Still wrong I fear...
What do you make of all this?
Thank you very much again!
Daniel

4. Feb 20, 2013

### danielakkerma

Sorry, I am going to have to bump this...
Has anyone got an idea?

5. Feb 20, 2013

### haruspex

That would be true if the water were going straight out to atmospheric pressure there. Applying Bernoulli to the transition from surface of water to start of pipe: v12/2+p1/ρ=gh. However, I get 12N, not 6.

6. Feb 21, 2013

### danielakkerma

Interesting, could you elaborate on that?
I thought that given that the water was still in the beginning, it was simply a question of converting the "weight" of the water to Dynamic pressure.
It's just like water flowing out of a tank, with a hole in one side.
How did you arrive at 12N?(that's pretty close! I may have missed a factor of 1/2, so I'd like to trace your answer, if you would indulge me)...
Thanks again,
I think we're making progress here¸
Daniel

Last edited: Feb 21, 2013
7. Feb 21, 2013

### haruspex

You have to be careful to consider pressure and motion at the same location. If you consider a point at the depth of the pipe but on the opposite side, it will be at the pressure given by the depth, but it won't be moving. If you consider a point at the start of the exit pipe, it will be moving, but its pressure will be lower (as given by Bernoulli). At the end of the exit pipe, we're back to atmospheric and moving faster still. Again, you can set up a Bernoulli equation. So you have two equations and three unknowns (pone pressure, two speeds). Your third equation is that the rate of flow of mass is constant along the pipe.

8. Feb 22, 2013

### danielakkerma

Haru,
Thanks for your detailed responses, and I'm sorry if I'm stretching this out for far longer than it should be. I very much value your patience with me here.
The problem is, I don't quite see what you mean:
For a small fluid parcel with volume dV, positioned inside the draining pipe(and initially at rest), one could construct the following Bernoulli equality:
$$P_1 = P_2+\frac{\rho {v_2}^2}{2}$$(No change in height, along the funnelling tube).
So basically, all that's left to be determined is P_1-P_2.
But here's the deal:
P_1(to the "left" of the fluid, towards the basin), has to be: P_atm + P_fluid column:
P_atm + ρgh(there are no other sources of pressure, that I'm aware of).
To the right:
P_atm(the nozzles opens up to air)+0(since the fluid to the right is taken to be still "at rest").
P_atm cancels out.
So where does all this lead?
I would be very grateful if you could show your own derivation, as I am somewhat stuck!
Still, I am very thankful for all your help,
And hoping for some breakthrough,
Daniel

9. Feb 22, 2013

### haruspex

But that is not a feasible combination. If it is inside the pipe it must already be moving. You have to think of the entire tank as part of the flow.
If water enters the pipe at speed V1, pressure P1 (above atmospheric), area S1, and leaves at speed V2, pressure P2 (= atmospheric, i.e. 0), area S2, we have:
gh = V12/2+P1/ρ = V22/2
R (mass flow rate) = V1S1 = V2S2
Whence V2 = √(2gh), V1 = √(2gh)S2/S1, P1 = ρgh(1-S22/S12)
To get the backthrust from the pipe, one way is to compare the applied force at the start of the pipe, P1S1, with the resulting mass*acceleration, ρR(V2-V1). Take the difference of those and you should get 6N (I made a mistake before).

10. Feb 22, 2013

### rude man

I look at it as follows:
It's the horizontal component of force that matters. So dF = pdS since the area under question is the difference in areas between S+dS and S = dS = 2πr dr, r = radius.

I also figure that according to Bernoulli applied to the tank's surface and the final outlet point 2 is (atmospheric pressure disregarded at both ends)
0 = - ρgh + ρv22/2 → v22 = 2ρgh so working backwards into the pipe,
ρv2(x)/2 + p(x) = ρv22/2 where v(x = 0) = v1.

Combined with S(x)v(x) = S2v2 you can now integrate dF from x = 0 to x = L, L = length of tapered section. S(x) = πr2(x).
I have not done the numbers.

(May be of interest: if you assume an abrupt change in pipe area from 3 cm2 to 1c2, the force works out to 8N. That is very easy to compute & I was hoping it would come out to 6!)

11. Feb 22, 2013

### haruspex

How the cross section depends on distance along the pipe is unknown, so I'd prefer a solution that makes no assumption about that. If you assume it's a cone then S(x) = A(x+B)2.

12. Feb 22, 2013

### rude man

If no such assumption needs to be made then I would think that I'm entitled to include the limiting case of abrupt aperture transition, in which case I'm fairly confident of my solution of 8N.

