Q : 8    In a hot water heating system, there is a cylindrical pipe of length \small 28\hspace{1mm}m and diameter \small 5\hspace{1mm}cm. Find the total radiating surface in the system.
 

Answers (1)

Given,

Length of the cylindrical pipe = l = \small 28\hspace{1mm}m

Diameter =  \small d = 5\hspace{1mm}cm = 0.05\ m

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius r and length l = 2\pi r l

\therefore CSA of this pipe = 

= 2\times\frac{22}{7}\times\frac{0.05}{2}\times28

\\ = 22\times0.05\times4 \\ = 4.4\ m^2

Therefore, the total radiating surface of the system is 4.4\ m^2

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