# 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

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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

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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

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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

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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.

#### MrsZee

Yes, but the question asks about the force after the level has dropped very little.
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level

#### Chestermiller

Mentor
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level
This is not correct. The exit pressure from the pipe is zero gauge. Moreover, the fluid is being accelerated horizontally, so the tank is exerting a force to accelerate the fluid, and the fluid is exerting a corresponding reaction force on the tank. The OP's result is spot on, and captures this force correctly.

#### haruspex

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2018 Award
which is the force when the plug is released; which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe level
No. That would be the force before the plug is removed.
The questioner wants us to consider the force when a more-or-less steady state has been achieved, on the assumption that the level has not declined much.
In practice, of course, the flow increases swiftly from 0, reaches a maximum, then declines as the level in the tank drops.
Usually, with this question, we are not even required to worry about the rate at which the level drops, but here we are advised to do so. Using Bernoulli’s equation and the incompressibility of water we get that the force is $\frac{2gh\rho a}{1-(\frac aA)^2}$.

#### MrsZee

potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations

#### haruspex

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Gold Member
2018 Award
potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations
Note the wording: "the necessary force to keep the container still when the plug has been removed."
The most reasonable interpretation is to find the maximum force that occurs subsequent to the unplugging. Your approach yields $\rho gha$. As I stated in post #15, Bernoulli's equation gives a little more than double that. However, I regard the denominator there as bogus. A full analysis (if I remember correctly) gives a peak value a bit less than $2\rho gha$.

#### Chestermiller

Mentor
potato-potato
the pressure before the plug is released is the one to create the force of the flow as the plug is released; split seconds; after the plug is released everything starts moving because of this force

we can agree to disagree
btw, the question is will the system trip over, or will it get to the steady state you are waiting to happen to start calculations
You omitted the part of the force necessary for accelerating the fluid horizontally, from zero horizontal velocity at the top of the tank to the horizontal velocity in the exit pipe. Before the plug is removed, the fluid is not being accelerated horizontally. This is the reason the correct answer is twice as high as your (incorrect) answer. The OP and haruspex analyzed the problem correctly. For more details on how to properly do this kind of calculation, see Bird, Stewart, and Lightfoot, Transport Phenomena, section on Macroscopic Momentum Balances.

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

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