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 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 v = \sqrt{2gh} [/tex].

<br /> <br /> Square the whole thing and it starts to look familiar.<br /> <br /> V^2 = 2gh<br /> <br /> 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
P+ \rho gh + \frac{1}{2}\rho v^2.
What does the other side of the equation look like?

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