Thread: Draining a spherical tank filled with water?

Hi
I have recently been presented with this prolem:

Suppose that a spherical tank (whose diameter is 4m) is filled to a depth of 4m at time t = 0 and is draining through a hole in the bottom. Find the depth of the fluid in terms of time. Assume the constant in Torricelli’s law is k = 0.01

I initially did apply Torricelli's Law (A(x)*dh/dt= -k*square root(h), but upon trying to determine the cross-section area of Spherical tank I had trouble comming up with the area since, as the water drains the radius of the circle is to change.

Would someone please help

You want to use:

Hey prosri29.

One suggestion is to use the volume integration method which rotates a curve around an axis to get the volume.

Basically if you know the volume that has been lost at particular time, you can solve for a particular x0 by solving V = Integral (x0 to 1) pi*f(x)^2dx and what this represents is some volume from some point x0 onwards to the right to the end of the sphere (i.e. the volume that is lost).

The x0 will correspond to the "height" at which the water level is at (you may have to adjust this height if you are rotating the circle where the circle is centred at the origin), but the idea is the same.

So in short, you have an expression for volume at time t (or volume loss) and then you can use the above to get the height given that particular volume loss (or current volume).

Hi mark

Thanks for the reply but can you go through the steps for how you formulated the area. And isnt it suppose to be A(h)=pi(2^2-(2-h)^2)

The two forms are equivalent, since .

I considered the circle centered at (2,0) of radius 2:

I understand but,
when applying the area formula to Torricelli’s law I get:
A(x)*dh/dt=-k Sqrt(h)

pi(2^2-(h-2)^2)* dh/dt=-0.01*sqrt(h)
(4-(h^2-4h+4))/sqrt(h) dh=-0.01/pi dt
(h^2-4h)/sqrt(h) dh=-0.01/pi dt

Implictly intergrating both sides we get:

int(h^2-4h)/sqrt dh=int(-0.01/pi)dt

2/5*h^(5/2)-8/3*h^(3/2) +c = -(0.01/pi)t +d

My question now is how do you represent this as h(t)????

I understand but,
when applying the area formula to Torricelli’s law I get:
A(x)*dh/dt=-k Sqrt(h)

pi(2^2-(h-2)^2)* dh/dt=-0.01*sqrt(h)
(4-(h^2-4h+4))/sqrt(h) dh=-0.01/pi dt
(h^2-4h)/sqrt(h) dh=-0.01/pi dt

Implictly intergrating both sides we get:

int(h^2-4h)/sqrt dh=int(-0.01/pi)dt

2/5*h^(5/2)-8/3*h^(3/2) +c = -(0.01/pi)t +d

My question now is how do you represent this as h(t)????

You can solve for t, but not for h, as far as I can see.

Understood

cheers MarkFL2 and chiro for your contributions