Heat & Mass Transfer

Chapter 2
Steady-State Heat Conduction

Bachelors Program in Mechnical Engineering
Heat & Mass Transfer • Bachelor

Steady-State Heat Conduction

Introduction

Heat conduction is the transfer of thermal energy through a stationary medium caused by a temperature gradient. In mechanical engineering it governs the performance of walls, pipes, pressure vessels, thermal shields, electronic packages, heat exchangers, turbine components, refrigeration equipment, and energy-efficient buildings. This chapter develops steady, one-dimensional conduction models for plane walls, cylinders, and spheres, then extends them to composite systems, thermal contact resistance, insulation, fins, and internally heat-generating solids.

A steady-state condition exists when temperature at every location is independent of time, although temperature may vary with position. Under steady conditions, the energy stored within the body does not change. Unless heat is generated internally, the heat rate entering a control volume equals the heat rate leaving it.

Learning Objectives

After studying this chapter, you should be able to:

Apply Fourier’s law to plane, cylindrical, and spherical geometries.
Construct and interpret thermal-resistance networks.
Analyze composite walls and thermal contact resistance.
Explain the critical radius of insulation.
Determine heat transfer and temperature distributions for common fin configurations.
Analyze steady conduction with uniform volumetric heat generation.
Select appropriate models, identify assumptions, and assess practical limitations.

1. Fourier’s Law and Thermal Resistance

Fourier’s law states that heat flows in the direction of decreasing temperature. In one dimension,

\dot Q_x = -k A \frac{dT}{dx}

where \dot Q_x is the heat-transfer rate (W), k is thermal conductivity (W·m⁻¹·K⁻¹), A is the area normal to heat flow (m²), and dT/dx is the temperature gradient (K·m⁻¹). The negative sign expresses heat flow from high to low temperature.

For constant k and a plane wall of thickness L with temperatures T₁ and T₂,

\dot Q = \frac{kA(T₁-T₂)}{L}

The corresponding conduction resistance is

R_cond = \frac{L}{kA}, \qquad \dot Q = \frac{T₁-T₂}{R_cond}

Thermal resistance has units K·W⁻¹. This analogy resembles Ohm’s law: temperature difference is analogous to voltage, heat rate to current, and thermal resistance to electrical resistance. The analogy is useful, but it does not eliminate the need to check geometry and boundary conditions.

For a general one-dimensional, variable-area path with constant k,

R_cond=\int_x₁^x₂\frac{dx}{kA(x)}, \qquad \dot Q=\frac{T₁-T₂}{R_cond}

This expression is the basis for cylindrical and spherical conduction.

Assumptions commonly used

The basic resistance equations generally assume steady state, one-dimensional heat flow, known boundary temperatures, negligible radiation within the solid, no internal heat generation, and constant or suitably averaged thermal conductivity. A model is reliable only when these assumptions are compatible with the physical system.

2. Plane-Wall Conduction

For a plane wall with uniform area A and thickness L, the temperature profile is linear when k is constant and there is no heat generation:

T(x)=T₁-\frac{T₁-T₂}{L}x

The heat flux is heat rate per unit area:

q''=\frac{\dot Q}{A}=k\frac{T₁-T₂}{L}

where q″ has units W·m⁻². If thermal conductivity depends strongly on temperature, the integrated form is more appropriate:

\dot Q=\frac{A}{L}\int_T₂^T₁k(T)\,dT

For a multilayer plane wall, layers are in series. If layer i has thickness Lᵢ and conductivity kᵢ,

R_wall=\sum_i\frac{L_i}{k_iA}

and \dot Q equals the overall temperature difference divided by this total resistance.

Convection boundaries

When a fluid contacts a wall, the surface-to-fluid heat transfer is modeled by Newton’s law of cooling:

\dot Q=hA(T_s-T_\infty), \qquad R_conv=\frac{1}{hA}

Here h is the convection coefficient (W·m⁻²·K⁻¹), T_s is the surface temperature, and T∞ is the bulk-fluid temperature. For a wall between two fluids,

R_total=\frac{1}{h₁A}+\sum_i\frac{L_i}{k_iA}+\frac{1}{h₂A}

The heat rate is \dot Q=(T∞,1−T∞,2)/R_total. Temperature drops across each resistance are proportional to that resistance.

