Question

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.
Find​

Answer 1

Answer:

Diameter = 5 cm

Radius = Diameter ÷2=5cm÷2=2.5cm

length of pipe =28m=2800cm

Total Radiating Surface of pipe = Curved surface area + 2 × Area of circular crossection

=2πrh+2πr

2

=2πr(h+r)=2×

7

22

×2.5(2800+2.5)=44039.28cm

2

Answer 2

Question:-

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?

Answer:-

Given:-

$r = \frac{5}{2} = 2.5cm \\ in \: m \: = \frac{2.5}{100} \\ = 0.025m$

$h = 28m$

$total \: radiating \: surface \: in \: the \: system = total \: surface \: area \: of \: the \: \\cylinder. \\ \longrightarrow 2 \: \pi \: r \: (h + r) \\ \longrightarrow 2 \times \frac{22}{7} \times 0.025 \times (28 + 0.025)m ^{2} \\ \longrightarrow \frac{44 \times 0025 \times 28.025}{7} \\ \longrightarrow 4.4m ^{2} (approx)$