Sinking a cylinder with varying hole sizes

lloydthebartender

Problem Statement
A cylinder, open on one end and sealed on the other except for a small hole of varying radii. It will be placed in a bucket of water from the side of the hole, and the time taken for the cylinder to completely submerged will be measured.
Relevant Equations
$F_{cylinder}=F_{gravity}-F_{buoyant}-F_{drag}$
$F_{ext}+v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}=m_{total}v$
I'm trying to write up some theory for this experiment I'm carrying out. I'll be changing the radius of the hole at the bottom of the cylinder, and I think the time taken for the cylinder to completely submerge is inversely proportional to the size of the hole. Problem is that I'm stuck at deriving this, so any help is much appreciated!

First I have equated all forces acting on the cylinder, and simplified it:
$F_{cylinder}=F_{gravity}-F_{buoyant}-F_{drag}$
$F_{cylinder}=m_{cylinder}g-\frac{1}{2}C\rho v^2A$

I found the force of the water onto the cylinder and then equated to the net force:
$F_{water}=\frac{\mathrm{d} p_{water}}{\mathrm{d} x}$
$F_{net}=m_{cylinder}g - \frac{1}{2}C\rho v^2A + \frac{\mathrm{d} p_{water}}{\mathrm{d} x}$

Then I derived and used the variable-mass equation:
$F_{ext}+v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}=m_{total}v$

I equated F_{ext} to F_{net}:
$m_{cylinder}g - \frac{1}{2}C\rho v^2A + \frac{\mathrm{d} p_{water}}{\mathrm{d} x}=m_{total}v-v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}$

And then I'm stuck and I have no idea what to do next...if there is anything missing please tell me. The issue is that the mass of water, as well as the downwards velocity of the cylinder, is changing at a rate I do not know.

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haruspex

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Problem Statement: A cylinder, open on one end and sealed on the other except for a small hole of varying radii. It will be placed in a bucket of water from the side of the hole, and the time taken for the cylinder to completely submerged will be measured.
Relevant Equations: $F_{cylinder}=F_{gravity}-F_{buoyant}-F_{drag}$
$F_{ext}+v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}=m_{total}v$

I'm trying to write up some theory for this experiment I'm carrying out. I'll be changing the radius of the hole at the bottom of the cylinder, and I think the time taken for the cylinder to completely submerge is inversely proportional to the size of the hole. Problem is that I'm stuck at deriving this, so any help is much appreciated!

First I have equated all forces acting on the cylinder, and simplified it:
$F_{cylinder}=F_{gravity}-F_{buoyant}-F_{drag}$
$F_{cylinder}=m_{cylinder}g-\frac{1}{2}C\rho v^2A$

I found the force of the water onto the cylinder and then equated to the net force:
$F_{water}=\frac{\mathrm{d} p_{water}}{\mathrm{d} x}$
$F_{net}=m_{cylinder}g - \frac{1}{2}C\rho v^2A + \frac{\mathrm{d} p_{water}}{\mathrm{d} x}$

Then I derived and used the variable-mass equation:
$F_{ext}+v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}=m_{total}v$

I equated F_{ext} to F_{net}:
$m_{cylinder}g - \frac{1}{2}C\rho v^2A + \frac{\mathrm{d} p_{water}}{\mathrm{d} x}=m_{total}v-v_{rel}\frac{\mathrm{d} m_{total}}{\mathrm{d} x}$

And then I'm stuck and I have no idea what to do next...if there is anything missing please tell me. The issue is that the mass of water, as well as the downwards velocity of the cylinder, is changing at a rate I do not know.
You have not defined your variables, but it looks like your v is the rate of descent of the cylinder and the drag term is the external drag on it. That equation would be right if the cylinder were completely open at both ends.

But the hole is small, so it will sink slowly. That makes drag linear, not quadratic.
Also, the level inside the cylinder will be less than outside. This means there will be buoyancy and the resulting pressure difference is what drives the flow of water into the cylinder, subject to quadratic drag. These parts of the equation system are probably a lot more important than any external drag.

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lloydthebartender

$v$ is the rate of descent of the cylinder
$u$ is the rate of inflow (which I take as constant)
$v_{rel}$ is $u-v$ in other words the difference in velocity
$A$ is the area of exposure between the base and the water, i.e. the area of the base minus the area of the hole.

$F_{net} = F_{gravity} - F_{buoyant} + F_{drag}$
$F_{drag} = -bv$

Do you mind clarifying this?
Also, the level inside the cylinder will be less than outside. This means there will be buoyancy and the resulting pressure difference is what drives the flow of water into the cylinder, subject to quadratic drag. These parts of the equation system are probably a lot more important than any external drag.
As I understand the buoyancy is $F_{b} = \rho*g*V$ where V is the volume of the displaced liquid. So does this value increase at a quadratic rate? If so what equation would I use?

