Question

A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 13 ft, what percentage of the total capacity is being used? (Round your answer to one decimal place.)

Answer 1

Answer:

$\%A_F=77.335\%$

Explanation:

Given:

• diameter of tank, $d=14\ ft$

• level of the tank filled in its horizontal position, $h=13\ ft$

• Now refer the schematic that show water with blue colour.

The triangle ORQ is symmetric about OS as it comes from center O on the cord QR at S.

$RO=QO=7\ ft$ (∵ radius of the cylinder)

$RS=\sqrt{RO^2-OS^2}$

$RS=3.6055\ ft$

Now the area of triangle ORQ:

$A_t=\frac{1}{2}\times QR\times OS$

$A_t=RS\times OS$

$A_t=3.6055\times 6$

$A_t=21.6333\ ft^2$

Now the angle ROS:

$\cos\theta=\frac{OS}{OR}$

$\theta=\cos^{-1}(\frac{6}{7} )$

$\theta=31.0027^{\circ}$

Therefore the reflex angle ROQ:

$\rangle ROQ=360^{\circ}-\theta^{\circ}$

$\rangle ROQ=360^{\circ}-31.0027^{\circ}$

$\rangle ROQ=328.9973^{\circ}$

Now the area of sector ROQPR:

We have the area of full circle, $A=\pi.r^2$

where:

r = radius of the circle

hence for sector:

$A_S=\pi\times 7^2\times \frac{328.9973}{360}$

$A_S=140.6811\ ft^2$

Now the cross sectional area filled with water:

$A_F=A_S+A_t$

$A_F=140.6811-21.6333$

$A_F=119.0478\ ft^2$

Total cross sectional area of tank:

$A=\pi.r^2$

$A=\pi\times 7^2$

$A=153.9380\ ft^2$

Now the percentage of total capacity used:

$\%A_F=\frac{A_F}{A}\times 100\%$

$\%A_F=\frac{119.0478}{153.9380} \times 100$

$\%A_F=77.335\%$