Chapter 3: One-Dimensional Steady-State Conduction

Source files used: late Heat Transfer (05), Heat Transfer (06), Heat Transfer (07), Heat Transfer (08), Heat Transfer (09); textbook Chapter 3 as support. No worked examples are included.

1. Big Picture

Chapter 3 is where conduction becomes an engineering tool. Chapter 2 taught how to pose conduction problems with differential equations. Chapter 3 shows that many steady, one-dimensional conduction problems can be reduced to thermal resistance networks.

Biddle’s basic idea is:

$$\boxed{\text{Heat flow behaves like current through resistors.}}$$

Temperature difference acts like voltage difference. Heat rate acts like current. Thermal resistance controls how much heat flows.

$$\dot{q}=\frac{\Delta T}{R_{\text{total}}}$$

This chapter also covers fins. Fins are extended surfaces used to increase convection area. Biddle motivates them with radiator/electronics examples: if you cannot change the fluid or temperature difference much, increasing surface area is often the practical engineering move.

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2. Core Ideas

2.1 Resistance Method

For steady, one-dimensional conduction with no generation:

$$\dot{q}=\frac{T_{\text{hot}}-T_{\text{cold}}}{R}$$

This works because the same heat rate passes through each layer in series. If $100\ \text{W}$ enters layer A, $100\ \text{W}$ leaves layer A and enters layer B, assuming steady state and no generation.

Thermal resistance units:

$$[R]=\text{K/W or }^\circ\text{C/W}$$

The resistance method is valid when:

• heat flow is essentially one-dimensional,
• steady state applies,
• properties are constant or represented by average values,
• no internal generation appears in the resistance segment,
• temperatures are uniform over each surface/node.

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2.2 Series and Parallel Networks

Series: same heat rate through each resistance.

$$R_{\text{total}}=R_1+R_2+R_3+\cdots$$ $$\dot{q}=\frac{T_1-T_2}{R_{\text{total}}}$$

Parallel: same temperature difference across multiple paths, heat rates split.

$$\dot{q}_{\text{total}}=q_1+q_2+\cdots$$ $$\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots$$

Use parallel networks when heat can take multiple paths between the same two temperature nodes.

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2.3 Temperature Drop Logic

Biddle emphasizes this intuition with composite walls:

$$q''=k\left|\frac{dT}{dx}\right|$$

If two materials carry the same heat flux, the material with lower $k$ must have a larger temperature gradient. So insulation has a large temperature drop; copper has a small temperature drop.

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2.4 Overall Heat Transfer Coefficient

The overall heat transfer coefficient $U$ is a shorthand for a whole network of conduction and convection resistances.

$$\dot{q}=UA\Delta T$$

But $UA$ is really:

$$UA=\frac{1}{R_{\text{total}}}$$

For tubes, the chosen area matters. You may define $U_i$ based on inside area or $U_o$ based on outside area. The product $UA$ stays tied to the same total resistance.

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2.5 Thermal Contact Resistance

Real surfaces are rough. When two materials touch, microscopic air gaps may exist. Air is a poor conductor, so the interface adds resistance.

Thermal contact resistance is inserted between two material resistances:

Material A resistance → contact resistance → Material B resistance

If a problem gives contact resistance, include it in the network.

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2.6 Fins / Extended Surfaces

Fins increase heat transfer by increasing surface area. They are used in radiators, electronics cooling, baseboard heaters, engine components, and many other systems.

Newton’s law says:

$$\dot{q}=hA(T_s-T_\infty)$$

If $h$, $T_s$, and $T_\infty$ are hard to change, increasing area $A$ is the practical option.

But fins are not isothermal. The fin is hottest near the base and cooler near the tip, so we need fin efficiency.

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3. Main Governing Equations and Formulas

3.1 Plane Wall Conduction Resistance

$$R_{\text{cond}}=\frac{L}{kA}$$ $$\dot{q}=\frac{T_1-T_2}{L/(kA)}=kA\frac{T_1-T_2}{L}$$

Use for steady 1D conduction through a flat wall.

Assumptions:

• steady,
• 1D,
• constant $k$,
• no generation.

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3.2 Cylindrical Wall Conduction Resistance

For radial conduction through a cylinder:

$$R_{cond,cyl}=\frac{\ln(r_2/r_1)}{2\pi kL}$$ $$\dot{q}=\frac{T_1-T_2}{\ln(r_2/r_1)/(2\pi kL)}$$

where:

• $r_1$: inner radius,
• $r_2$: outer radius,
• $L$: cylinder length.

