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Speed of water leaving trough

  • #1
**Below is the question. No need to solve it for me. Providing me with an equation would be fine. Thanks**

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is 0.14 m below the level of the water that is in the tank.

The acceleration of gravity is 9.81 m/s^2.

If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole?

Assume that the trough is large enough that the velocity of the water at the top is zero. Answer in units of m/s.
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
Try Bernoulli.
 
  • #3
So...

pressure + 1/2 * density * velocity^2 + density * acceleration * elevation = constant


I believe that density crosses out.

One question is what is the pressure equal to?
 
  • #4
ok i think its 100,000 Pascals. When I plug everything in I get 447.2

Somehow that doesn't seem correct though
 
  • #5
Doc Al
Mentor
44,882
1,129
You need to compare a point at the water surface to one at the hole. Both are open to the atmosphere.
 
  • #6
You need to compare a point at the water surface to one at the hole. Both are open to the atmosphere.
Could you explain this further?
 
  • #7
Could someone possibly walk me through this. Sound a little more difficult than I'm used to.
 
  • #8
alphysicist
Homework Helper
2,238
1
Hi European Sens,

Bernoulli's equation indicates that

[tex]
P + \frac{1}{2} \rho v^2 + \rho g h
[/tex]

is constant along a flow. So the first step is to compute that quantity for the top of the trough, and then separately compute it at the hole. What do you get for those two quantities? Since the quantity is constant, what would you do next?
 

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