Question

Cylindrical wall of length 2m and inner , outer radius are (50 cm, 60 cm) and the thermal conduction coefficient of its material is 20 W/m.K and the termprature of the surrouding medium is 15 C and the h = 25 w/mk

find dQ/dt total

Answer 1

What is the temperature on the other side of the wall?
Can you define h?

Answer 2

h = convective heat transfer coefficient
You can find the temperature on the other side
using Fourier equation:

dQ/dt = -kA dTheta/dx

k = 20 W/m.k (given)




I am stuck on finding the surface area of the shape (A) in heat conviction.
is it

2pi R L or ( 2piRL + 2piR^2 - 2pir^2 )

Answer 3

I think that, here we have to use cylindrical coordinates, so the formula above can be rewritten as below:
\[\huge \frac{{dQ}}{{dt}} = h4\pi {r^2}\frac{{dT}}{{dr}}\]

Answer 4

If the pipe is cylindrical, it should read:
\(\large \dfrac{{dQ}}{{dt}} = -\lambda \;2\pi rl\dfrac{{dT}}{{dr}}\)

@TrojanPoem:
h and \(\lambda\) have different units:
h in W.m\(^{-2}\).K\(^{-1}\)
\(\lambda\) in W.m\(^{-1}\).K\(^{-1}\)

Answer 5

I persist in saying that the inner temperature has to be known in order to find out the thermal power going across the walls of the pipe ;-)

Answer 6

A good approximation is to ignore the curvature of the pipe.
Then thermal resistances are easier to work out.
\(R_{thermal}=\dfrac{1}{2\pi R_{1/2}l}(\dfrac{1}{h}+\dfrac{R_2-R_1}{\lambda}+\dfrac{1}{h})\) where \(R_{1/2}\) is a mean value of \(R_1\) and \(R_2\).

Answer 7

Then \(\large \dfrac{{dQ}}{{dt}} = \dfrac{{\Delta T}}{{R_{thermal}}}\)

Answer 8

@Vincent-Lyon.Fr , Totally correct.

Answer 9

my formula is wrong, since I was thinking about a thermal dispersion by means of a spherical object

Answer 10

@Michele_Laino , why should we use 2 pi r l as the surface Area ?