Heat Transfer · Chapter 3
0%
Heat Transfer · Chapter Three

Steady Heat Conduction

At steady state, a thermal-resistance network turns geometry, materials, interfaces, and fluid films into an auditable map of heat flow.

The analogy is powerful only when the heat paths and areas match the physical system. We develop series and parallel networks, curved walls, critical insulation radius, fins, and shape factors with explicit applicability checks.

Learning objectives

  • Construct steady one-dimensional thermal-resistance networks from physical heat paths.
  • Compute interface temperatures and identify dominant resistances.
  • Select plane, cylindrical, spherical, contact, and convection resistances with correct areas.
  • Judge when insulation or fins increase useful performance.
  • Audit networks by node balances, units, limits, and correlation/property validity.
3–1

Steady Heat Conduction in Plane Walls

For steady one-dimensional, source-free heat flow, each element obeys \(\dot Q=\Delta T/R\). Series elements carry the same rate; parallel paths share the same node-to-node temperature difference. This requires well-defined approximately isothermal nodes.

Thermal Resistance Analogy

Steady heat flow obeys a relation identical in form to Ohm's law: the driving temperature difference equals the heat rate times a thermal resistance. Resistances in series add; resistances in parallel combine by reciprocals — just like electrical circuits.

\[ \dot{Q} = \frac{\Delta T}{R}, \qquad R_{\text{series}} = \sum R_i \]

Conduction and convection resistances
\[ R_{\text{wall}} = \frac{L}{kA}, \qquad R_{\text{conv}} = \frac{1}{hA} \]
Radiation resistance
\[ R_{\text{rad}} = \frac{1}{h_{\text{rad}}A}, \qquad h_{\text{rad}} = \varepsilon\sigma (T_s^2 + T_{\text{surr}}^2)(T_s + T_{\text{surr}}) \]
Overall heat transfer coefficient
\[ \dot{Q} = U A\,\Delta T_{\text{overall}}, \qquad \frac{1}{UA} = R_{\text{total}} \]

Physical meaning

A larger resistance consumes a larger fraction of the total temperature drop. That observation locates the most effective design change before detailed optimization.

EXAMPLE 1Cold-room composite panel
Problem. A 6.0 m² cold-room panel separates 28°C warehouse air from −4°C room air. From warm to cold it has \(h_o=12\), 1.0-mm steel (\(k=45\)), 80-mm foam (\(k=0.028\)), and \(h_i=7.0\), in SI units. Find the heat gain and both foam-face temperatures.

Strategy

Steady heat follows one series path through two films and two layers, so use the minimum closing set: total resistance, rate, then node drops.

Givens

  • \(A=6.0\) m²; \(T_{\infty,o}=28^\circ\)C, \(T_{\infty,i}=-4^\circ\)C; supplied \(h,L,k\).

Physical/model interpretation

One-dimensional steady conduction, perfect contact, no radiation or thermal bridges; positive heat is into the cold room.

Step-by-step solution
1. Build resistances. \(R_o=1/(12\cdot6)=0.01389\), \(R_{st}=0.001/(45\cdot6)=3.70\times10^{-6}\), \(R_f=0.080/(0.028\cdot6)=0.47619\), \(R_i=1/(7\cdot6)=0.02381\) K/W.
2. Total and rate. \(R_t=0.51389\ \mathrm{K/W}\); \(\dot Q=(28-(-4))/R_t=62.27\) W.
3. Warm foam face. \(T_{f,w}=28-\dot Q(R_o+R_{st})=27.13^\circ\)C.
4. Cold foam face. \(T_{f,c}=T_{f,w}-\dot QR_f=-2.53^\circ\)C.

Unit/sign/limit checks

Units: The overall relation gives K/(K/W) = W, while every node drop \(\dot QR_i\) gives W·K/W = K, matching the calculated temperatures.

Signs: Temperatures decrease monotonically from 28°C to −4°C in the declared inward heat direction, and every resistance consumes a positive temperature drop.

Limit: The foam takes 29.66 K of the 32 K total drop because it supplies 92.7% of \(R_t\); removing foam would therefore increase heat gain sharply.

Final answer: \(\boxed{\dot Q=62.3\ \mathrm W,\ T_{f,w}=27.1^\circ C,\ T_{f,c}=-2.53^\circ C}\)

Engineering insight

Steel resistance is negligible; fasteners and joints, omitted here, may dominate an actual panel and require thermal-bridge testing.

3–2

Thermal Contact Resistance

Nominally touching solids meet at asperities, constricting heat flow and leaving low-conductivity gaps. \(h_c\) is an empirical joint property, not a material constant: pressure, roughness, flatness, oxidation, and interface material matter.

Thermal Contact Conductance, \(h_c\)

The resistance \(R_c=1/(h_cA)\) creates a finite temperature jump \(\dot QR_c\). Use conductance data matching the surface finish, pressure, temperature, and interstitial medium; extrapolation can be unsafe.