Thank you

haruspex

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u is the rate of inflow (which I take as constant)
It will vary, but it should suffice to consider a steady state.
I assume this is the linear velocity relative to the cylinder.
$v_{rel}$ is ... the difference in velocity
I don't see how that is interesting. Think about how the level of the water within the cylinder will change.
A is the area of exposure between the base and the water, i.e. the area of the base minus the area of the hole.
Again, I do not see the relevance. The flow within the hole is quite different from that up the sides of the cylinder.
The area of the hole, however, is highly relevant.
buoyancy is $F_{b} = \rho*g*V$ where V is the volume of the displaced liquid
How does that volume relate to a) the extent by which the cylinder has descended and b) the volume of water which has flowed into the cylinder?

Have you drawn yourself a diagram of the partially descended cylinder, showing the water inside?

lloydthebartender

It will vary, but it should suffice to consider a steady state.
I assume this is the linear velocity relative to the cylinder.

I don't see how that is interesting. Think about how the level of the water within the cylinder will change.

Again, I do not see the relevance. The flow within the hole is quite different from that up the sides of the cylinder.
The area of the hole, however, is highly relevant.

How does that volume relate to a) the extent by which the cylinder has descended and b) the volume of water which has flowed into the cylinder?

Have you drawn yourself a diagram of the partially descended cylinder, showing the water inside?
This isn't the greatest drawing but here it is:

I'm measuring the time taken for the cylinder to completely submerge from the surface level.
How does that volume relate to a) the extent by which the cylinder has descended and b) the volume of water which has flowed into the cylinder?
I suppose since the value of $V$ increases from 0 to the volume of the cylinder, I could denote that in terms of time and in-flow? That way I would apply Bernoulli's equation before and after the hole...Or I could even use the continuity equation $A_{1}v_{1}=A_{2}v_{2}$ where $A_{1}$ is the small hole and $A_{2}$ is the cross-sectional area within the cylinder, in order to find $v_{2}$ which would be the rate at which the water inside the cylinder rises.

Perhaps I'm misunderstanding something because I have been trying to solve this in terms of the forces which apply on to the cylinder.

haruspex

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This isn't the greatest drawing
The situation you need to analyse is steady state descent, so half submerged, say. Show the internal and external water levels.
No need for perspective; a side view will do.
the value of V increases from 0 to the volume of the cylinder
Yes, except that things will change quite suddenly when the water outside overtops the cylinder. That will be before the cylinder is full.
I would apply Bernoulli's equation before and after the hole.
Yes!
use the continuity equation
That too.
solve this in terms of the forces which apply on to the cylinder.
Ok, but as I posted you should probably concentrate on steady state, which means forces would be in equilibrium. I think Bernoulli, Archimedes and continuity will get you a long way.

lloydthebartender

The situation you need to analyse is steady state descent, so half submerged, say. Show the internal and external water levels.
No need for perspective; a side view will do.

Yes, except that things will change quite suddenly when the water outside overtops the cylinder. That will be before the cylinder is full.

Yes!

That too.

Ok, but as I posted you should probably concentrate on steady state, which means forces would be in equilibrium. I think Bernoulli, Archimedes and continuity will get you a long way.
Ok great.

I know that:
$\frac{1}{2}\rho u^{2}+pgz_{1}+P_{1}=\frac{1}{2}\rho v^{2}+pgz_{2}+P_{2}$, where
$ρ$ is fluid density, $u$ is the rate of in-flow, $v$ is the rate of descent of the cylinder, $z_{1}$ is the height of the bottom of the cylinder from ground level, $z_{2}$ is the height of the surface of the liquid within the cylinder from ground level, $P_{1}$ is pressure at the small hole and $P_{2}$ is pressure surface of the water inside the cylinder.

Also:
$A_{1}u=A_{2}v$ where
$A_{1}$ is the size of the small hole $\pi R^2$ and $A_{2}$ the cross-sectional area of the cylinder $\pi r^2$.
So I can write the first equation:
$\frac{1}{2}\rho u^{2}+pgz_{1}+P_{1}=\frac{1}{2}\rho (\frac{A_{2}u}{A_{1}})^{2}+pgz_{2}+P_{2}$

At time TT where the cylinder is completely submerged it will look like this:

At time $T$, $V$ is equal to the volume of the cylinder. In the end, I need a relation between T and R...the volumetric flow rate of in-flow is:
$Q=uA_{1}$
Therefore:
$V=uA_{1}T$
$u=\frac{V}{A_{1}t}$
Plugging back into Bernoulli:
$\frac{1}{2}\rho (\frac{V}{A_{1}t})^{2}+pgz_{1}+P_{1}=\frac{1}{2}\rho (\frac{A_{2}u}{A_{1}})^{2}+pgz_{2}+P_{2}$
And I suppose I rearrange this to get a relation between TT and RR?

