Steady Heat Conduction

Page 1: Steady Heat Conduction Overview

• Scope: Fundamentals of steady one-dimensional heat conduction, thermal resistance networks, contact resistance, extended surfaces (fins), and multidimensional conduction shape factors.

• Instructor: Prof. J. L. Lage

Page 2: Chapter Core Topics

• Thermal resistance and the direct-current (DC) circuit analogy

• Critical radius of insulation for cylindrical and spherical bodies

• Thermal contact resistance and methods for reduction

• Extended surfaces (fins): governing equations, boundary conditions, efficiency, and effectiveness

• Multidimensional steady conduction shape factors

Page 3: Thermal Resistance & DC Circuit Analogy for Plane Walls

• Assumptions: Steady state ($\frac{d()}{dt} = 0$), one-dimensional heat diffusion, no energy generation ($\dot{e}_g = 0$), constant properties.

• Analogy to Ohm's law ($I = \frac{E_1 - E_2}{R_e}$):   $\text{Flow} = \frac{\text{Driving Potential}}{\text{Thermal Resistance}}$

• Fourier's law of conduction across a plane wall of thickness $L$ and surface area $A$:   $\dot{Q}_{\text{cond, wall}} = k A \frac{T_1 - T_2}{L} = \frac{T_1 - T_2}{R_{\text{wall}}}$

• Conduction thermal resistance:   R_{\text{wall}} = \frac{L}{k A}\; (^\n\circ\text{C/W})

Page 4: Convection Thermal Resistance

• Newton's law of cooling:   $\dot{Q}_{\text{conv}} = h A_s (T_s - T_\infty) = \frac{T_s - T_\infty}{R_{\text{conv}}}$

• Convection resistance:   R_{\text{conv}} = \frac{1}{h A_s}\; (^\n\circ\text{C/W})

• Limiting condition: As $h \rightarrow \infty$, $R_{\text{conv}} \rightarrow 0$, yielding $T_s \approx T_\infty$.

• Biot number ($Bi$) criterion for lumped capacitance:   $Bi = \frac{h L_c}{k} = \frac{R_{\text{cond}}}{R_{\text{conv}}} < 0.1$

Page 5: Radiation Thermal Resistance

• Stefan-Boltzmann law linearized with respect to a radiation heat transfer coefficient ($h_{\text{rad}}$):   $\dot{Q}_{\text{rad}} = \varepsilon \sigma A_s (T_s^4 - T_{\text{surr}}^4) = h_{\text{rad}} A_s (T_s - T_{\text{surr}}) = \frac{T_s - T_{\text{surr}}}{R_{\text{rad}}}$

• Radiation heat transfer coefficient:   $h_{\text{rad}} = \varepsilon \sigma (T_s^2 + T_{\text{surr}}^2)(T_s + T_{\text{surr}})\; (\text{W/m}^2\cdot\text{K})$

• Radiation resistance:   $R_{\text{rad}} = \frac{1}{h_{\text{rad}} A_s}\; (\text{K/W})$

• Non-linearity: $R_{\text{rad}}$ is strongly temperature-dependent, $f(T)$.

Page 6: Thermal Resistance Networks in Series

• For a plane wall exposed to convection on both sides:   $\dot{Q} = \frac{T_{\infty 1} - T_1}{R_{\text{conv, 1}}} = \frac{T_1 - T_2}{R_{\text{wall}}} = \frac{T_2 - T_{\infty 2}}{R_{\text{conv, 2}}} = \frac{T_{\infty 1} - T_{\infty 2}}{R_{\text{total}}}$

• Total thermal resistance in series:   R_{\text{total}} = R_{\text{conv, 1}} + R_{\text{wall}} + R_{\text{conv, 2}} = \frac{1}{h_1 A} + \frac{L}{k A} + \frac{1}{h_2 A}\; (^\n\circ\text{C/W})

Page 7: Composite Vertical Planes (Series Layers)

• Total thermal resistance of multi-layer walls in series:   $R_{\text{total}} = R_{\text{conv, 1}} + R_{\text{wall, 1}} + R_{\text{wall, 2}} + R_{\text{conv, 2}} = \frac{1}{h_1 A} + \frac{L_1}{k_1 A} + \frac{L_2}{k_2 A} + \frac{1}{h_2 A}$