3. Cylindrical Conduction

For radial conduction through a cylindrical shell, the area changes with radius: A(r)=2πrL. Integrating Fourier’s law gives

\dot Q=\frac{2\pi kL(T₁-T₂)}{\ln(r₂/r₁)}

and

R_cyl=\frac{\ln(r₂/r₁)}{2\pi kL}

where r₁ and r₂ are inner and outer radii and L is cylinder length. The temperature distribution is logarithmic:

T(r)=T₁-\frac{T₁-T₂}{\ln(r₂/r₁)}\ln\left(\frac{r}{r₁}\right)

For a multilayer cylindrical wall,

R_cyl,total=\sum_i\frac{\ln(r_i+1/r_i)}{2\pi k_iL}

Convection resistances at the inner and outer surfaces are 1/(hᵢ2πrᵢL) and 1/(hₒ2πrₒL), respectively. Areas must be evaluated at the actual surface radius; using one mean area can introduce error for thick walls.

4. Spherical Conduction

For radial conduction through a spherical shell, A(r)=4πr². Integration gives

\dot Q=4\pi k\frac{T₁-T₂}{(1/r₁)-(1/r₂)}

and

R_sph=\frac{1}{4\pi k}\left(\frac{1}{r₁}-\frac{1}{r₂}\right)

The temperature distribution is

T(r)=T₁-(T₁-T₂)\frac{(1/r₁)-(1/r)}{(1/r₁)-(1/r₂)}

Spherical resistance is important for tanks, cryogenic vessels, buried storage systems, and insulated spherical reactors.

5. Composite Walls and Contact Resistance

A composite system may contain several materials and fluid interfaces. Resistances in series carry the same heat rate; parallel paths experience the same temperature difference and divide the heat rate.

At an imperfect interface, microscopic roughness traps air and reduces the real contact area. The interface is represented by

R_contact=\frac{1}{h_cA_c}

where h_c is the thermal contact conductance (W·m⁻²·K⁻¹). Its value depends on surface roughness, contact pressure, material properties, oxidation, and interstitial material. The interface temperature jump is

\Delta T_contact=\dot Q R_contact

Contact resistance can be reduced by increasing clamping pressure, improving surface finish, using a compliant thermal interface material, or applying thermal grease where compatible with the application. It should not be confused with bulk resistance: it occurs at a boundary and may dominate thin, highly conductive assemblies.

Worked Example 1: Composite plane wall

A wall of area 10 m² consists of 0.10 m of concrete (k=1.4 W·m⁻¹·K⁻¹) and 0.05 m of insulation (k=0.05 W·m⁻¹·K⁻¹). The convection coefficients on the two sides are 10 and 25 W·m⁻²·K⁻¹. Fluid temperatures are 20 °C and −10 °C. Find the heat-transfer rate.

The total resistance is

R_t=\frac{1}{10(10)}+\frac{0.10}{1.4(10)}+\frac{0.05}{0.05(10)}+\frac{1}{25(10)}

Thus R_t=0.010+0.00714+0.100+0.004=0.12114 K·W⁻¹. The temperature difference is 30 K, so

\dot Q=\frac{30}{0.12114}=247.6\ W

The insulation accounts for most of the resistance and therefore most of the temperature drop. This illustrates why adding a material with low k is effective.

6. Critical Radius of Insulation

Insulation normally reduces heat transfer, but for small cylinders or spheres it initially increases the outer area available for convection. The competition is between added conduction resistance and reduced convection resistance.

For a cylinder with inner radius r₁ and insulation conductivity k, the total resistance per length is

R=\frac{\ln(r₂/r₁)}{2\pi kL}+\frac{1}{h(2\pi r₂L)}

Differentiating with respect to r₂ gives the critical radius:

r_crit,cyl=\frac{k}{h}

For a sphere,

r_crit,sph=\frac{2k}{h}

If the existing radius is greater than the critical radius, adding insulation decreases heat transfer immediately. If it is smaller, heat transfer may initially increase until the critical radius is reached. In most building walls the radius concept is irrelevant, while it can matter for thin wires, small pipes, electrical cables, and biomedical probes. The critical radius is an idealized result based on constant h and one-dimensional conduction.

