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Velocity of water out of reservoir.

by John09
Tags: reservoir, velocity, water
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John09
#1
Aug12-11, 06:17 PM
P: 2
I have a cylindrical tank and I know that the efflux speed is proportional to the square root of the depth of the hole from the surface. So u=k sqrt(w). I need to algebraically determine the constant or k in that situation. Has anyone got any ideas as to how I should approach this? I was thinking that I could try and find the acceleration in i and j components and integrate it for velocity but didn't get far.

Thanks for any help.
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cjl
#2
Aug12-11, 06:32 PM
P: 1,008
If you're ignoring the viscous effects, the efflux speed will be such that the dynamic pressure is equal to the static pressure just inside the hole. Dynamic pressure is 1/2*rho*v2, so rearranging for v, we can get that v = sqrt(2*p/rho). Since the pressure in a tank is simply from hydrostatic equilibrium (P = rho*g*h), we can plug in for P:

v = sqrt(2*rho*g*h/rho) = sqrt(2*g*h).

So, your constant is sqrt(2g).
John09
#3
Aug12-11, 07:04 PM
P: 2
Hm I don't think taking pressure into account is necessary as it is not part of our coursework.

CDCraig123
#4
Aug14-11, 01:23 AM
P: 32
Velocity of water out of reservoir.

If you want to know the velocity in fluid dynamics you need to know 2 things. Volume and pressure to find velocity. Unless you can invent some kinda new math cjl is right.


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