Question

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter
5 cm. Find the total radiating surface in the system.
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Answer 1

Given:-

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

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To find:-

• Total radiating surface in the system.

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Solution:-

• The cylinder pipe radiates from the curved surface and not from the ends.

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So,

• we need to find curved surface of cylinder.

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Here,

• Length (l) = 28 m
• Diameter (d) = 5 cm.

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Then,

• Radius of circular end of pipe:-

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⇛ r = 5/2 cm

⇛ r = 2.5 cm

⇛ r = 2.5 × 1/100

⇛ r = 0.025 m

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☆Total radiating surface in the system = CSA of cylinder pipe

⇛ 2πrh

⇛ 2 × 22/7 × 0.025 × 28

⇛ 4.4 m²

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Hence,

• the total radiating surface in the system is 4.4 m².

Answer 2

Given :

Height (h) of cylindrical pipe Length of cylindrical pipe = 28 m

Radius (r) of circular end of pipe

$= \frac{5}{2} \: cm \: = 2.5 \: cm = 0.0025$

We know that,

CSA of cylindrical pipe =

$2\pi \: rh$

$= 2 \times \frac{22}{7 } \times 0.0025 \times 28m {}^{2}$

$= 4.4m {}^{2}$

Therefore, the area of the radiating surface of the system is

$4.4 \: cm {}^{2}$