Question

In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22/7)​

Answer 1

Answer:

Here's your answer

Step-by-step explanation:

$We \: need \: to \: find \: CSA \: of \: Cylinder \\ Height \: of \: Cylinder(h) = 28m \\ Radius \: of \: circular \: end \: of \: pipe = \frac{5}{2} cm \\ = 2.5cm \\ Total \: Radiating \: Surface \: of \: pipe \\ = CSA + 2 × Area \: of \: circular \: crossection \\ =2πrh+ {2πr}^{2} \\ =2πr(h+r) \\ 2 \times \frac{22}{7} \times 2.5(2800+2.5) \\ {=44039.28cm}^{2}$

Answer 2

$\bf{\dag}\:{\underline{\underline{\sf{Question\::}}}}$

• In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Assume π = 22/7)

$\bf{\dag}\:{\underline{\underline{\sf{Answer\::}}}}$

• Total radiating surface in the system is 4.4 m².

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$\bf{\dag}\:{\underline{\underline{\sf{Given\::}}}}$

• Length of cylindrical pipe = 28 cm
• Diameter of cylindrical pipe = 5 cm

$\bf{\dag}\:{\underline{\underline{\sf{To\:Find\::}}}}$

• Total radiating surface in the system?

$\bf{\dag}\:{\underline{\underline{\sf{Solution\::}}}}$

☯ Firstly let's find the radius of a cylindrical pipe. We clearly know that;

$\quad\bf{\dag}\:{\underline{\boxed{\sf{\red{Radius = Diameter\:\div\:2}}}}}$

$\underline{\sf{\bigstar\:Putting\:all\:values\::}}$

$\\ :\implies\:\sf Radius = 5\:\div\:2$

$\\ :\implies\:\sf Radius = \dfrac{5}{2}\:cm$

☯ Converting radius into meters. We clearly know that;

$\quad\bf{\dag}\:{\underline{\boxed{\sf{\pink{1\:m = 100\:cm}}}}}$

Therefore,

$\\ :\implies\:\sf Radius\:in\:meters = \dfrac{5}{2}\:\div\:100$

$\\ :\implies\:\sf Radius\:in\:meters = \dfrac{5}{2}\:\times\:\dfrac{1}{100}$

$\\ :\implies\:\sf Radius\:in\:meters = \dfrac{5\:\times\:1}{2\:\times\:100}$

$\\ :\implies\:\sf Radius\:in\:meters = {\cancel{\dfrac{5}{200}}}$

$\\ :\implies\:\sf Radius\:in\:meters = \dfrac{1}{40}\:m$

$\underline{\underline{\boxed{\sf{\therefore\:Radius\:of\:pipe\:is\:\bf{\dfrac{1}{40}\:m}}}}}$

☯ Now, we have to find it's total radiating surface i.e, curved surface area (CSA). We clearly know that,

$\quad\bf{\dag}\:{\underline{\boxed{\sf{\purple{CSA_{(cylinder)} = 2\pi rh}}}}}$

Where,

• r denotes radius of cylinder
• h denotes height of cylinder

We have,

• Radius of pipe (r) = 5/40 m
• Height of pipe (h) = 28 m

$\underline{\sf{\bigstar\:Putting\:all\:values\::}}$

$\\ :\implies\:\sf CSA_{(cylinder)} = 2\:\times\:\dfrac{\cancel{22}}{\cancel{7}}\:\times\:\dfrac{1}{\cancel{40}}\:\times\:\cancel{28}$

$\\ :\implies\:\sf CSA_{(cylinder)} = \cancel{2}\:\times\:\dfrac{11}{1}\:\times\:\dfrac{1}{\cancel{20}}\:\times\:4$

$\\ :\implies\:\sf CSA_{(cylinder)} = \dfrac{1\:\times\:11\:\times\:4}{10}$

$\\ :\implies\:\sf CSA_{(cylinder)} = \dfrac{11\:\times\:4}{10}$

$\\ :\implies\:\sf CSA_{(cylinder)} = \dfrac{44}{10}$

$\\ :\implies\:\sf CSA_{(cylinder)} = 4.4\:m^2$

$\underline{\underline{\boxed{\sf{\therefore\:Total\:radiating\:surface\:is\:\bf{4.4\:m^2}}}}}$

$\:$

$\bf{\dag}\:{\underline{\underline{\sf{\rm{I}\sf{mportant\:Formulae\::}}}}}$

↠ TSA of cube = 6a²

↠ CSA of cube = 4a²

↠ Volume of cube = a³

↠ TSA of cuboid = 2(lb + bh + hl)

↠ CSA of cuboid = 2(l + b)h

↠ Volume of cuboid = l × b × h

↠ TSA of cylinder = 2πr(r + h)

↠ CSA of cylinder = 2πrh

↠ Volume of cylinder = πr²h

↠ Volume of hollow cylinder = πh(R²-r²)

↠ TSA of cone = πr(l + r)

↠ CSA of cone = πrl

↠ Volume of cone = 1/3 πr²h

↠ SA of sphere = 4πr²

↠ Volume of sphere = 4/3 πr³

↠ TSA of hemisphere = 3πr²

↠ CSA of hemisphere = 2πr²

↠ Volume of hemisphere = 2/3 πr³

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