Basic (?) calc problem with flow out of a hole in a bucket

[krebs]

Homework Statement

(This isn't homework; someone is insisting that the graph of height of a leaking water column/time is a parabola, which makes no logical sense to me, but I want to prove it)
You have a bucket (5 gal) filled to the top with water, and a small hole in the bottom. Find the general shape of the graph describing the height of the water column with respect to time (as well as its derivatives). I'm ignoring all constants, such as the diameter of the hole.

Homework Equations

Q = flow rate
H = height of water in bucket
A = cross section of hole
g = gravity
Q = A √(2gH) ⇒ Q ∝ √H

The Attempt at a Solution

The way I tried to solve it is to say that ΔH/Δt ∝ -Q ∝ -√H. I'm a little rusty on my calculus, but I think this is correct... I used Wolfram to try and solve it. It gives me the result: H = ¼(-2ct + c² + t²). A parabola! What?? It doesn't make any sense! H is always, physically, positive, and thus -√H is always negative and so H/t should always have a negative slope. A parabolic solution implies that the bucket refills itself by magic. Did I break a math rule, or is Wolfram just totally wrong this time?

If the formula was Q = -H, then the solution would be H = e^-t; I expect a very similar solution here. Why does the square root mess it up so much?

http://www.wolframalpha.com/input/?i=dy/dx+=+-(y^.5)

 

[Chestermiller]

$$\frac{dH}{dt}=-k\sqrt{H}$$
H=H₀ at t = 0.
Do you know how to solve this differential equation, subject to the prescribed initial condition, without using wolframalpha?

 

[krebs]

$$\frac{dH}{dt}=-k\sqrt{H}$$
H=H₀ at t = 0.
Do you know how to solve this differential equation, subject to the prescribed initial condition, without using wolframalpha?

I do not, but here is my best shot:

$$\frac{dH}{\sqrt{H}}=-k{dt}$$

This simplifies (?) to: (does dH integrate to 1, or H?)

$$2\sqrt{H} = \frac{-k^2{t}}{2}$$

So $$H = \frac{k^4 t^2}{16} + H_0$$

It's been a while since I did any calculus. I can't believe that it could be a parabola...

 

[Chestermiller]

You need to go back and review calculus.
$$\sqrt{H}=\sqrt{H_0}-\frac{kt}{2}$$
So,
$$H=H_0-\sqrt{H_0}kt+\left(\frac{kt}{2}\right)^2$$

 

[krebs]

You need to go back and review calculus.
$$\sqrt{H}=\sqrt{H_0}-\frac{kt}{2}$$
So,
$$H=H_0-\sqrt{H_0}kt+\left(\frac{kt}{2}\right)^2$$

I don't understand how you got the first step... would you mind giving an explanation? If your math is right, I don't understand how it can be a parabola. Does that not imply the bucket refills itself after a given amount of time?

 

[HallsofIvy]

From [itex]\frac{dH}{\sqrt{H}}= H^{-1/2}dH= -kdt[/itex] integrating both sides (and using [itex]\int H^{-1/2}dH= \frac{1}{-1/2+ 1}H^{-1/2+ 1}[/itex]) [itex]2H^{1/2}= -kt+ c[/itex] where where "C" is the "constant of integration". Squaring both sides [itex]H= (-kt+ C)^2/4[/itex]. If [itex]H(0)= H_0[/itex] then [itex]H(0)= C^2/4[/itex] so [itex]C= 2\sqrt{H_0}[/itex]. [tex]H= \frac{(2\sqrt{H_0}- kt)^2}{4}[/tex].

The graph of that is a parabola opening downward with vertex at [itex](0, H_0)[/itex]. As t increases, H decreases until H= 0 at which point, because there is no more water in the bucket, the whole analysis no longer applies. H increases only for t< 0 where, again, the analysis does not apply because we assumed from the start that the hole only opens when t= 0.