Contact resistance
\[ R_c = \frac{1}{h_c A_c}, \qquad \Delta T_{\text{interface}} = \dot{Q}\,R_c \]
Audit cueA temperature discontinuity inside a homogeneous solid is suspicious; at a characterized interface it is the expected signature of contact resistance.
3–3

Generalized Thermal Resistance Networks

Side-by-side materials form parallel paths only if they connect the same two temperature nodes. Shared convection films can create lateral coupling; the simple network then becomes an approximation whose quality worsens near path boundaries.

Parallel resistances
\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \]
T1 R1 R2 R3 R4 T2
Figure 3–1A mixed network: series resistances \(R_1\) and \(R_4\) bracket a parallel pair \(R_2 \parallel R_3\), modeling a composite wall with side-by-side materials.
3–4

Heat Conduction in Cylinders and Spheres

Radial area varies as \(2\pi rL\) for a cylinder and \(4\pi r^2\) for a sphere. Integrating Fourier’s law therefore gives logarithmic and inverse-radius resistances; inserting an arbitrary area into \(L/kA\) is not generally valid.

Cylindrical shell resistance
\[ R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L} \]
Spherical shell resistance
\[ R_{\text{sph}} = \frac{r_2 - r_1}{4\pi k\, r_1 r_2} \]
EXAMPLE 2Condensation-risk check on an insulated line
Problem. A 10-m chilled-process line has insulation inner radius 25 mm, outer radius 55 mm, \(k=0.036\ \mathrm{W/(m\,K)}\). The pipe/insulation interface is at 5°C and ambient air is 30°C with \(h=9.0\ \mathrm{W/(m^2K)}\). Find inward heat leak and outer-surface temperature; compare with a 19°C dew point.

Strategy

Radial cylindrical conduction and outside convection are series elements; the surface node follows from the convection drop.

Givens

  • \(r_1=0.025\) m, \(r_2=0.055\) m, \(L=10\) m, \(k=0.036\), \(h=9.0\), \(T_\infty=30^\circ\)C, \(T_1=5^\circ\)C.

Physical/model interpretation

Steady radial 1-D model with constant \(k\), negligible pipe/contact resistance and radiation. Positive heat points inward.

Step-by-step solution
1. Conduction resistance. \(R_{ins}=\ln(0.055/0.025)/(2\pi(0.036)(10))=0.3486\ \mathrm{K/W}\).
2. Film resistance. \(R_o=1/[h(2\pi r_2L)]=1/[9(2\pi)(0.055)(10)]=0.03215\ \mathrm{K/W}\).
3. Heat leak. \(\dot Q=(30-5)/(0.3486+0.03215)=65.66\) W.
4. Surface. \(T_{s,o}=30-\dot QR_o=30-(65.66)(0.03215)=27.89^\circ\)C.

Unit/sign/limit checks

Units: Both resistances are K/W; result is W.

Signs: \(30>27.89>5^\circ\)C along inward heat flow.

Limit: If \(h\to\infty\), the outer surface approaches 30°C and film resistance vanishes.

Final answer: \(\boxed{\dot Q=65.7\ \mathrm W\ (6.57\ W/m),\ T_{s,o}=27.9^\circ C}\); no condensation is predicted because \(T_{s,o}>19^\circ\)C.

Engineering insight

Vapor-barrier seams and thermal bridges can be colder than this ideal surface; field condensation control needs humidity uncertainty and installation review.

3–5

Critical Radius of Insulation

For a cylinder, insulation raises conduction resistance while enlarged area lowers external convection resistance. Differentiating their sum yields \(r_{cr}=k/h\); this result assumes constant \(k,h\), cylindrical geometry, and an outside convection boundary.

Critical Radius of Insulation

Insulating a cylinder (or sphere) only reduces heat loss once the outer radius exceeds a critical value. Below it, added insulation actually increases heat transfer because the gain in exposed surface area outweighs the added conduction resistance.

\[ r_{\text{cr,cyl}} = \frac{k}{h}, \qquad r_{\text{cr,sph}} = \frac{2k}{h} \]

Key Idea The critical radius marks maximum heat transfer, not a recommended thickness. Fire, electrical, condensation, personnel-protection, and energy requirements usually control insulation selection.
3–6

Heat Transfer from Finned Surfaces

A fin adds area but is not isothermal: axial conduction supplies distributed convection. The classic equation assumes steady one-dimensional fin conduction, uniform \(h\), constant \(k\), negligible radiation, and no generation.

The Fin Equation

An element balance yields \(d^2\theta/dx^2-m^2\theta=0\), where \(\theta=T-T_\infty\) and \(m^2=hp/(kA_c)\). The tip boundary selects the solution; an adiabatic-tip approximation is best for a slender fin whose tip area is small.

