What is the Volume of Water Leaving a Tank Through a Rectangular Opening?

AI Thread Summary
The discussion centers on calculating the volume of water leaving a tank through a rectangular opening using Bernoulli's law and the continuity equation. The formula derived is 2/3(L)√(2g(h1³-h2³)). Participants suggest breaking the opening into small horizontal sections for integration and emphasize the importance of understanding the speed of water flow. One user initially struggles with the integration process but eventually clarifies their confusion regarding height variables. The conversation concludes with a successful application of the formula after resolving the misunderstanding.
juanitotruan77
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Homework Statement



5. A rectangular opening on the side of a water tanks has a width of L. The upper part of the opening it's in a height of h1 underneath the water y and the bottom part it's in h2 . prove that the volume of water that leaves the tank is given by :
2/3(L)√(2g(h1³-h2³))

Homework Equations


Bernoulli's law
e2dd757ae8209ce6fed45947ad9ad43b.png

Continuity equation

d8ceecc84d9efdc2ac536c90630b71c2.png

Volumetric flux
ce0b80a445b4435285d84193ddf63b32.png

both V and v are Speed.

The Attempt at a Solution


No idea so far.

Homework Statement


Homework Equations


The Attempt at a Solution

 
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hi juanitotruan77! welcome to pf! :smile:

imagine the opening is divided into tiny horizontal openings of height dh, and integrate from h1 to h2

what do you get? :wink:
 
I think i get it, i'll try it. But I'm not very familiar with using integrals. Can you explain me how?

thanks, btw.
 
try first

see how far you get :wink:
 
i'm trying, but i don't know how to get to torricelli's law from bernoulli's in a rectangular area.
 
if dh is small enough, you can regard the pressure as constant

so find the speed v from Bernoulli's equation
 
i can't, i don't understand where i have to put the differential of h. I think i should ask a teacher tomorrow.
 
the rectangle is length L, and starts at height h, and finishes at height h + dh

you can regard the pressure as being constant, = ρgh

find v, then multiply by the area (L*dh) to get the flow :smile:
 
oh, that sounds reasonable, thanks :D
 
  • #10
well, i tried, but i couldn't make it work.
 
  • #11
(just got up :zzz:)

show us what you did :smile:

hint: 2/3(L)√(2g(h1³-h2³)) = (L)√(2g) times [(2/3)h13/2 - (2/3)h23/2] :wink:
 
  • #12
Got it, bro, i already finished, i think i was cofused with the h terms. I used y terms instead. it's quite easy now that i did it. lol. Thanks.
 
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