# Sink time of a cap with varying hole diameters

#### Jefffff

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)

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 cm2. 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
Px=Fx / Ax

Px+(1/2)ρvx2ghx = Py+(1/2)ρvy2ghy

Axvx=Ayvy

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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#### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

#### mfb

Mentor
t = A*x^B
This is not an exponential formula (which would need B^x).
What is x, what are the units?
How did you fit the data?
Can you show your raw data?

It has no pressure (except for atmospheric pressure...?)
As everything is relative to atmospheric pressure anyway - right.
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?
I would expect a nearly constant sinking rate as well, but some things will lead to deviations.
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?
Do absolute pressure values matter? Does the difference change?
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.
An ideal flow into nothing will continue to accelerate a bit behind the hole, and form a smaller stream. That reduces the mass flow a bit. I guess an even larger effect here is the pool of surrounding water that quickly forms. Your water is moving through this other water. How does that impact your flow rate?

#### Jefffff

Here's the raw data. I've done a lot of research and experimenting. I know that forces play a key role in the equation. The buoyant force plays a role. I'm going to assume that it varies linearly with time for this scenario. It starts at zero, and reaches a maximum when the entire cap has submerged. With each trial, I can find the buoyant force at any given time t then.

#### mfb

Mentor
Buoyant force and mass of the filled cap increase in the same way.

I think the interaction of inflowing water with water already in the cap is the key point.

Your time uncertainty looks very small.

#### JBA

Gold Member
Since the density of the cap material is greater than that of water, once the base of the cap is fully submerged only weight and orifice flow are effective and buoyancy should no longer be a part of the issue. It would be interesting to see a profile of sink rate vs depth to see if there is a transition point at the point that the cap end becomes fully submerged. I am thinking that the time to reach that transition point would be longer for the smaller orifice and therefore the buoyancy might have a more significant effect on the sink rate for those orifices. Just a passing thought.

#### mfb

Mentor
@JBA: Time is stopped once the cap is surrounded by water, the time to sink to the bottom is not included:
and timed how long it would take for each to fully submerge

#### JBA

Gold Member
I am talking about is the point at which only the base of the cap becomes fully submerged not the entire cab. and cap becomes only a sinking ring rather than a floating buoyant shell.

#### Jefffff

The time uncertainty was very small, as my trials were at the most a few hundred milliseconds apart. I was also pretty surprised by that part.

To describe the interaction of inflowing water with water already in the cap, I would apply bernoulli's principle to the water under the hole and the water already in the cap, right?

In which case it would be P1=P2+(1/2)ρv2+ρgh

P1 is Fnet divided by area. Fnet is going to be Fgrav - Fbuoy.

Assuming that Fbuoy varies linearly with time (the sinking ring as opposed to the sinking shell, for smaller diametres, Fbuoy should play a much more significant role)

Would P2 be equal to the weight of the water above the hole divided by the area of the hole, (and does not vary linearly with time I guess) or would we ignore P2? And for ρgh , would the change in height be referring to the small height the water travels up (just through the thickness of the hole, 2.2 mm) or would we say that the water is going all the way up to the surface of the growing pool in the cap, in which case change in height also varies with time?

#### mfb

Mentor
To describe the interaction of inflowing water with water already in the cap, I would apply bernoulli's principle to the water under the hole and the water already in the cap, right?
That doesn't work. It would require nicely separated flows in the cap. You do not have those, that is the point.
I don't see an analytic approach to this point.

#### JBA

Gold Member
At this point, I now think you can discount my prior concern about the effect of the bottom of the cup.

First, just a quick note, the P1 and P2 pressure terms in the Bernoulli equation should both be expressed in terms of water head (ρgh) and for your case in determining flow rate the equation reduces to (1/2)ρv2 = ρgh outside cap - ρgh inside cap.

In actuality, the pressure differential across the cap hole is not related to the diameter of the cup bottom; however, the weight of the cap does determine how deep the cup will will initially sit in the water and therefore determine the differential in water head (depth) between the outside of the cap/hole and the inside of the cap/hole. The difference in pressure across the hole will be equal to the difference in water head (depth) outside the cup vs the water head (depth) inside the cup.

