Heat Transfer — Conduction
Conduction is the primary mode of heat transfer in solids and stationary fluids. In GATE, this topic tests your understanding of Fourier’s law, thermal resistance networks, transient conduction, and the Biot/Fourier numbers. Typically 3–6 marks per year appear from this topic, often combined with convection in compound heat transfer problems.
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Fourier’s Law of Conduction $$q = -k \frac{dT}{dx}$$
Where $q$ = heat flux (W/m²), $k$ = thermal conductivity, $dT/dx$ = temperature gradient (negative because heat flows down gradient).
Thermal Conductivity $k$:
- Metals: 10–400 W/m·K (copper ~385, steel ~45)
- Non-metals: 0.02–10 W/m·K (insulation ~0.02–0.05)
- Gases: 0.01–0.1 W/m·K
Plane Wall Conduction (Steady State) $$q = \frac{\Delta T}{R_{th}} = \frac{T_1 - T_2}{L/kA}$$
Where $R_{th} = L/(kA)$ is thermal resistance.
Thermal Resistance Network:
- Series: $R_{eq} = R_1 + R_2 + R_3 + \ldots$
- Parallel: $1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 + \ldots$
Biot Number: $Bi = \frac{hL_c}{k}$ — Ratio of internal conduction resistance to external convection resistance.
- $Bi < 0.1$: Temperature inside body is nearly uniform → Lumped capacitance valid
Fourier Number: $Fo = \frac{\alpha t}{L_c^2}$ — Dimensionless time; ratio of heat conduction rate to heat storage rate.
⚡ Exam Tip: For unsteady problems, always check $Bi < 0.1$ before using lumped capacitance. If $Bi > 0.1$, use Heisler charts or analytical solutions.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Fourier’s Law — The Conduction Equation
Heat conduction occurs due to lattice vibration (phonons) and free electron transport in solids. Fourier’s law is the constitutive equation:
$$\vec{q} = -k \nabla T$$
For one-dimensional steady conduction: $$\frac{d}{dx}\left(k \frac{dT}{dx}\right) = 0$$
When $k$ is constant: $\frac{d^2T}{dx^2} = 0$ → linear temperature profile.
Thermal Conductivity — Physical Interpretation
Metals: High $k$ due to free electrons. Copper (385 W/m·K) > Aluminum (237) > Steel (45).
Insulators: Low $k$ due to air pockets and porous structure. Glass wool, rock wool.
Effect of temperature: $k$ may increase (metals) or decrease (insulators, gases) with temperature. Use mean temperature $k_m = (k_1 + k_2)/2$ for linear variation.
Steady State Conduction — Extended Surfaces (Fins)
Fins increase heat transfer surface area. Fin equation (long fin with insulated tip): $$\frac{d^2T}{dx^2} - m^2(T - T_\infty) = 0$$ Where $m = \sqrt{hP/(kA_c)}$
Fin heat transfer: $$q_{fin} = \sqrt{hPkA_c} \cdot (T_0 - T_\infty) \cdot \tanh(mL)$$
Fin efficiency: $\eta_{fin} = \frac{q_{fin}}{q_{fin,ideal}} = \frac{\tanh(mL)}{mL}$
Where $q_{fin,ideal} = hA_{fin}(T_0 - T_\infty)$, $A_{fin} = PL$ (surface area of fin).
Fin Effectiveness
$$\epsilon_{fin} = \frac{q_{with,fin}}{q_{without,fin}} = \frac{\sqrt{hPkA_c} \cdot \tanh(mL)}{hA_c}$$
Use fins when: $\epsilon > 2$ or when convection coefficient $h$ is small (natural convection, gases).
Cylindrical Geometry — Hollow Cylinder
For a hollow cylinder with inner radius $r_i$, outer radius $r_o$, and length $L$: $$q = \frac{2\pi kL(T_i - T_o)}{\ln(r_o/r_i)}$$
Critical insulation radius: $r_{crit} = k_{insulation}/h$ — adding insulation beyond this radius reduces heat transfer (important for pipe insulation).
Spherical Geometry — Hollow Sphere
$$q = \frac{4\pi k(T_i - T_o)}{(1/r_i) - (1/r_o)}$$
Transient Conduction
Lumped Capacitance Method ($Bi < 0.1$)
When internal conduction resistance is negligible (small $Bi$), temperature is spatially uniform: $$\frac{T - T_\infty}{T_i - T_\infty} = e^{-(hA/\rho Vc) \cdot t}$$
Time constant: $\tau = \rho VC/(hA) = 1/(hA/\rho VC)$
Semi-Infinite Solid
For sudden exposure to convection: $$T(x,t) - T_\infty = (T_i - T_\infty) \cdot \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$
Where $\alpha = k/(\rho c)$ is thermal diffusivity.
