Originally posted by XanthosNZ
So PBE6, explain why you think it's wrong.
Your solution makes perfect sense if you assume the outflow to be constant and independent of the height in the tank. However, Bernoulli's equation models this situation a little better than that, and gives some interesting restrictions (which are actually quite intuitive when you see them).
(This derivation is a little long, but I hope it's fairly clear and straightforward):
Setting up Bernoulli's equation for one outlet, where point 1 is the top of the tank and point 2 is on the free jet at the outlet, we have:
P1/gamma + v1^2/(2g) + h1 = P2/gamma + v2^2/(2g) + h2
P1, P2 are pressures
v1, v2 are flow velocities
h1, h2 are heights
gamma is the specific gravity of water = density * gravity
g is the gravitational constant
We set h2 equal to 0 as a reference height. Now, assuming both water surfaces are exposed to the atmosphere, and the height of the tank is not several miles high so atmospheric pressure is constant, we can cancel P1/gamma and P2/gamma. Also, assuming that the water at the top of the tank is not moving quickly compared to the jet, we can assume v1 = 0. After simplifying, we are left with:
h1 = v2^2/(2g)
Solving for v2 we get:
v2 = (2g*h1)^0.5
By the continuity equation, the flowrate from the outlet is equal to the flow velocity v2 times the outlet area A, so:
Qout = A*v2 = A*(2g*h1)^0.5
Assuming a cylindrical tank, the volume of water in the tank V is the height of the water h1 times the cross-sectional area R, so:
Qout = A*[2g*(V/R)]^0.5 = k*(V^0.5) for simplicity's sake.
Now that we have an expression for the flowrate from the outlet as a function of the volume in the tank, we can set up a differential equation to solve for the volume in the tank as a function of time.
change in volume = flow in - flow out
dV/dt = Qin - n*Qout (...n outlets, all assumed to flow at the same rate)
dV/dt = Qin - n*k*(V^0.5)
This is a seperable differential equation. Seperating both parts we get:
dV/(Qin - n*k*(V^0.5)) = dt
Integrating both sides, we get:
(1/nk)^2 * [2*(V0^0.5 - V^0.5) + 2*Qin*LN[(Qin - nk*V0^0.5)/(Qin - nk*V^0.5)]] = t
The interesting thing here is that if Qin = nk*V^0.5, or V = (Qin/nk)^2, then the LN part of this equation will be undefined. Values of V less than (Qin/nk)^2 give negative values for t, which are clearly not possible. The physical explanation for this is that as the volume in the tank is reduced, there is not enough driving force (hydrostatic pressure in this case) to evacuate the fluid in the tank at a faster rate than it's coming in. In effect, it is not possible to empty the tank so long as water in flowing in. I used a cylindrical tank as the basis for this proof for simplicity, but this principle applies to any shaped tank.