# Homework Help: Liquid flowing in a tube

1. Apr 15, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

A horizontal tube of uniform cross sectional area 'A' is bent in the form of U. If the liquid of density 'ρ' enters and leaves the tube with velocity 'v' then how much external force is required to hold the bend stationary ?

2. Relevant equations

3. The attempt at a solution

I think the bend in the tube provides centripetal force to the fluid flowing through it .As a result the fluid also exerts force on the bend. But then radius of the curve is not given.

In 1 sec liquid of mass ρAv is flowing through any cross section of the tube.

I would be grateful if somebody could help me with the problem.

2. Apr 15, 2014

### TSny

You can work the problem using centripetal force, but there's an easier way.
momentum

3. Apr 15, 2014

### Tanya Sharma

The change in momentum/sec 2ρAv2 of the liquid is the force by the tube on the liquid.Same force is applied by liquid on the tube .So external force F = 2ρAv2 needs to be applied , in opposite direction to the force by liquid to keep the tube at rest .

Is it correct ?

Last edited: Apr 15, 2014
4. Apr 15, 2014

### TSny

Looks good!

5. Apr 15, 2014

### Tanya Sharma

Thanks TSny !

There is a similar problem in which I am having difficulty .

Q . Water flows with a velocity v along a rubber tube having the form of a ring. The radius of the ring is R and diameter of the tube is d (d<<R). Find the force with which the rubber tube is stretched.(ρ is the density of water).

My reasoning : The mass/sec of water flowing through any cross section of the tube is given by ρπd2v/4 .

Not sure what to do next .

6. Apr 15, 2014

### TSny

Can you relate a ring to a U-tube?

7. Apr 15, 2014

### Tanya Sharma

Three similarities I can think

1) Liquid goes in liquid comes out.
2) In U tube a part is curved whereas a ring is completely circular.
3) Change of momentum/sec of liquid is same in both the cases i.e 2ρAv2

Last edited: Apr 15, 2014
8. Apr 15, 2014

### TSny

For what part of the circumference of the ring does this expression apply?

9. Apr 15, 2014

### Tanya Sharma

Approx. 100% of the circumference .Almost a full circle .

10. Apr 15, 2014

### TSny

No. Think about where the factor of 2 comes from in 2ρAv2 in the result for the U-tube.

11. Apr 15, 2014

### Tanya Sharma

Factor of 2 comes from a reversal of momentum of the fluid which had entered the curved part.

12. Apr 15, 2014

### TSny

Yes. How much of the circumference corresponds to reversal of momentum?

13. Apr 15, 2014

### Tanya Sharma

Can't assign a numeric value ,but a very small fraction of the circumference .

14. Apr 15, 2014

### TSny

Consider a cross section of the ring and the direction of momentum flowing through it. How far around the ring do you need to go before you reach a second cross section where the direction of momentum is reversed relative to the first cross section?

15. Apr 15, 2014

### Tanya Sharma

The distance around the ring is ∏R i.e half the circumference.

16. Apr 15, 2014

### TSny

OK. Draw a free body diagram for the portion of the rubber tube representing half a circumference. Include the total force due to the motion of the fluid in this section of the tube as well as any other forces acting on this section of the tube.

17. Apr 15, 2014

### Tanya Sharma

In the attachment I have shown the left half of the tube.

The net force due to the fluid is towards left having magnitude 2ρAv2. Since it is in equilibrium an equal amount of force is exerted by right half of the rubber tube.

#### Attached Files:

• ###### ring.GIF
File size:
2.1 KB
Views:
110
18. Apr 15, 2014

### TSny

OK. Can you be more specific about the force exerted by the right half? At what point(s) of the left half does the force of the right half act?

19. Apr 15, 2014

### Tanya Sharma

The force on the left half is exerted on the circumference of two circular end points of the tube (in red). It's exaggerated view is shown towards right . The blue lines indicate the force on the circumference .The right half exerts force through the two end points .

#### Attached Files:

• ###### ring.GIF
File size:
4.1 KB
Views:
110
20. Apr 15, 2014

### TSny

OK. How would you relate the question "Find the force with which the rubber tube is stretched" to the force on a circular cross section (i.e., one of your blue forces in your figure)?

21. Apr 15, 2014

### Tanya Sharma

I am not entirely convinced but I will give a try

The force with which left half of the rubber tube is stretched = Sum of the forces on the two circular cross sections = 2(force on a circular cross section)

22. Apr 15, 2014

### TSny

So, it comes down to interpreting the question. I tend to think that the question is asking for the "tension" force in the rubber tube created by the moving fluid. If you take a cross section of the tube, then the tension is the total force on the circumference of the tube at the cross section, as shown in your figure. If that's what the problem is asking for, then you should be able to get the answer easily from your free-body diagram.

23. Apr 15, 2014

### Tanya Sharma

Is post#21 right ?

Force on a circular cross section = 1/2(Force on the left half of the tube) = ρAv2 .

Is it correct ?

Okay

I think this is the crux . But ,sorry I didn't understand .Could you please elaborate

Last edited: Apr 15, 2014
24. Apr 15, 2014

### TSny

I would say that's right. But I think the force they want is the force on a circular cross section.

Yes. I think that's the answer to the question.

I was just trying to say that my interpretation of the question is to calculate the force on a circular cross section, which is what you have done. Hope I haven't misinterpreted the question.

25. Apr 15, 2014

### Tanya Sharma

Okay...so our goal was to find the force at any cross section of the tube and for that we chose half the circumference as it is easy to find the force on it with the help of change of momentum of the liquid . Right ?

Now after doing the problem ,I am not able to relate this to the OP i.e the case of a U tube.In the OP we applied a similar approach but we were not calculating the force at any cross section ,rather the net force on the U tube .

How does the same concept fit in both the cases ?