 

[krebs]

From [itex]\frac{dH}{\sqrt{H}}= H^{-1/2}dH= -kdt[/itex] integrating both sides (and using [itex]\int H^{-1/2}dH= \frac{1}{-1/2+ 1}H^{-1/2+ 1}[/itex]) [itex]2H^{1/2}= -kt+ c[/itex] where where "C" is the "constant of integration". Squaring both sides [itex]H= (-kt+ C)^2/4[/itex]. If [itex]H(0)= H_0[/itex] then [itex]H(0)= C^2/4[/itex] so [itex]C= 2\sqrt{H_0}[/itex]. [tex]H= \frac{(2\sqrt{H_0}- kt)^2}{4}[/tex].

The graph of that is a parabola opening downward with vertex at [itex](0, H_0)[/itex]. As t increases, H decreases until H= 0 at which point, because there is no more water in the bucket, the whole analysis no longer applies. H increases only for t< 0 where, again, the analysis does not apply because we assumed from the start that the hole only opens when t= 0.

Thanks for the step by step. I was able to follow it through to completion. However, isn't $$H= \frac{(2\sqrt{H_0}- kt)^2}{4}$$ an upward opening parabola? I also still don't understand how we can get a parabola out of this... Are you absolutely sure your math is correct; or, is it possible that Q = √H is an incorrect assumption to start with?

Just thinking about it intuitively as $$\frac{dH}{dt}=-k\sqrt{H}$$, at large values of H, the slope of H/t is sharply negative, while as the graph approaches the x-axis from the left, the slope approaches zero. It follows that lim t→∞ = 0. lim t→-∞ = ∞. This also makes sense realistically, as an infinitely tall pillar of water would take an infinite amount of time to empty regardless of the high rate of exit through the hole at the bottom. Everything I have ever experienced tells me this should be a graph resembling e^x, and there are many systems with which I am familiar, such as capacitors, that have an e^x curve and which empty at a rate proportional to their "charge". I don't see how having a fractional power derivative can cause such strange things to happen...

 

[Chestermiller]

HallsofIvy's result is exactly the same as the one I gave in post #4, so I guess my math must be right. How is it possible that you don't recognize my 2nd equation in post #4 as a parabola? Apparently you need some brushing up on algebra also.

The initial rate of decrease of H is ##-k\sqrt{H_0}##, so H decreases rapidly with time as you anticipated. And the rate of decrease of H gets smaller as time progresses. But the height H does reach zero after a finite time. Your experience with the height decreasing exponentially with x is based on systems where the rate of change of height is proportional to H, rather than proportional to ##\sqrt{H}##. So this system does not behave like a capacitor. Contrary to your intuition, having the fractional power derivative does not cause anything strange to happen; you are the only one for whom the behavior seems strange.

Chet

 

[krebs]

HallsofIvy's result is exactly the same as the one I gave in post #4, so I guess my math must be right. How is it possible that you don't recognize my 2nd equation in post #4 as a parabola? Apparently you need some brushing up on algebra also.

The initial rate of decrease of H is ##-k\sqrt{H_0}##, so H decreases rapidly with time as you anticipated. And the rate of decrease of H gets smaller as time progresses. But the height H does reach zero after a finite time. Your experience with the height decreasing exponentially with x is based on systems where the rate of change of height is proportional to H, rather than proportional to ##\sqrt{H}##. So this system does not behave like a capacitor. Contrary to your intuition, having the fractional power derivative does not cause anything strange to happen; you are the only one for whom the behavior seems strange.

Chet

What makes you think I didn't recognize it as a parabola? I said very clearly that I didn't understand how, even if the math is correct, the solution could be parabolic, and my reasoning is still logical if ultimately flawed. I followed your math, and I don't recognize any errors, though solving much beyond simple derivatives is difficult for me. The fact remains that the parabolas both of your results obtained open upwards H ∝ T². And yet you claim there is nothing at all to be confused about. The only explanation I can see is that the left half of the parabola is what we are using, and we have to ignore the right half of it.