7. Extended Surfaces (Fins)

Fins increase heat transfer by enlarging the exposed surface area. They are used on air-cooled engines, heat sinks, radiators, condensers, and electronic devices. Heat conducts from the base along the fin and is then transferred by convection to the surroundings.

For a straight fin of uniform cross-section with constant k and h, define excess temperature θ=T−T∞. The one-dimensional fin equation is

\frac{d²\theta}{dx²}-m²\theta=0, \qquad m=\sqrt{\frac{hP}{kA_c}}

where P is the fin perimeter, A_c is its cross-sectional area, and m has units m⁻¹. The derivation assumes steady state, uniform properties, negligible radiation, one-dimensional conduction, and uniform h.

For an infinitely long fin, θ approaches zero at large x and

\dot Q_f=\sqrt{hPkA_c}\,(T_b-T_\infty)

For a fin with an insulated (adiabatic) tip and length L,

\dot Q_f=\sqrt{hPkA_c}\,(T_b-T_\infty)\tanh(mL)

For a convective tip, the exact expression is

\dot Q_f=\sqrt{hPkA_c}\,(T_b-T_\infty)\frac{\sinh(mL)+(h/mk)\cosh(mL)}{\cosh(mL)+(h/mk)\sinh(mL)}

A useful approximate treatment replaces the actual length by corrected length L_c=L+A_tip/P and then uses the insulated-tip expression. This is appropriate when tip convection is modest.

Fin efficiency is the ratio of actual fin heat transfer to the heat transfer if the entire fin were at base temperature:

\eta_f=\frac{\dot Q_f}{hA_f(T_b-T_\infty)}

For a straight fin with an insulated tip,

\eta_f=\frac{\tanh(mL)}{mL}

Fin effectiveness compares fin heat transfer with heat transfer from the base area without a fin:

\varepsilon_f=\frac{\dot Q_f}{hA_b(T_b-T_\infty)}

A fin is generally worthwhile when effectiveness is substantially greater than 1. High conductivity, a large exposed perimeter, and a high convection coefficient improve fin performance, but increasing length eventually gives diminishing returns because the distal portion approaches ambient temperature.

8. Steady Conduction with Heat Generation

Internal heat generation occurs in electrical conductors, nuclear fuel, chemical reactors, batteries, and exothermic materials. For constant properties, the steady heat equation is

\nabla²T+\frac{\dot q_g}{k}=0

where \dot q_g is volumetric heat generation (W·m⁻³). The source term changes the temperature profile from linear to curved.

Plane wall with uniform generation

For a wall of half-thickness L, symmetry at x=0, and surface temperature T_s at x=±L,

T(x)=T_s+\frac{\dot q_g}{2k}(L²-x²)

The maximum temperature occurs at the center:

T_max=T_s+\frac{\dot q_gL²}{2k}

The heat flux at a surface is \dot q_gL, because half the generated heat leaves through each surface.

Long solid cylinder

For a cylinder of radius R with surface temperature T_s,

T(r)=T_s+\frac{\dot q_g}{4k}(R²-r²), \qquad T_max=T_s+\frac{\dot q_gR²}{4k}

Solid sphere

For a sphere of radius R,

T(r)=T_s+\frac{\dot q_g}{6k}(R²-r²), \qquad T_max=T_s+\frac{\dot q_gR²}{6k}

These expressions require uniform generation, constant k, and a prescribed surface temperature. If the surface is convective, the surface temperature must first be found from the total generated heat and the convection resistance.

Worked Example 2: Heat-generating cylinder

A long ceramic cylinder has radius 0.02 m, k=2 W·m⁻¹·K⁻¹, and uniform generation \dot q_g=1.0×10⁶ W·m⁻³. Its surface is maintained at 80 °C. The center temperature is

T_max=80+\frac{(1.0\times10⁶)(0.02)²}{4(2)}

Since (0.02)²=0.0004 m², the temperature rise is 50 K. Therefore T_max=130 °C. The result should be checked against material limits and any temperature dependence of k.

