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A cistern which has a leak on the bottom is filled in $1$ hr. Had there been no leak, it could have been filled in $40$ minutes. If the cistern is full, at what time will the leakage empty it?

My attempt

Let the cistern hold $x$l of water.

If there is leak,

To fill $x$l of water, it takes $1$hr.

If there is no leak,

To fill $x$l of water, it takes $40$ minutes.

Now, how should I move on?

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  • $\begingroup$ Try to find out rate of leaking and rate of filling in terms of $x$ in the unit of l/hr. $\endgroup$ – Brian Cheung May 4 '16 at 16:17
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    $\begingroup$ You say "Had there been no leak, it could not have been filled in 40 minutes." Oh? $\endgroup$ – user21820 May 4 '16 at 16:22
  • $\begingroup$ Inflow $x$. Outflow $kx$. Hence capacity $40x$. Time to fill with leak is $60$, so $60x(1-k)=40x$, so $k=\frac{1}{3}$. Hence time for leak to empty cistern is $\frac{40x}{kx}=120$ or 2 hrs. $\endgroup$ – almagest May 4 '16 at 16:33
  • $\begingroup$ Any accurate answer has to use the differential equation and some additional assumptions like mine, because the difference between 40min and 1hour is significant, which implies that the leak is not that small, which means that it leaks at a significantly varying rate, faster when it's fuller. $\endgroup$ – user21820 May 4 '16 at 16:57
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    $\begingroup$ @user21820 Yeah, that's probably not plausible if the cistern is made of cement or cast iron. Maybe if it's made of some soft plastic. (But this was never really a serious answer in the first place!) $\endgroup$ – David K May 5 '16 at 13:11
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Let $t$ be time and $x$ be the ratio of the cistern filled with water. Assume that the cistern has a constant horizontal cross-section, which implies that the rate of leaking is proportional to the square-root of the amount of water by Torricelli's law, since the leak is at the bottom, and the amount is proportional to the water depth. Also assume that the cistern is filled via a tap at a constant rate.

The differential equation for the leaking cistern is then:

$\frac{dx}{dt} = a - c\sqrt{x}$.

where $a = \frac{3}{2}$ is the rate of tap flow, since it takes $\frac{2}{3}$ hours to fill the cistern when there is no leak, and $c$ is some positive constant that is larger for larger leaks.

We know that throughout the process of filling the cistern $a - c\sqrt{x} > 0$, otherwise it can never be filled. Thus during that process we have:

  $\frac{1}{a-c\sqrt{x}} \frac{dx}{dt} = 1$.

  $\int \frac{1}{a-c\sqrt{x}} \ dx = \int \frac{1}{a-c\sqrt{x}} \frac{dx}{dt} \ dt = t + k$ for some constant $k$.

We find that:

  $\int \frac{1}{a-c\sqrt{x}} \ dx = \int \left( - \frac{1}{c\sqrt{x}} + \frac{a}{c\sqrt{x}(a-c\sqrt{x})} \right) \ dx = - \frac{2}{c} \sqrt{x} - \frac{2a}{c^2} \ln(a-c\sqrt{x})$.

Therefore:

$- \frac{2}{c} \sqrt{x} - \frac{2a}{c^2} \ln(a-c\sqrt{x}) = t + m$ for some constant $m$.

Now we know that at $t = 0$ we have $x = 0$, therefore $m = - \frac{2a}{c^2} \ln(a)$.

Also at $t = 1$ (hour) we have $x = 1$ (full tank), therefore $- \frac{2}{c} - \frac{2a}{c^2} \ln(a-c) = 1 - \frac{2a}{c^2} \ln(a)$. Since $a = \frac{3}{2}$, we can numerically solve for $c$:

$c \approx 0.7114$.

If the leaky cistern starts full and the tap is off, it will run dry according to:

$\frac{dx}{dt} = -c\sqrt{x}$.

Which we can solve easily:

  $\int \frac{1}{2\sqrt{x}} \ dx = \int -\frac{c}{2} \ dt$.

  $\sqrt{x} = 1 - \frac{c}{2} t$.   [since at $t=0$ we have $x=1$]

  $x = ( 1 - \frac{c}{2} t )^2$.

Therefore the answer is:

The full leaky cistern will run dry at $t = \frac{2}{c} \approx 2.81$ hours.

An interesting feature of filling such leaky cisterns with a tap is that they can only be filled up to a certain limit, at which the rate of tap flow is equal to the rate of leak. That limit is at most $(\frac{a}{c})^2$, because if $x$ ever reaches that then $\frac{dx}{dt} = 0$ and so $x$ will not increase anymore. The fact that $x$ really tends to that limit can be seen by observing that $\ln(a-c\sqrt{x})$ must tend to $\infty$ as $t \to \infty$. This means that the tap flow rate $a$ must be at least $c$ otherwise it will be unable to fill the cistern.

Note

The previous version of this answer was using the assumption that leak rate is proportional to pressure at leak, which after almagest's comment and some online searching I now believe is not physically correct for water. I think it might be valid for very tiny leaks or if the cistern is leaking because the material is permeable, in which case viscosity has a large effect and so the Hagen–Poiseuille equation would imply my assumption. But if the cistern is quite big then the tap flow rate has to be reasonably high to be able to fill it up in 1 hour if there is no leak, so 20 min difference if there is a leak implies that the leak is actually not so tiny, which means that perhaps viscosity is not a dominant factor, and Torricelli's law would be the relevant one.

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  • $\begingroup$ So you are saying the answer is about 8 hours? $\endgroup$ – N.S.JOHN May 4 '16 at 17:44
  • $\begingroup$ @N.S.JOHN: Yes if you consider it empty at 0.1% full. I chose that value since it is relatively reasonable for most ordinary cistern shapes. $\endgroup$ – user21820 May 5 '16 at 2:41
  • $\begingroup$ @N.S.JOHN: I've changed the model used in my answer. In this model there is a well-defined time at which the filled leaky cistern runs dry, which is nearly 3 hours, significantly more than the 2 hours given by the naive solution. $\endgroup$ – user21820 May 5 '16 at 11:06
  • $\begingroup$ The interesting thing about this answer is that unless the problem in question came from a physics textbook (in which case I would have expected a better description of the nature of the "leak"), this is almost surely not the answer that was intended, even though it is more realistic than the intended (naive) answer. In my opinion, this shows that the question is a bad one, and the person who wrote it should have chosen some other "realistic" instantiation of their relative rates problem using phenomena that they actually understood and that actually worked as desired. $\endgroup$ – David K May 5 '16 at 11:33
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    $\begingroup$ After all this effort (apparently correct, too--I checked), it seems wrong that you would get no credit for it. At the very least it shows how much fun we can have finding plausible solutions to a poorly-conceived word problem. +1 $\endgroup$ – David K May 5 '16 at 13:09
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Let $a$ the outflow rate and $b$ the inflow rate in $\frac{l}{min}$. In 60 minutes the cistern is filled. The equation is

$(-a+b)\cdot 60= c \quad(\text{1 cistern})$

If there is no leak the cistern is filled in 40 minutes.

$b\cdot 40= c \quad(\text{1 cistern})$

$b=\frac{c}{40}$

The first equation becomes

$(-a+\frac{c}{40})\cdot 60= c $

Solving the equation you´ll get $a=\frac{1}{120}c$. That means that in one minute $\frac{1}{120}$ of the content will be drained. Hence in 2 hours the cistern is completely empty.

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  • $\begingroup$ Unfortunately, this answer is wrong by nearly an hour, even though it is clearly the kind of answer expected by the question-setter. $\endgroup$ – user21820 May 5 '16 at 11:28
  • $\begingroup$ This still gets my vote for best answer (so far). There seems to be an unfortunate tendency for writers of lower-level math textbooks to insert very badly-conceived problems into them. What to do about that? In the context of having to answer the question on homework or in an exam, guessing the author's intent (even if it's wrong-headed) is the main thing. If there is some effective way to push back against the tendency to ask such poor questions, I'd like to hear it. Feynman tried but gave up due to the apparent intractability of the problem. $\endgroup$ – David K May 5 '16 at 11:46
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Quickly, $$V(t)=\int_0^t f(s) - \delta(s)\mathrm{d}s.$$ If $f(s) = f$ and $\delta(s)=\delta$ for $s\geq0$, then $$ V^T = f-\delta$$ $$V^T=V(1)=\int_0^1 f(s) \mathrm{d}s- \int_0^1\delta(s)\mathrm{d}s = f-\delta$$ If $f(s) = f$ and $\delta(s)=0$ for $s\geq0$, then $$ V^T = V(2/3)=\int_0^{2/3} f(s) \mathrm{d}s- \int_0^{2/3}\delta(s)\mathrm{d}s = \frac{2}{3}f$$

Now, $$V(u) =0= V^T - \int_0^{u}\delta(s)\mathrm{d}s \implies 0 =V^T-u\delta$$ Using, the relations obtained before, we can say $$\delta=\frac{1}{3}f \qquad \text{and} \qquad V^T = \frac{2}{3}f.$$ Therefore $$u\delta=V^T \implies u\frac{1}{3}f=\frac{2}{3}f \implies u=2.$$ The time unit used was, of course, hours. So, if the cistern is full, if the input flow is stopped, it will get empty in 2 hours.

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Take Volume =$V$, Rate of filling =$x$, Rate of leaking=$l $

We need $\frac{V}{l}$

$\frac{V}{x}=40$

$\frac{V}{x-l}=60$

$\frac{x-l}{x}=\frac{2}{3}$

$l=\frac{x}{3}$

$\frac{V}{x}.\frac {x}{l}=\frac{V}{l}=120$

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    $\begingroup$ Don´t panic. The OP recognize your answer without any extra hints. $\endgroup$ – callculus May 4 '16 at 17:00

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