• Heat transfer rate across the system:   $\dot{Q} = \frac{T_{\infty 1} - T_{\infty 2}}{R_{\text{total}}}$

• Evaluation of intermediate interface temperatures:   $\dot{Q} = \frac{T_i - T_j}{R_{\text{total}, i-j}}$   $T_1 = T_{\infty 1} - \dot{Q} R_{\text{conv, 1}}$   $T_2 = T_{\infty 1} - \dot{Q} (R_{\text{conv, 1}} + R_{\text{wall, 1}})$   $T_3 = T_{\infty 2} + \dot{Q} R_{\text{conv, 2}}$

Page 8: Composite Horizontal Planes (Parallel Layers)

• Parallel heat conduction through adjacent parallel layers 1 and 2:   $\dot{Q} = \dot{Q}_1 + \dot{Q}_2 = \frac{T_1 - T_2}{R_1} + \frac{T_1 - T_2}{R_2} = \frac{T_1 - T_2}{R_{\text{total}}}$

• Equivalent resistance:   $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \rightarrow R_{\text{total}} = \frac{R_1 R_2}{R_1 + R_2}$

• Note: At steady state, the temperature profile $T(x)$ along each parallel path depends on geometry and boundary temperatures, while the total heat flow depends on the individual thermal conductivities.

Page 9: Combined Series-Parallel Systems

• One-dimensional approximation assumption: Interfaces perpendicular to the $x$-direction are isothermal surfaces.

• Total thermal resistance for parallel sections (1 and 2) combined with series sections (layer 3 and convection):   $R_{\text{total}} = R_{12} + R_3 + R_{\text{conv}} = \frac{R_1 R_2}{R_1 + R_2} + R_3 + R_{\text{conv}}$

• Overall heat transfer rate:   $\dot{Q} = \frac{T_1 - T_\infty}{R_{\text{total}}}$

Page 10: Radial Systems – Cylindrical Layers

• Steady 1D radial governing equation:   $\frac{d}{dr}\left(r \frac{dT}{dr}\right) = 0$

• Conduction resistance of a cylindrical layer between radii $r_1$ and $r_2$ with length $L$:   $R_{\text{cyl}} = \frac{\ln(r_2 / r_1)}{2 \pi L k}$

• Total resistance with inner and outer convection:   $R_{\text{total}} = \frac{1}{2 \pi r_1 L h_1} + \frac{\ln(r_2 / r_1)}{2 \pi L k} + \frac{1}{2 \pi r_2 L h_2}$

• For multi-layer composite cylinders:   $R_{\text{total}} = \frac{1}{h_1 A_1} + \sum_{i=1}^{n} \frac{\ln(r_{i+1} / r_i)}{2 \pi L k_i} + \frac{1}{h_2 A_{n+1}}$

Page 11: Radial Systems – Spherical Layers

• Steady 1D radial governing equation:   $\frac{d}{dr}\left(r^2 \frac{dT}{dr}\right) = 0 \quad \text{or} \quad r \frac{d^2 T}{dr^2} + 2 \frac{dT}{dr} = 0$

• Conduction resistance of a spherical shell between radii $r_1$ and $r_2$:   $R_{\text{sph}} = \frac{r_2 - r_1}{4 \pi r_1 r_2 k}$

• Rate of heat conduction through the shell:   $\dot{Q}_{\text{cond, sph}} = \frac{T_1 - T_2}{R_{\text{sph}}}$

Page 12: Problem 3-23 Double-Pane Window Specification

• Geometry: Window area $A = 1.0\,\text{m} \times 1.5\,\text{m} = 1.5\,\text{m}^2$.

• Layers: Two glass layers ($L_g = 0.004\,\text{m}$, $k_g = 0.78\,\text{W/m}\cdot\text{K}$) separated by an air gap ($L_a = 0.005\,\text{m}$, $k_a = 0.025\,\text{W/m}\cdot\text{K}$).

