# Homework Help: Water Flowing through U-tubes (Velocity Given Cross-Sectional Areas)

1. Jan 28, 2013

### SignaturePF

1. The problem statement, all variables and given/known data
See question 23

A' = 1/2A

2. Relevant equations
Av = A'v'

3. The attempt at a solution
I thought this was just a simple Av = A'v' problem
which would lead to v' = 2v. But there is apparently more to it, as the answer is √2v.
It could have to do with the momentum because F_net = 0 so it is conserved.
In that case,
mv = mv but I'm confused about how to use that to solve the problem.

2. Jan 28, 2013

### SignaturePF

Any thoughts?

3. Jan 28, 2013

### Staff: Mentor

Use conservation of momentum. In terms of the fluid density, the area, and the velocity v, what is the rate of change of momentum of the fluid passing through the U tube on the left? Same question for the U tube on the right.

4. Jan 28, 2013

### SignaturePF

momentum = mv
dp/dt = m/t(v)
dp/dt = (ρAv)v
dp/dt = ρAv^2 - This is the left

Right:
dp/dt = pA'v'^2
A' = .5A
dp/dt = ρAv'^2/2

The two rate changes are equal ( net forces are zero).
v^2 = v'^2/2
2v^2 = v'^2
v' = √2v

This is the right answer. Thanks.

Just a question; from where did you get the motivation to use rate change of momentum?

5. Jan 28, 2013

### Staff: Mentor

The problem said that the assembly of u tubes is in equilibrium. That means that the forces exerted by the fluids on the u tubes must be equal. The force exerted by each fluid on each u tube is equal to the force exerted by the u tube on the fluid. The latter is equal to the rate of change of momentum of the fluid.

6. Jan 28, 2013

### SignaturePF

Ok I see where you were going.
Equlibrium → ƩF = 0 → FL = FR
F = dp/dt
dp/dt = mv/dt
Given, that v is constant, m is changing
dp/dt = dm/dt*v
dm/dt = ρAv
dp/dt = ρAv^2
And that's it right.

I'm just making sure so I don't end up memorizing scenarios.

7. Jan 28, 2013

### Staff: Mentor

Yes. Well done. I hold in very high regard a student like you who focuses on fundamentals. You can look forward to a bright future.

Chet