Heisler Charts — Finite Bodies
For $Bi > 0.1$, use Heisler/Grober charts:
- Find Fourier number $Fo = \alpha t/L_c^2$
- Find Biot numbers for each dimension
- Read dimensionless temperature $\theta^* = (T - T_\infty)/(T_i - T_\infty)$
⚡ Common Mistake: Confusing $L_c$ (characteristic length) with actual thickness. For a slab: $L_c = V/A = L/2$ for one-sided convection.
Thermal Resistance in Composite Walls
For a composite wall with $n$ layers:
$$R_{total} = \frac{1}{h_1A} + \sum_{i=1}^{n}\frac{L_i}{k_iA} + \frac{1}{h_2A}$$
$$q = \frac{\Delta T_{overall}}{R_{total}}$$
⚡ GATE Tip: For cylindrical pipes, resistance is $\ln(r_o/r_i)/(2\pi kL)$ — don’t forget the $\ln$ term!
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Exact Solution — Transient Conduction
Transient Heat Equation
For one-dimensional conduction with constant properties: $$\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}$$
Boundary conditions:
- Insulated surface: $\partial T/\partial x = 0$ at $x=0$
- Convection: $-k \partial T/\partial x = h(T - T_\infty)$ at $x=L$
Separation of Variables Solution
For a slab initially at $T_i$, suddenly exposed to convection at $x=0$ with insulated $x=L$: $$T(x,t) - T_\infty = \sum_{n=1}^{\infty} A_n \cos(\lambda_n x) e^{-\alpha \lambda_n^2 t}$$
Where eigenvalues $\lambda_n$ satisfy $\tan(\lambda_n L) = Bi/\lambda_n L$.
Infinite and Semi-Infinite Bodies
Infinite solid: Only valid for times such that thermal penetration hasn’t reached boundaries. $$T - T_\infty = (T_i - T_\infty) \cdot \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$
Semi-infinite solid with convection: $$\frac{T - T_\infty}{T_i - T_\infty} = 1 - \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right) + \frac{h\sqrt{\alpha t}}{k} \cdot \text{ierfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$
Contact Resistance
When two solids are joined, imperfect contact creates a temperature discontinuity: $$R_{contact} = \frac{\Delta T}{q}$$
Reducing contact resistance: Increase contact pressure, use thermal interface materials (thermal grease, indium foil).
Thermal Diffusivity — Physical Meaning
$$\alpha = \frac{k}{\rho c}$$
| Material | $\alpha$ (m²/s) |
|---|---|
| Copper | $1.1 \times 10^{-4}$ |
| Aluminum | $9.7 \times 10^{-5}$ |
| Steel | $1.2 \times 10^{-5}$ |
| Air | $2.2 \times 10^{-5}$ |
| Water | $1.4 \times 10^{-7}$ |
High $\alpha$ → heat spreads quickly (copper). Low $\alpha$ → heat stays localized (water is slow).
Dimensionless Numbers — Summary
| Number | Formula | Physical Significance |
|---|---|---|
| Biot $Bi$ | $hL_c/k$ | Internal conduction / External convection resistance |
| Fourier $Fo$ | $\alpha t/L_c^2$ | Dimensionless time |
| Fourier-Biot | $Fo \cdot Bi = ht/k\rho c$ | Combined parameter |
| Thermal resistance | $R_{th} = \Delta T/q$ | Resistance to heat flow |
Example Problem
GATE 2022 (ME) Style: A large steel plate (k = 45 W/m·K, ρ = 7800 kg/m³, c = 500 J/kg·K) of thickness 10 cm, initially at 200°C, is suddenly cooled by convection with air at 30°C (h = 100 W/m²·K). Find the time for the center to reach 100°C.
Solution approach:
- $L_c = V/A = 0.05$ m (half-thickness)
- $Bi = hL_c/k = 100 \times 0.05/45 = 0.111$ → borderline, use Heisler or lumped
- Since $Bi \approx 0.1$, use lumped: $\theta/\theta_i = e^{-hAt/\rho Vc}$
- $t = -\ln(70/170) \times \rho Vc/(hA) = 0.88 \times 7800 \times 0.05 \times 500 / 100 \approx 1715$ s ≈ 28.5 minutes
Check: $Bi = 0.111 > 0.1$ — using exact Heisler chart would give slightly different value. For GATE, state your assumption.
Fins — Extended Analysis
Fin with Convective Tip (not insulated)
$$q_{fin} = \sqrt{hPkA_c} \cdot (T_0 - T_\infty) \cdot \frac{\sinh(mL) + (h/mk)\cosh(mL)}{\cosh(mL) + (h/mk)\sinh(mL)}$$
Fin Array Effectiveness
$$q_{total} = q_{unfinned} + N \cdot q_{fin} \cdot \eta_{fin}$$
Where $N$ = number of fins, $\eta_{fin}$ = fin efficiency.
Previous Year GATE Pattern
| Year | Topic Focus | Marks |
|---|---|---|
| 2023 | Thermal resistance, composite wall | 3 |
| 2022 | Transient conduction, Bi number | 5 |
| 2021 | Fin efficiency, extended surfaces | 2 |
| 2020 | Cylindrical conduction, critical radius | 3 |
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