9.1 Conduction, Convection, and Radiation Fundamentals

9.1 Conduction, Convection, and Radiation Fundamentals

Heat transfer is thermal energy in transit due to a spatial temperature difference. The study of heat transfer differs from thermodynamics: while thermodynamics deals with the end states and the quantity of energy required to transition between them, heat transfer predicts the rate at which energy is exchanged and the temperature distribution within a system. For chemical engineers, mastering the three fundamental modes of heat transfer—conduction, convection, and radiation—is essential for designing reactors, separators, and utility systems, and is a major focus of the FE Chemical exam.

Conduction & Fourier's Law

Conduction is the transfer of energy from more energetic particles of a substance to adjacent, less energetic particles due to interactions between them. In gases and liquids, conduction is due to the collisions and diffusion of molecules during their random motion. In solids, conduction is attributed to lattice vibrations (phonons) and the translation of free electrons.

The fundamental rate equation for conduction is Fourier’s Law. In its general vector form, it states that the heat flux is proportional to the negative gradient of temperature:

$\vec{q}'' = -k \nabla T$

where $\vec{q}''$ is the heat flux vector ($\text{W/m}^2$), and $k$ is the thermal conductivity ($\text{W/(m·K)}$), which is a transport property of the material. For steady-state, one-dimensional heat conduction in a Cartesian coordinate system (e.g., through a plane wall of thickness $L$ and surface area $A$), Fourier's Law integrates to:

$\dot{Q} = \frac{kA(T_1 - T_2)}{L}$

where $\dot{Q}$ is the heat transfer rate ($\text{W}$), and $T_1$ and $T_2$ are the temperatures at the boundaries ($T_1 > T_2$).

Cylindrical Geometries

For radial conduction through a cylindrical wall (such as pipe walls or pipe insulation), the heat transfer area changes with radius ($A = 2\pi r L$). Integrating Fourier's Law in cylindrical coordinates yields:

$\dot{Q} = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}$

where $L$ is the length of the cylinder, and $r_1$ and $r_2$ are the inner and outer radii, respectively.

Thermal Resistance

By analogy to Ohm's law ($I = V/R$), we can define a thermal resistance for conduction ($R_{cond}$):

• Plane wall conduction resistance: $R_{cond} = \frac{L}{kA}$
• Cylindrical shell conduction resistance: $R_{cond, cyl} = \frac{\ln(r_2/r_1)}{2\pi k L}$

Convection & Newton's Law of Cooling

Convection heat transfer is comprised of two mechanisms: energy transfer due to random molecular motion (diffusion) and energy transfer due to the bulk, or macroscopic, motion of the fluid (advection). When a fluid moves over a surface at a different temperature, the heat transfer is governed by Newton's Law of Cooling:

$\dot{Q} = h A (T_s - T_\infty)$

where $h$ is the convective heat transfer coefficient ($\text{W/(m}^2\text{·K)}$), $T_s$ is the surface temperature, and $T_\infty$ is the bulk fluid temperature. The coefficient $h$ is not a thermodynamic property of the fluid; instead, it depends on surface geometry, the nature of fluid motion, fluid properties, and bulk fluid velocity.

Convection is broadly classified into two categories:

1. Forced Convection: The flow is caused by external means, such as a pump, fan, or atmospheric winds.
2. Natural (Free) Convection: The flow is induced by buoyancy forces within the fluid, which arise from density differences caused by temperature variations.

Convective Dimensionless Numbers

To characterize convective heat transfer and scale experimental data, chemical engineers rely on several key dimensionless groups:

For forced convection, correlations typically take the form $Nu = f(Re, Pr)$. For natural convection, correlations take the form $Nu = f(Gr, Pr)$ or $Nu = f(Ra)$.

Radiation & Stefan-Boltzmann Law

Radiation is energy emitted by matter that is at a non-zero temperature. Unlike conduction and convection, radiation does not require the presence of a material medium and propagates at the speed of light via electromagnetic waves.