Use for pipes, tubes, cylindrical insulation, and rods with radial heat flow.

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3.3 Spherical Shell Conduction Resistance

$$R_{cond,sph}=\frac{1}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$$

Use for radial conduction through spherical shells.

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3.4 Convection Resistance

$$R_{\text{conv}}=\frac{1}{hA}$$ $$\dot{q}=\frac{T_s-T_\infty}{1/(hA)}=hA(T_s-T_\infty)$$

Use when a surface exchanges heat with a fluid.

Area must match the surface where convection occurs.

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3.5 Composite Wall in Series

$$\dot{q}=\frac{T_{\infty,1}-T_{\infty,2}}{\frac{1}{h_1A}+\frac{L_A}{k_AA}+\frac{L_B}{k_BA}+\frac{1}{h_2A}}$$

Use when fluids exist on both sides of a composite wall.

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3.6 Cylindrical Tube with Inside and Outside Convection

$$\dot{q}=\frac{T_{\infty,i}-T_{\infty,o}}{\frac{1}{h_iA_i}+\frac{\ln(r_o/r_i)}{2\pi kL}+\frac{1}{h_oA_o}}$$

where:

$$A_i=2\pi r_iL,\qquad A_o=2\pi r_oL$$

Use for heat transfer through pipe/tube walls.

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3.7 Overall Heat Transfer Coefficient

$$\dot{q}=UA\Delta T$$ $$\frac{1}{UA}=R_{\text{total}}$$

For a tube:

$$\frac{1}{U_iA_i}=\frac{1}{h_iA_i}+\frac{\ln(r_o/r_i)}{2\pi kL}+\frac{1}{h_oA_o}$$

or

$$\frac{1}{U_oA_o}=\frac{1}{h_iA_i}+\frac{\ln(r_o/r_i)}{2\pi kL}+\frac{1}{h_oA_o}$$

The numerical values of $U_i$ and $U_o$ differ because the areas differ. The total conductance $UA$ represents the same total resistance.

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3.8 Thermal Contact Resistance

If contact resistance per unit area is given as $R_{t,c}''$:

$$R_{t,c}=\frac{R_{t,c}''}{A}$$

Temperature drop across contact:

$$\Delta T_{contact}=qR_{t,c}$$

Use only if the problem gives or requires contact resistance.

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3.9 Plane Wall with Uniform Heat Generation

For a slab of half-thickness $L$, symmetry at $x=0$, surfaces at $x=\pm L$, surface temperature $T_s$:

$$T(x)=T_s+\frac{\dot{q}}{2k}(L^2-x^2)$$

Maximum temperature at the center:

$$T_{\text{max}}=T_s+\frac{\dot{q} L^2}{2k}$$

Surface heat flux:

$$q_s''=\dot{q} L$$

Use for internally heated plane walls with symmetric cooling.

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3.10 Solid Cylinder with Uniform Heat Generation

For a solid cylinder radius $r_o$ and surface temperature $T_s$:

$$T(r)=T_s+\frac{\dot{q}}{4k}(r_o^2-r^2)$$

Maximum temperature at center:

$$T_{\text{max}}=T_s+\frac{\dot{q} r_o^2}{4k}$$

Heat generated per unit length:

$$q'=\dot{q}\pi r_o^2$$

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3.11 Fin Temperature Variable

$$\theta=T-T_\infty$$

Base excess temperature:

$$\theta_b=T_b-T_\infty$$

Fin parameter:

$$m=\sqrt{\frac{hP}{kA_c}}$$

where:

• $P$: fin perimeter exposed to convection,
• $A_c$: fin cross-sectional area,
• $k$: fin material conductivity,
• $h$: convection coefficient.

Fin differential equation for a uniform cross-section fin:

$$\frac{d^2\theta}{dx^2}-m^2\theta=0$$

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3.12 Fin Heat Rate — Adiabatic Tip

$$q_f=\sqrt{hPkA_c}\,\theta_b\tanh(mL)$$

Use when the fin tip is modeled as insulated or by symmetry.

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3.13 Fin Heat Rate — Convective Tip

$$q_f=M\frac{\sinh(mL)+\frac{h}{mk}\cosh(mL)}{\cosh(mL)+\frac{h}{mk}\sinh(mL)}$$

where:

$$M=\sqrt{hPkA_c}\,\theta_b$$

Use when convection from the tip is included directly.

