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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. How much room will be left in the floor area of the silo when the pile is 60 ft high? How fast is the floor area of the pile growing at the height?

Solution

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Answered 2 years ago
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When the pile is 60ft high, the radius of the pile is 1.560=901.5\cdot 60=90 ft

Since the radius of the silo is 200 ft,

The floor area left=π(radius of the silo)2π(radius of the pile)2\text{The floor area left} = \pi\left(\text{radius of the silo}\right)^2-\pi\left(\text{radius of the pile}\right)^2

The floor area left=π(200)2π(90)2=31900π square feet\text{The floor area left} = \pi\left(200\right)^2-\pi\left(90\right)^2=31900\pi\text{ square feet}

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