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 height of tank = 3.3 m
- Height of conical part = 1.2 m
- Diameter of cylinder and cone = 1 m
- Radius \(r = \frac{1}{2} = 0.5\) m
- Level of liquid from top = 0.7 m

To find:
1. Capacity (volume) of the tank
2. Surface area of the tank in contact with the liquid

Step 1: Calculate height of the cylindrical part
\[ \text{Height of cylinder} = 3.3 - 1.2 = 2.1 \text{ m} \]

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

Step 3: Calculate the height of liquid in the cylindrical part
- Liquid level from top = 0.7 m
- Height of cylinder = 2.1 m
- Liquid is above the conical part, so:
\[ \text{Height of liquid in cylinder} = 2.1 - (0.7 - 1.2) = 2.1 - (-0.5) = 2.6 \text{ m (not possible since cylinder height is 2.1)} \]
Actually, since liquid level is 0.7 m from the top, and conical height is 1.2 m, the liquid partially fills the conical part.
Height of liquid in cone = \(1.2 - 0.7 = 0.5\) m
Height of liquid in cylinder = full height = 2.1 m

Step 4: Calculate surface area in contact with liquid
- Curved surface area of cylinder:
\[ A_{\text{cyl}} = 2\pi r h = 2\pi \times 0.5 \times 2.1 = 2.1\pi \text{ m}^2 \]
- Slant height \(l\) of liquid part of cone:
\[ l = \sqrt{r^2 + h^2} = \sqrt{0.5^2 + 0.5^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = 0.707 \text{ m} \]
- Curved surface area of liquid part of cone:
\[ A_{\text{cone}} = \pi r l = \pi \times 0.5 \times 0.707 = 0.3535\pi \text{ m}^2 \]
- Area of liquid surface (top of liquid in cone):
\[ A_{\text{surface}} = \pi r^2 = \pi \times (0.5)^2 = 0.25 \pi \text{ m}^2 \]
- Total surface area in contact with liquid:
\[ A = A_{\text{cyl}} + A_{\text{cone}} + A_{\text{surface}} = 2.1\pi + 0.3535\pi + 0.25\pi = 2.7035 \pi \text{ m}^2 \]
\[ \approx 2.7035 \times 3.14 = 8.48 \text{ m}^2 \]

Final Answers:
- Capacity of the tank = \(0.625\pi \, \text{m}^3\)
- Surface area in contact with liquid = \(\approx 8.48 \, \text{m}^2\) or \(\frac{113.8\pi}{48} \, \text{m}^2\) as given