13. Feb 22, 2013

### haruspex

You certainly have to assume that the pipe is long in comparison to its width; otherwise flow effects within the tank near the exit become important. Although this is not explicitly stated, it is clear from the accompanying diagram. But I don't see how you get 8N with an abrupt change. Wouldn't you get ρgh(S1-S2), which I calculate as about 9N?

14. Feb 22, 2013

### Staff: Mentor

I solved this problem independently, making use of an overall momentum balance as the final step, and got:
$$F=\rho ghS_1(1-\frac{S_2}{S_1})^2$$ which gives a force of 6 N. As haruspex indicated, it was unnecessary to integrate along the length of the taper. I might also mention that haruspex hinted at using an overall momentum balance, but no one seemed to be very receptive to this idea. Too bad. It seemed obvious to me that this is what was needed.

Of course, this analysis only applies to an ideal (inviscid) fluid.

Chet

Last edited: Feb 23, 2013
15. Feb 23, 2013

### rude man

F = p1(S1 - S2)
p1 = (ρ/2)(v22 - v12) (Bernoulli along the pipe)
v22/2 = gh (Bernoulli from top of tank to S2)
v1 = v2S2/S1
F = 4.01e4(3e-4 - 1e-4) = 8

CM, does momentum transfer give F = 6 or F = 8 or ?? for an abrupt transition from S1 to S2 do you think?

Last edited: Feb 23, 2013
16. Feb 23, 2013

### danielakkerma

Guys, everyone, thank you so much!
I've indeed gleaned a great of useful info here, and finally got to the bottom of this.
Haruspex, a special thanks goes out to you, for getting the ball rolling; but everyone, you're all terrific!
Thank you all!

17. Feb 23, 2013

### rude man

I got the same thing using ∫Fdt = Δmomentum (can't use p for momentum here! ).
Now to figure out why my abrupt-junction calculation misfired ...

18. Feb 23, 2013

### Staff: Mentor

The overall momentum balance in the axial direction (on the fluid) gives:

$$P_1S_1-P_2S_2-F=\rho v_2^2S_2-\rho v_1^2S_1$$

with $P_2=0$ and with $\rho v_2S_2=\rho v_1S_1$

For an ideal (inviscid) fluid, this gives a value of F = 6N, irrespective of the transition from S1 to S2, provided only that there is enough pipe length following the transition for the streamlines to straighten out again before the fluid exits the pipe.

19. Feb 23, 2013

### haruspex

I'm pretty sure it would be for the reason I gave - it's hard to know the pressure distribution in the approach to an abrupt change because it depends on details of the flow. But I can't be certain without seeing your calculation.

20. Feb 23, 2013

### Staff: Mentor

Yes. you are right about that. One can't get the details of the distribution of pressure at the wall without solving the differential flow equations. But that isn't necessary for this problem, since the overall momentum balance delivers the overall integrated force at the boundary, which is all that is asked for. What the overall momentum balance does tell you, however, is that the details of how the channel is tapered are not relevant to the value of the overall force. Of course, if the fluid has viscosity, these results are modified, and the channel taper has at least some effect, since mechanical energy is no longer conserved (i.e., the Bernoulli equation is no longer exact).

21. Feb 23, 2013

### rude man

See post 15.
But I think my problem was that for an abrupt junction you can't assume streamlines so Bernoulli is not applicable.

22. Feb 23, 2013

### Staff: Mentor

Hi Rude Man,

Who says that you can't assume streamlines for an abrupt contraction? My understanding is that, as long as the fluid is ideal, Bernoulli applies. Even with viscous flow, you can assume streamlines.

Chet

23. Feb 24, 2013

### rude man

I would think that, as the fluid bangs up against the abrupt junction, a streamline up to that point is destroyed.

OK then help me figure out what is wrong with my post 15!

24. Feb 24, 2013

### Staff: Mentor

Hi Rude Man,

No way. The streamline along the wall makes a 90 degree turn at the junction and heads downword. As long as the flow is steady and the continuity equation applies, streamlines can be identified. In 2D cartesian coordinates, for example,

$$\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}=0$$

and you can write:

$$v_x=\frac{\partial \psi}{\partial y}$$
$$v_y=-\frac{\partial \psi}{\partial x}$$
where ψ is the stream function. This satisfies the continuity equation identically.

The thing that's wrong with post #15 is the force (momentum) balance.

Chet

25. Feb 24, 2013

### rude man

Yes I know but would still like to know what is wrong with my reasoning. It's simply finding p1 by Bernoulli from just to the left of the S1/S2 junction to the outlet, Bernoulli again from the top of the tank to the outlet, then saying F = p1(S1 - S2).