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Water is leaking out of an inverted conical tank at a rate of $0.0068\,\rm m^3/min$. At the same time water is being pumped into the tank at a constant rate. The tank has height $14\,\rm m$ and the diameter at the top is $5.5\,\rm m$. If the water level is rising at a rate of $0.24\,\rm m/min$ when the height of the water is $3\,\rm m$, find the rate at which water is being pumped into the tank.

I tried to solve, but I don't know what the appropriate symbols would be to use in this situations. I got to

$$V' = \frac{2}{3}\pi r r'h + \frac{1}{3}\pi r^2 h',$$

but I don't know what to do next. If anyone could provide a step by step explanation I'd be grateful.

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Let $h(t)$ be the height of the water at time $t$, and let $r(t)$ be the radius of the surface of the water at time $t$; from similar triangles we know that

$$\frac{h(t)}{r(t)}=\frac{14}{2.75}=\frac{56}{11}\;.$$

We could solve this for either $r(t)$ or $h(t)$ in terms of the other, but notice that we’re told $h'(t)$ at a particular moment, and we’re not told anything about an specific value of $r(t)$. This suggests that we’d be better off working in terms of $h(t)$, so we’ll solve for $r(t)$ and get $$r(t)=\frac{11}{56}h(t)\;.$$

At time $t$ the volume $V(t)$ of water in the tank is the volume of a right circular cone with height $h(t)$ and base radius $r(t)$, which is given by

$$V(t)=\frac13\pi r(t)^2h(t)=\frac{\pi}3\left(\frac{11}{56}h(t)\right)^2h(t)=\frac{121\pi}{9408}h(t)^3\;.$$

Then

$$V'(t)=\frac{121\pi}{3136}h(t)^2h'(t)\;.$$

We’re told that $h'(t)=0.24$ when $h(t)=3$; if we call that moment time $t_0$, we have

$$V'(t_0)=\frac{121\pi}{3136}\cdot3^2\cdot0.24=\frac{3267\pi}{39200}\text{ m}^3/\text{min}\;.$$

Now let $v$ be the rate in cubic metres per minute at which water is being pumped into the tank. Taking into account both the inflow and the leakage, we know that at all times

$$V'(t)=v-0.0068\text{ m}^3/\text{min}\;.$$

In particular, at time $t_0$ we have

$$v-0.0068=\frac{3267\pi}{39200}\;,$$

which is completely straightforward to solve for $v$.

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You are fine up to $V'=\frac 23 \pi rhr'+\frac 13 \pi r^2h'$. The fact that the tank is conical says that $\frac rh=\frac {2.75}{14}$ or $h=\frac {56}{11}r$, which also implies $h'=\frac {56}{11}r'$. You can insert this to get $V'$, which is the net increase rate in volume at the given conditions. Then if you get some units for the $0.0068$ you can add that to $V'$ to get the inflow rate.

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I think the simplest way to understand this problem is to recognize that

$$V^{'}_{total} = V^{'}_{in} - V^{'}_{out}$$, or in alternative notation,

$$\frac {dV_{total}}{dt} = \frac {d_{pump}}{dt}-\frac {d_{leak}}{dt}$$

Once you recognize that, everything else is straight forward.

Let's use $\frac {dp}{dt}$ for the rate of change due to the pump and $\frac {dl}{dt}$ for the rate of change due to the leak. So,

$$\frac {dV}{dt} = \frac {dp}{dt} - \frac {dl}{dt}$$

Simplify the Volume Equation

Now, proceed in the normal way. $V = \frac {1}{3}\pi r^2h$ and we can eliminate $r$ to simplify the equation. $tan~\theta= \frac {r}{h}$ or $r=h~tan~\theta$. (This comes from similar triangles and $\theta$ is constant.) Thus, our volume equation becomes $V=\frac{1}{3}\pi (h~tan~\theta)^2h$. We know $tan~\theta$. It is $\frac {2.75}{14} = \frac {11}{56}$. The volume equation therefore is $V=\frac{1}{3}\pi(\frac{11h}{56})^2h = \frac {1}{3}\pi \frac {121h^2}{3136}h = \frac{121\pi h^3}{9408}$

Differentiate

$$V= \frac {121\pi h^3}{9408}$$ $$\frac {dV}{dt} = 3\cdot \frac {121\pi h^2}{9408}\frac{dh}{dt}$$ $$\frac {dV}{dt} = \frac {121\pi h^2}{3136}\frac{dh}{dt}$$

Make Substitutions

Recalling that $\frac {dV}{dt} = \frac {dp}{dt} - \frac {dl}{dt}$, our derivative becomes

$$\frac {dp}{dt} - \frac {dl}{dt} =\frac {121\pi h^2}{3136}\frac{dh}{dt}$$

We know all of the values except $\frac {dp}{dt}$ and we can solve for that:

$$\frac {dp}{dt} - 0.0068 = \frac {121\pi}{3136}\cdot3^2\cdot0.24$$ $$\frac {dp}{dt} = \frac {121\pi}{3136}\cdot3^2\cdot0.24 + 0.0068 $$ $$\frac {dp}{dt}\approx 0.2686~m^3/min$$

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