I came here to try and better my understanding of a difficult subject which I regrettably had few opportunities to study. The elitist insults are really unnecessarily and antagonistic.

 

[Mark44]

I do not, but here is my best shot:

$$\frac{dH}{\sqrt{H}}=-k{dt}$$

This simplifies (?) to: (does dH integrate to 1, or H?)

$$2\sqrt{H} = \frac{-k^2{t}}{2}$$

So $$H = \frac{k^4 t^2}{16} + H_0$$

You integrated the left side correctly, but not the right side. ##\int -k dt = -kt + C##, not ##\frac{-k^2 t}{2}##. The differential dt on the right side indicates that t is the variable of integration, so -k is just a constant.

It's been a while since I did any calculus. I can't believe that it could be a parabola...

Nor do I. The height of water in the leaking bucket can't possibly be increasing.

The flaw in your reasoning, I believe, is your initial assumption that dH/dt is proportional to ##\sqrt{H}##.

Squaring both sides [itex]H= (-kt+ C)^2/4[/itex]. If [itex]H(0)= H_0[/itex] then [itex]H(0)= C^2/4[/itex] so [itex]C= 2\sqrt{H_0}[/itex]. [tex]H= \frac{(2\sqrt{H_0}- kt)^2}{4}[/tex].

The graph of that is a parabola opening downward with vertex at [itex](0, H_0)[/itex].

That is incorrect - the parabola opens upward, which suggests that modeling this situation by ##dH/dt = k\sqrt{H}## doesn't work.

Thanks for the step by step. I was able to follow it through to completion. However, isn't $$H= \frac{(2\sqrt{H_0}- kt)^2}{4}$$ an upward opening parabola?

Yes.

I also still don't understand how we can get a parabola out of this... Are you absolutely sure your math is correct; or, is it possible that Q = √H is an incorrect assumption to start with?

That's how it seems to me.

 

[krebs]

You integrated the left side correctly, but not the right side. ##\int -k dt = -kt + C##, not ##\frac{-k^2 t}{2}##. The differential dt on the right side indicates that t is the variable of integration, so -k is just a constant.

Nor do I. The height of water in the leaking bucket can't possibly be increasing.

The flaw in your reasoning, I believe, is your initial assumption that dH/dt is proportional to ##\sqrt{H}##.That is incorrect - the parabola opens upward, which suggests that modeling this situation by ##dH/dt = k\sqrt{H}## doesn't work.
Yes.
That's how it seems to me.

Thanks for the response. I can see my mistake now with the calculation. I'm trying to learn calculus from an old textbook, and it is tough sometimes; thank you for explaining it. I looked around online, and someone saying that $$k$$ varies with Q in the formula, so I think you're right about the assumption being wrong about the flow rate formula.

 

[HallsofIvy]

What makes you think I didn't recognize it as a parabola? I said very clearly that I didn't understand how, even if the math is correct, the solution could be parabolic, and my reasoning is still logical if ultimately flawed. I followed your math, and I don't recognize any errors, though solving much beyond simple derivatives is difficult for me. The fact remains that the parabolas both of your results obtained open upwards H ∝ T². And yet you claim there is nothing at all to be confused about. The only explanation I can see is that the left half of the parabola is what we are using, and we have to ignore the right half of it.

I came here to try and better my understanding of a difficult subject which I regrettably had few opportunities to study. The elitist insults are really unnecessarily and antagonistic.

The point I was trying to make was that the equation doesn't know anything about the "meanings" of the variables ("t is time" etc.) IF you go back before t= 0 then the equation appears to be showing water going into the bucket at the same rate at which it is going out of the bucket for t greater than 0. It is up to the person setting up the problem to recognize that the equation only works for x greater than equal to 0.

 

[Chestermiller]

I came here to try and better my understanding of a difficult subject which I regrettably had few opportunities to study. The elitist insults are really unnecessarily and antagonistic.

