Question

In a hot water heating system, there is a cylindrical pipe of length 21 m and diameter 5 cm. Find the total radiating surface in the system.​

Answer 1

Step-by-step explanation:

Given:

H=28m, R=

2

5

=2.5cm =0.025m (

100

2.5

)

⇒CSA of pipe (cylindrical)=2πrh

=2×

7

22

×0.025×28m

2

=4.4m

2

Answer 2

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h = 28m

2r=5cm

$\therefore$ r = $\large\dfrac{5}{2}$ cm = $\large\dfrac{5}{2×100}$

= $\large\dfrac{5}{200}$ m = $\large\dfrac{1}{40}$ m

$\therefore$ Total radiating surface in the system

= ${\small{\boxed{\sf{\pink{2πrh}}}}}$

= 2 × $\large\dfrac{27}{7}$ × $\large\dfrac{1}{40}$ × 28 = ${\underline{\underline{4.4m²}}}$

The area of the radiating surface of system is 4.4 m²

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