• Temperatures and coefficients: $T_{\infty 1} = 20\,^\circ\text{C}$, $h_i = 40\,\text{W/m}^2\cdot\text{K}$, $T_{\infty 2} = -20\,^\circ\text{C}$, $h_o = 20\,\text{W/m}^2\cdot\text{K}$.

• Objectives: Find steady heat loss rate $\dot{Q}$ and the temperature drop across the largest thermal resistance.

Page 13: Solution to Problem 3-23

• Total thermal resistance per unit area:   $R_{\text{total}} = \frac{1}{h_i A} + \frac{L_g}{k_g A} + \frac{L_a}{k_a A} + \frac{L_g}{k_g A} + \frac{1}{h_o A}$   $R_{\text{total}} = \frac{0.025 + 0.000513 + 0.2 + 0.000513 + 0.05}{1.5} = 0.1840\,^\circ\text{C/W}$

• Heat loss rate:   $\dot{Q} = \frac{T_{\infty 1} - T_{\infty 2}}{R_{\text{total}}} = 210\,\text{W}$

• The largest resistance is the air gap ($R_a = \frac{L_a}{k_a A} = 0.1333\,^\circ\text{C/W}$, normalized $0.2 / A$):   $\Delta T_a = \dot{Q} R_a = (210\,\text{W}) \frac{0.005\,\text{m}}{(0.025\,\text{W/m}\cdot\text{K})(1.5\,\text{m}^2)} = 28\,^\circ\text{C}$

Page 14: Problem 3-29 Rear Window Defogger

• Data: Thickness $L = 5\,\text{mm}$, $k = 1.2\,\text{W/m}\cdot\text{K}$, uniform heating flux $\dot{q}_h = 1300\,\text{W/m}^2$.

• Interior air: $T_{\infty, i} = 22\,^\circ\text{C}$, $h_i = 15\,\text{W/m}^2\cdot\text{K}$.

• Exterior air: $T_{\infty, o} = -5\,^\circ\text{C}$, $h_o = 100\,\text{W/m}^2\cdot\text{K}$.

• Nodal conservation at heating element node ($T_1$):   $\dot{q}_h = \dot{q}_{\text{inside}} + \dot{q}_{\text{outside}} = \frac{T_1 - T_{\infty, i}}{R_i} + \frac{T_1 - T_{\infty, o}}{R_{\text{win}} + R_o}$

Page 15: Problem 3-51 Copper-Epoxy Sandwich Setup

• Configuration: $1\,\text{mm}$ copper plate ($k = 386\,\text{W/m}\cdot\text{K}$) between two $5\,\text{mm}$ epoxy boards ($k = 0.26\,\text{W/m}\cdot\text{K}$).

• Size: $15\,\text{cm} \times 20\,\text{cm}$.

• Thermal contact conductance: $h_c = 6000\,\text{W/m}^2\cdot\text{K}$ on both interfaces.

• Objective: Determine percentage error in total thermal resistance if contact conductance is omitted.

Page 16: Solution to Problem 3-51

• Resistances for unit area ($A = 1\,\text{m}^2$):   $R_{\text{contact}} = \frac{1}{h_c A} = 0.00017\,^\circ\text{C/W}$   $R_{\text{plate}} = \frac{L}{k A} = \frac{0.001}{(386)(1)} = 2.6 \times 10^{-6}\,^\circ\text{C/W}$   $R_{\text{epoxy}} = \frac{L}{k A} = \frac{0.005}{(0.26)(1)} = 0.01923\,^\circ\text{C/W}$

• Total resistance:   $R_{\text{total}} = 2 R_{\text{contact}} + R_{\text{plate}} + 2 R_{\text{epoxy}} = 0.03880\,^\circ\text{C/W}$

• Percentage error:   $\%\text{Error} = \frac{2 R_{\text{contact}}}{R_{\text{total}}} \times 100 = \frac{2 (0.00017)}{0.03880} \times 100 = 0.88\%$   (The contact resistance is negligible).

Page 17: Problem 3-66 Building Wall with Support Members

• Steel support members: $k = 50\,\text{W/m}\cdot\text{K}$, thickness $t_{23} = 8\,\text{cm}$, height $L_B = 0.5\,\text{cm}$.

