Question

Fermentation tanks are designed in the form of a cylinder mounted on a cone as shown below:

The total height of the tank is 3.3 m and the height of the conical part is 1.2 m. The diameter of the cylindrical as well as the conical part is 1 m. Find the capacity of the tank. If the level of liquid in the tank is 0.7 m from the top, find the surface area of the tank in contact with liquid.

Hint:

Use the formulas for volume and lateral surface area of cylinders and cones: \( V_{\text{cyl}} = \pi r^2 h \), \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h \), \( A_{\text{lateral cone}} = \pi r l \), \( A_{\text{lateral cyl}} = 2\pi r h \)

Answer 1

Given:
- Total tank height = 3.3 m
- Cone height = 1.2 m
- Cylinder and cone diameter = 1 m
- Radius \(r = \frac{1}{2} = 0.5\) m
- Liquid level from top = 0.7 m

To find:
1. Tank capacity (volume)
2. Liquid-contact surface area

Step 1: Cylinder height calculation
\[\text{Cylinder height} = 3.3 - 1.2 = 2.1 \text{ m}\]

Step 2: Tank volume calculation (cylinder + cone)
- Cylinder volume:
\[V_{\text{cyl}} = \pi r^2 h = \pi (0.5)^2 \times 2.1 = \pi \times 0.25 \times 2.1 = 0.525\pi \text{ m}^3\]
- Cone volume:
\[V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (0.5)^2 \times 1.2 = \frac{1}{3} \pi \times 0.25 \times 1.2 = 0.1\pi \text{ m}^3\]
- Total volume:
\[V = V_{\text{cyl}} + V_{\text{cone}} = 0.525\pi + 0.1\pi = 0.625\pi \text{ m}^3\]

Step 3: Liquid height calculations
- Liquid level from top = 0.7 m
- Cylinder height = 2.1 m
- Liquid partially fills the cone:
Height of liquid in cone = \(1.2 - 0.7 = 0.5\) m
Height of liquid in cylinder = 2.1 m

Step 4: Surface area calculation
- Cylinder curved surface area:
\[A_{\text{cyl}} = 2\pi r h = 2\pi \times 0.5 \times 2.1 = 2.1\pi \text{ m}^2\]
- Slant height \(l\) of liquid part of cone:
\[l = \sqrt{r^2 + h^2} = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = 0.707 \text{ m}\]
- Cone curved surface area (liquid part):
\[A_{\text{cone}} = \pi r l = \pi \times 0.5 \times 0.707 = 0.3535\pi \text{ m}^2\]
- Liquid surface area (top of cone):
\[A_{\text{surface}} = \pi r^2 = \pi \times (0.5)^2 = 0.25 \pi \text{ m}^2\]
- Total liquid-contact surface area:
\[A = A_{\text{cyl}} + A_{\text{cone}} + A_{\text{surface}} = 2.1\pi + 0.3535\pi + 0.25\pi = 2.7035 \pi \text{ m}^2\]
\[\approx 2.7035 \times 3.14 = 8.48 \text{ m}^2\]

Final Answers:
- Tank capacity = \(0.625\pi \, \text{m}^3\)
- Liquid-contact surface area = \(\approx 8.48 \, \text{m}^2\) or \(\frac{113.8\pi}{48} \, \text{m}^2\) as given