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title: Critical radius of insulation for a cylinder
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# Critical radius of insulation for a cylinder
Unit 1, Lecture 5

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## Critical radius of insulation for a cylinder

- Course: Heat and Mass Transfer (DI05019071)
- Unit 1: Conduction
- Lecture 5: Critical radius of insulation for a cylinder

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Welcome. Today we look at radial systems, specifically cylinders. We will uncover an interesting paradox where adding insulation doesn't always insulate.
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## Lecture Agenda

- 1. Conduction in Radial Systems (Cylinders)
- 2. Thermal Resistance of a Cylinder
- 3. Adding Insulation: Plane Wall vs. Cylinder
- 4. The Competing Effects in a Cylinder
- 5. Definition of Critical Radius
- 6. Mathematical Derivation of $r_c$
- 7. Practical Applications

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We'll start by defining thermal resistance for a cylinder, which differs from a plane wall because the area is not constant. Then we'll derive the critical radius.
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## Conduction through a Hollow Cylinder

- Consider a hollow cylinder (pipe) of length $L$, inner radius $r_1$, outer radius $r_2$.
- Heat flows radially outwards.
- Fourier's Law in cylindrical coordinates (1D radial):
- $$Q = -kA \frac{dT}{dr}$$
- Important: The area $A$ is not constant! It varies with radius $r$.
- Surface area at radius $r$: $A = 2\pi r L$

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Unlike a flat wall where area is constant, heat flowing outward through a pipe passes through an ever-increasing surface area. This changes the math.
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## Thermal Resistance of a Cylinder

- Substitute $A = 2\pi r L$ into Fourier's Law:
- $$Q = -k(2\pi r L) \frac{dT}{dr}$$
- Separate variables and integrate from $r_1$ to $r_2$ and $T_1$ to $T_2$:
- $$Q \int_{r_1}^{r_2} \frac{dr}{r} = -2\pi k L \int_{T_1}^{T_2} dT$$
- $$Q = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}$$
- Thermal Resistance: $R_{cyl} = \frac{\ln(r_2/r_1)}{2\pi k L}$

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The integration of 1/r yields a natural logarithm. Therefore, the thermal resistance of a cylinder is logarithmic, not linear.
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## Adding Insulation: Plane Wall vs Cylinder

- Plane Wall:
- Adding insulation increases thickness ($L$).
- Conduction resistance ($L/kA$) increases.
- Outer area ($A$) stays constant, so convection resistance ($1/hA$) is constant.
- Result: Total resistance ALWAYS increases. Heat loss ALWAYS decreases.
- Cylinder:
- Adding insulation increases outer radius ($r_2$).

<!--
For a flat wall, adding insulation is straightforward. More insulation equals less heat loss, always. But a cylinder is a different story.
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## The Competing Effects in a Cylinder

- When you add insulation to a pipe (increasing outer radius $r_0$):
- 1. Conduction Resistance ($R_{cond} = \frac{\ln(r_0/r_i)}{2\pi k L}$) INCREASES.
-    - Tends to decrease heat transfer.
- 2. Convection Resistance ($R_{conv} = \frac{1}{h(2\pi r_0 L)}$) DECREASES (because outer surface area $A = 2\pi r_0 L$ increases).
-    - Tends to increase heat transfer.
- Which effect dominates?

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This is the core concept. Adding insulation makes the conductive path harder, but it also provides a much larger surface area for the surrounding air to cool. These two effects fight each other.
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## Total Thermal Resistance

- Consider a pipe covered with insulation of outer radius $r$.
- Total Resistance from pipe surface to ambient air:
- $$R_{total} = R_{insulation} + R_{convection}$$
- $$R_{total} = \frac{\ln(r/r_i)}{2\pi k L} + \frac{1}{h(2\pi r L)}$$
- Where $k$ is insulation conductivity and $h$ is outside convection coefficient.
- Heat transfer $Q = \frac{\Delta T}{R_{total}}$.

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To find out what happens to heat transfer, we need to see how the total resistance changes as the outer radius 'r' changes.
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## Finding the Critical Radius

- To find the radius where heat transfer is maximum, we minimize total resistance.
- Differentiate $R_{total}$ with respect to $r$ and equate to zero:
- $$\frac{dR_{total}}{dr} = \frac{1}{2\pi k L r} - \frac{1}{h 2\pi L r^2} = 0$$
- $$\frac{1}{k r} = \frac{1}{h r^2}$$
- Solving for $r$, we get the Critical Radius ($r_c$):
- $$r_c = \frac{k}{h}$$

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By using basic calculus to find the minimum of the resistance function, we arrive at a remarkably simple formula. The critical radius depends only on the insulation's thermal conductivity and the external convection coefficient.
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## Understanding the Critical Radius Graph

- Plot of Heat Transfer ($Q$) vs. Outer Radius ($r$):
- - The curve starts at bare pipe radius $r_i$.
- - It rises to a maximum peak at $r = r_c$.
- - It then continuously decreases for $r > r_c$.
- Behavior:
- If $r_i < r_c$: Adding insulation initially INCREASES heat loss up to $r_c$.
- If $r_i \ge r_c$: Adding insulation DECREASES heat loss immediately.

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This graph is crucial. It shows that if your pipe is very thin (smaller than the critical radius), wrapping it in insulation will actually make it lose more heat until it reaches the critical thickness.
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## Physical Significance

- Why does $Q$ increase initially if $r_i < r_c$?
- Because for small radii, the increase in surface area (decreasing convection resistance) overwhelms the added thermal resistance of the insulation.
- Beyond $r_c$, the added conduction resistance becomes dominant, and heat loss drops.
- Maximum heat loss occurs exactly at $r = r_c$.

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At very small diameters, the geometric effect of increasing surface area is much stronger than the insulating effect of the material.
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## Application: Electrical Cables

- Goal: Maximize heat dissipation (prevent overheating).
- Electrical wires are very thin (radius $r_i$ is very small, usually $r_i < r_c$).
- Adding plastic/rubber insulation ($k$ is small, but $h$ for natural convection is also small, making $r_c = k/h$ larger than the wire).
- Result: The plastic coating on wires actually HELPS cool the wire while providing electrical safety!

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This is a perfect example where we want to operate below the critical radius. The plastic sheath on your phone charger cable helps it stay cool by increasing its surface area.
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## Application: Steam Pipes

- Goal: Minimize heat loss (save energy).
- Steam pipes are large (radius $r_i$ is large, usually $r_i > r_c$).
- Since the bare pipe is already larger than the critical radius, ANY insulation added will immediately decrease heat loss.
- Result: Insulating large pipes is always effective for conserving heat.

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For industrial pipes, the radius is almost always larger than critical radius. So we don't have to worry about accidentally increasing heat loss; insulation will work as intended.
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## Summary & Next Lecture

- Thermal resistance of a cylinder is logarithmic: $R = \ln(r_2/r_1) / (2\pi k L)$.
- Critical radius of insulation for a cylinder is $r_c = k/h$.
- If $r < r_c$, adding insulation increases heat transfer.
- Next time: We will wrap up Unit 1 with Overall Heat Transfer Coefficient analysis for cylindrical systems and look at extended surfaces (Fins).

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To summarize, the critical radius is a fascinating phenomenon governed by the formula k/h. It dictates whether insulation will act as a blanket or a radiator.
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