9. Applied Case: Insulated Hot-Water Pipe

Background

Consider a steel hot-water pipe in a plant carrying water at 120 °C. The pipe has inner radius 0.025 m, outer steel radius 0.030 m, and length 12 m. It is insulated to an outer radius of 0.080 m. The insulation conductivity is 0.045 W·m⁻¹·K⁻¹, steel conductivity is 45 W·m⁻¹·K⁻¹, the internal convection coefficient is 500 W·m⁻²·K⁻¹, and the external coefficient is 8 W·m⁻²·K⁻¹. Ambient temperature is 25 °C.

Analysis

The resistances are

R_i=\frac{1}{h_i2\pi r_iL}

R_steel=\frac{\ln(0.030/0.025)}{2\pi(45)(12)}

R_ins=\frac{\ln(0.080/0.030)}{2\pi(0.045)(12)}

R_o=\frac{1}{h_o2\pi r_oL}

Numerically, R_i≈0.00212 K·W⁻¹, R_steel≈0.000052 K·W⁻¹, R_ins≈0.0306 K·W⁻¹, and R_o≈0.0207 K·W⁻¹. Therefore R_total≈0.0535 K·W⁻¹ and

\dot Q=\frac{120-25}{0.0535}\approx1776\ W

The insulation and external convection resistances dominate. The critical radius is k_ins/h_o=0.045/8=0.0056 m, which is less than the existing steel radius. Thus insulation reduces heat loss from the start. In practice, the design would also consider insulation temperature rating, fire performance, moisture ingress, mechanical damage, personnel protection, and maintenance access.

Learning points

The cylindrical model is preferable to a plane-wall approximation because the radius ratio is substantial. Resistance ranking identifies where design changes matter. Increasing steel conductivity would have negligible effect, whereas improving insulation or reducing external convection would have a much larger influence.

10. Design Practice, Limitations, and Contemporary Engineering

Good thermal analysis begins with a physical sketch, clearly defined boundaries, consistent SI units, and an explicit statement of assumptions. Choose plane, cylindrical, or spherical geometry according to the dominant heat-flow path. Use thermal-resistance networks only when the heat flow is approximately one-dimensional and the interfaces can be represented by lumped resistances.

Important limitations include multidimensional spreading, temperature-dependent properties, radiation, natural-convection variation, moisture in insulation, transient operation, and uncertain contact conductance. Computational tools such as finite-element and finite-volume software are useful when geometry or boundary conditions are complex, but they do not replace conservation laws, mesh verification, or experimental validation. Engineers should perform energy balances, inspect temperature fields for physical plausibility, and conduct sensitivity studies on uncertain k, h, and contact resistance values.

Modern applications increasingly combine conduction models with digital sensors and data-driven monitoring. AI may help estimate degradation or infer effective thermal properties from operating data, but training data quality, extrapolation risk, explainability, cybersecurity, and safety-critical validation must be addressed. Thermal designs must also comply with applicable pressure-vessel, electrical, fire, building, and occupational-safety standards.

Chapter Summary

Steady conduction is governed by Fourier’s law and energy conservation. Plane walls have linear temperature profiles for constant k and no generation, while cylindrical and spherical systems have logarithmic and reciprocal-radius distributions because their areas vary with position. Thermal resistance provides a powerful engineering method: series resistances carry the same heat rate, while parallel paths share the same temperature difference. Composite walls require bulk and contact resistances to be considered together. The critical radius explains why insulation can initially increase heat loss from very small cylinders or spheres. Fins enhance convection by increasing area, but their temperature decreases away from the base, so efficiency and effectiveness must be distinguished. Uniform internal generation produces parabolic temperature profiles and centerline maximum temperatures. Reliable design requires appropriate geometry, valid assumptions, unit consistency, resistance ranking, and verification against material and safety limits.

Key Terms

Fourier’s law: Relation between heat flux and temperature gradient.

