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title: Numerical Examples on Extended Surfaces
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# Numerical Examples on Extended Surfaces
Unit 1, Lecture 15

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## Numerical Examples on Extended Surfaces

- Course: Heat and Mass Transfer (DI05019071)
- Unit 1: Conduction
- Lecture 15: Problem Solving

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Welcome. Today is all about application. We will take the formulas we've derived over the last few lectures and apply them to solve numerical problems.
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## Lecture Agenda

- 1. Problem Solving Strategy
- 2. Numerical 1: Infinitely Long Fin Application
- 3. Numerical 2: Evaluating Fin Efficiency
- 4. Numerical 3: Justifying Fin Addition (Effectiveness)

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We will start with a general strategy, then move through three distinct problems: an infinite fin, an efficiency calculation, and an effectiveness evaluation.
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## Problem Solving Strategy

- Step 1: Identify geometry (Rectangular, Pin) and extract L, w, t, D.
- Step 2: Calculate Area (Ac) and Perimeter (P).
- Step 3: Calculate the fin parameter 'm' = √(hP/kAc).
- Step 4: Determine the appropriate boundary condition (Infinite, Insulated tip, etc.).
- Step 5: Apply the correct formula for Q_fin or η_fin.

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Always start by calculating Ac and P. Once you have 'm', the rest is just selecting the right formula based on the tip condition.
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## Numerical 1: Problem Statement

- A long copper rod (k = 398 W/mK) of 5 mm diameter extends from a surface at 120°C.
- The ambient air is at 25°C with a convection coefficient of 20 W/m²K.
- Assuming the rod is infinitely long, determine the heat transfer rate from the rod.

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This is a classic infinite fin problem using a cylindrical pin fin geometry.
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## Numerical 1: Step 1 & 2 (Geometry)

- Diameter D = 0.005 m
- Perimeter P = π * D = π * 0.005 = 0.0157 m
- Cross-sectional Area Ac = (π/4) * D² = (π/4) * (0.005)² = 1.963 × 10⁻⁵ m²
- Base excess temp θb = Tb - T∞ = 120 - 25 = 95°C

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First, we convert everything to standard SI units and calculate the geometrical properties of the rod.
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## Numerical 1: Step 3 & 5 (m & Q_fin)

- Calculate m = √(hP / kAc)
- m = √( (20 * 0.0157) / (398 * 1.963×10⁻⁵) ) = √( 0.314 / 0.00781 ) ≈ 6.34 m⁻¹
- For infinite fin: Q_fin = √(hPkAc) * θb
- Alternatively: Q_fin = k * Ac * m * θb
- Q_fin = 398 * 1.963×10⁻⁵ * 6.34 * 95 ≈ 4.70 W

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We calculate m, and then use either of the equivalent infinite fin heat transfer equations. The rod dissipates 4.7 Watts.
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## Numerical 2: Problem Statement

- A rectangular aluminum fin (k = 200 W/mK) is 3 mm thick and 7.5 cm long.
- It is attached to a wall at 300°C. Ambient air is 50°C with h = 10 W/m²K.
- Determine the fin efficiency assuming an insulated tip.

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Now let's look at efficiency for a finite rectangular fin. Assume unit width (w = 1m) for calculations if not given.
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## Numerical 2: Geometry and 'm'

- Assume width w = 1 m. Thickness t = 0.003 m, L = 0.075 m.
- P = 2*(w+t) ≈ 2w = 2 m (since t << w)
- Ac = w*t = 1 * 0.003 = 0.003 m²
- m = √(hP / kAc) = √( (10 * 2) / (200 * 0.003) ) = √(20 / 0.6) = √33.33 ≈ 5.77 m⁻¹
- mL = 5.77 * 0.075 = 0.433

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For thin rectangular fins, the perimeter is dominated by the width, so we approximate P as 2w. We calculate the dimensionless parameter mL.
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## Numerical 2: Calculating Efficiency

- Efficiency formula for insulated tip: η_fin = tanh(mL) / mL
- η_fin = tanh(0.433) / 0.433
- tanh(0.433) ≈ 0.407
- η_fin = 0.407 / 0.433 ≈ 0.94 or 94%

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Using the insulated tip formula, we find the fin is 94% efficient. This means it transfers 94% of the heat it would if the entire fin was at 300 degrees.
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## Numerical 3: Effectiveness Check

- For the same fin in Numerical 2, what is the effectiveness?
- ε_fin = (A_fin / Ac) * η_fin
- A_fin = P * L = 2 * 0.075 = 0.15 m²
- Ac = 0.003 m²
- ε_fin = (0.15 / 0.003) * 0.94 = 50 * 0.94 = 47
- Since 47 >> 2, the use of this fin is highly justified.

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Even though efficiency is 94%, because the fin increases the surface area by a factor of 50, the effectiveness is massive. It's a very good design.
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## Summary

- Systematically break down problems: Geometry -> m parameter -> formula.
- Be careful with units, especially mm vs. m.
- For thin rectangular fins, Perimeter P ≈ 2w.
- Efficiency η and Effectiveness ε are easily linked by the area ratio.

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The key to mastering these problems is a methodical approach and being meticulous with your unit conversions.
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## Next Lecture Preview

- Topic: Unit 2 - Convection
- We are wrapping up Conduction.
- Next, we move to the fluid dynamics of heat transfer: Newton's Law of Cooling in depth.

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This concludes our unit on Conduction. In the next lecture, we transition to Unit 2, where we focus entirely on Convection and the boundary layer.
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