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3.14 Infinite Fin Approximation

$$q_f=\sqrt{hPkA_c}\,\theta_b$$

Use only when the fin is long enough. Biddle corrected the criterion in lecture: do not use the transient-conduction value by mistake. The infinite-fin model is acceptable only for sufficiently large $mL$, commonly around $mL\ge 2.65$ for the table criterion he used.

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3.15 Corrected Length Approximation

For many straight fins:

$$L_c=L+\frac{A_c}{P}$$

Then approximate a convective-tip fin as an adiabatic-tip fin with corrected length:

$$q_f\approx\sqrt{hPkA_c}\,\theta_b\tanh(mL_c)$$

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3.16 Fin Efficiency

$$\eta_f=\frac{q_f}{hA_f\theta_b}$$

where $hA_f\theta_b$ is the heat transfer if the whole fin were at base temperature.

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3.17 Overall Surface Efficiency

$$\eta_o=1-\frac{A_f}{A_t}(1-\eta_f)$$

Total heat transfer from a finned surface:

$$q_t=\eta_o hA_t\theta_b$$

where:

• $A_f$: total fin area,
• $A_t$: total exposed area including fins and exposed base.

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4. Problem-Solving Workflow

4.1 Resistance Network Problems

1. Draw the physical heat path.
2. Identify temperature nodes: fluid temperatures, surface temperatures, interfaces.
3. Identify each resistance: convection, conduction, contact, radiation if included approximately.
4. Decide series or parallel.
5. Compute total resistance.
6. Use $\dot{q}=\Delta T/R_{\text{total}}$.
7. Find intermediate temperatures from $\Delta T=qR$.
8. Check whether heat flow direction makes sense.

4.2 Finned Surface Problems

1. Identify fin geometry: rectangular, pin, circumferential, etc.
2. Determine $A_c$, $P$, $A_f$, and $L$ or $L_c$.
3. Find material $k$ and convection coefficient $h$.
4. Compute $m=\sqrt{hP/(kA_c)}$.
5. Choose tip condition: convective, adiabatic, prescribed temperature, or infinite.
6. Compute $q_f$ or $\eta_f$.
7. If multiple fins, combine with exposed base area using $\eta_o$ or direct summation.

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5. Decision Rules / Decision Trees

5.1 Resistance Network Validity

Steady state? AND one-dimensional path? AND no internal generation in the resistance layer? AND uniform temperatures over nodes? → resistance network is appropriate.

If generation exists inside the layer: → use generation equations or differential equation, not a simple $L/(kA)$ resistance alone.

5.2 Plane Wall vs Cylinder vs Sphere

Flat wall, constant area? → $R=L/(kA)$

Radial pipe/tube conduction? → $R=\ln(r_o/r_i)/(2\pi kL)$

Radial spherical shell? → $R=(1/r_1-1/r_2)/(4\pi k)$

5.3 Series vs Parallel

Same heat rate through each layer? → series

Heat splits into multiple paths between same temperature nodes? → parallel

5.4 Fin Tip Decision

Tip convection included explicitly? → convective-tip formula

Tip heat loss negligible or symmetry plane? → adiabatic-tip formula

Very long fin, $mL$ sufficiently large? → infinite-fin formula

Quick engineering approximation for convective tip? → corrected length with adiabatic-tip formula

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6. Important Tables / Correlations Needed

6.1 Thermal Resistances

Case	Resistance
Plane wall	$L/(kA)$
Cylinder	$\ln(r_2/r_1)/(2\pi kL)$
Sphere	$(1/r_1-1/r_2)/(4\pi k)$
Convection	$1/(hA)$
Contact	$R_{t,c}''/A$

6.2 Fin Geometry Quantities

Fin Type	$A_c$	$P$
Rectangular fin, width $w$, thickness $t$	$wt$	approximately $2(w+t)$
Circular pin fin, diameter $D$	$\pi D^2/4$	$\pi D$

Use the actual geometry stated in the problem.

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7. Key Takeaways

• Thermal resistance networks are the main tool for steady 1D conduction.
• Same heat rate flows through series resistances.
• Temperature drops are larger across larger resistances.
• Low-$k$ materials produce large temperature gradients.
• For cylinders, area changes with radius, so the resistance is logarithmic.
• $U$ is just shorthand for all resistances combined.
• Contact resistance can matter when surfaces are rough or poorly bonded.
• Fins increase area but are not isothermal.
• Fin efficiency tells how effectively the fin area is used.
• Always check whether the infinite-fin approximation is valid before using it.