Two variable physics problem

I’m struggling with a classic physics problem:

We have a can filled with water. The can has a hole on the bottom. We measure how long it takes for the water to drain out and try to write an equation that predicts this time based on the water height and hole diameter.

So the variables are:
t = time the water takes to drain
h = height of the water column before the drain begins
d = diameter of the hole

I analyzed the data and got a few equations but I can only account for two out of the three variables. I’m not sure that I know how to get all 3 in the same equation.

For instance, I know that:
t=35.261*(1/d)-31.134 at h = 8
and that:
t=8.495*sqrt(h) - 1.005 at d = .635

At first I thought that since they both equalled t, I could say t^2 = the product of the two equations. However, this produces inaccurate results because the coefficients of the two equations only hold true in those specific circumstances where I make one of the variables constant.

So there is my dilemma. I can find the time as a function of diameter, or as a function of height, but I can’t account for both variables at once. I’ve never solved a problem like this before. What do I do?

Seems like it would be somewhat easier if you didn’t have those constants in the equations. Where did they come from anyway? It doesn’t seem like they should be there. For example:

t=35.261*(1/d)-31.134 at h = 8

So your equation is saying if the diameter is 2 units, it takes -13.5035 seconds (or whatever) for the water to drain? That’s pretty damn fast!

By the way, if it’s not obvious, I was talking about the constants being subtracted off in each equation; the 31.134 in the first equation, and the 1.005 in the second. I don’t think there should be any.

I’m thinking of a barrel or garbage can. Something cylindrical. Then the time T for the water to flow would be directly proportional to the height H and inversely proportional to the square of the diameter D of the hole in the bottom.

Greater height means more water and more time to drain. Bigger hole means greater flow of water and less time to drain.

So T = k * H/D[sup]2[/sup].

True only if you’re assuming choked flow through the orifice, and that the vena contracta diameter is a constant percentage of the orifice diameter. Otherwise, flow rate is proportional to the square root of pressure, which is proportional to water height.

Then, flow rate Q = k * sqrt(H) / D^2
and volume V = integral from 0 to t ( k * sqrt(H) / D^2 )

this is a classic problem which requires differential calculus. The equivalent electrical is the discharge of a capacitor through a resistor. The answer is an exponential function base en the number e. There is now way of resolving this without resorting to differential equations.

Not quite, since you can’t get a choked-flow situation (flow rate [current] is at a maximum, increasing pressure [voltage] doesn’t further increase flow rate) through a resistor. If you think of this as an engineering problem, rather than physics, then a choked-flow approximation may be good enough. If the hole is small compared to the container, then flow will be choked most of the way and the earlier-posted, simple equation will do.

Outrider, your equations are inconsistant. At h = 8, d = 0.635, your two equations should agree, but I get t=24.395 with the first, and t=23.022 with the second.

If they were consistant, you could solve for a single equation by assuming the equation was separable in the two variable h and d:

t(h,d) = C*f(d)*g(h)

where C = t(h=8,d=0.635), f(d=0.635) = 1, g(h=8) = 1.

Then

C * f(d) * g(8) = 35.261*(1/d)-31.134
C * f(0.635) * g(d) = 8.495*sqrt(h) - 1.005

Using g(8) = f(0.635) = 1, these become

f(d) = (35.261*(1/d)-31.134) / C
g(h) = (8.495*sqrt(h) - 1.005) / C

since your equations are inconsistant at h=8, d = 0.635, about the best you can do is pick C = (24.395+23.022)/2

Also, there’s no guarantee the equation is separable, but at least you’ll have a reasonable guess for a starting point.

On further reflection, when solving for f(d), use the value for C you got for that equation: C = 24.395. When solving for g(h), use C = 23.022.

Then use the average value for C to get t(h,d):

t(h,d) = 0.5*(24.395+23.022)*f(d)*g(h)

ElvisL1ves, I haven’t a clue what you are talking about. The problem of discharging a capacitor through a resistor is in chapter 1 of electronics 101. It is about the simplest problem you can imagine. The problem can only be resolved using differential equations because you have a circular reference: the current depends on the voltage which, in turn depends on the integral of the current.

The solution, as any electrical engineer can tell you is:

V = Vo * e^(-t/(r*c))

e is the number 2.718282 and rc is called the time constant of the circuit.

So, if you start with a capacitor charged to 100 volts it will discharge exponentially never really reaching 0 volts.



t/(rc)  Volt

  0.0  100.00
  0.1   90.48
  0.2   81.87
  0.3   74.08
  0.4   67.03
  0.5   60.65
  0.6   54.88
  0.7   49.66
  0.8   44.93
  0.9   40.66
  1.0   36.79
  1.5   22.31
  2.0   13.53
  3.0    4.98
  4.0    1.83
  5.0    0.67
  6.0    0.25
  7.0    0.09
  8.0    0.03
  9.0    0.01
 10.0    0.00


you can see it discharges very rapidly at first and much slower later and in a time equal to 5 or 6 time constants it can be considered discharged.

In the first period it loses 63.21% of its voltage. In the following period it loses 63.21% of what’s left and so on. You can see it will never fully discharge.

So I reiterate: There is no way of resolving this problem without resorting to differential calculus (as you can easily see if you look in any such text). If you find a way of resolving this problem without resorting to differential calculus, you will have a great place among the mathematicians of history.

Where did everybody go? I only said “differential calculus” and everybody disappears like suddenly they remembered some urgent chores they had to do…

Sailor, I think he’s talking about a situation in which the flow is only proportional to sqrt pressure until a certain point, at which the flow hits a maximum and doesn’t increase further. He’s saying that in a capacitor/resistor, there is no such practical restriction.

This is some kind of fluid dynamics thing I’m not too familiar with.

In the case you are describing, you are correct, but I think Elvis has a point about ‘choked flow’.

from Perspectives on Safety Relief Valve Sizing for Two Phase Flow

well, I doubt the water coming out of the can would rech sonic velocity. At any rate, if the flow changes with the level and the level changes with how much water has escaped, there is a circular reference that only differential calculus can resolve. but it seems everybody disappeared into the woodwork. What happened?