We know that it takes 2.5 hours to empty the tank with 10 valves open. Over the 2.5 hours the sum of the water in the tank initially and the water going into the tank will equal the amount of water leaving the tank. Let's set that up as an equation.
w + 2.5x = 2.5*10y
We also know that it takes 5.5 hours to empty the tank with 6 valves open. Let's set that up as an equation.
w + 5.5x = 5.5*6y
I don't like decimals so let's multiply both equations by 2:
2w+5x=50y
2w+11x=66y
To solve the problem we need to know the relationship between w and x, so lets solve for y in the first equation and subsitute in the the second:
y=(2w+5x)/50.
Substituting this in the second equation...
2w+11x=66*(2w+5x)/50
100w+550x=132w+330x
220x=32w
w=220x/32
w=6.875x
So 6.875 times the input rate equals the capacity of the tank. Thus it would take 6.875 hours to fill the tank.
Andrew N wrote it to add his two cents on this problem, arguing in real life the rate at which the water leaves will depend on the water pressure in the tank. Here are his comments.
Hi Michael,
First off, I'd like to say that you've done a fine job with your math problems website. The problems are interesting, and the solutions are quite thorough and easy to understand for the most part. This is no easy feat. I also love your gambling website, the Wizard of Odds. You taught me how to play blackjack! Thanks for that.
Now, problem 163 about emptying the water tanks has a physical wrinkle which in missing from the problem statement: if water is constantly entering the tank, there is no way the tank can be emptied. This is because the flow rate through the outlets is related to the height of the water in the tank. I can prove this assertion as follows.
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
where:
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 flow rate out 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 flow rate out of 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.
Just my 2 cents.
Thanks,
Andrew N
Michael Shackleford, ASA - June 22, 2001