Diameter of hole vs time to submerge

In summary: The cap will fall into the water if you drop it from an initial position where it is completely out of the water.
  • #1
jackkk
6
0
Hey everyone, I'm new to physics forums and I would really like some help with my physics report. For my high school physics topic I have chosen an experiment where holes with differing diameters are drilled into the centre of the bottom of PVC pipe caps and the time taken for the each cap to fully submerge below the water line is recorded. Could someone help me with my overall understanding of the topic and how it may relate to Bernoulli's principle?

1. Homework Statement

I've conducted the experiment and this is what I know:
When the cap is dropped, the force weight is slightly larger than the buoyancy force and PVC is more dense than water. The buoyancy force upwards will result in water rushing through opening of the hole in the bottom of the cap. As the water rushes into the cap, the cap becomes heavier and consequentially the buoyant force is slightly smaller than the force of gravity on the cap because the PVC is more dense. Increasing the diameter of the hole size results in a higher flow rate as the larger the diameter of the hole is, the more liquid that will flow through it, that would otherwise be motionless at the bottom of the cap. Furthermore, smaller the hole is, the ratio of water touching the inside wall of the hole (and therefore at a zero velocity due to the no slip condition) to the water actually rushing through increases.

Questions:
Why does a higher flow rate decrease fluid pressure and how does this apply?
What is the relationship between the liquid outside the cap and the liquid inside the cap? note: the water level is higher outside than inside
Will the pressure of the water going through the hole change?
Will the pressure due to the growing column of water on top play a role?
 
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  • #2
Hard to answer this one without breaking forum rules and actually telling you how to solve it .

Let's have a go and see how we progress :

Draw me a picture of the cap sitting partly submerged in a large area deep pond of water . Represent the cap by a piece of vertical axis tube with a flat bottom .

Show where the level of water inside the cap is likely to be relative to the pond surface level . Just use your intuition .

Label the drawing and show any dimensions that you think relevant .

Show any pressures and forces that you think may be acting .

Once we have a good picture we can move on .
 
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  • #3
Best to work on the general problem first before introducing actual values . So use symbols like p1 , Fv , h2 etc for your annotations .
 
  • #4
FullSizeRender 23.jpg
 

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  • #6
I should have mentioned the bottom of the cap is a dome.

I know that pressure is a dependant on depth from surface of water. Therefore, the pressure at a point in water inside the cap will be less than the pressure at a point of the same height outside of the cap (as in height from ground). So am I correct in saying that P2 is at a distance of H2 from the surface of the water as demonstrated in the diagram? This would tell us that the their is a greater force acting on the water particles outside the hole compared to inside the hole, and this is what is causing the water to rush in.
Thus, if the hole was larger, the pressures would be the same (initially) and therefore a larger hole simply allows for more fluid to flow into the cup?

Are these conclusions valid?
 
  • #7
Nidum said:
Best to work on the general problem first before introducing actual values . So use symbols like p1 , Fv , h2 etc for your annotations .
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/87/87877-9884520a80438f91f0e849866f95459b.jpg
 

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  • #8
(1) When you say 'drop the cap' what exactly do you mean ?

(a) From an initial position completely out of the water so that cap has an initial velocity when it hits the water .
(b) From an initial position where the lowest points on the cap just touch the water surface and cap has zero initial velocity .

(2) What diameters are the holes relative to the cap diameter ? (Roughly)

(c) Large .
(d) Small .
(e) Very small .

(3) In your experiments is the descent of the cap into the water stable ? It looks like it could wobble or even fall over in the initial stages of descent .
 
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  • #9
Nidum said:
(1) When you say 'drop the cap' what exactly do you mean ?

(a) From an initial position completely out of the water so that cap has an initial velocity when it hits the water .
(b) From an initial position where the lowest points on the cap just touch the water surface and cap has zero initial velocity .

(2) What diameters are the holes relative to the cap diameter ? (Roughly)

(c) Large .
(d) Small .
(e) Very small .

(3) In your experiments is the descent of the cap into the water stable ? It looks like it could wobble or even fall over in the initial stages of descent .

1)
The cap is dropped from an initial position completely out of the water so that the cap has an initial velocity. The height from which the cap is dropped stays constant for each cap however, due to splashing and water escaping the beaker, the caps used last had a longer distance to fall and therefore a higher initial velocity. I know that this effects the distance the cap moves into the water and therefore affects the force and initial flow rate (due to a larger pressure difference). As I measured the height between the bottom of the cap and the water level for each test, I can not only calculate the initial velocity of each cap upon impact, but I can also calculate an estimated impact force for each cap as I filmed each trial and know the distance each cap reaches into the water before stopping. These are things I can talk about when it comes to error in experimental design as the best design would have been to have the cap already floating in water having the hole plugged up, and then releasing the plug.

2)
The area of the bottom of each cap is approximately 0.0025m^2 (based on area of a circle), and the holes range from 6.835x10^-6m^2 to 1.297x10^-4m^2.

3)
Due to the drop there is bobbing up and down of the cap in water. Whilst the cap is actually sinking it is very stable (other than vertical bobbing) due to the dome shape of the bottom (I believe?).
 

1. What is the relationship between the diameter of a hole and the time it takes to submerge?

The diameter of a hole and the time it takes to submerge are inversely proportional. This means that as the diameter of the hole increases, the time to submerge decreases. Similarly, as the diameter decreases, the time to submerge increases.

2. Is there a specific formula that relates the diameter of a hole to the time it takes to submerge?

Yes, there is a formula known as Torricelli's Law that relates the diameter of a hole to the time it takes to submerge. It states that the time to submerge is equal to the square root of 2 times the length of the hole divided by the acceleration due to gravity.

3. How does the depth of the water affect the relationship between the diameter of a hole and the time to submerge?

The depth of the water does not have a significant impact on the relationship between the diameter of a hole and the time to submerge. As long as the water is deep enough to fully submerge the object, the relationship remains the same.

4. Are there any other factors that can affect the time it takes for an object to submerge through a hole?

Yes, there are other factors that can affect the time it takes for an object to submerge through a hole. These include the shape of the object, the density of the object, and the viscosity of the liquid the object is being submerged in.

5. How can the relationship between diameter of a hole and time to submerge be applied in real life situations?

This relationship can be applied in various situations, such as calculating the time it takes for a sink or bathtub to drain, or determining the size of a hole needed for a specific amount of liquid to drain within a certain time frame. It can also be used in engineering and physics experiments to study fluid dynamics.

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