You're absolutely right. I totally apologize. Thanks for the reminder that I need to be more patient. It's just that, sometimes I get a little defensive when someone questions whether I did the math correctly. Sorry again.

Chet

 

[Ray Vickson]

I do not, but here is my best shot:

$$\frac{dH}{\sqrt{H}}=-k{dt}$$

This simplifies (?) to: (does dH integrate to 1, or H?)

$$2\sqrt{H} = \frac{-k^2{t}}{2}$$

So $$H = \frac{k^4 t^2}{16} + H_0$$

It's been a while since I did any calculus. I can't believe that it could be a parabola...

Thanks for the response. I can see my mistake now with the calculation. I'm trying to learn calculus from an old textbook, and it is tough sometimes; thank you for explaining it. I looked around online, and someone saying that $$k$$ varies with Q in the formula, so I think you're right about the assumption being wrong about the flow rate formula.

I don't think there is anything much wrong with the ##Q## formula. Where you are in error is in assuming that your differential equation applies outside a certain time interval. For example, if the hole had been plugged by a cork that was removed at time t = 0, then whatever happened for t < 0 would not follow the differential equation you wrote. And, of course, once the bucket is empty, that's it---no more flow. So, a more correct differential equation would be:
[tex] \frac{dH}{dt} = \begin{cases}
?, & t < 0 \\
-k \sqrt{H}, & 0 \leq t \leq T \\
0,& t > T
\end{cases}
[/tex]
where ##T## is the smallest positive time at which ##H = 0##.

Now that is a pretty long-winded way of writing things (but is 100% correct); a more usual way would be to say that ##dH/dt = -k \sqrt{H}## on ##[0,T]##, where ##T## is as stated before.

 

[krebs]

I don't think there is anything much wrong with the ##Q## formula. Where you are in error is in assuming that your differential equation applies outside a certain time interval. For example, if the hole had been plugged by a cork that was removed at time t = 0, then whatever happened for t < 0 would not follow the differential equation you wrote. And, of course, once the bucket is empty, that's it---no more flow. So, a more correct differential equation would be:
[tex] \frac{dH}{dt} = \begin{cases}
?, & t < 0 \\
-k \sqrt{H}, & 0 \leq t \leq T \\
0,& t > T
\end{cases}
[/tex]
where ##T## is the smallest positive time at which ##H = 0##.

Now that is a pretty long-winded way of writing things (but is 100% correct); a more usual way would be to say that ##dH/dt = -k \sqrt{H}## on ##[0,T]##, where ##T## is as stated before.

Wow, this finally makes sense! $$t<0$$ makes intuitive sense because we are choosing $$t=0$$ at an arbitrary water column height, $$H_0$$, so the fact that it continues into the negative t is acceptable. Furthermore, once H = 0, the slope is $$-\sqrt{0}$$, thus dx/dt = 0 , t > T. Not being a calculus whiz, does anyone know why the integral comes out as a parabola? If the slope of the integral is its derivative, shouldn't it both graphs hit 0 at t = T, and remain flat thereon out?

Math is weird.

You're absolutely right. I totally apologize. Thanks for the reminder that I need to be more patient. It's just that, sometimes I get a little defensive when someone questions whether I did the math correctly. Sorry again.

Chet

Thanks for the apology. I meant no disrespect in questioning your math, I just was looking for an explanation because it doesn't make sense from my limited understanding. No harm done.

 

[Chestermiller]

Not being a calculus whiz, does anyone know why the integral comes out as a parabola? If the slope of the integral is its derivative, shouldn't it both graphs hit 0 at t = T, and remain flat thereon out?

I don't think there is an answer to the question of why the solution comes out to be a parabola. That's just a characteristic of the solution to the mathematical equations for this problem.

