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. • You can solve this with differentiation to find the optimal thickness. A good exercise for students. Nov 25, 2021 at 9:48

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

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()

• Before I try to think upon the answer, I need to be sure we're on the same page. I'm referring to a case where let's say $r_{cr}=1cm$ and we first chose some value of $r_2$, say $r_2=0.5cm$, then plot a graph between Q and r (i.e. we keep on adding the insulation and noting the heat transfer rate). Then again we chose a different value of $r_2= 1cm=r_{cr}$ say, then plot a graph between Q and r. Then again we chose a different value of $r2=2cm$ then plot a graph between Q and r. I wanted to know this variation. How will these graphs look like. Are we on the same page? Nov 25, 2021 at 10:43
• Somewhat like this (not sure if the graph is right, but just to show the idea)- drive.google.com/file/d/1rNe0gxauEablx7j53OQWOJNgQ-WHcd-a/… Nov 25, 2021 at 10:54
• @HarshitRajput I dug my textbook, and it wasn't as I remember. Please check to see if I am making some gross mistake in my derivation. I also put in the code in python to create the plot.
– NMech
Nov 25, 2021 at 17:03