• Insulation: $k = 0.03\,\text{W/m}\cdot\text{K}$, thickness $t_{23} = 8\,\text{cm}$, height $L_A = 60\,\text{cm}$.

• Gypsum inner board: $k = 0.5\,\text{W/m}\cdot\text{K}$, thickness $t_{12} = 1\,\text{cm}$.

• Brick outer wall: $k = 1.0\,\text{W/m}\cdot\text{K}$, thickness $t_{34} = 10\,\text{cm}$.

• Boundary temperatures: $T_1 = 20\,^\circ\text{C}$, $T_4 = 35\,^\circ\text{C}$.

Page 18: Problem 3-66 Domain and Thermal Modeling

• Unit section height: $L_A + L_B = 60 + 0.5 = 60.5\,\text{cm} = 0.605\,\text{m}$.

• Wall section consists of:

  • Gypsum board in series ($R_{12}$

  • Parallel combination of insulation ($R_{23a}$) and steel stud ($R_{23b}$)

  • Brick layer in series ($R_{34}$

Page 19: Solution to Problem 3-66

• Unit area resistances ($1\,\text{m}^2$):   $R_{12} = \frac{t_{12}}{k_{12}} = \frac{0.01}{0.5} = 0.02\,\text{m}^2\cdot^\circ\text{C/W}$   $R_{23a} = t_{23} \frac{L_a + L_b}{k_{23a} L_a} = (0.08) \frac{0.6 + 0.005}{(0.03)(0.6)} = 2.645\,\text{m}^2\cdot^\circ\text{C/W}$   $R_{23b} = t_{23} \frac{L_a + L_b}{k_{23b} L_b} = (0.08) \frac{0.6 + 0.005}{(50)(0.005)} = 1.32 \times 10^{-5}\,\text{m}^2\cdot^\circ\text{C/W}$   $R_{34} = \frac{t_{34}}{k_{34}} = \frac{0.1}{1.0} = 0.1\,\text{m}^2\cdot^\circ\text{C/W}$

• Combined total resistance:   $R_{\text{total}} = R_{12} + \left(\frac{R_{23a} R_{23b}}{R_{23a} + R_{23b}}\right) + R_{34} = 0.02 + 1.32 \times 10^{-5} + 0.1 = 0.120\,\text{m}^2\cdot^\circ\text{C/W}$

• Heat flux:   $\dot{q} = \frac{T_4 - T_1}{R_{\text{total}}} = \frac{(35 - 20)\,^\circ\text{C}}{0.120\,\text{m}^2\cdot^\circ\text{C/W}} = 125\,\text{W/m}^2$

Page 20: Critical Radius of Insulation

• Adding insulation increases conduction resistance but decreases outer surface convection resistance due to increased surface area.

• Total resistance as a function of outer radius $r_2$:   $R_{\text{total}} = \frac{\ln(r_2 / r_1)}{2 \pi k L} + \frac{1}{2 \pi r_2 L h}$

• Extreme value condition ($\frac{dR_{\text{total}}}{dr_2} = 0$):   $r_{\text{cr, cylinder}} = \frac{k}{h}$   $r_{\text{cr, sphere}} = \frac{2 k}{h}$

• Significance: For $r_1 < r_{\text{cr}}$, adding insulation increases heat loss up to $r_2 = r_{\text{cr}}$ where $\dot{Q}$ is maximum.

Page 21: Effective Radius of Insulation

• Defined as the outer radius $r_{\text{eff}}$ where heat transfer equals that of the bare cylinder ($\dot{Q} = \dot{Q}_{\text{bare}}$).

• Balance equation:   $\frac{\ln(r_{\text{eff}} / r_1)}{k} + \frac{1}{h r_{\text{eff}}} = \frac{1}{h r_1}$

• In terms of critical radius:   $\ln\left(\frac{r_{\text{eff}}}{r_1}\right) + \frac{r_{\text{cr}}}{r_{\text{eff}}} = \frac{r_{\text{cr}}}{r_1}$

Page 22: Problem 3-83 Insulated Wire Specification

• Wire: Diameter $D_1 = 2.2\,\text{mm}$ ($r_1 = 1.1\,\text{mm}$), length $L = 10\,\text{m}$.

