Solving Water Storage Tank Problem with Bernoulli's Equation

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Homework Help Overview

The problem involves a water storage tank that is open to the air, with a height of 1 meter. A hole is made at the center of the tank's wall, and the question is about determining the speed at which water exits the tank using Bernoulli's equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's equation and question the interpretation of the final height in the equation. There is confusion regarding why the final height is considered to be 0.5 meters instead of 0 meters.

Discussion Status

Some participants have provided insights into the misunderstanding of the problem setup, particularly regarding the height at which the hole is located. There is recognition of the need to clarify the problem statement to align with the assumptions made in the calculations.

Contextual Notes

There is an indication that the problem may have been misinterpreted, leading to confusion about the correct height to use in the calculations. The original poster acknowledges a misunderstanding of the question's wording.

VanessaN
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Homework Statement



A water storage tank is open to air on the top and has a height of 1 m. If the tank is completely full and a hole is made at the center of the wall of the tank, how fast will water exit the tank?

Homework Equations



Pressure is the same as atmospheric pressure because the tank is open to the air. Also, linear flow speed at the surface is essentially zero, so...
Bernoulli's equation can be simplified to:

rho * g * h1 = 1/2 * rho * v2^2 + rho * g * h2

The Attempt at a Solution


[/B]
rho= density
g= 10 m/ s^2
height 1 (or height initial)= 1 m
Here is where I don't understand the solution...
height 2 is apparently = 0.5 m, but this is not in the question stem...

rho * g * h1 = 1/2 * rho * v2^2 + rho * g * h2

My attempt at solving for v2 ultimately comes out to be:
v2= sq rt 2g (h1-h2)= sq rt 2* 10 m/s * (1 m - 0) = sq rt 20 meters

But the correct way to solve was:
v2= sq rt 2g (h1-h2)= sq rt 2* 10 m/s * (1 m - 0.5 m) = sq rt 10 meters

So, I guess my question is why would the final height (height 2) be 0.5 m?
Or, is there any other way to get the correct answer of sq rt 10 meters?
 
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VanessaN said:

Homework Statement



A water storage tank is open to air on the top and has a height of 1 m. If the tank is completely full and a hole is made at the center of the wall of the tank, how fast will water exit the tank?

Homework Equations



Pressure is the same as atmospheric pressure because the tank is open to the air. Also, linear flow speed at the surface is essentially zero, so...
Bernoulli's equation can be simplified to:

rho * g * h1 = 1/2 * rho * v2^2 + rho * g * h2

The Attempt at a Solution



rho * g * h1 = 1/2 * rho * v2^2 + rho * g * h2

solving for v2, which ultimately comes out to be:
v2= sq rt 2g (h1-h2)= sq rt 2* 10 m/s * (1 m - 0) = sq rt 20

I did not get the correct answer. The correct answer is apparently sq rt 10. Can somebody please help me figure out where I went wrong? Thanks in advance!:woot:
Your post is hard to follow, since you don't define the variables used in your equations nor do you list their values. How do you expect anyone to help? :oops: :rolleyes: o_O

In any event, have you heard of Torricelli's Law?

http://en.wikipedia.org/wiki/Torricelli's_law
 
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They asked for the case where the hole is half way down the wall of the tank, not at the bottom of the tank.

Chet
 
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Chestermiller said:
They asked for the case where the hole is half way down the wall of the tank, not at the bottom of the tank.

Chet

ahhh, the case where i make a problem much more difficult by not understanding the question... the word "center" should have tipped me off to the fact that the final height would be 0.5!

Thank you for your help i appreciate it!
 

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