Heat Transfer · Chapter 2
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Heat Transfer · Chapter Two

Heat Conduction Equation

A temperature field is the record left by diffusion, storage, and internal generation. The conduction equation is their local energy balance.

We build that balance, choose a coordinate system and boundary conditions, and audit simplified solutions before trusting them.

Learning objectives

  • Classify a conduction field by dimensionality, time dependence, and generation.
  • Derive and simplify the conduction equation without deleting physically active terms.
  • Translate thermal descriptions into temperature, flux, convection, symmetry, and initial conditions.
  • Solve steady one-dimensional constant- and variable-property fields.
  • Verify solutions using units, boundary conditions, extrema, and global energy conservation.
2–1

Introduction

A general field is \(T(x,y,z,t)\). “Steady” means every fixed point has \(\partial T/\partial t=0\), not that temperature is spatially uniform. A one-dimensional model is justified when geometry and boundary conditions make gradients in the omitted directions negligible, usually away from edges.

Field, flux, source, and storage

Fourier’s vector law is \(\mathbf q''=-k\nabla T\), where \(\mathbf q''\) is W/m². A volumetric source \(\dot q''' \,[\mathrm{W/m^3}]\) adds energy throughout matter; \(\rho c_p\,\partial T/\partial t\) stores it. Conductivity \(k\) should be evaluated over the actual material temperature range, not automatically at room temperature.

Total heat generation in a volume
\[ \dot Q_{\rm gen}=\int_V\dot q'''\,dV=\dot q'''V\quad\text{only when }\dot q'''\text{ is uniform}. \]

Physical meaning

The gradient sets direction and flux; curvature reveals whether distributed generation or transient storage is reshaping the field.

2–2

One-Dimensional Heat Conduction Equation

For a differential element, conduction in minus conduction out plus generation equals storage. Dividing by volume and taking the limit produces the field equation. Geometry enters through the divergence operator because flow area changes with position.

Plane wall (constant \(k\))
\[ \frac{\partial^2 T}{\partial x^2}+\frac{\dot q'''}{k}=\frac{1}{\alpha}\frac{\partial T}{\partial t},\qquad \alpha=\frac{k}{\rho c_p}. \]
Long cylinder (radial)
\[ \frac1r\frac{\partial}{\partial r}\!\left(r\frac{\partial T}{\partial r}\right)+\frac{\dot q'''}k=\frac1\alpha\frac{\partial T}{\partial t}. \]
Sphere (radial)
\[ \frac1{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial T}{\partial r}\right)+\frac{\dot q'''}k=\frac1\alpha\frac{\partial T}{\partial t}. \]
These forms assume isotropic, constant \(k,\rho,c_p\). With variable \(k\), retain \(\nabla\!\cdot(k\nabla T)\).
Key Idea The three equations differ only in the conduction (divergence) term, which carries the geometry. Cylindrical and spherical forms gain \(1/r\) and \(1/r^2\) factors because the cross-sectional area grows with radius.
2–3

General Heat Conduction Equation

The safest general statement is \(\nabla\!\cdot(k\nabla T)+\dot q'''=\rho c_p\,\partial T/\partial t\). Constant isotropic \(k\) permits division by \(k\):

Three-dimensional, constant \(k\) (rectangular)
\[ \nabla^2T+\frac{\dot q'''}k=\frac1\alpha\frac{\partial T}{\partial t}. \]

Named Special Cases

Steady with a source is Poisson’s equation; steady and source-free is Laplace’s equation; transient and source-free is the diffusion equation. Terms may be removed only after a physical argument, not because the resulting equation is familiar.

Laplace Equation (steady, no generation)

When conduction is steady and there is no internal heat generation, the temperature field satisfies \(\nabla^2 T = 0\). Its solutions are harmonic functions, with extrema occurring only on the boundaries of the region.

\[ \nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} = 0 \]

2–4

Boundary and Initial Conditions

The differential equation describes a family of fields. Boundary conditions describe how the body meets its surroundings; a transient model also requires \(T(\mathbf x,0)\). The outward-normal convention prevents sign errors.

Specified Temperature (Dirichlet)

Prescribed temperature: \(T=T_s\). It represents a sufficiently large thermal reservoir or controlled boundary.

Specified Heat Flux (Neumann)

With outward normal \(n\), heat leaving is \(-k\,\partial T/\partial n=q_s''\). Adiabatic and symmetry boundaries have zero normal gradient, but symmetry does not imply the entire body is insulated.

Convection Boundary Condition

At a convection boundary, \(-k\,\partial T/\partial n=h(T_s-T_\infty)\). Here \(h\) comes from an applicable correlation or measurement and fluid properties are normally evaluated at the correlation’s stated reference temperature.

Radiation & Interface Conditions

Radiation is nonlinear: \(-k\,\partial T/\partial n=\varepsilon\sigma(T_s^4-T_{\rm surr}^4)\), with absolute temperatures. Perfect interfaces have continuous temperature and flux; contact resistance permits a temperature jump.