Regarding your question about "shouldn't both graphs hit 0 at t = T, and remain flat thereon out?" Please note that this is a physics problem, and the mathematics is just a tool for helping to quantify what is going on. But the physics must dominate. The physics is the master and the mathematics is the slave. And the physics says that, once the tank is empty, liquid doesn't spontaneously start backing up into the tank from the hole that it just discharged through under gravity. So the mathematics must obey, and not apply beyond t = T.

Chet

 

[krebs]

I don't think there is anything much wrong with the ##Q## formula. Where you are in error is in assuming that your differential equation applies outside a certain time interval. For example, if the hole had been plugged by a cork that was removed at time t = 0, then whatever happened for t < 0 would not follow the differential equation you wrote. And, of course, once the bucket is empty, that's it---no more flow. So, a more correct differential equation would be:
[tex] \frac{dH}{dt} = \begin{cases}
?, & t < 0 \\
-k \sqrt{H}, & 0 \leq t \leq T \\
0,& t > T
\end{cases}
[/tex]
where ##T## is the smallest positive time at which ##H = 0##.

Now that is a pretty long-winded way of writing things (but is 100% correct); a more usual way would be to say that ##dH/dt = -k \sqrt{H}## on ##[0,T]##, where ##T## is as stated before.

I don't think there is an answer to the question of why the solution comes out to be a parabola. That's just a characteristic of the solution to the mathematical equations for this problem.

Regarding your question about "shouldn't both graphs hit 0 at t = T, and remain flat thereon out?" Please note that this is a physics problem, and the mathematics is just a tool for helping to quantify what is going on. But the physics must dominate. The physics is the master and the mathematics is the slave. And the physics says that, once the tank is empty, liquid doesn't spontaneously start backing up into the tank from the hole that it just discharged through under gravity. So the mathematics must obey, and not apply beyond t = T.

Chet

I've worked it some more and I think I fully understand the problem now. The function H(t) is continuous and doesn't have any vertical tangents, cusps or corners, yet for some reason its derivative is actually a piece-wise function. I didn't realize that this was possible, and I don't really know why. I graphed it if anyone is interested: Red is H/t, blue is dH/dt https://www.desmos.com/calculator/yzwdcws2kr

Of course you can also take the derivative of our newly found formula, set it equal to $$-k\sqrt{H}$$ and get the same exact parabola, showing that the derivative is correct. The weirdness with the derivative makes me seriously doubt if $$Q = k\sqrt{2gH}$$ is accurate when $$ H \rightarrow 0 $$.

 

[Ray Vickson]

I've worked it some more and I think I fully understand the problem now. The function H(t) is continuous and doesn't have any vertical tangents, cusps or corners, yet for some reason its derivative is actually a piece-wise function. I didn't realize that this was possible, and I don't really know why. I graphed it if anyone is interested: Red is H/t, blue is dH/dt https://www.desmos.com/calculator/yzwdcws2kr

Of course you can also take the derivative of our newly found formula, set it equal to $$-k\sqrt{H}$$ and get the same exact parabola, showing that the derivative is correct. The weirdness with the derivative makes me seriously doubt if $$Q = k\sqrt{2gH}$$ is accurate when $$ H \rightarrow 0 $$.

If you doubt the accuracy of the formula when ##H \to 0##, can you tell us which Law of Physics you think is being violated by the original differential equation?

Anyway, about no cusps or corners: no, the function ##H(t)## (outside the region ##0 < t < T##) can (and sometimes does) have a cusp at ##t = 0##. For example, if the bucket was filled to height ##H_0## for all ##t < 0## (before a cork plugging the hole was removed at time 0), then we would have:
[tex] H(t) = \begin{cases} H_0 & t < 0 \\
\frac{1}{4} (2\sqrt{H_0}- kt)^2& 0 \leq t \leq 2 \sqrt{H_0}/k \\
0 & t > 2 \sqrt{H_0}/k
\end{cases}
[/tex]
This has a cusp at ##t = 0##. Other scenarios for ##t < 0## might very well give you a different behavior slightly to the left of ##t = 0##, but the fact remains that this is all fantasy: we are given absolutely NO information about what happened before ##t = 0##, so have no basis at all for arguing one way or another. However, things are very different around the time ##T = 2 \sqrt{H_0}/2##; there, the actual physics of the problem dictates that the water height remains at ##H = 0## thereafter.