Last edited:

haruspex

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Ok great.

I know that:
12ρu2+ρgz1+P1=12ρv2+ρgz2+P212ρu2+ρgz1+P1=12ρv2+ρgz2+P2, where
ρρ is fluid density, uu is the rate of in-flow, vv is the rate of descent of the cylinder, z1z1 is the height of the bottom of the cylinder from ground level, z2z2 is the height of the surface of the liquid within the cylinder from ground level, P1P1 is pressure at the small hole and P2P2 is pressure surface of the water inside the cylinder.

Also:
A1u=A2vA1u=A2v, where
A1isA1is the size of the small hole πR2πR2 and A2isA2is the cross-sectional area of the cylinder πr2πr2.
So I can write the first equation:
12ρu2+ρgz1+P1=12ρ(R2ur2)2+ρgz2+P212ρu2+ρgz1+P1=12ρ(R2ur2)2+ρgz2+P2

At time TT where the cylinder is completely submerged it will look like this:
View attachment 243868

At time TT, VV is equal to the volume of the cylinder. In the end, I need a relation between T and R...the volumetric flow rate of in-flow is:
Q=uA1Q=uA1
Therefore:
V=uA1TV=uA1T
u=VA1Tu=VA1T
Plugging back into Bernoulli:
12ρ(VA1T)2+ρgz1+P1=12ρ(R2ur2)2+ρgz2+P212ρ(VA1T)2+ρgz1+P1=12ρ(R2ur2)2+ρgz2+P2
And I suppose I rearrange this to get a relation between TT and RR?

Edit: Rearranging I got
$R=T^\frac{1}{2}(\frac{C_{1}z_{2}-C_{2}z_{2}+C_{3}}{V_{2}C_{4}})^{\frac{1}{4}}$
This is very hard decipher. Please repost with the LaTeX working properly.
The z terms in Bernoulli refer to heights from some common level. The water just below the hole is effectively at the same height as the water just above it. The pressure difference vertically across the hole is accounted for by the P1 and P2 terms.
P2P2 is pressure surface of the water inside the cylinder.
P2 should be the pressure just above the hole, not at the surface of the water.

At time TT, VV is equal to the volume of the cylinder.
No. As I wrote, things will change swiftly when the outside level reaches the top of the cylinder. No need to consider a time beyond that.

lloydthebartender

This is very hard decipher. Please repost with the LaTeX working properly.

The z terms in Bernoulli refer to heights from some common level. The water just below the hole is effectively at the same height as the water just above it. The pressure difference vertically across the hole is accounted for by the P1 and P2 terms.

P2 should be the pressure just above the hole, not at the surface of the water.

No. As I wrote, things will change swiftly when the outside level reaches the top of the cylinder. No need to consider a time beyond that.
I think the formatting got messed up as I was editing it, sorry. It should be fixed now.

haruspex

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I think the formatting got messed up as I was editing it, sorry. It should be fixed now.
Much better, thanks. But my other comments still stand.

haruspex

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A couple more thoughts...
Since we are assuming steady state, z1-z2 will be constant.
The depth to the bottom of the tank is irrelevant, so instead create a variable for the depth of the bottom of the cylinder. What is the relationship between that and v?

lloydthebartender

The depth to the bottom of the tank is irrelevant, so instead create a variable for the depth of the bottom of the cylinder. What is the relationship between that and v?
Is this between the surface of the water inside the cylinder and the bottom of the cylinder?

I just realized...isn't the Bernoulli equation for the speed of the water, and not of the cylinder? If I were to take two points, inside and just above the hole at the bottom, all I'm doing is finding the speed of the water, and not of the velocity of the descending cylinder itself. I still don't see how the Bernoulli equation can account for this??? This is why I had the $F_{net}$ equation in the first place...to find the downwards force and then find the velocity from that (I think).

haruspex

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Is this between the surface of the water inside the cylinder and the bottom of the cylinder?
It can be depth below the outside surface or depth of water within the cylinder. Create another variable for the difference between them.
isn't the Bernoulli equation for the speed of the water, and not of the cylinder?
Yes, of course, but you need this.
This is why I had the FnetFnetF_{net} equation in the first place...to find the downwards force and then find the velocity from that
But you are not going to be able to find it without considering the flow of water into the hole.

lloydthebartender

It can be depth below the outside surface or depth of water within the cylinder. Create another variable for the difference between them.

Yes, of course, but you need this.

