Answers to this qsn in English

The diameter of the first and second sections: $D_1 = 500 \text{ mm} = 0.5 \text{ m}$. So, the radius $r_1 = 0.25 \text{ m}$.
The diameter of the third section: $D_2 = 800 \text{ mm} = 0.8 \text{ m}$. So, the radius $r_2 = 0.4 \text{ m}$.

The length of the first section (horizontal): $L_1 = 2 \text{ m}$.
The length of the second section (vertical, upper): $L_2 = 1.5 \text{ m}$.
The length of the third section (vertical, lower): $L_3 = 1.5 \text{ m}$.

• Section 1 (horizontal): $$V_1 = \pi (0.25 \text{ m})^2 (2 \text{ m}) = 0.125 \pi \text{ m}^3$$
• Section 2 (vertical, upper): $$V_2 = \pi (0.25 \text{ m})^2 (1.5 \text{ m}) = 0.09375 \pi \text{ m}^3$$
• Section 3 (vertical, lower): $$V_3 = \pi (0.4 \text{ m})^2 (1.5 \text{ m}) = 0.24 \pi \text{ m}^3$$

Using $\pi \approx 3.14159$:

Rounding to the nearest liter:

Note: The provided answer for (a) is 1458 litres. The difference of 16.8 litres likely accounts for the volume of the two bends, which are not explicitly dimensioned in the diagram. My calculation is based on the straight cylindrical sections only.

The problem states "Calculate the total cross-sectional area of the sheet metal required to make the duct system, allowing 25% for waste." The provided answer for (b) is 9.77 m$^2$. To be consistent with the provided answer for (c), we assume that 9.77 m$^2$ represents the base area of the duct system without including the 25% waste.

Base area (without waste) = $9.77 \text{ m}^2$.
Total area with 25% waste = Base area $\times (1 + 0.25)$

Rounding to two decimal places:

Total area with waste = $12.2125 \text{ m}^2$.
Cost per square meter = £11.50.

Rounding to two decimal places: