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Question

A planning engineer for a new alum plant must present some estimates to his company regarding the capacity of a silo designed to store bauxite ore until it is processed into alum. The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder 100 ft high with a radius of 200 ft. The conveyor carries ore at a rate of 60,000 πft3/h\pi \mathrm{ft}^{3} / \mathrm{h} and the ore maintains a conical shape whose radius is 1.5 times its height. If, at a certain time t, the pile is 60 ft high, how long will it take for the pile to reach the top of the silo?

Solution

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It is given that the ore is forming a conical shape, we know that the volume of a cone is given by

V=13πr2hV=\dfrac{1}{3}\pi r^2h

It is given that r=1.5hr=1.5h

V=13π(1.5h)2hV=\dfrac{1}{3}\pi\left(1.5h\right)^2h

V=13π(2.25h2)hV=\dfrac{1}{3}\pi\left(2.25h^2\right)h

V=0.75πh3V=0.75\pi h^3

Differentiate both sides with respect to time

dVdt=0.75πd(h3)dt\dfrac{dV}{dt}=0.75\pi \dfrac{d\left(h^3\right)}{dt}

The conveyor is bringing the ore at the rate of 60,000π\pi ft3/^3/h, this is dVdt\dfrac{dV}{dt}

60000π=0.75πd(h3)dhdhdtCHAIN RULE60000\pi=0.75\pi \underbrace{\dfrac{d\left(h^3\right)}{dh}\cdot\dfrac{dh}{dt}}_{\text{\color{#c34632}CHAIN RULE}}

60000π=0.75π3h2dhdt60000\pi=0.75\pi\cdot3h^2\cdot\dfrac{dh}{dt}

60000π=2.25πh2dhdt60000\pi=2.25\pi h^2\cdot\dfrac{dh}{dt}

Rewrite the equation in the differential form as shown below:

60000πdt=2.25πh2dh60000\pi\hspace{1mm}dt=2.25\pi h^2\hspace{1mm}dh

Integrate both sides

60000πdt=2.25πh2dh\int 60000\pi\hspace{1mm}dt=\int 2.25\pi h^2\hspace{1mm}dh

60000πt=2.25πh2+12+1+C60000\pi t=2.25\pi \dfrac{h^{2+1}}{2+1}+C

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