Calculating Centripetal Force in a Bent Tube

[Tanya Sharma]

Homework Statement

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 ?

Homework Equations

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.

 

[TSny]

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

Spoiler

momentum

 

[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 ?

 

[TSny]

Looks good!

 

[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 .

 

[TSny]

Can you relate a ring to a U-tube?

 

[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

 

[TSny]

3) Change of momentum/sec of liquid is same in both the cases i.e 2ρAv²

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

 

[Tanya Sharma]

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

 

[TSny]

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

 

[Tanya Sharma]

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

 

[TSny]

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

 

[Tanya Sharma]

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

 

[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?

 

[Tanya Sharma]

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?

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

 

[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.

 

[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.

 

[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?

 

[Tanya Sharma]

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?

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 .

 

[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)?

 

[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)

 

[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.

 

[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 ?

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.

Okay

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.

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

 

[TSny]

Is post#21 right ?

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

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

Is it correct ?

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

But ,sorry I didn't understand .Could you please elaborate .

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.

 

[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 ?

 

[TSny]

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 ?

Yes.

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 ?

The curved portion of the U-tube may be considered a semicircle. In solving the U-tube question, you found the force that the fluid exerts on the curved, semicircular portion of the U-tube. So, this is also the force that the fluid exerts on a semicircle of the rubber tube. That was the purpose of relating the circular rubber tube question to the U-tube question.

Once you have the net force on a semicircular portion of the rubber tube, you can deduce the tension force at a cross section.

 

[Tanya Sharma]

Excellent ! Thank you very much TSny.

OK...enough of physics

By the way , "Thanks ∞ " looks very impressive

In between I didn't realize I crossed 1000 posts

 

[TSny]

In between I didn't realize I crossed 1000 posts

Congrats! you're now an old timer.

 

[Tanya Sharma]

Hello TSny

I have come up with another method to solve the problem in post#5 ,so I would like to discuss it with you .

I feel this problem is similar to finding the tension T in a ring of mass M and length L rotating with a constant angular velocity ω.If we consider a differential element dθ of the ring and do a force balance ,then we arrive at the result i.e $T= \frac{MLω^2}{4\pi^2}$ .We will use this result in our problem .

In this problem $M = \frac{ρ(\pi d^2)(2\pi R)}{4} = \frac{ρ\pi^2d^2R}{2} ,L = 2\pi R ,ω = \frac{v}{R}$

$$ T = \frac{ρ\pi d^2v^2}{4} = ρAv^2 $$ .This is same result as we obtained in post#23

What is your opinion ?

 

[TSny]

That looks good. Nice work.