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tandoorichicken
Feb5-04, 10:13 PM
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
Feb6-04, 06:52 AM
Please show us what you have to work with and what you have tried on this problem.

ShawnD
Feb6-04, 06:58 AM
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.

V^2 = 2gh

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
Feb6-04, 10:34 AM
Good! A good solid hint without completely solving the problem!

himanshu121
Feb6-04, 02:18 PM
Write the Bernoulli's Theorem at the Point just inside the tank and just outside the tank.

tandoorichicken
Feb9-04, 07:13 PM
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}