Initial condition (transient problems)
\[ T(x,0) = T_i(x) \]
2–5

Solution of Steady One-Dimensional Problems

For a steady, constant-\(k\), source-free plane wall, \(d^2T/dx^2=0\); two integrations and two independent boundary conditions produce a linear profile. A cylindrical profile is logarithmic because radial area grows.

Linear temperature profile in a plane wall
\[ T(x) = T_1 + (T_2 - T_1)\frac{x}{L}, \qquad \dot{Q} = kA\frac{T_1 - T_2}{L} \]
Key Idea In steady 1-D conduction with no generation, the heat rate \(\dot{Q}\) is the same through every cross-section. In a plane wall the profile is linear; in a cylinder it is logarithmic; in a sphere it varies as \(1/r\).
EXAMPLE 1Temperature inside a sensor window
Problem. A 12-mm polymer optical window has \(k=0.24\ \mathrm{W/(m\,K)}\), area \(0.018\ \mathrm{m^2}\), and faces maintained at 68°C and 32°C. Find \(T\) 4 mm from the hot face and the signed heat rate in the positive \(x\)-direction.

Strategy

Fixed face temperatures, constant properties, steady one-dimensional plane conduction imply the source-free wall equation.

Givens

  • \(L=0.012\) m, \(x=0.004\) m, \(A=0.018\) m², \(k=0.24\) W/(m·K), \(T(0)=68^\circ\)C, \(T(L)=32^\circ\)C.

Physical/model interpretation

Edge effects and generation are neglected; \(+x\) points from hot to cold. Constant area makes the profile linear.

Step-by-step solution
1. Simplify. \(d^2T/dx^2=0\), so \(T=C_1x+C_2\).
2. Apply boundaries. \(C_2=68^\circ\mathrm C\), \(C_1=(32-68)/0.012=-3000\ \mathrm{K/m}\).
3. Evaluate temperature. \(T(0.004)=68-3000(0.004)=56.0^\circ\mathrm C\).
4. Evaluate rate. \(\dot Q_x=-kA\,dT/dx=-(0.24)(0.018)(-3000)=12.96\ \mathrm W\).

Unit/sign/limit checks

Units: In \(-kA\,dT/dx\), [W/(m·K)]m²(K/m) reduces to W; the computed 12.96 therefore has heat-rate rather than heat-flux units.

Signs: The gradient is negative while Fourier’s law carries a minus sign, yielding a positive rate in \(+x\), consistent with heat flowing from 68°C toward 32°C.

Limit: Since \(x/L=1/3\), 56°C is exactly one-third through the 36 K face-to-face drop; it also lies between both prescribed boundary temperatures.

Final answer: \(\boxed{T=56.0^\circ\mathrm C,\ \dot Q_x=+13.0\ \mathrm W}\)

Engineering insight

The interior temperature follows directly from position only because \(k\) and area are constant; a graded material would curve this profile.

2–6

Heat Generation in a Solid

Uniform generation produces a concave-down temperature profile in a plane wall cooled at its surfaces. Symmetry fixes zero slope at the midplane, while a surface condition fixes the temperature level.

Surface temperature with uniform generation
\[ T_s=T_\infty+\frac{\dot q'''V}{hA_s} \]
Maximum (center) temperature rise
\[ T_0-T_s=\frac{\dot q'''L^2}{2k}\ {\rm(wall)},\quad\frac{\dot q'''r_o^2}{4k}\ {\rm(cylinder)},\quad\frac{\dot q'''r_o^2}{6k}\ {\rm(sphere)}. \]
\(L\) is the wall half-thickness; \(r_o\) is the outer radius. The peak occurs at the center / midplane.
EXAMPLE 2Peak temperature in a battery module slab
Problem. A symmetric 40-mm-thick battery module is modeled as a slab with \(k=1.6\ \mathrm{W/(m\,K)}\) and uniform generation \(1.8\times10^4\ \mathrm{W/m^3}\). Both faces convect to 24°C coolant with \(h=55\ \mathrm{W/(m^2K)}\). Find surface and midplane temperatures.

Strategy

Symmetry reduces the domain to half-thickness \(L=0.020\) m; global generation first determines surface flux, then the parabolic solution gives the peak.

Givens

  • \(L=0.020\) m, \(\dot q'''=1.8\times10^4\) W/m³, \(k=1.6\) W/(m·K), \(h=55\) W/(m²·K), \(T_\infty=24^\circ\)C.

Physical/model interpretation

Steady 1-D constant-property conduction; identical cooling on both faces; contact and edge effects omitted.

Step-by-step solution
1. Surface flux. Per unit face area, \(q_s''=\dot q'''L=(1.8\times10^4)(0.020)=360\ \mathrm{W/m^2}\).
2. Surface condition. \(T_s-T_\infty=q_s''/h=360/55=6.545\) K, so \(T_s=30.55^\circ\)C.
3. Interior rise. \(T_0-T_s=\dot q'''L^2/(2k)=18000(0.020)^2/(3.2)=2.25\) K.
4. Peak. \(T_0=30.55+2.25=32.80^\circ\)C.

