025 Review Problem - Time required to fill a reservoir of water

Problem 25
A reservoir 10 ft. deep is in the form of a frustum of inverted square pyramid with bases of 100 and 90 ft. on a side respectively. How long will it require an inlet pipe to fill the reservoir if the water pours in at the rate of 200 gal. per min.? (One gal. = 231 cu. in.)

Solution 25

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Reservoir capacity
$V = \frac{1}{3}(A_1 + A_2 + \sqrt{A_1A_2})h$

$V = \frac{1}{3}[ \, 100^2 + 90^2 + \sqrt{100^2(90^2)} \, ](10)$

$V = 90,333.33 ~ \text{ft.}^3$

$V = 90,333.33 ~ \text{ft.}^3 \times \left( \dfrac{12 ~ \text{in.}}{1 ~ \text{ft.}} \right)^3$

$V = 156,096,000 ~ \text{in.}^3$

$V = 156,096,000 ~ \text{in.}^3 \times \dfrac{1 ~ \text{gal.}}{231 \text{in.}^3}$

$V = 675,740.2597 ~ \text{gal.}$

Volume = Discharge × time
$V = Qt$

$675,740.2597 = 200t$

$t = 3,378.7 ~ \text{min.}$

$t = 3,378.7 ~ \text{min.} \times \dfrac{1 ~ \text{hr.}}{60 ~ \text{min.}}$

$t = 56.312 ~ \text{hrs.}$ answer

025 Review Problem - Time required to fill a reservoir of water

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