 

[krebs]

If you doubt the accuracy of the formula when ##H \to 0##, can you tell us which Law of Physics you think is being violated by the original differential equation?

Anyway, about no cusps or corners: no, the function ##H(t)## (outside the region ##0 < t < T##) can (and sometimes does) have a cusp at ##t = 0##. For example, if the bucket was filled to height ##H_0## for all ##t < 0## (before a cork plugging the hole was removed at time 0), then we would have:
[tex] H(t) = \begin{cases} H_0 & t < 0 \\
\frac{1}{4} (2\sqrt{H_0}- kt)^2& 0 \leq t \leq 2 \sqrt{H_0}/k \\
0 & t > 2 \sqrt{H_0}/k
\end{cases}
[/tex]
This has a cusp at ##t = 0##. Other scenarios for ##t < 0## might very well give you a different behavior slightly to the left of ##t = 0##, but the fact remains that this is all fantasy: we are given absolutely NO information about what happened before ##t = 0##, so have no basis at all for arguing one way or another. However, things are very different around the time ##T = 2 \sqrt{H_0}/2##; there, the actual physics of the problem dictates that the water height remains at ##H = 0## thereafter.

Yes, if you have a cork, then you'll have a cusp at t=0; I suppose you must be right. I guess expecting this to be an e^-x is sort of akin to saying that if I poured out a glass of water, it would take an infinite amount of time for the last drop to hit the ground, which is nonsense. Obviously k changes during a real-world experiment especially when H is very small (partially open hole in the bucket). When I did the actual experiment, the graph was fitted in the form of e^-x, but I can see both graphs look very similar, especially with the noise of real data.

Thanks for the help, everyone.

 

[Ray Vickson]

Yes, if you have a cork, then you'll have a cusp at t=0; I suppose you must be right. I guess expecting this to be an e^-x is sort of akin to saying that if I poured out a glass of water, it would take an infinite amount of time for the last drop to hit the ground, which is nonsense. Obviously k changes during a real-world experiment especially when H is very small (partially open hole in the bucket). When I did the actual experiment, the graph was fitted in the form of e^-x, but I can see both graphs look very similar, especially with the noise of real data.

Thanks for the help, everyone.

You say "obviously, k changes during a real-world experiment...". That is not obvious at all, and I wonder why you make the claim.

Of course, in the real world, when we get down to the molecular scale (for example, when there is a film one-molecule thick left in the bottom of the barrel) we cannot continue to use macroscopic, lab-scale equations and to ignore such phenomena as evaporation and the like. However, I think you are referring to more "lab-scale" phenomena, and seem to take issue with the basic laws governing such matters. I wonder what your evidence is. (Note that inevitable random errors infect any experiment, and measuring levels and rates to, say, 7 decimal places is out of the question, so I would not believe a claim based on such errors occurring.)

 

[krebs]

You say "obviously, k changes during a real-world experiment...". That is not obvious at all, and I wonder why you make the claim.

Of course, in the real world, when we get down to the molecular scale (for example, when there is a film one-molecule thick left in the bottom of the barrel) we cannot continue to use macroscopic, lab-scale equations and to ignore such phenomena as evaporation and the like. However, I think you are referring to more "lab-scale" phenomena, and seem to take issue with the basic laws governing such matters. I wonder what your evidence is. (Note that inevitable random errors infect any experiment, and measuring levels and rates to, say, 7 decimal places is out of the question, so I would not believe a claim based on such errors occurring.)

k changes at the point where h < d, where d is the diameter of the hole, since the effective size of the aperture is now smaller. Your measurements would then appear more e^-x like because the curve would take much longer to reach the x-axis for the very last leg of its journey. I don't mean to beat a dead horse.