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title: Heat Flow Through a Rectangular Fin
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# Heat Flow Through a Rectangular Fin
Unit 1, Lecture 12

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## Heat Flow Through a Rectangular Fin

- Course: Heat and Mass Transfer (DI05019071)
- Unit 1: Conduction
- Lecture 12: Deriving the Fin Equation

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Welcome. Today we will establish the mathematical foundation for fin analysis by deriving the governing differential equation.
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## Lecture Agenda

- 1. Standard Assumptions for Fin Analysis
- 2. The Rectangular Fin Geometry
- 3. Energy Balance on a Differential Element
- 4. Derivation of the Governing Equation
- 5. The Fin Parameter (m)

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We will start with the assumptions to simplify our model, then apply the principle of conservation of energy to a small element of the fin.
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## Assumptions in Fin Analysis

- 1. Steady-state heat conduction.
- 2. One-dimensional heat flow (temperature varies only along the length 'x').
- 3. Constant thermal conductivity (k) of the fin material.
- 4. Uniform convection coefficient (h) over the entire fin surface.
- 5. No internal heat generation.
- 6. Negligible radiation (or combined into an effective 'h').

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To make the mathematics tractable, we assume 1D flow. Since the fin is thin, temperature variation across its thickness is negligible compared to along its length.
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## Geometry of a Rectangular Fin

- Consider a straight rectangular fin:
- Length = L, Width = w, Thickness = t.
- Cross-sectional Area (Ac) = w * t.
- Perimeter (P) = 2 * (w + t).
- Base temperature = Tb, Ambient temperature = T∞.

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Here is our setup. The fin extends in the x-direction. We'll focus on a tiny slice of the fin of thickness dx.
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## Energy Balance Formulation

- Take a differential element of thickness 'dx' at distance 'x'.
- Energy In (Conduction): Qx
- Energy Out (Conduction): Qx+dx
- Energy Out (Convection): Qconv
- Steady State Balance: Qx = Qx+dx + Qconv

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By conservation of energy, the heat conducted into the element must equal the heat conducted out plus the heat convected away from its surface.
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## Applying Fourier's and Newton's Laws

- From Fourier's Law:
- Qx = -k * Ac * (dT/dx)
- Qx+dx = Qx + (d/dx)(Qx) * dx
- From Newton's Law of Cooling:
- Qconv = h * dAs * (T - T∞)
- Where surface area dAs = P * dx (Perimeter * thickness).

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We substitute the rate equations into our energy balance. Notice that the convective area is the perimeter times dx.
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## Deriving the Differential Equation

- Substitute into balance: Qx = [Qx + (d/dx)(Qx) * dx] + h*P*dx*(T - T∞)
- Simplify: -(d/dx)(Qx) * dx = h*P*dx*(T - T∞)
- Divide by dx: -(d/dx)(-k*Ac * dT/dx) = h*P*(T - T∞)
- k * Ac * (d²T/dx²) = h*P*(T - T∞)

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After rearranging and substituting Fourier's law, we arrive at a second-order differential equation for temperature.
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## The General Governing Equation

- Rearranging the terms:
- d²T/dx² - (h*P / k*Ac) * (T - T∞) = 0
- To simplify, we introduce the excess temperature θ = T - T∞.
- Since T∞ is constant, dθ/dx = dT/dx.
- Equation becomes: d²θ/dx² - (h*P / k*Ac) * θ = 0

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Defining the excess temperature theta makes the differential equation homogeneous and easier to solve.
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## The Fin Parameter (m)

- Let m² = h*P / k*Ac
- The parameter 'm' has units of m⁻¹.
- The equation simplifies to:
- d²θ/dx² - m²θ = 0
- This is a second-order, linear, homogeneous ordinary differential equation.

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The term m combines all the geometric and thermal properties of the fin into a single constant. It determines how rapidly temperature drops along the fin.
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## General Solution of the Equation

- The general solution for d²θ/dx² - m²θ = 0 is:
- θ(x) = C1 * e^(mx) + C2 * e^(-mx)
- Where C1 and C2 are constants of integration.
- To find C1 and C2, we need two boundary conditions.

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This exponential solution form tells us how temperature decays along the fin. We need specific physical conditions at the base and the tip to evaluate C1 and C2.
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## Summary

- Assumed 1D steady-state heat conduction.
- Derived energy balance on a small dx element.
- Obtained the governing equation: d²θ/dx² - m²θ = 0.
- The fin parameter m relies on geometry and material properties.

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We now have the general mathematical framework. The next step is applying boundary conditions based on the fin's length.
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## Next Lecture Preview

- Topic: Analysis of an Infinitely Long Fin
- We will apply specific boundary conditions to solve for C1 and C2.
- We'll evaluate the heat transfer rate for a very long fin.

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Next lecture, we will look at the simplest case: a fin that is so long that its tip temperature reaches the ambient fluid temperature.
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