Question

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

Answer 1

Answer:

44039.28cm^2

Step-by-step explanation:

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)

=2×22/7×2.5(2800+2.5)cm^2

=44039.28cm^2

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Answer 2

Answer:-

$\red{\bigstar}$ Total radiating surface in the system $\large\leadsto\boxed{\tt\purple{44039.28 \: cm^2}}$

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• Given:-

• Length of cylindrical pipe = 28m

• Diameter of cylindrical pipe = 5cm

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• To Find:-

• Total radiating surface in system.

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• Solution:-

Given that,

Length of cylindrical pipe is 28m

➪ $\sf 28 \times 100 \: cm$

➪ $\bf 2800 \: cm$

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Also,

Diameter of cylindrical pipe is 5cm

➪ $\sf Radius = \dfrac{Diameter}{2}$

➪ $\bf Radius = \dfrac{5}{2} cm$

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Now,

We know,

$\pink{\bigstar}$ $\underline{\boxed{\bf\red{C.S.A = 2 \pi rh}}}$

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• Substituting in the Formula:-

➪ $\sf C.S.A = 2 \times \dfrac{22}{7} \times \dfrac{5}{2} \times 2800$

➪ $\sf C.S.A = 22 \times 5 \times 400$

➪ $\sf C.S.A = 110 \times 400$

★ ${\bf\green{C.S.A = 44000 \: cm^2}}$

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Now,

✯ Total Radiating surface = C.S.A of Cylinder + Area of circular ends

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➪ $\sf 44000 + 2 \pi r^2$

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➪ $\sf 44000 + 2 \times \dfrac{22}{7} \times \dfrac{5}{2} \times \dfrac{5}{2}$

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➪ $\sf 44000 + \dfrac{11 \times 25}{7}$

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➪ $\sf 44000 + \dfrac{275}{7}$

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➪ $\sf 44000 + 39.2857$

★ $\large{\bf\pink{44039.28 \: cm^2}}$

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Therefore, the total radiating surface in the system is 44039.28 cm².