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- Surface Areas and Volumes - Exercise 13.1
- Surface Areas and Volumes - Exercise 13.2
- Surface Areas and Volumes - Exercise 13.3
- Surface Areas and Volumes - Exercise 13.4
- Surface Areas and Volumes - Exercise 13.5
- Surface Areas and Volumes - Exercise 13.6
- Surface Areas and Volumes - Exercise 13.7
- Surface Areas and Volumes - Exercise 13.8
- Surface Areas and Volumes - Exercise 13.9

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

the total radiating surface in the system. (Assume π=22/7)

Answer:

Height of cylindrical pipe = Length of cylindrical pipe = 28 m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5 cm = 0.025 m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2 x 22/7 x 0.025 x 28 \mathrm{m}^{2}

=4.4 \mathrm{m}^{2}

The area of the radiating surface of the system is 4.4 \mathrm{m}^{2}

"hello students welcome to lino q a video
session
i am saf your match tutor and question
for today is
in a hot water heating system there is a
cylindrical pipe
of length 28 meter and diameter 5
centimeter
find the total radiating surface in the
system
assume phi is equal to 22 by 7
height of the cylindrical pipe will be
equal to the length of the cylindrical
pipe
which is 28 meter it is given over here
that length of the cylindrical pipe is
28 meter therefore this is the height
so
and that is equal to let us say it is
h
radius of the circular end of the pipe
can be found now with the help of the
diameter
we know that the radius is equal to
half of the diameter so r is equal to
radius
diameter of pipe
divided by 2
pi upon 2 so that is
2.5 centimeter
every unit over here is in meter so let
us convert this 2.5 centimeter
in meter so this will give us 0.025
meter now curved surface area of the
cylindrical pipe is 2pi
rh pi into radius into height
so it can be found by substituting
values of r and h
y value over here is 22 by 7 r is
0.025 and h is
28 simplify it
and you will get the answer 4.4 meter
square
so the area of the radiating surface of
the system is
4.4 meter square
this is our final answer if you have any
queries
you can drop it in our comment section
and subscribe to lido for more such q a
thank you for watching
"

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