• Cover: Plastic thickness $t = 1\,\text{mm}$ ($r_2 = 2.1\,\text{mm}$), $k = 0.15\,\text{W/m}\cdot\text{K}$.

• Electrical input: $I = 13\,\text{A}$, $V = 8\,\text{V}$.

• Convection: $T_\infty = 30\,^\circ\text{C}$, $h = 24\,\text{W/m}^2\cdot\text{K}$.

• Objectives: Find interface temperature $T_1$ and determine the effect of doubling plastic thickness.

Page 23: Solution to Problem 3-83

• Heat dissipation rate:   $\dot{Q} = V I = (8\,\text{V})(13\,\text{A}) = 104\,\text{W}$

• Thermal resistances:   $R_{\text{conv}} = \frac{1}{h (2 \pi r_2 L)} = \frac{1}{(24)[\pi (0.0042)(10)]} = 0.3158\,^\circ\text{C/W}$   $R_{\text{plastic}} = \frac{\ln(r_2 / r_1)}{2 \pi k L} = \frac{\ln(2.1 / 1.1)}{2 \pi (0.15)(10)} = 0.0686\,^\circ\text{C/W}$   $R_{\text{total}} = 0.3844\,^\circ\text{C/W}$

• Interface temperature:   $T_1 = T_\infty + \dot{Q} R_{\text{total}} = 30 + (104)(0.3844) = 70.0\,^\circ\text{C}$

• Critical radius:   $r_{\text{cr}} = \frac{k}{h} = \frac{0.15}{24} = 0.00625\,\text{m} = 6.25\,\text{mm}$

• Conclusion: Doubling thickness yields $r_2 = 3.1\,\text{mm} < r_{\text{cr}}$. Heat loss increases, decreasing interface temperature.

Page 24: Overall Heat Transfer Coefficient ($U$)

• Definition for combined systems:   $\dot{Q} = U A \Delta T = \frac{\Delta T}{R_{\text{total}}}$

• Equivalence:   $U A = \frac{1}{R_{\text{total}}}\; (\text{W/K})$

• Context: Highly useful for steady-state heat exchanger design.

Page 25: Thermal Contact Resistance

• Phenomenon: Actual physical contact between surfaces occurs at discrete microscopic high points; remaining space is occupied by interstitial voids.

• Temperature discontinuity $\Delta T_{\text{interface}}$ arises across the joint:   $R_c = \frac{\Delta T_{\text{interface}}}{\dot{Q}/A} = \frac{1}{h_c}\; (\text{m}^2\cdot^\circ\text{C/W})$   where $h_c$ is the thermal contact conductance.

Page 26: Interfacial Conductance & Contact Resistance Reduction

• Contact conductance ($h_c$) for aluminum plates ($10\,\mu\text{m}$ roughness, $1\,\text{atm}$):

  • Air: $3640\,\text{W/m}^2\cdot\text{K}$

  • Helium: $9520\,\text{W/m}^2\cdot\text{K}$

  • Hydrogen: $13{,}900\,\text{W/m}^2\cdot\text{K}$

  • Silicone oil: $19{,}000\,\text{W/m}^2\cdot\text{K}$

  • Glycerin: $37{,}700\,\text{W/m}^2\cdot\text{K}$

• Methods to minimize $R_c$:

  • Thermal grease (e.g., silicone oil)

  • High-conductivity interstitial gas (He, $H_2$)

  • Soft metallic foils (tin, silver, copper, nickel, aluminum)

  • Increasing joint contact pressure

Page 27: Finned Surfaces Principle

• Newton's cooling rate $\dot{Q} = h A_s (T_s - T_\infty)$ is enhanced either by increasing $h$ or surface area $A_s$.

• Fins enhance heat dissipation by extending convection area into the surrounding fluid.

