The cylindrical tank with vertical circular cross section of height 14 ft. and
water level of 12 ft. holds a considerable percentage of its capacity.
The percentage of the tank being used is approximately 91.2%
Reasons:
The given parameters are;
Diameter of the tank = 14 ft.
Depth of water in the tank = 12 ft.
Required:
The percentage of total capacity being used
Solution:
Angle subtended by the segment of the tank without water, θ, is given as
follows;
[tex]\theta = 2 \times arccos\left(\dfrac{5}{7} \right)[/tex]
Cross sectional area of the top part of the tank not being used for water is
therefore;
[tex]A_{top} = 7^2 \cdot \left ( \dfrac{2 \times arccos\left(\dfrac{5}{7} \right) \times \pi }{360^{\circ}} - \dfrac{sin\left(2 \times arccos\left(\dfrac{5}{7} \right) \right)}{2} \right) \approx 13.49[/tex]
Volume of the top part of the tank, [tex]V_{top}[/tex] = [tex]A_{top}[/tex] × L
Where;
L = The length of the tank
∴ [tex]V_{top}[/tex] ≈ 13.49·L
Volume of the entire tank, V = π·r²·L
∴ V = 7² × π × L = 49·π·L
Capacity (volume) of the tank being used, [tex]V_{in \, use}[/tex] = V - [tex]V_{top}[/tex]
∴ [tex]V_{in \, use}[/tex] = 49·π·L - 13.49·L = (49·π - 13.49)·L
Percentage of the capacity being used, is therefore;
[tex]Percentage \ use = \dfrac{V_{in \, use}}{V} \times 100[/tex]
Therefore;
[tex]Percentage \ use = \dfrac{ (49\cdot \pi - 13.49)\cdot L}{49\cdot \pi\cdot L} \times 100 = \dfrac{ (49\cdot \pi - 13.49)}{49\cdot \pi} \times 100 \approx 91.2\%[/tex]
The percentage of the tank being used is approximately 91.2%.
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