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title: Heat Conduction through Cylinders
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# Heat Conduction through Cylinders
Unit 1, Lecture 8

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## Heat Conduction through Cylinders

- Course: Heat and Mass Transfer (DI05019071)
- Unit 1: Conduction
- Lecture 8: Heat Conduction through Cylinders

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Welcome. Today we move from flat walls to curved surfaces, specifically cylinders.
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## Lecture Agenda

- 1. 1D Conduction in a Cylinder
- 2. Fourier's Law in Radial Direction
- 3. Temperature Distribution (Derivation)
- 4. Logarithmic Temperature Profile
- 5. Thermal Resistance of a Cylinder
- 6. Composite Cylindrical Walls

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We will derive the equations for radial heat flow and see how they differ from the plane wall.
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## 1D Conduction in a Cylinder

- Many applications involve cylindrical geometries: pipes, tubes, wires.
- If a pipe is long and temperatures depend only on radius (r), it's 1D radial conduction.
- Unlike a plane wall, the heat transfer area A = 2πrL changes with radius r.

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The key difference here is that as heat moves outward, the area it flows through increases.
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## Fourier's Law in Radial Direction

- Fourier's Law: qr = -k * A * (dT/dr)
- Substitute A = 2πrL:
- qr = -k * (2πrL) * (dT/dr)
- At steady state, the heat transfer rate qr is constant.

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We use Fourier's law again, but replace the constant area with the radial area formula.
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## Temperature Distribution (Derivation)

- Separate variables and integrate from r1 to r2:
- qr * ∫(1/r) dr = -k * 2πL * ∫dT
- qr * ln(r2/r1) = k * 2πL * (T1 - T2)
- qr = 2πLk * (T1 - T2) / ln(r2/r1)

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Integrating 1/r gives us a natural logarithm. This is the fundamental equation for radial heat transfer.
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## Logarithmic Temperature Profile

- Solving for T(r) gives:
- T(r) = T1 - (T1 - T2) * [ln(r/r1) / ln(r2/r1)]
- The temperature profile is logarithmic, not linear.
- The slope dT/dr decreases as r increases because area increases.

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Because the area gets larger as you move outward, the temperature doesn't need to drop as fast to push the same amount of heat.
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## Thermal Resistance of a Cylinder

- Rearranging the heat rate equation to qr = ΔT / R_th:
- R_th_cyl = ln(r2/r1) / (2πLk)
- This is the radial thermal resistance.
- It depends on the ratio of radii, not just the thickness.

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This is our new building block for thermal circuits involving pipes and tubes.
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## Composite Cylindrical Walls

- Example: A pipe covered with a layer of insulation.
- Inner pipe (r1 to r2, k_pipe) and insulation (r2 to r3, k_ins).
- Just like plane walls, these resistances are in series.
- Total Resistance R_total = R_pipe + R_insulation

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When we insulate a pipe, we are adding a cylindrical resistance in series.
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## Series Circuit for Cylinders

- R_total = [ln(r2/r1) / 2πLk_pipe] + [ln(r3/r2) / 2πLk_ins]
- qr = (T_inner - T_outer) / R_total
- Often, the pipe resistance is very small compared to the insulation and can be neglected for metals.

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The math looks a bit more complex with the logs, but the series circuit principle is identical to the plane wall.
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## Example Problem Introduction

- A steel pipe (k=50 W/mK) with inner radius 50 mm, outer 60 mm.
- Covered with 40 mm of insulation (k=0.05 W/mK).
- Inner surface is 200°C, outer surface is 30°C.
- Calculate heat loss per meter of pipe length (L=1m).

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Let's try calculating the heat loss for a typical insulated steam line.
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## Example Problem Solution

- r1=0.05m, r2=0.06m, r3=0.10m.
- R_pipe = ln(0.06/0.05) / (2π*1*50) = 0.00058 K/W
- R_ins = ln(0.10/0.06) / (2π*1*0.05) = 1.625 K/W
- R_total = 1.6256 K/W
- q/L = (200 - 30) / 1.6256 = 104.6 W/m

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As expected, the metal pipe offers almost no resistance compared to the thick layer of insulation.
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## Summary

- Radial conduction has a non-constant area.
- The temperature profile in a cylinder is logarithmic.
- Cylindrical thermal resistance is ln(r2/r1) / (2πLk).
- Composite cylinders can be analyzed using series thermal circuits.

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Remember the logarithmic nature of cylindrical resistance; it's a fundamental difference from flat walls.
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## Next Lecture Preview

- Topic: Critical Radius of Insulation
- Does adding insulation to a pipe ALWAYS decrease heat loss?
- We'll explore a counter-intuitive phenomenon regarding cylinders and convection.

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Next time, we will ask a surprising question: Can adding insulation actually increase heat loss? The answer is yes, under certain conditions.
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