Cylindrical tank w/ water flowing out of small tube near bottom

[Linus Pauling]

1. A cylindrical open tank needs cleaning. The tank is filled with water to a height meter, so you decide to empty it by letting the water flow steadily from an opening at the side of the tank, located near the bottom. The cross-sectional area of the tank is square meters, while that of the opening is square meters



2. I've got a bunch of calculations below. I need help with setting up and solving the final integral.



3. A_1 = area of surface, v_1 = velocity of flow at suraface
A_2 = area of small tube, v_2 = velocity of flow out of tube

A_1 >> A_2

V' = A_2 * v_2 = rate of discharge

V' = A_1 * dh/dt, relates rate of discharge to drop in height of liquid

v_2 = sqrt[2gh + v_1^2]

v_1 = v_2 *(A_2/A_1)

v_1 = sqrt[(2gh) / (1-(A_2/A_1)^2)]

dh/dt = rate of change of height of water = -A_2/A_1 * sqrt(2gh)

Ok, so now the strategy is to solve a separable first-order ODE like so:

dy/dx = f(x)g(y)
dy/g(y) = f(x)dx, and integrate both sides.

So in my case, these are the integrals I obtain:

dh/dt = -A_2/A_1 * sqrt[2gh]
Plus in g = 9.81 and obtain:

dh/dt = -4.23*(A_2/A_1)*sqrt(h)

So I get the following integrals:

h^(-0.5)dh = -4.23*(A_2/A_1)dt with h_0 and 0.5h_0 as the limits of integration. I have solved this a couple different ways and obtained incorrect answers.

Can someone point out what the correct integral I need to solve is?

Thanks!

 

[Linus Pauling]

Any ideas?

 

[Delphi51]

What is the question?
I'm having trouble following your work - why do you say
v_2 = sqrt[2gh + v_1^2]
It seems to say that if v_1 was zero, v_2 would be nonzero which doesn't make sense. Then it looks like you dropped the v_1^2 to get
dh/dt = = -A_2/A_1 * sqrt(2gh)
Actually that does make sense. What did you find from this that wasn't correct? I see you can get h as a function of time.