When will the barrel become half empty?

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SUMMARY

The discussion centers on calculating the time it takes for a barrel with a volume of 200 liters and a height of 1 meter to become half-empty through a 1 cm² hole at the bottom. The relevant equations include Bernoulli's principle, Torricelli's law, and the continuity equation. The key steps involve correlating the sinking velocity of the liquid level with the velocity of the exiting water, leading to a differential equation that can be solved through integration. The solution requires understanding fluid dynamics principles and applying them correctly to derive the time to half-empty.

PREREQUISITES
  • Understanding of Bernoulli's principle
  • Familiarity with Torricelli's law
  • Basic knowledge of differential equations
  • Concept of fluid continuity equation
NEXT STEPS
  • Study the application of Bernoulli's principle in fluid dynamics
  • Learn how to derive and solve differential equations related to fluid flow
  • Explore the implications of Torricelli's law in real-world scenarios
  • Investigate the continuity equation and its applications in fluid mechanics
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Students studying fluid dynamics, engineers working with fluid systems, and anyone interested in solving practical physics problems involving fluid flow.

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Homework Statement


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I'm not a native english speaker, so I apologize if my explanations are a bit awkward. I do not have a solution to this problem and I'd be grateful if someone could check this/help. I also have absolutely no idea how to even begin. Asked my colleagues, none of them know. I'm pretty desperate.

There's a barrel filled with water. The volume of water inside is V=200L. Its height is 1m.
We decide to make a hole on the bottom of the barrel. The size of the hole is 1cm2.

How much time will it take for the barrel to become half-empty (when will the height drop to 1/2 of the initial height)?

Homework Equations


What I believe is relevant here is:

Applying Bernoulli's principle:
p1+qgh1+1/2qv1^2=p2+qgh2+1/2qv2^2

Toricelli (this is derived from Bernoulli's principle in this case from what I understand):
v=sqrt(2*g*h)

Continuity equation:
S1v1=s2v2

The Attempt at a Solution


I got used to solving problems that include manometers, tubes of different sizes etc., but with a simple problem like this, I'm not sure where to begin.

First I did basic conversions.

V=200*10-3m3
S=1*10-4m2

My first thought was writing down:

h1=1/2*h2

and proceed from there, but I didn't get anything useful, since none of the equations I've written down took the size of the small hole into consideration.

t=v*s would seem fine, but since velocity isn't constant, I need a differential equation, ending up with an integral I cannot solve nor does it seem like a way to solve this (please correct me if I'm wrong).

I tried writing down the Bernoulli equation I've written above and it just leads to v=sqrt(2gh). Again, I'm not getting anything. Tried S1*v1=S2*v2, but since v1 is close to zero, I can't get anything from this equation either.
 
Physics news on Phys.org
1) In a first step you have to correlate the sinking velocity of the liquid level (##dh/dt##), with the velocity of the exiting water (with Torricelli)
2) By separation of the variables ##h## and ##t## and integrating (which is quite easy to write down, but not that easy to do), you obtain an initial value problem, which you can solve by inserting the height of the level at ##t=0##

But I suggest you start with 1)
 

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