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Hi guys, this is my first time on the forum, and I was wondering if you could help me out with this problem. I have regrettably chosen a topic for a high school lab assignment that was quite complex compared to what was expected. For my investigation, I've drilled holes of varying size into identical PVC pipe caps (cylindrical in shape), and timed how long it would take for each to fully submerge after being gently placed into water at approximately room temperature. I've obtained the data points and it forms a nearly perfect exponential correlation. The line of best fit is the equation:

t = A*x^B

A: 716.3 +/- 29.06

B: -1.750 +/- 0.02683

I know that Bernoulli's equation fits into all of this. I will look at the difference in pressure just inside the hole versus on the outer sides. Using the continuity equation, I can see the change in volume over time, and once the volume of the cap is completely filled the cap will submerge. (I also noticed that surface tension is a factor, the water tends to pool over the rim before finally overcoming surface tension)

The diameter of a cap is 8.9 cm. The frontal surface area (the area in contact with the water) is 63.05 cm

P

P

I'm going to list down things that I believe are at work in my equations:

The first pressure P1 will be equal to the force of the cap pressing down on the water, or 0.066*9.8 divided by the area that the cap occupies, 0.0063 metres squared. This works out to be 102.7 Pa.

The water will have zero initial velocity and zero initial gravitational potential energy per unit volume, only the initial pressure energy due to the weight of the cap.

The water as it goes through the hole will have gained a minuscule amount of gravitational potential energy ( the hole is 2.4mm thick) and gained kinetic energy per unit volume. It has no pressure (except for atmospheric pressure...?) , P2 = 0 because it is rushing up and out of the cap. This only occurs for the initial moment of the cap being placed in the water. However, as water rushes into the cap, the cap becomes heavier and the buoyant force is always slightly smaller than the force of gravity on the cap because the plastic is slightly denser... causing the cap to continue sinking at more or less the same rate? (This is what I observed in the lab - the water seemed to fill the cap at a uniform rate) Will the pressure of the water going through the hole still be zero? Or will the pressure due to the growing column of water on top play a role? I need to be able to determine how the velocity of water through the hole varies with time, and plug this into the continuity equation to see the mass flow rate over time, the value of time after which the area under the curve exceeds the volume of the cap is the time at which the cap sinks?

Using the information, I calculated an exit velocity of 0.45 m/s

Taking the smallest diametre hole, radius 2 mm, and doing A*v gives me a volume flow rate of 5.6565 * 10^-6 metres cubed per second. The volume of the cap is 0.000225 m^3 so it would take 39.78 seconds for the smallest cap to fill? This is 23 seconds off from my observed time of 63 seconds, so I believe this definitely involves calculus or more complex equations. I haven't considered atmospheric pressure, how it impacts my experiment is unclear to me. I'd imagine that since it is pushing from both the bottom and the top, it cancels out.

Any help would be greatly appreciated.. I haven't learned any concepts of fluid dynamics in class yet, this is all off the textbook and internet.

t = A*x^B

A: 716.3 +/- 29.06

B: -1.750 +/- 0.02683

I know that Bernoulli's equation fits into all of this. I will look at the difference in pressure just inside the hole versus on the outer sides. Using the continuity equation, I can see the change in volume over time, and once the volume of the cap is completely filled the cap will submerge. (I also noticed that surface tension is a factor, the water tends to pool over the rim before finally overcoming surface tension)

**1. Homework Statement**The diameter of a cap is 8.9 cm. The frontal surface area (the area in contact with the water) is 63.05 cm

^{2}. The temperature of the water was kept constant at 24 degrees Celsius. The volume is approximately 225 mL and the mass is around 66 grams. The holes range from 4mm to 20 mm, and the ratio of area of the hole to the frontal SA becomes more significant towards the 20mm end. The hole has a thickness of 2.4 mm and is drilled into the exact centre of the cap. The density of water is 1000 kg/m cubed.**2. Homework Equations**P

_{x}=F_{x}/ A_{x}P

_{x}+(1/2)ρv_{x}^{2}+ρ*gh*= P_{x}_{y}+(1/2)ρv_{y}^{2}+ρ*gh*

A_{y}A

_{x}v_{x}=*A*_{y}v_{y}**3. The Attempt at a Solution**

I'm going to list down things that I believe are at work in my equations:

The first pressure P1 will be equal to the force of the cap pressing down on the water, or 0.066*9.8 divided by the area that the cap occupies, 0.0063 metres squared. This works out to be 102.7 Pa.

The water will have zero initial velocity and zero initial gravitational potential energy per unit volume, only the initial pressure energy due to the weight of the cap.

The water as it goes through the hole will have gained a minuscule amount of gravitational potential energy ( the hole is 2.4mm thick) and gained kinetic energy per unit volume. It has no pressure (except for atmospheric pressure...?) , P2 = 0 because it is rushing up and out of the cap. This only occurs for the initial moment of the cap being placed in the water. However, as water rushes into the cap, the cap becomes heavier and the buoyant force is always slightly smaller than the force of gravity on the cap because the plastic is slightly denser... causing the cap to continue sinking at more or less the same rate? (This is what I observed in the lab - the water seemed to fill the cap at a uniform rate) Will the pressure of the water going through the hole still be zero? Or will the pressure due to the growing column of water on top play a role? I need to be able to determine how the velocity of water through the hole varies with time, and plug this into the continuity equation to see the mass flow rate over time, the value of time after which the area under the curve exceeds the volume of the cap is the time at which the cap sinks?

Using the information, I calculated an exit velocity of 0.45 m/s

Taking the smallest diametre hole, radius 2 mm, and doing A*v gives me a volume flow rate of 5.6565 * 10^-6 metres cubed per second. The volume of the cap is 0.000225 m^3 so it would take 39.78 seconds for the smallest cap to fill? This is 23 seconds off from my observed time of 63 seconds, so I believe this definitely involves calculus or more complex equations. I haven't considered atmospheric pressure, how it impacts my experiment is unclear to me. I'd imagine that since it is pushing from both the bottom and the top, it cancels out.

Any help would be greatly appreciated.. I haven't learned any concepts of fluid dynamics in class yet, this is all off the textbook and internet.

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