Interactive Lecture
Conduction, Convection, and Radiation Heat Transfer
by Dr Arnaldo Delli Carri
Heat transfer is the movement of thermal energy from one object or material to another due to a temperature difference. It is a fundamental concept in engineering, governing everything from building insulation to electronic cooling systems. Heat naturally flows from regions of higher temperature to regions of lower temperature until thermal equilibrium is reached.
Heat moves by three mechanisms — conduction (direct contact), convection (fluid motion), and radiation (electromagnetic waves). Most engineering problems combine two or three of them at once.
Conduction is heat transfer through a solid material or between materials in direct contact. It occurs when faster-moving molecules (at higher temperature) collide with slower-moving molecules (at lower temperature), transferring kinetic energy.
Fourier's Law of Heat Conduction:
\[Q = -kA\frac{dT}{dx}\]
For steady-state conduction through a plane wall:
\[Q = \frac{kA(T_1 - T_2)}{L}\]
Where:
Dividing through by area gives the heat flux, the heat rate per unit area:
\[q = \frac{Q}{A} \text{ (W/m}^2\text{)}\]
Thermal Conductivity: Different materials have vastly different thermal conductivities:
Thermal Resistance (Conduction):
\[R_{cond} = \frac{L}{kA}\]
Using thermal resistance analogy (similar to electrical resistance):
\[Q = \frac{\Delta T}{R_{total}}\]
Convection is heat transfer between a solid surface and a moving fluid (liquid or gas). It combines conduction within the fluid and the bulk motion of the fluid.
Newton's Law of Cooling:
\[Q = hA(T_s - T_\infty)\]
Where:
Types of Convection:
Thermal Resistance (Convection):
\[R_{conv} = \frac{1}{hA}\]
Radiation is heat transfer through electromagnetic waves, requiring no medium. All objects emit thermal radiation proportional to the fourth power of their absolute temperature.
Stefan-Boltzmann Law:
\[Q = \varepsilon \sigma A T^4\]
For radiation exchange between two surfaces:
\[Q = \varepsilon \sigma A (T_A^4 - T_B^4)\]
Where:
Emissivity Values:
In most engineering applications, heat transfer occurs through multiple modes simultaneously. For example, heat loss from a building involves conduction through the wall, convection at the surfaces, and possibly radiation.
Overall Heat Transfer Coefficient:
For a composite wall with convection at both surfaces:
\[Q = UA\Delta T\]
Where \(U\) is the overall heat transfer coefficient:
\[\frac{1}{UA} = R_{total} = R_{conv,1} + R_{cond,1} + R_{cond,2} + \cdots + R_{conv,2}\]
\[\frac{1}{U} = \frac{1}{h_1} + \frac{L_1}{k_1} + \frac{L_2}{k_2} + \cdots + \frac{1}{h_2}\]
Explore how different materials, thicknesses, and boundary conditions affect heat transfer. Observe temperature distribution and heat flux through a composite wall.
In "All Three Modes", the sim treats the combined conduction/convection path and the radiation path as parallel routes between the same two boundary temperatures: \(Q_{total} = Q_{cond+conv} + Q_{rad}\), each driven by the full \(\Delta T\). The infobox reports an equivalent resistance \(R = \Delta T / Q_{total}\) — a lumped value, not a series network with a separate radiation resistance.
