Cylinder with a horizontal tube (Fluid dynamics problem)

In summary: The force when the plug is released is ##F=\frac{mΔv}{Δt}=ρ*v_2*A_2*Δv## which will be the product of the cross sectional area of the pipe multiplied with the pressure at the pipe...
  • #1
ValeForce46
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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.]
Relevant Equations
Bernoulli's Principle: ##p+ρgz+\frac{1}{2}ρv^2=constant##
1567707951747.png

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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  • #2
ValeForce46 said:
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.
 
  • #3
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.
 
  • #4
haruspex said:
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:
  • #5
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?
 
  • #6
Chestermiller said:
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...
 
  • #7
ValeForce46 said:
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.
 
  • #8
ValeForce46 said:
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
 
  • #9
ValeForce46 said:
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.
 
  • #10
MrsZee said:
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.
 
  • #11
haruspex said:
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
 
  • #12
ValeForce46 said:
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.
 
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  • #13
haruspex said:
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
 
  • #14
MrsZee said:
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.
 
  • #15
MrsZee said:
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}##.
 
  • #16
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
 
  • #17
MrsZee said:
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##.
 
  • #18
MrsZee said:
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.
 
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Related to Cylinder with a horizontal tube (Fluid dynamics problem)

1. What is a "cylinder with a horizontal tube" in fluid dynamics?

A cylinder with a horizontal tube is a common geometry used in fluid dynamics experiments and simulations. It consists of a cylindrical body with a straight tube attached to the side, typically at a right angle. This setup allows for the study of fluid flow in and around the cylinder and tube, which can have important applications in various industries such as aerodynamics and hydrodynamics.

2. What is the purpose of studying fluid dynamics in a cylinder with a horizontal tube?

The study of fluid dynamics in a cylinder with a horizontal tube can provide valuable insights into the behavior of fluids in real-world scenarios. This setup allows researchers to investigate phenomena such as drag and lift forces, vortex shedding, and turbulence, which are crucial in understanding the flow of fluids around and through objects.

3. How is the flow of fluid in a cylinder with a horizontal tube affected by different parameters?

The flow of fluid in a cylinder with a horizontal tube can be affected by various parameters such as the Reynolds number, the diameter and length of the cylinder and tube, the fluid viscosity, and the velocity of the fluid. These parameters can influence the flow regime, pressure distribution, and drag force on the cylinder, which are important factors to consider in fluid dynamics studies.

4. What types of experiments can be conducted using a cylinder with a horizontal tube?

A cylinder with a horizontal tube is a versatile setup that can be used for a wide range of fluid dynamics experiments. Some common experiments include flow visualization, pressure measurement, and force measurement. Researchers can also use this setup to study the effects of various flow control techniques, such as the use of flow control devices or surface modifications, to improve fluid flow around the cylinder and tube.

5. How does the flow of fluid in a cylinder with a horizontal tube compare to other geometries?

The flow of fluid in a cylinder with a horizontal tube differs from other geometries, such as a plain cylinder or a cylinder with a vertical tube, due to the additional interaction between the cylinder and the tube. This interaction can significantly affect the flow behavior, and it is essential to consider when choosing the appropriate geometry for a particular fluid dynamics study.

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