But you are not going to be able to find it without considering the flow of water into the hole.
So the diagram would look something like this; I am to find the relation between $v$, the rate of water just above the hole and $h$.

Well since $h$ diminishes to 0, and speed=dist/time, I can write it as $v=\frac{h}{T}$? I would then use the Bernoulli equation above and place $v=\frac{h}{T}$? and $u=\frac{V}{A_{1}T}$?

Sorry I know you're helping but I'm terribly thick in the skull sometimes.

haruspex

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since h diminishes to 0,
Why would it do that?

As I wrote, try assuming steady state, i.e. the cylinder descends at a constant rate. What does that say about the relationship between the buoyancy force and the weight of the (empty) cylinder?
In terms of your diagram in post #14, what is the force of buoyancy?
What do you conclude about h?

lloydthebartender

As I wrote, try assuming steady state, i.e. the cylinder descends at a constant rate. What does that say about the relationship between the buoyancy force and the weight of the (empty) cylinder?
Does this mean the weight equals the sum of drag and buoyancy?
In terms of your diagram in post #14, what is the force of buoyancy?
Since $F_{buoyancy}$ is proportional to the volume of water displaced...in this case, it would be the cross-sectional area of the cylinder multiplied by the height of the water within the cylinder (H), so $\pi r^2 H$.
$F_{weight} = F_{buoyancy} + F_{drag}$
$(m_{water}+m_{cylinder})g = \rho g\pi r^2H - bv^2$?

haruspex

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Does this mean the weight equals the sum of drag and buoyancy?
Yes, but let's see if we can get a result ignoring drag for now.
multiplied by the height of the water within the cylinder
Did you mean that? What is this H? Is it different from h in the diagram?

lloydthebartender

Did you mean that? What is this H? Is it different from h in the diagram?
Yes sorry I think I misunderstood; I was calculating the volume of water inside of the cylinder, and H is just the height of the water inside the cylinder from the base of the cylinder.
I need buoyancy so instead, I would calculate the volume of water displaced; in the case of my diagram in #14, the volume of water displaced is the volume of the cylinder itself under the water surface?

haruspex

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Yes sorry I think I misunderstood; I was calculating the volume of water inside of the cylinder, and H is just the height of the water inside the cylinder from the base of the cylinder.
I need buoyancy so instead, I would calculate the volume of water displaced; in the case of my diagram in #14, the volume of water displaced is the volume of the cylinder itself under the water surface?
Right, assuming you mean the surface outside the cylinder, but there is also water in the cylinder, so that much is not really being displaced. The displaced volume is the airspace.

lloydthebartender

Right, assuming you mean the surface outside the cylinder, but there is also water in the cylinder, so that much is not really being displaced. The displaced volume is the airspace.
So this value of volume being displaced is $\pi r^2h$, $h$ being from the diagram from #14?

haruspex

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So this value of volume being displaced is $\pi r^2h$, $h$ being from the diagram from #14?
Yes. Can you answer my other questions in post #15?

lloydthebartender

Yes. Can you answer my other questions in post #15?
As I wrote, try assuming steady state, i.e. the cylinder descends at a constant rate. What does that say about the relationship between the buoyancy force and the weight of the (empty) cylinder?
$F_{buoyancy} = F_{gravity}$
In terms of your diagram in post #14, what is the force of buoyancy?
$F_{buoyancy} = \rho g (\pi r^2h)$
What do you conclude about h?
$h=\frac{m}{\rho \pi r^2}$
So $h$ is proportional to the inverse square of the radius of the cylinder?

haruspex

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$F_{buoyancy} = F_{gravity}$

$F_{buoyancy} = \rho g (\pi r^2h)$

$h=\frac{m}{\rho \pi r^2}$
So $h$ is proportional to the inverse square of the radius of the cylinder?
Yes, but the important point for now is that h is constant.

It's going to make the algebra a little easier if we just write A for the cross-sectional area of the cylinder and a for that of the hole (instead of having to write $\pi r^2$ everywhere).

Next, say the depth of the bottom of the cylinder at some instant is y (below the outside water level) and the velocity of water entering the hole, relative to the hole, is v.
What differential equation relates A, a, v and y?

lloydthebartender

Next, say the depth of the bottom of the cylinder at some instant is y (below the outside water level) and the velocity of water entering the hole, relative to the hole, is v.

Since the rate at which the water rises is $v$ so $\int vdt=y-h$? I know $h=\frac{m}{\rho A}$ so $\int vdt=y-\frac{m}{\rho A}$?

haruspex

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Since the rate at which the water rises is $v$
No, I defined v as the linear speed of the water, relative to the hole, as it flows through the hole. In time dt, what volume is that?

"Sinking a cylinder with varying hole sizes"

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