Thermal conductivity: Material property measuring ability to conduct heat.

Heat flux: Heat-transfer rate per unit area.

Thermal resistance: Ratio of temperature difference to heat-transfer rate.

Convection resistance: Resistance associated with fluid-to-surface heat transfer.

Composite wall: Wall made of two or more thermal layers.

Thermal contact resistance: Resistance caused by imperfect contact between solids.

Cylindrical conduction: Radial conduction through a cylindrical geometry.

Spherical conduction: Radial conduction through a spherical geometry.

Critical radius: Radius at which insulation produces maximum heat transfer.

Fin: Extended surface used to increase convective heat transfer.

Fin efficiency: Actual fin heat transfer divided by ideal isothermal-fin transfer.

Fin effectiveness: Fin heat transfer compared with un finned base transfer.

Volumetric heat generation: Heat generated per unit volume of material.

Steady state: Condition in which temperature does not vary with time.

One-dimensional conduction: Model in which temperature varies significantly in one coordinate only.

Boundary condition: Thermal condition imposed at a surface or symmetry location.

Temperature gradient: Spatial rate of temperature change.

Review Questions

Short Answer Questions

State Fourier’s law and explain the significance of its negative sign.
Why is cylindrical conduction resistance logarithmic rather than linear?
Distinguish thermal contact resistance from bulk conduction resistance.
Define fin efficiency and fin effectiveness.
Where is the maximum temperature located in a uniformly heat-generating symmetric solid?

Long Answer Questions

Derive the conduction resistance of a cylindrical shell from Fourier’s law.
Explain how a composite wall with convection boundaries is represented by a resistance network.
Derive the critical radius of insulation for a cylinder and discuss its practical meaning.
Develop the temperature distribution for a plane wall with uniform heat generation.
Explain the assumptions behind the one-dimensional fin equation and discuss when they may fail.

Analytical / Application Questions

A multilayer cylindrical pipe has known dimensions, conductivities, and convection coefficients. Construct its complete resistance network and determine the heat loss and each interface temperature.
Compare two fin designs with equal material volume but different lengths and thicknesses. Use fin efficiency and effectiveness to recommend a design.
A battery module generates heat internally and is cooled by convection. Determine the maximum temperature using a heat-generation model, then assess whether changing conductivity, radius, or convection coefficient gives the greatest reduction.

Multiple Choice Questions with Answers

Question 1

For a plane wall with constant thermal conductivity and no heat generation, the temperature profile is:

A. Linear

B. Logarithmic

C. Reciprocal

D. Exponential

Correct Answer: A. Linear

Question 2

The conduction resistance of a cylindrical wall is:

A. L/(kA)

B. ln(r₂/r₁)/(2πkL)

C. (1/r₁−1/r₂)/(4πk)

D. 1/(hA)

Correct Answer: B. ln(r₂/r₁)/(2πkL)

Question 3

A thermal contact resistance primarily represents:

A. Radiation through a vacuum

B. Resistance caused by imperfect solid-to-solid contact

C. Fluid friction in a pipe

D. Heat generation inside a solid

Correct Answer: B. Resistance caused by imperfect solid-to-solid contact

Question 4

The critical radius of insulation for a cylinder is:

A. h/k

B. 2k/h

C. k/h

D. kh

Correct Answer: C. k/h

Question 5

For a symmetric plane wall with uniform heat generation, the maximum temperature occurs:

A. At either surface

B. At the center plane

C. Outside the wall

D. At one-quarter thickness only

Correct Answer: B. At the center plane

Question 6

If two thermal resistances are in series, they carry:

A. The same heat-transfer rate

B. The same temperature difference

C. Zero heat transfer

D. Opposite heat rates

Correct Answer: A. The same heat-transfer rate

Question 7

For a straight fin with an insulated tip, fin efficiency is:

A. mL/tanh(mL)

B. tanh(mL)/(mL)

C. hP/(kA_c)

D. kA_c/(hP)

Correct Answer: B. tanh(mL)/(mL)

Question 8

The critical radius of insulation for a sphere is:

A. k/h