In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.
Question:

In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Length of the cylindrical pipe, h = 28 m

Radius of the cylindrical pipe, $r=\frac{5}{2}=2.5 \mathrm{~cm}=0.025 \mathrm{~m} \quad(1 \mathrm{~m}=100 \mathrm{~cm})$

∴ Total radiating surface in the system

= Curved surface area of the cylindrical pipe

$=2 \pi r h$

$=2 \times \frac{22}{7} \times 0.025 \times 28$

$=4.4 \mathrm{~m}^{2}$

Thus, the total radiating surface in the system is 4.4 m2.