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Is this solution correct?

What I know is that the volume of the tank is $V = \pi r^2 h$, where r and h are in meter. Water is drained by a rate of $2,7\frac{m^3}{min}$. How fast does the water level decrease in this tank?

So if I set $H(t)$ to be the height after $t$ minutes, the volume should be $V(t) = (\pi r^2)H(t) \Rightarrow H(t) = \frac{V(t)}{\pi r^2}$.

This gives $H'(t) = \frac{V'(t)}{\pi r^2} = -2,7\frac{m^3}{\pi r^2 t}$, but I'm not sure if this is the correct solution. Could anyone tell me if any of my steps are incorrect, and if so, how do I correct them?

Thanks in advance

Frank Vel
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  • So $r$ is the tank radius, right? Is there the nozzle radius as well? – Aaron Maroja Oct 21 '14 at 00:01
  • @AaronMaroja no radius for the nozzle. – Frank Vel Oct 21 '14 at 09:29
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    If $V$ is a volume of water in the tank and water is being drained out, then $V$ decreases, so $V^\prime$ should be negative. And so $H^\prime$. The exact calculations depend, however, on a tank position. Is its axis vertical or horizontal? Or may be inclined somewhat...? – CiaPan Oct 21 '14 at 09:41
  • @CiaPan The axis is vertical. And yes it's supposed to be negative... – Frank Vel Oct 21 '14 at 10:24
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    OK, so once you replace $t$ with $\text{min}$ (a minute) in $-2,7\frac{m^3}{\pi r^2 t}$ your answer will be correct. You might also convert a minute to seconds to get your answer in SI units. – CiaPan Oct 21 '14 at 10:29

1 Answers1

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In one minute, the water volume decreases by $2.7$ m$^3$. With a section of $\pi r^2$, measuring $r$ in meters, this change of volume corresponds to a reduction in height by $\frac{2.7}{\pi r^2}$ meters per minute.

Anatoly
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  • why wouldn't it be $\frac{2.7}{\pi r^2}$ cubic meters per minute? otherwise the answer would be $k\times\frac{1}{meter\times minute}$ – Frank Vel Oct 21 '14 at 09:35
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    You're asking for a reduction in height, fvel, and height is measured in meters, not cubic meters. – Gerry Myerson Oct 21 '14 at 09:52
  • The numerator $2.7$ is measured in cubic meters per minute. The denominator $\pi r^2$ is measured in square meters. The ratio then gives meters per minute, which is the linear variation in height. – Anatoly Oct 21 '14 at 10:04
  • @GerryMyerson Yes, but $\frac{2.7}{\pi r^2}$ meters per minute $= \frac{2.7}{\pi r^2}\times\frac{meter}{minute}$, which wouldn't cancel out the $r$ meters in $r^2$, since $r$ is measured in meters. Only $\frac{2.7 m^3}{\pi r^2}$ would give an answer in meters, since $\frac{m^3}{r^2}$ would give a meter in the numerator? – Frank Vel Oct 21 '14 at 10:31
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    When you write "$\frac{2.7}{\pi r^2}$ meter per minute", clearly specifying that $r$ is measured in meters (this specification is necessary, otherwise the formula is incorrect), it is implicit that the meaning of $r$ in the denominator includes only its numerical value (the final result of the ratio in terms of dimensional measures is already specified as meters per minute). – Anatoly Oct 21 '14 at 11:31