Velocity of water out of reservoir.

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SUMMARY

The velocity of water exiting a cylindrical tank is determined by the equation v = sqrt(2gh), where v represents the efflux speed, g is the acceleration due to gravity, and h is the depth of the water. The constant k in the equation u = k sqrt(w) is established as sqrt(2g) when considering hydrostatic pressure. The discussion emphasizes that dynamic pressure equates to static pressure at the hole, leading to this relationship. Viscous effects are deemed negligible for this analysis, focusing solely on hydrostatic equilibrium.

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  • Knowledge of gravitational acceleration (g)
  • Basic algebra for manipulating equations
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This discussion is beneficial for students and professionals in physics, engineering, and fluid dynamics, particularly those interested in understanding fluid flow and pressure relationships in cylindrical tanks.

John09
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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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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).
 
Hm I don't think taking pressure into account is necessary as it is not part of our coursework.
 
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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