Rate of Flow of water out of can

[pacelweb]

In my Further Maths class today...

1. Produce a function which can b used to predict the depth of water in a can with a hole in it
The experiment, conducted by a friend and I, gave results of the height left from times 0s to 1010s. At t=0, depth (h) = 0.125m

Homework Equations

Torricelli's equation $$v=\sqrt{2gh}$$

The Attempt at a Solution

We can deduce by inspection of a graph plotted of the data that the rate of flow out of the can is proportional to the height left (Note this is a decay, hence the negative):
$$\frac{dh}{dt}\propto{-h}$$

$$\frac{dh}{dt}=-kt$$

Solving the differential equation we get:
$$h=Ae^{-kt}$$

Substituting t=0, A=0.125m
$$h=0.125e^{-kt}$$

Attempting to find k, and using autograph, I substituted different values of t and their corresponding values of h from my results. However, I found that t was different in each case- the higher the value of t, the higher value of k; or graphically, using a lower value or k, the function fits the data at low values of t, for higher values of k the function is a better fit at higher t values.
I cannot figure out in my mind what physical property(ies) k represents. There was a hint from my teacher that velocity may play a part, and I have looked at Torricelli's $$v=\sqrt{}2gh$$ but cannot link into how this would help me.

Could anyone suggest what k may actually represent?

P.S sorry if this should be in Introductory Physics, but as we have done it in our maths lesson I naturally put it in here...
Many thanks, Philip Wiseman

 

[Mark44]

I'm confused. Don't you mean dh/dt = -kh rather than dt/dh = -kt? The first equation suggests that the rate of change of height is proportional to the height of water remaining in the can. I can't think of any physical meaning for the second equation.

Since you are having trouble fitting your data to your solution curve, it might be that your basic differential equation doesn't fit reality. Torrecelli's equation says that velocity is proportional to h^1/2, and we can infer from that, that the rate of change of volume is likewise proportional to h^1/2. This would say that dV/dt = -kh^1/2. To get this in terms of h, notice that the volume of water in the can depends on the fixed radius of the can and the height of water, so dV/dt is proportional to dh/dt, hence dh/dt = -Kh^1/2, where K is different from k, and both are positive.

 

[pacelweb]

Thanks, I have now changed it to dh/dt in stead of dt/dh.
Would I then solve to find h, giving 2h^1/2=-Kt + c ?

 

[Mark44]

Yes, so h = (-2Kt + C')^2, where C' = 2C. And you know that at t = 0, h = .125 m, so you can solve for C', which should be sqrt(.125).

Now, if you can fiddle with K so that your data fits this curve, then you'll have a good idea you're on the track that your teacher suggested in the hint.

As for a physical meaning for K, I would think it's probably related to the size and shape of the outlet hole.

 

[pacelweb]

Thanks very much. I will give it a go in autograph tomorrow.
I went in today to find that I was the only one who'd even attempted it!