## Fluid flow

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].

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 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.

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 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 $$v = \sqrt{2gh}$$