| Concept | Equation | Description |
|---|---|---|
| Fourier's Law (Conduction) | \(Q = -kA\frac{dT}{dx}\) | Heat conduction rate |
| Conduction (Plane Wall) | \(Q = \frac{kA(T_1 - T_2)}{L}\) | Steady-state heat transfer through wall |
| Thermal Resistance (Cond.) | \(R_{cond} = \frac{L}{kA}\) | Conduction thermal resistance |
| Newton's Law of Cooling | \(Q = hA(T_s - T_\infty)\) | Convective heat transfer |
| Thermal Resistance (Conv.) | \(R_{conv} = \frac{1}{hA}\) | Convection thermal resistance |
| Stefan-Boltzmann Law | \(Q = \varepsilon\sigma A(T_A^4 - T_B^4)\) | Radiation heat transfer |
| Overall Heat Transfer | \(Q = UA\Delta T\) | Combined heat transfer modes |
| Overall HT Coefficient | \(\frac{1}{U} = \frac{1}{h_1} + \frac{L_1}{k_1} + \frac{L_2}{k_2} + \cdots + \frac{1}{h_2}\) | Series thermal resistances (multi-layer) |
| Thermal Diffusivity | \(\alpha = \frac{k}{\rho c_p}\) | Rate of temperature change |
| Biot Number | \(Bi = \frac{hL_c}{k}\) | Ratio of internal to surface resistance |
A concrete wall (k = 1.4 W/m·K) is 20 cm thick. The inside surface temperature is 25°C and the outside surface temperature is 5°C. Calculate:
Show Solution
Solution:
Given:
(a) Heat flux:
Using Fourier's law:
\[q = \frac{k(T_1 - T_2)}{L} = \frac{1.4 \times (25 - 5)}{0.2}\]
\[q = \frac{1.4 \times 20}{0.2} = \frac{28}{0.2} = 140 \text{ W/m}^2\]
(b) Total heat transfer:
\[Q = q \times A = 140 \times 15 = 2100 \text{ W} = 2.1 \text{ kW}\]
Answers:
A hot plate at 150°C is cooled by air at 25°C flowing over it. The convection heat transfer coefficient is h = 20 W/m²·K. If the plate area is 0.5 m², calculate:
Show Solution
Solution:
Given:
(a) Heat transfer rate:
Using Newton's law of cooling:
\[Q = hA(T_s - T_\infty) = 20 \times 0.5 \times (150 - 25)\]
\[Q = 10 \times 125 = 1250 \text{ W} = 1.25 \text{ kW}\]
(b) Convective thermal resistance:
\[R_{conv} = \frac{1}{hA} = \frac{1}{20 \times 0.5} = \frac{1}{10} = 0.1 \text{ K/W}\]
Verification: \(Q = \frac{\Delta T}{R} = \frac{125}{0.1} = 1250 \text{ W}\) ✓
Answers:
A composite wall consists of:
Inside convection: h₁ = 10 W/m²·K, T∞,1 = 22°C
Outside convection: h₂ = 25 W/m²·K, T∞,2 = -5°C
Calculate:
Show Solution
Solution:
Given:
(a) Overall heat transfer coefficient:
\[\frac{1}{U} = \frac{1}{h_1} + \frac{L_1}{k_1} + \frac{L_2}{k_2} + \frac{1}{h_2}\]
\[\frac{1}{U} = \frac{1}{10} + \frac{0.15}{0.7} + \frac{0.05}{0.04} + \frac{1}{25}\]
\[\frac{1}{U} = 0.1 + 0.214 + 1.25 + 0.04 = 1.604 \text{ m}^2\text{·K/W}\]
\[U = \frac{1}{1.604} = 0.623 \text{ W/m}^2\text{·K}\]
(b) Heat flux:
\[q = U(T_{\infty,1} - T_{\infty,2}) = 0.623 \times (22 - (-5)) = 0.623 \times 27 = 16.8 \text{ W/m}^2\]
(c) Interface temperature:
From inside to interface:
\[q = \frac{T_{\infty,1} - T_{interface}}{R_1 + R_{brick}}\]
Where \(R_1 = 0.1\) and \(R_{brick} = 0.214\)
\[16.8 = \frac{22 - T_{interface}}{0.1 + 0.214}\]
\[16.8 = \frac{22 - T_{interface}}{0.314}\]
\[22 - T_{interface} = 16.8 \times 0.314 = 5.28\]
\[T_{interface} = 22 - 5.28 = 16.72°C\]
Answers:
A radiator surface at 80°C (ε = 0.9, A = 2 m²) radiates to surroundings at 20°C. Calculate the net radiative heat transfer rate.
Show Solution
Solution:
Given:
Stefan-Boltzmann requires absolute temperature, so convert to Kelvin first:
\[T_A = 80 + 273.15 = 353.15 \text{ K}, \qquad T_B = 20 + 273.15 = 293.15 \text{ K}\]
\[Q = \varepsilon \sigma A (T_A^4 - T_B^4)\]
\[T_A^4 = 353.15^4 = 1.555 \times 10^{10} \text{ K}^4, \qquad T_B^4 = 293.15^4 = 7.385 \times 10^{9} \text{ K}^4\]
\[Q = 0.9 \times 5.67 \times 10^{-8} \times 2 \times (1.555 \times 10^{10} - 7.385 \times 10^{9})\]
\[Q = 0.9 \times 5.67 \times 10^{-8} \times 2 \times 8.169 \times 10^{9} = 833.7 \text{ W}\]
Answer: Q ≈ 834 W (0.834 kW)