Thermal Resistance and Critical Radius of Insulation

There is an analogy between thermal and electrical resistance concept. The analog of  heat flow rate $\dot{Q}$ is current (I) and the analog of temperature difference is voltage difference.

we know that $I = \frac{ΔV}{R}$     similarly,     $\dot{Q} = \frac{ΔT}{Rth}$

Conduction Resistance

• Slab of thickness L

By Fourier equation,

\[ \dot{Q} = \frac{kAΔT}{L}\]    

If we compare with \( \dot{Q} = \frac{T1- T2}{Rth}\)

$\displaystyle Rth = \frac{L}{kA}$

k is thermal conductivity 

A is Area of cross section

• Hallow Cylinder

Assumptions:

1. Steady state, 1D heat flow in radial direction

2. k is uniform

r1 and r2 are inner and outer radius of cylinder.

T1 and T2 inner and out surface temperature.

L is the length of cylinder.

heat flow rate $\displaystyle \dot{Q} =- k A \frac{dT}{dr}$

or,     $\displaystyle \dot{Q} =- k 2 \pi rL \frac{dT}{dr}$

$\displaystyle \ \frac{\dot{Q} dr}{r} =- k 2 \pi L dT$

integrating from $ T = T_1$ at $ r = r_1$ to  $ T = T_2$ at $ r = r_2$ yields,

$\displaystyle \dot{Q}\ln(r_2/r_1) = 2 \pi kL (T_2-T_1)$

or,    $\displaystyle \dot{Q} = 2 \pi kL\frac{T_2-T_1}{\ln(r_2/r_1)}$

thus, thermal resistance      $\displaystyle R = \frac{\ln(r_2 / r_1)}{2\pi kL}$

• Hallow Sphere

Assumptions:

1. Steady state, 1D heat flow only in radial direction    

2. k is uniform

r1 and r2 are inner and outer radius of sphere.

T1 and T2 inner and out surface temperature.

heat flow rate $\displaystyle \dot{Q} =- k A \frac{dT}{dr}$

or,     $\displaystyle \dot{Q} =- k 4 \pi r^2 \frac{dT}{dr}$

$\displaystyle \ \frac{\dot{Q} dr}{r^2} =- k 4 \pi dT$

integrating from $ T = T_1$ at $ r = r_1$ to  $ T = T_2$ at $ r = r_2$ yields,

$\displaystyle \dot{Q}\frac{(r_2 - r_1)}{r_1 r_2} = 4 \pi k (T_2-T_1)$

or,    $\displaystyle \dot{Q} = 4 \pi k \frac{(T_2-T_1) r_1 r_2}{(r_2 - r_1)}$

thus thermal resistance      $\displaystyle R = \frac{(r_2 - r_1)}{4\pi k r_1 r_2}$

Convection Thermal Resistance

By Newton's law of cooling,  $\displaystyle \dot{Q} = h A(T_w - T_\infty)$

or, $\displaystyle \dot{Q} = \frac{(T_w - T_\infty) }{\frac{1 }{h A }}$   

thus thermal resistance in convection is      $\displaystyle Rconv = \frac{1}{h A}$

Critical Radius of Insulation

Assumptions:

1. Steady state condition

2. 1D heat flow only in radial direction     

3. Negligible thermal resistance due to cylinder wall

As we add insulation, intially thermal resistance (R) 

decreases since r2 increases that will increases

heat transfer rate. At some critical thickness $ r_{cr} $ thermal resistance increases again and heat transfer is reduced. 

To find $ r_{cr} $, set total thermal resistance  $\displaystyle  \frac{dR}{dr} = 0. $

$\displaystyle R = \frac{1}{2\pi r L h} +  \frac{\ln(r/r_1)}{2\pi k L}$

$\displaystyle \frac{dR}{dr} =  \frac{1}{2\pi k r L} - \frac{1}{2\pi r^2 L h} = 0$

thus, Critical radius $\displaystyle r_{cr} = \frac{k}{ h}$

similarly for sphere,  $\displaystyle r_{cr} = \frac{2k}{ h}$

• For insulation thickness less than that $ r_{cr} $ heat loss increases with increasing r and after $ r_{cr} $ heat loss decreases with increasing r.

Insulation Critical Radius