Speed of Water Emerging from Large Tank: [itex] \sqrt{2gh} [/tex]

[tandoorichicken]

A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex] v = \sqrt{2gh} [/tex].

 

[HallsofIvy]

Please show us what you have to work with and what you have tried on this problem.

 

[ShawnD]

Originally posted by tandoorichicken
A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex] v = \sqrt{2gh} [/tex].

Square the whole thing and it starts to look familiar.

[tex]V^2 = 2gh[/tex]

Remember that gravity applies a downward force on the fluid and that a force on any side of a fluid is applied to ALL sides of the fluid.

 

[HallsofIvy]

Good! A good solid hint without completely solving the problem!

 

[himanshu121]

Write the Bernoulli's Theorem at the Point just inside the tank and just outside the tank.

 

[tandoorichicken]

ummmmm...
If I use Bernoulli's Equation, one side of it says
[tex] P+ \rho gh + \frac{1}{2}\rho v^2 [/tex].
What does the other side of the equation look like?

If the other side is 0,
then
[tex] P = \rho gh [/tex]
[tex]2\rho gh = -\frac{1}{2}\rho v^2 [/tex]
But since h is going below the surface the negative is expected and can be ignored. Then-
[tex] 2\rho gh = \frac{1}{2}\rho v^2 [/tex]
The density of water is 1 g/cm^3.
[tex] 2gh = \frac{1}{2} v^2 [/tex]
[tex]4gh = v^2 [/tex]
But then I get
[tex] v = 2\sqrt{gh} [/tex], not [tex] v = \sqrt{2gh} [/tex]