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
Answers
Answered by
98
total radiating surface will be CSA
=2πrh
=2×22/7×0.05/2×28
=22×0.05×4
=4.4m sq
=2πrh
=2×22/7×0.05/2×28
=22×0.05×4
=4.4m sq
Answered by
69
Answer:
Step-by-step explanation:
Given:
length/height = 28 m
diameter = 5 cm = 0.05 m
area of Total radiation surface is CSA ( Curved surface area )
CSA = 2πrh
= 2 × 22/7 × 0.05/2 × 28
= 22 × 0.05 × 4
= 4.4 m²
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