When you first place the cup into the water it will be empty and will sink to a certain depth based upon the weight of the cup; then, as water flows through the hole and the cup sinks the water head outside the cup increases with the water depth and water depth inside the cup increases at the same rate based upon the rate of flow through the hole in the cup. As a result, the rate of sinking of the cup, once placed in the water, will be directly proportional to the rate of water flow through the hole; and, therefore, the differential pressure at the hole should be constant as the cup sinks because the differential water height between the water outside the cap and inside the cap will remain same as the cup sinks to depth at which the outside water reaches the top edge of the cup. A which point the sinking rate will increase rapidly as the water starts filling the cup both through the hole and over the cup rim.

While this may appear to be an indeterminate situation to analyze; in your case, it is made a little easier because the change in water depth outside the cap as ii slowly submerges is very small and therefore the outside head pressure on the hole can be assumed to constant; and,the time between the initial placing of the cap in the water and the point at which it sinks should be equal to the weight of water (density x volume) in the cup at the point at which it sinks / rate of flow (lb/sec) through the hole. If we assume that flow rate is linear with the change in P1 and P2 (a reasonable assumption based upon Bernoulli's equation), then, over the distance of Hmax - Ho. this flow should be equal to the average flow rate at 1/2 of the H max at which water begins to spill over the the cap top lip. Unfortunately, establishing an equation for the time factor for time in which water is is both flowing through the hole and the top edge of the cap until the cap sinks is beyond imagining.

PS A more elegant method of doing the above average flow rate time calculation would be calculate the total flow based upon an integral equation where mass flow rate ρv^2/2 = P1 (initial submerged cap depth * ρg) - P2 with P2 = an integral ∫ of H (the water height in the cup) from H=0 to Hmax (the height of the water inside the cap at point that water begins to spill over the top edge of the cup), if you want to try and pursue that route.

I suspect you may have further questions about this; and, if so, don't be hesitant to ask them. Even after some 20 years of working with orifice and fluid flow applications It has taken considerable thought on my part to come to the conclusions I have presented above

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#### mfb

Mentor
If we assume that flow rate is linear with the change in P1 and P2 (a reasonable assumption based upon Bernoulli's equation)
I don't think so. It would require a constant velocity. Initially the whole pressure difference can be used to accelerate the water moving through the hole. Later, we have to accelerate a lot of water around the water stream due to "friction" - this will need energy and slow down the water.
then, over the distance of Hmax - Ho. this flow should be equal to the average flow rate at 1/2 of the H max at which water begins to spill over the the cap top lip.
I don't get where the factor 2 comes from. The pressure difference is constant.

#### JBA

Gold Member
The pressure difference is not constant because the backpressure is 0 at the start with no water in the cap and then continuously increases as the water level inside the cap rises to the top inside edge of the cup. Of course, by that point some top outside water will have occurred; so, this analysis is only valid for fill times up to the point just prior to the initiation of the top edge spill over.

If you want to verify whether or not the mass flow rate decreases linearly with increasing inside water depth then simply make a graph of mass flow rate vs. inside water depth with a constant outside water pressure to see the result. If it is not linear then you will have to go the integral method; or, the multi interval element to element averages solution route, which if the slope variations in the curve are minor and you use sufficiently small element intervals will be functionally close to an integration solution.

One item I should have also discussed in my last post is orifice flow coefficients which adjust for the fact that the shape of your hole effects the actual effective flow area for the fluid passage. Assuming you used a drilled hole with a clean square edge entry the the generally accepted coefficient for that type of orifice is 0.60 (ie, the area of the stream of water passing through the hole is only 60% of the actual hole area); and, edge effects on the discharge side of the hole can reduce the effective flow area even lower due to induced discharge turbulence that increases the effective backpressure on the orifice. As a result if your calculated times are uniformly low, then I suggest you start by using the 0.60 modifier (and if necessary adjust it even lower) on your flow rate calculations to see if that adjustment improves the correlation between your test times and calculated times.

#### mfb

Mentor
The pressure difference is not constant because the backpressure is 0 at the start with no water in the cap and then continuously increases as the water level inside the cap rises to the top inside edge of the cup.
Both pressure inside and outside will increase in the same way if the cap sinks down.
edge effects on the discharge side of the hole can reduce the effective flow area even lower due to induced discharge turbulence that increases the effective backpressure on the orifice
That is the point I was mentioning in the previous posts.

JBA

#### Jefffff

Hi guys and thanks so much for your help!