Heat rate, insulated-tip (or corrected-length) fin
\[ \dot{Q}_{\text{fin}} = \sqrt{h p k A_c}\,(T_b - T_\infty)\tanh(aL_c), \qquad a = \sqrt{\frac{hp}{kA_c}} \]
Fin efficiency and effectiveness
\[ \eta_{\text{fin}} = \frac{\dot{Q}_{\text{fin}}}{\dot{Q}_{\text{max}}} = \frac{\dot{Q}_{\text{fin}}}{h A_{\text{fin}}(T_b - T_\infty)}, \qquad \varepsilon_{\text{fin}} = \frac{\dot{Q}_{\text{fin}}}{h A_c (T_b - T_\infty)} \]
Efficiency compares the fin to an ideal isothermal fin; effectiveness compares finned to unfinned. Fins are worthwhile when \(\varepsilon_{\text{fin}} \gtrsim 2\).
EXAMPLE 3Screening an inverter heat-sink fin
Problem. An aluminum plate fin is 40 mm long, 30 mm wide, and 2.0 mm thick, with \(k=180\ \mathrm{W/(m\,K)}\). Its base is 85°C in 35°C air; a validated duct test gives uniform \(h=28\ \mathrm{W/(m^2K)}\). Using an adiabatic tip, find fin heat rate, efficiency, and effectiveness relative to its footprint.

Strategy

A straight constant-section fin with supplied uniform \(h\) matches the insulated-tip fin solution; calculate \(mL\) before judging performance.

Givens

  • \(L=0.040\) m, \(w=0.030\) m, \(t=0.002\) m, \(k=180\), \(h=28\), \(\theta_b=50\) K.

Physical/model interpretation

\(A_c=wt\), \(p=2(w+t)\); steady 1-D conduction, negligible radiation/contact loss, adiabatic tip.

Step-by-step solution
1. Geometry. \(A_c=6.00\times10^{-5}\ \mathrm{m^2}\), \(p=0.064\) m, and convecting area \(A_f=pL=0.00256\ \mathrm{m^2}\).
2. Fin parameter. \(m=\sqrt{hp/(kA_c)}=\sqrt{28(0.064)/[180(6.0\times10^{-5})]}=12.88\ \mathrm{m^{-1}}\); \(mL=0.515\).
3. Heat rate. \(\dot Q_f=\sqrt{hpkA_c}\theta_b\tanh(mL)=\sqrt{28(0.064)(180)(6.0\times10^{-5})}(50)\tanh(0.515)=3.30\) W.
4. Metrics. \(\eta_f=\tanh(mL)/(mL)=0.914\); \(\varepsilon_f=\dot Q_f/[hA_c\theta_b]=3.30/[28(6.0\times10^{-5})(50)]=39.3\).

Unit/sign/limit checks

Units: In \(\sqrt{hpkA_c}\theta\), the quantity under the root has units W²/K², so multiplication by the 50 K excess gives W; efficiency and effectiveness are dimensionless ratios.

Signs: With \(\theta_b=T_b-T_\infty=+50\) K, the positive 3.30 W rate leaves the hotter base; reversing the temperature excess would reverse the heat-rate sign.

Limit: \(0<\eta_f=0.914<1\), and the independent check \(\eta_fhA_f\theta_b=3.28\) W agrees within rounding; as \(L\to0\), fin heat transfer tends to zero.

Final answer: \(\boxed{\dot Q_f=3.30\ \mathrm W,\ \eta_f=0.914,\ \varepsilon_f=39.3}\)

Engineering insight

The fin is thermally effective, but an array also changes airflow and \(h\); spacing, pressure drop, base spreading, extrusion tolerances, and test correlation remain design checks.

3–7

Heat Transfer in Common Configurations

Tabulated shape factors compress a particular steady, source-free, constant-\(k\) multidimensional boundary-value problem into \(\dot Q=Sk\Delta T\). Use one only when its geometry and listed restrictions—such as burial depth and length—match.

Conduction shape factor
\[ \dot{Q} = S\,k\,(T_1 - T_2) \]
\(S\) (units of length) depends only on geometry. Example: a pipe of length \(L\) buried at depth \(z\) gives \( S = \dfrac{2\pi L}{\cosh^{-1}(2z/D)} \).

Problem-solving strategy

  1. Sketch heat paths and define isothermal nodes and positive heat direction.
  2. Select geometry-specific resistances and evaluate properties at representative layer temperatures, iterating if needed.
  3. Combine only true series or parallel elements; retain contact and film resistances when evidence supports them.
  4. Solve heat rate first, then recover node temperatures from \(\Delta T_i=\dot Q_iR_i\).
  5. Audit units, node balances, temperature ordering, limiting resistance, and every correlation or shape-factor range.

Chapter recap

  • Networks are energy balances arranged around physical paths and isothermal nodes.
  • Curved geometry, interfaces, and fins require their own validated resistance models.
  • Dominant-resistance and limiting-case checks turn arithmetic into engineering judgment.

Before moving on

  • Can you draw and label a mixed network before calculating?
  • Can you recover every interface temperature and verify each node balance?
  • Can you defend property temperatures and model validity ranges?
← Chapter 2 · Conduction Equation Chapter 4 · Transient Conduction →