How Does Water Level Change Over Time in a Draining Cylindrical Tank?

[uniidiot]

Homework Statement

A tank is in a shape of a cylinder with a circular cross-section of area A. Initially the depth of water in the tank (the head) is h0. At time t = 0 water is allowed to leave the tank ffrom a valve at the bottom. The rate at which the water leaves the tank is proportional to the head of the water at that instant; the constant of proportionality, k, is related to the discharge coefficient or the tank.

Derive an expression for the head of water in the tank at any time t after it is allowed to empty.

How long does it take for the tank to lose 1/2 its water? How long does it take for the tank to empty.

Homework Equations

dy/dx = -ky

y = Ce^(-kt)

The Attempt at a Solution

dh/dt = -k(h - h0)

h(time) = (h - h0)e^(-kt)

ln ((h - h0) / h(time)) = kt

and this is where i get stuck that's even if I've got the first bit right?

any help would be great!

:)

 

[neutrino]

Read the question carefully, it says that the rate at which water leaves the tank is proportional to the height of the water at the the time.

 

[uniidiot]

ok so:

the rate at which water leaves the tank = -kh

i'm a little confused about teh "discharge coefficient"

 

[neutrino]

ok so:

the rate at which water leaves the tank = -kh

So what does "the rate at which water leaves the tank" actually mean?

i'm a little confused about teh "discharge coefficient"

Just a bunch of fancy words to make the problem look interesting. If the complete problem reads as you have stated it here, then there's nothing to be confused about.

 

[uniidiot]

So what does "the rate at which water leaves the tank" actually mean?

the speed at which the height of the water decreases

well the diffrence of the height from the initial value is h - h0

so the rate of change is

dh / dt = (h - h0) / (t - t0)

?

 

[uniidiot]

Ok i think i might have cracked the 1st bit now,

dh/dt = -kh

h = Ce^-kt

ln h = -kt + A
but i don't understand how to work out how long it will take to empty 1/2 of the water, there are no figures to work with?thanks for your help so far neutrino.

 

[neutrino]

the speed at which the height of the water decreases

Not exactly. The answer is in the title of this thread!

well the diffrence of the height from the initial value is h - h0

so the rate of change is

dh / dt = (h - h0) / (t - t0)

?

At any instant the rate at which the volume decreases is proportional to the height of the water at that instant. That would give you,

\frac{dV}{dt} = -kh