# Cylinder with a horizontal tube (Fluid dynamics problem)

#### ValeForce46

Homework Statement
A container is made up of a vertical cylinder, diameter $D=9.0 cm$, on which is placed an horizontal tube, diameter $d=3.0 cm$, with axis at a distance $l=5.0 cm$ from the bottom of the cylinder. The other end of the horizontal tube is plugged and the container is filled with water until the height of $h=50 cm$ (see picture). Assuming that the surface is perfectly smooth (no friction), determine the value of the necessary force to keep the container still when the plug has been removed. [Advice: the lowering speed of the column of water should NOT be neglected.]
Homework Equations
Bernoulli's Principle: $p+ρgz+\frac{1}{2}ρv^2=constant$

I'll call $v_2$ the flow speed on the horizontal tube and $v_1$ the lowering speed of the column.
Even if the lowering speed isn't negligible, the problem says "...when the plug has been removed", so can I consider the height unchanged?
In order to find the force, I need the pressure when the plug is off. Is it $p'=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2$ ?
First thing I know is $v_1*A_1=v_2*A_2$ ($A_1$ and $A_2$ are the two section).
Second thing I know is $p_{atm}+\frac{1}{2}ρv_2^2=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_1^2$ thanks to Bernoulli's principle.
So I can find $v_2$.
The force should be
$$F=p'*A=[p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2]*(\frac{D}{2})^2*π$$
Is something wrong?

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#### haruspex

Homework Helper
Gold Member
2018 Award
Is it $p'=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2$ ?
I see no basis for such an equation. Can you justify it?

Having found v2, think about rate of change of momentum.

#### MrsZee

I am not sure about the accuracy of the EQ haruspex is pointing to.
But for sure you need to start with the pressure when the plug is ON.
After unplugging, everything will decrease with the time ( the pressure and both velocities ) , until there is no liquid going through pipe.

For the stability of the system, velocity out of the pipe in the un[lugging moment can be the only reason for container to NOT stay stable.

#### ValeForce46

I see no basis for such an equation. Can you justify it?
I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?

Last edited:

#### Chestermiller

Mentor
You can't consider the height unchanged. If the flow out the pipe is v2, what is the downward flow rate in the vertical tube v1 (given that the fluid can be considered incompressible)? From this, what does Bernoulli equation give you?

#### ValeForce46

You can't consider the height unchanged. If the flow out the pipe is v2, what is the downward flow rate in the vertical tube v1 (given that the fluid can be considered incompressible)? From this, what does Bernoulli equation give you?
The downward flow rate should be $v_1=v_2*\frac{A_2}{A_1}$
Bernoulli's principle gives me $p_{atm}+ρg(h-l-x)+\frac{1}{2}ρv_1^2=p_{atm}+\frac{1}{2}ρv_2^2$ if I can't consider the height unchanged...

#### Chestermiller

Mentor
The downward flow rate should be $v_1=v_2*\frac{A_2}{A_1}$
Bernoulli's principle gives me $p_{atm}+ρg(h-l-x)+\frac{1}{2}ρv_1^2=p_{atm}+\frac{1}{2}ρv_2^2$ if I can't consider the height unchanged...
They're only (tacitly) asking for the force F at short time when h has not changed significantly yet.

Also, in this problem, to get the force, you would find it easier to work in terms of gauge pressures and not absolute pressures.

#### MrsZee

I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?
the flow is not constant; it will decrease as the fluid drains

#### haruspex

Homework Helper
Gold Member
2018 Award
I thought that was the pressure to a certain height $(h-l)$ and flow speed $v_2$.
How do I find the force, having found v2? $F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv$ because flow rate is constant?
Sorry, in my cutting and pasting I picked out the wrong equation. The one I see no justification for is
$$F=p'*A=[p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2]*(\frac{D}{2})^2*π$$
As I posted, when you have found v2, think about horizontal momentum.

But as @Chestermiller points out, your Bernoulli equation needs to allow for the downward velocity at the top of the tank.
It is a bit of a fudge. On the one hand they want you to treat the level of water in the tank as unchanged, on the other, not to treat it as unchanging. I.e. pretend that it instantly transits from static to steady state flow.

#### haruspex

Homework Helper
Gold Member
2018 Award
the flow is not constant; it will decrease as the fluid drains
Yes, but the question asks about the force after the level has dropped very little.

#### ValeForce46

As I posted, when you have found v2, think about horizontal momentum.
Why is this wrong? $F=\frac{mΔv}{Δt}=ρ∗v_2∗A_2∗Δv=ρ*A_2*v_2^2$
v1 is vertical so it doesn't affect horizontal momentum

#### haruspex

Homework Helper
Gold Member
2018 Award
Why is this wrong? $F=\frac{mΔv}{Δt}=ρ∗v_2∗A_2∗Δv=ρ*A_2*v_2^2$
v1 is vertical so it doesn't affect horizontal momentum
I didn't say that was wrong, but I omitted to say it was right.
You have all the pieces.

"Cylinder with a horizontal tube (Fluid dynamics problem)"

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