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Rate of fill of a cone-shaped fluid container

  • #1
127
0
An engineer came to me with the following problem.
I'm not convinced the problem is correct - but I've probably just overlooked something, since I haven't done any vaguely practical problems in years.

Anyway, here it is.

Homework Statement



[PLAIN]http://img526.imageshack.us/img526/1911/54308613.png [Broken]

Homework Equations



Volume of a cone, I'm guessing trivial applications of chain rule in 1D. Some integration (integrating factor at worst? Probably separable.)

The Attempt at a Solution


Putting pi*7.5^2*30/3 (should be the volume of that cone) into google gives 1,700 or so.
So suppose the cone is almost full - then the out-flow greatly dwarfs the in-flow.
So the cone can never fill up, surely?
Indeed, the liquid level should just sit at a stable equilibrium at 400 units of liquid.

What have I missed?

The ridiculous non-SI units don't help as well.

Is the answer complex? (even pure imaginary?)
 
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Answers and Replies

  • #2
33,508
5,193
I'm having problems figuring out which parts of the problem you're doing. Are you asking about part c? The rate at which water is entering the tank is 600 ft3/min, and the rate out is 1.5 * 562.5[itex]\pi[/itex] ft3/min, which as you have noticed, is much larger than the rate of inflow.

Most likely there is a typo in the problem - maybe they switched the inflow and outflow rates or got the numbers wrong somehow.

I don't see any stable equilibrium of 400 units, as you said. Every bit of water that comes in will go right out, so the tank will never accumulate any depth of water.
 
  • #3
127
0
dV/dt := 600 - 1.5V

So dV/dt = 0 at V=400.

Some water must accumulate in the tank. Intuitively, because V=0 (initially) gives dV/dt > 0 (and we can integrate a strict inequality happily).

Maybe it's not a stable equilibrium, but won't all scenarios always end up at V=400? Like some kind of attractive point.

And yes, I apologize, I forgot to mention I was talking about part c).
(The rest is trivial, but I should have said so)

At first I thought they'd switched the inflow for outflow, but then I thought it'd be reasonable to set students a problem where the outflow depends on the weight of fluid in the container in some simple way (such that they can actually solve stuff analytically). Must be a typo of some other flavour, then.

Thanks.
 
  • #4
33,508
5,193
dV/dt := 600 - 1.5V
OK, I totally misinterpreted what they meant by V as the volume of the tank, not the volume of water in the tank at time t.
So dV/dt = 0 at V=400.

Some water must accumulate in the tank. Intuitively, because V=0 (initially) gives dV/dt > 0 (and we can integrate a strict inequality happily).

Maybe it's not a stable equilibrium, but won't all scenarios always end up at V=400? Like some kind of attractive point.

And yes, I apologize, I forgot to mention I was talking about part c).
(The rest is trivial, but I should have said so)

At first I thought they'd switched the inflow for outflow, but then I thought it'd be reasonable to set students a problem where the outflow depends on the weight of fluid in the container in some simple way (such that they can actually solve stuff analytically). Must be a typo of some other flavour, then.
Solving the DE + initial condition, I get V(t) = 400 - 600e-3t/2

As t gets larger and larger, the exponential term decreases to 0, so yes, there is an equilibrium value.
 

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