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation and acts as a perfect emitter. The emissive power of a blackbody is given by the Stefan-Boltzmann Law:

$E_b = \sigma T^4$

where $T$ is the absolute temperature ($\text{K}$ or $R$) and $\sigma$ is the Stefan-Boltzmann constant:

• $\sigma = 5.670 \times 10^{-8} \text{ W/(m}^2\text{·K}^4\text{)}$ in SI units
• $\sigma = 0.1714 \times 10^{-8} \text{ Btu/(h·ft}^2\text{·R}^4\text{)}$ in USCS units

Real surfaces emit less energy than a blackbody at the same temperature. This is accounted for by the emissivity ($\epsilon$), a dimensionless surface property ($0 \le \epsilon \le 1$):

$E = \epsilon \sigma T^4$

View Factors

When analyzing radiation exchange between multiple surfaces, we must account for their orientation relative to one another. The view factor (or shape factor), $F_{ij}$, is defined as the fraction of the radiation leaving surface $i$ that is directly intercepted by surface $j$.

Two fundamental rules govern view factors in an enclosure of $N$ surfaces:

1. Reciprocity Relation: $A_i F_{ij} = A_j F_{ji}$
2. Summation Rule: $\sum_{j=1}^N F_{ij} = 1$

For a flat or convex surface, none of the radiation leaving the surface can strike itself, so $F_{ii} = 0$. For a concave surface, some radiation strikes the surface itself, so $F_{ii} > 0$.

Worked Examples

Worked Example 1: Cylindrical Heat Conduction

A carbon steel pipe ($k = 50 \text{ W/(m·K)}$) with an inner radius $r_1 = 0.04 \text{ m}$ and an outer radius $r_2 = 0.05 \text{ m}$ is covered with a layer of fiberglass insulation ($k = 0.04 \text{ W/(m·K)}$) of thickness $0.03 \text{ m}$ ($r_3 = 0.08 \text{ m}$). The inner surface of the steel pipe is maintained at $220^\circ\text{C}$, and the outer surface of the insulation is at $40^\circ\text{C}$. Calculate the rate of heat loss per meter of pipe length.

Step 1: Identify the thermal resistances in series.

The heat must conduct radially through the pipe wall and then through the insulation. We calculate the resistances per unit length ($R'$):

$R'_{cond, pipe} = \frac{\ln(r_2/r_1)}{2\pi k_{pipe}} = \frac{\ln(0.05 / 0.04)}{2\pi \times 50} = \frac{0.2231}{314.16} \approx 0.00071 \text{ m}\cdot\text{K/W}$ $R'_{cond, ins} = \frac{\ln(r_3/r_2)}{2\pi k_{ins}} = \frac{\ln(0.08 / 0.05)}{2\pi \times 0.04} = \frac{0.4700}{0.2513} \approx 1.870 \text{ m}\cdot\text{K/W}$

Step 2: Sum the resistances to find the total thermal resistance per unit length.

$R'_{total} = R'_{cond, pipe} + R'_{cond, ins} = 0.00071 + 1.870 = 1.8707 \text{ m}\cdot\text{K/W}$

(Note that the thermal resistance of the metal pipe is negligible compared to the insulation layer, which is typical in industrial piping).

Step 3: Apply the radial heat rate equation.

$q' = \frac{T_{inner} - T_{outer}}{R'_{total}} = \frac{220 - 40}{1.8707} \approx 96.2 \text{ W/m}$

The rate of heat loss per meter of pipe length is $96.2 \text{ W/m}$.

Worked Example 2: View Factor Summation and Reciprocity

A long, triangular duct has an equilateral cross-section. Surface 1 is the bottom flat plate, Surface 2 is the left plate, and Surface 3 is the right plate. Find all nine view factors $F_{ij}$ assuming the duct is infinitely long.

Step 1: Identify self-view factors.

Since all three surfaces are flat plates, none of them can see themselves:

$F_{11} = F_{22} = F_{33} = 0$

Step 2: Apply the summation rule for Surface 1.

$F_{11} + F_{12} + F_{13} = 1 \implies F_{12} + F_{13} = 1$

By symmetry of the equilateral shape, the fraction of radiation going to Surface 2 must equal that going to Surface 3:

$F_{12} = F_{13} = 0.5$

Step 3: Determine the remaining view factors by symmetry.

Because the cross-section is an equilateral triangle and the duct is symmetric:

$F_{21} = F_{23} = 0.5$ $F_{31} = F_{32} = 0.5$

Thus, the complete matrix of view factors is:

$F = \begin{bmatrix} 0 \& 0.5 \& 0.5 \\ 0.5 \& 0 \& 0.5 \\ 0.5 \& 0.5 \& 0 \end{bmatrix}$