Unit/sign/limit checks

Units: \([\mathrm{W/m^3}]\,\mathrm{m^2}/[\mathrm{W/(m\,K)}]=\mathrm K\), and \(q_s''/h\) gives [W/m²]/[W/(m²·K)] = K.

Signs: Positive internal generation requires outward surface flux and produces the physically necessary ordering \(T_0>T_s>T_\infty\).

Limit: As \(\dot q'''\to0\), surface and center temperatures both approach coolant temperature; increasing \(k\) reduces only the center-to-surface rise.

Final answer: \(\boxed{T_s=30.5^\circ\mathrm C,\ T_{\max}=32.8^\circ\mathrm C}\)

Engineering insight

Most rise occurs across the coolant film (6.55 K), so increasing \(h\) helps more than a modest conductivity improvement; pack validation must still include contact nonuniformity.

2–7

Variable Thermal Conductivity, \(k(T)\)

When \(k\) varies appreciably, Fourier’s law integrates in temperature. The exact average below preserves the steady plane-wall rate; evaluating \(k\) at mean temperature is exact only for linear \(k(T)\).

Average thermal conductivity
\[ k_{\text{avg}} = \frac{1}{T_2 - T_1}\int_{T_1}^{T_2} k(T)\,dT \]
Plane-wall heat rate with \(k(T)\)
\[ \dot{Q} = k_{\text{avg}}\,A\,\frac{T_1 - T_2}{L} = k_0\!\left(1 + \beta\,\frac{T_1+T_2}{2}\right) A\,\frac{T_1 - T_2}{L} \]
EXAMPLE 3Variable conductivity of a furnace sight-port liner
Problem. A 30-mm ceramic liner of area 0.060 m² has faces at 800°C and 200°C. Supplied test data are represented by \(k(T)=1.20+8.0\times10^{-4}T\) W/(m·K), with \(T\) in °C. Determine the steady heat rate and compare it with using room-temperature \(k=1.22\) W/(m·K).

Strategy

The wide temperature interval and explicit \(k(T)\) cue the integral conductivity model, not a room-temperature lookup.

Givens

  • \(L=0.030\) m, \(A=0.060\) m², \(T_h=800^\circ\)C, \(T_c=200^\circ\)C, supplied \(k(T)\).

Physical/model interpretation

Steady one-dimensional, source-free conduction; positive heat is hot-to-cold. The empirical property fit is assumed valid over 200–800°C.

Step-by-step solution
1. Average property. \(k_{\rm avg}=(T_h-T_c)^{-1}\int_{T_c}^{T_h}(1.20+0.0008T)dT\).
2. Integrate. Since the law is linear, \(k_{\rm avg}=1.20+0.0008(500)=1.60\ \mathrm{W/(m\,K)}\).
3. Rate. \(\dot Q=k_{\rm avg}A(T_h-T_c)/L=(1.60)(0.060)(600)/0.030=1920\) W.
4. Compare. \(\dot Q_{RT}=1.22(0.060)(600)/0.030=1464\) W, an underprediction of \((1920-1464)/1920=23.8\%\).

Unit/sign/limit checks

Units: \([\mathrm{W/(m\,K)}]\mathrm{m^2}\mathrm K/\mathrm m=\mathrm W\), so 1920 W is dimensionally consistent with a heat rate.

Signs: The declared positive direction is from the 800°C face to the 200°C face; the positive 1.92 kW result agrees with that hot-to-cold direction.

Limit: \(k_{\rm avg}=1.60\) W/(m·K) lies between endpoint values 1.36 and 1.84, and the rate would vanish as the face temperatures become equal.

Final answer: \(\boxed{\dot Q=1.92\ \mathrm{kW}}\); room-temperature \(k\) underpredicts by \(\boxed{23.8\%}\).

Engineering insight

Property reference temperature is a model decision. Before design use, verify the fitted range, material grade, aging, and multidimensional port-edge losses.

Problem-solving strategy

  1. Recognize field geometry, steady/transient behavior, and sources.
  2. Choose coordinates, positive heat direction, and the minimum faithful dimension.
  3. Write the unsimplified balance, then justify each deleted term and property approximation.
  4. Apply independent boundary/initial conditions and solve symbolically first.
  5. Audit every boundary, units, extrema, and global energy conservation.

Chapter recap

  • Start from local energy conservation; geometry determines the divergence term.
  • Boundary and initial conditions are part of the model, not afterthoughts.
  • Generation creates curvature and interior maxima; variable conductivity changes gradients.

Before moving on

  • Can you defend every retained and omitted term?
  • Can you check a solution against boundaries and a whole-body energy balance?
  • Can you state where each property value came from and its reference temperature?
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