Critical Radius of Insulation

Question

Consider a long pipe inside which a hot fluid flows. The pipe is exposed to ambient air which is at at some temperature $T_\infty$. Let the inner radius of the pipe is $r_1$ and the outer radius is $r_2$. Let the rate of heat transfer through the pipe when no insulation is added, be $\dot{Q_{bare}}$.

An insulation is now added on the pipe with outer radius of insulation represented by $r$. As $r$ increases from $r_2$, the heat transfer rate first increases up to a certain value of $r=r_{cr}$ (called the critical radius of insulation)and then starts to decrease, until finally the rate of heat transfer falls below $\dot{Q_{bare}}$.

I was interested in knowing what happens to heat transfer rate vs r graph as I keep on increasing $r_2$ by keeping $r_1$ constant.

This question came in my mind when I saw cases where, say, if the critical radius of insulation was 1cm then if $r_2 = 3cm$ the rate of heat transfer would always decrease, if we add an insulation.

Answer 1

According to Cenget (Heat Transfer a Practical Approach), the steady state heat rate in a cylindrical shape of large length L is given by:

$$\dot{Q}_{cond} = S k \Delta T$$ where:

• k is the conductivity coefficient (units: [W/(m.K )])
• $\Delta T = (T_i- T_o)$ is the temperature difference outside and inside (units K)
• S is conduction shape factor (units: m) which for a cylindrical shape of large L is equal to : $$S = \frac{2\pi L}{\ln\left(\frac{D_o}{D_i}\right)}$$

Apart from the conduction, you need to also consider the convection, which has the following formula:

$$\dot{Q}_{cond} = h_{air} A \Delta T_{conv}$$

Where:

• h is the convective heat transfercoefficient(units: [W/(m^2.K )])
• $\Delta T = (T_o - T_\infty)$ is the temperature difference outside and inside (units K)
• A is the area of heat transfer L is equal to $\pi D_o L$:

For the steady state, $\dot{Q}_{cond}= \dot{Q}_{conv}$ should be the same for convective and conductive heat transfer and it should also be equal for the inner side of the wall with temperature $T_i$ up to far away from the pipe where the air temperature is $T_{\infty}$. In that case the total heat resistance can be written as:

$$\dot{Q}_{total}= U\Delta T_t = U\Delta (T_i-T\infty)$$

Therefore we can write:

$$\frac{\dot{Q}_{total}}{U}= T_i-T\infty, \qquad \frac{\dot{Q}_{cond}}{\frac{2\pi L}{\ln\left(\frac{D_o}{D_i}\right)} k }= T_i-T_o,\qquad \frac{\dot{Q}_{conv}}{h_{air} \pi D_o L} = T_o-T_\infty$$

We can therfore write:

$$ T_i-T\infty= (T_i-T_o) + (T_o-T_\infty)$$

i.e.: $$\frac{\dot{Q}_{total}}{U} = \frac{\dot{Q}_{cond}}{\frac{2\pi k L}{\ln\left(\frac{D_o}{D_i}\right)} } + \frac{\dot{Q}_{conv}}{h_{air} \pi D_o L} $$

or

$$\frac{1}{U} = \frac{1}{\frac{2\pi k L}{\ln\left(\frac{D_o}{D_i}\right)} } + \frac{1}{h_{air} \pi D_o L} = \frac{\ln\left(\frac{D_o}{D_i}\right)}{{2\pi k L} } + \frac{1}{h_{air} \pi D_o L} $$

or finally:

$$U = \frac{1}{\frac{\ln\left(D_o/D_i\right)}{{2\pi k L} } + \frac{1}{h_{air} \pi D_o L} } = \frac{2\pi L}{\frac{\ln\left(D_o/D_i\right)}{{k} } + \frac{2}{h_{air} D_o } }$$

Therefore the total heat rate when the temperature inside the pipe wall $T_i$ and the air temperature is known $T_\infty$ can be expressed as:

$$\dot{Q}_t = \frac{2\pi L}{\frac{\ln\left(D_o/D_i\right)}{{k} } + \frac{2}{h_{air} D_o } } (T_i-T_\infty)=\frac{2\pi L k h_{air} D_o }{\ln\left(D_o/D_i\right)h_{air} D_o + 2k } (T_i-T_\infty)$$ Depending on the ratio of h, k ($D_i= 2cm$, $L=1m$) you get different values

h =1, k = 2	h =1, k = 0.5

So bottom line is that I shouldn't trust my memory at my age.

import matplotlib.pyplot as plt
import matplotlib.dates as mdates
import numpy as np

kins=0.5
hair=1
L=1
U = lambda r: 2*np.pi*L/ (np.log(r/1)/kins + 1/(hair*r))

r=np.linspace(1, 20,num= 10000)
plt.plot(r, U(r),'.')

plt.xlabel('$r_o/r_i$')
plt.ylabel('heat rate')
plt.xscale('log')
plt.yscale('log') 
plt.grid()