Question

Q. 6
A thin shell of thickness 't' and radius 'r' has thermal conductivity K.
The thermal resistance is
4xrPK
1
O
2ar K
none
1
Tr²K​

Answer 1

Given that,

Thickness = t

Radius = r

Thermal conductivity $k=\dfrac{K_{0}}{r^2}$

We need to calculate the thermal resistance

Using formula of the thermal resistance

$R= \dfrac{fl}{A}$

Here, $f =\dfrac{1}{k}$

$R= \dfrac{l}{kA}$

Where, l = thickness

A = area

k = thermal conductivity

Put the value into the formula

$R=\dfrac{r^2}{K_{0}}\dfrac{t}{4\pi r^2}$

$R=\dfrac{t}{4\pi K_{0}}$

Hence, The thermal resistance between the inner and outer surface for radial heat flow is $\dfrac{t}{4\pi K_{0}}$