Calculate time for water to leak completely

  • Thread starter Sewager
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  • #1
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Homework Statement


Given the dimensions of a cylinder container with a hole, the dimensions of the hole, and the volume of water, calculate the time for water inside the container to completely leak out.

Homework Equations


Bernoulli equation: P + 1/2pv^2 + pgh = constant https://en.wikipedia.org/wiki/Bernoulli's_principle
Continuity equation: A1V1 = A2V2 http://theory.uwinnipeg.ca/mod_tech/node65.html

The Attempt at a Solution


- Assume the water is ideal liquid
- Assume container has an open top, so pressure cancels

So far I am able to find the relationship between height of the water in the container and the velocity at the top of the water. Since both height, volume, and velocity changes at each instant, I think that integration needs to be used. Can someone enlighten me on how I should incorporate calculus in the equation?

See attachment for attempt

Thank you!
 

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Answers and Replies

  • #2
SteamKing
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Homework Statement


Given the dimensions of a cylinder container with a hole, the dimensions of the hole, and the volume of water, calculate the time for water inside the container to completely leak out.

Homework Equations


Bernoulli equation: P + 1/2pv^2 + pgh = constant https://en.wikipedia.org/wiki/Bernoulli's_principle
Continuity equation: A1V1 = A2V2 http://theory.uwinnipeg.ca/mod_tech/node65.html

The Attempt at a Solution


- Assume the water is ideal liquid
- Assume container has an open top, so pressure cancels

So far I am able to find the relationship between height of the water in the container and the velocity at the top of the water. Since both height, volume, and velocity changes at each instant, I think that integration needs to be used. Can someone enlighten me on how I should incorporate calculus in the equation?

See attachment for attempt

Thank you!
You need to figure out which variables change with time.
 
  • #3
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You need to figure out which variables change with time.
Both height and velocity change with time. Can I replace velocity with dh/dt?
 
  • #4
SteamKing
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Both height and velocity change with time. Can I replace velocity with dh/dt?
You can't replace velocity directly and completely with dh/dt, but you should be able to write how velocity changes with time, given dh/dt, since you know how velocity is affected by h.
 
  • #5
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You can't replace velocity directly and completely with dh/dt, but you should be able to write how velocity changes with time, given dh/dt, since you know how velocity is affected by h.
Can you elaborate a little bit more? Thank you for your time :)
 
  • #6
SteamKing
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If you analyze this problem using Toricelli's Law (which can be derived from Bernoulli's equation using certain simplifications), the velocity of the stream coming out the hole is
$$v = \sqrt{2g ⋅ h}$$
https://en.wikipedia.org/wiki/Torricelli's_law

Since the volume of fluid coming out is also proportional to the velocity, you should be able to use the formula for velocity to determine how velocity changes with time. What happens if you differentiate the velocity equation w.r.t. time?

I've given you some pretty substantial hints. It's time for you to start showing what you are doing with them to solve this problem.
 
  • #7
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If you analyze this problem using Toricelli's Law (which can be derived from Bernoulli's equation using certain simplifications), the velocity of the stream coming out the hole is
$$v = \sqrt{2g ⋅ h}$$
https://en.wikipedia.org/wiki/Torricelli's_law

Since the volume of fluid coming out is also proportional to the velocity, you should be able to use the formula for velocity to determine how velocity changes with time. What happens if you differentiate the velocity equation w.r.t. time?

I've given you some pretty substantial hints. It's time for you to start showing what you are doing with them to solve this problem.
Hey SteamKing,
After following your suggestions, I was able to find the relationship between velocity and time:

$$h = \frac{T-v \pi r^2t}{\pi R^2}$$
where T is the volume of the water, r is the radius of the hole, R is the radius of the container, and t is time

$$v = \sqrt{2g\frac{T-v \pi r^2t}{\pi R^2}}$$

How do I integrate if velocity is on both sides?

Really appreciate your help :c
 
  • #8
haruspex
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$$h = \frac{T-v \pi r^2t}{\pi R^2}$$
where T is the volume of the water, r is the radius of the hole, R is the radius of the container, and t is time
I assume T is the initial volume. To get that equation you have to assume v is constant, but it is not.
 

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