Before I saw your replies, I did a bit on my own and came up with this equation relating diametre of hole to sink time:

Where x is the sink time in seconds, D is the diametre of the hole, ΔT is the time it takes to go from just touching the surface to full submersion, (time it takes for Fbuoy to go from zero to max), t is the time in seconds, 0.645 is Fgrav, 6.221*10^-3 is the area of the bottom of the cap, and -21.56 is due to the difference in gravity potential energy from climbing the 2.2 mm through the orifice. The value 0.363 was taken by using the equation Fbuoymax=Vρg where the total volume of the plastic of the cap is 37 cm3. Calculating for the upper limit of the definite integral when the integrand = 2.35 * 10^-4 (the total volume of water needed to sink the cap). This equation is most likely incorrect, I just combined the information I knew. Substituting values for delta-T and D, I obtained the values for x:

4 mm - 56.279 s
7 mm - 17.745 s
12 mm - 5.7758 s
15.5 mm - 3.7186 s
20 mm - 2.079 s

This is much closer to my experimental values, but I know it has to be wrong since I can't use my experimental times to compute a theoretical time. lol.
As for the orifice coefficient, the holes for 4mm and 7mm are nearly perfect circles, with square edges. The holes for 12 mm - 20 mm are far from perfect, as I didn't have drill bits of that size and had to resort to cutting multiple holes and filing the holes down. So, the orifice coefficients are probably much lower for those (huge uncertainty for radius as well)

I'll have a crack at it with the method that JBA described, but some of the variables I didn't obtain in the experiment. The values I obtained above will not fit in the error margins, but with this kind of experiment I'd imagine there are many very complex factors involved, and it's basically impossible to get the theoretical variables and obtain the experimental value to any degree of accuracy.

#### JBA

Gold Member
Jefffff, You need to revise your attack because mbf, is absolutely correct about the differential pressure remaining constant based upon the initial internal and external water height differential (due to the weight of the cup) when the cap is initially placed in the water. Ironically, I made essentially that same statement in my earlier post about the level differential being constant as the cap submerged and then fell off the train when on my rewrite when I failed to remember that initial statement and that it clearly implies a constant differential pressure across the hole during the submersion.

Good catch mbf, thanks for catching that before I sent Jeff down the wrong track!

That obviously makes the analysis much simpler as is more inline with Jeff's derived equation

#### Jefffff

So, the differential pressure, P1-P2 = a constant?

And my equation would be (1/2)ρv2=P1-P2 ?

#### JBA

Gold Member
More specifically the equation for your case is: (1/2)ρv2 = ρgh, where h is the submerged depth of the cap bottom face when the cap is floating in the water (with no hole, of course); and, that h value will remain constant as water flows through the hole and the cap submerges. Using that h, the hole area and the total water volume for the full cap, you should be able to calculate a submerging time for each hole size.

While I am not sure how it might affect your calculation results because it is generally used in large piping systems, I am sending the below site address where you can see how a sudden enlargement K factor between each hole and the inside diameter of the cap might affect your flow times. With your flat bottom hole entry, none of the Pipe Reduction cases given apply for your situation so the section you should look at is: 3.6 Square Expansion, For Re1 < 4000. For the calculation of hL=K* V^2 / 2g, be sure to use mm - sec values for V and g so that hL will be in mm. The hL calculated for each orifice should be subtracted from your above determined h.

https://neutrium.net/fluid_flow/pressure-loss-from-fittings-expansion-and-reduction-in-pipe-size/

If you have any questions of course ask; and, keep me apprised of your progress.

#### Jefffff

I've done all my calculations and I'm now writing the evaluation and conclusion of my lab now. The equation has worked out nicely and all correlations seem to be on par. Thank you so much for your help!

#### Artem

Hello guys, i'm sorry for writing here after three years of thread inactivity, but i have a question about an equations. It remained unclear for how is the final equation looking like and i would want to know it, because i have literally the same question about sink time of a cap in my school physics tournament

#### mfb

Mentor
Post 15 has a formula (that I didn't check), but the overall conclusion was "it is complicated".

#### Artem

So, this formula in post 15 is a final equation for finding the sink time, but then what about JBA words that we can use (1／2)pv2 ＝ pgh to find all what we need

#### Artem

If I understood correctly than we do not need that formula from post 15, but i don't know how to find the hole size using JBA formula

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