Page 28: Governing Fin Equation

• Conservation of energy on differential volume element $\Delta x$:   $\dot{Q}_{\text{cond}, x} - \dot{Q}_{\text{cond}, x+\Delta x} - \dot{Q}_{\text{conv}} = 0$

• Differential form with uniform cross-section ($A_c$) and perimeter ($p$):   $\frac{d^2 \theta}{dx^2} - m^2 \theta = 0$   where:   $\theta(x) = T(x) - T_\infty$   $m^2 = \frac{h p}{k A_c}$

• General solution:   $\theta(x) = C_1 e^{mx} + C_2 e^{-mx}$

• Boundary condition at base ($x = 0$):   $\theta(0) = \theta_b = T_b - T_\infty$

Page 29: Fin Solution 1 – Infinitely Long Fin

• Tip condition ($L \rightarrow \infty$): $\theta(L) = 0$

• Temperature profile:   $\frac{T(x) - T_\infty}{T_b - T_\infty} = e^{-mx}$

• Heat transfer rate:   $\dot{Q}_{\text{long fin}} = \sqrt{h p k A_c} (T_b - T_\infty)$

• Condition for fin enhancement over bare area ($\dot{Q}_{\text{long fin}} > \dot{Q}_{\text{bare}}$):   $k > h \frac{A_c}{p}$

Page 30: Fin Solution 2 – Adiabatic Tip

• Tip condition ($x = L$): $\left.\frac{d\theta}{dx}\right|_{x=L} = 0$

• Temperature profile:   $\frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh[m(L - x)]}{\cosh(mL)}$

• Heat transfer rate:   $\dot{Q}_{\text{adiabatic}} = \sqrt{h p k A_c} (T_b - T_\infty) \tanh(mL)$

Page 31: Fin Solution 3 – Specified Tip Temperature

• Tip condition ($x = L$): $\theta(L) = \theta_L = T_L - T_\infty$

• Temperature profile:   $\frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\left(\frac{T_L - T_\infty}{T_b - T_\infty}\right) \sinh(mx) + \sinh[m(L - x)]}{\sinh(mL)}$

• Heat transfer rate:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty) \frac{\cosh(mL) - \left(\frac{T_L - T_\infty}{T_b - T_\infty}\right)}{\sinh(mL)}$

Page 32: Fin Solution 4 – Convection at Tip

• Tip condition ($x = L$): $-k \left.\frac{dT}{dx}\right|_{x=L} = h (T(L) - T_\infty)$

• Temperature distribution:   $\frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh[m(L - x)] + \left(\frac{h}{mk}\right) \sinh[m(L - x)]}{\cosh(mL) + \left(\frac{h}{mk}\right) \sinh(mL)}$

• Heat transfer rate:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty) \frac{\sinh(mL) + \left(\frac{h}{mk}\right) \cosh(mL)}{\cosh(mL) + \left(\frac{h}{mk}\right) \sinh(mL)}$

Page 33: Corrected Fin Length Approximation

• An adiabatic tip solution is adapted for tip convection by defining a corrected fin length ($L_c$):   $L_c = L + \frac{A_c}{p}$

• Rectangular fin ($A_c = w t$, $p \approx 2w$):   $L_c = L + \frac{t}{2}$

• Cylindrical fin ($A_c = \frac{\pi D^2}{4}$, $p = \pi D$):   $L_c = L + \frac{D}{4}$

Page 34: Fin Efficiency ($\eta_{\text{fin}}$

• Definition:   $\eta_{\text{fin}} = \frac{\dot{Q}_{\text{fin}}}{\dot{Q}_{\text{fin, max}}} = \frac{\dot{Q}_{\text{fin}}}{h A_{\text{fin}} (T_b - T_\infty)}$

• Long fin efficiency:   $\eta_{\text{long fin}} = \frac{1}{mL}$

• Adiabatic tip fin efficiency:   $\eta_{\text{fin}} = \frac{\tanh(mL)}{mL}$

Page 35: Efficiency of Straight Longitudinal Fins

• Charts express efficiency $\eta_{\text{fin}}$ as a function of $\xi = L_c^{3/2} (h / k A_p)^{1/2}$, where $A_p$ is the profile area:

  • Rectangular: $L_c = L + \frac{t}{2}$, $A_p = L_c t$

  • Triangular: $L_c = L$, $A_p = \frac{L_c t}{2}$

  • Parabolic: $L_c = L$, $A_p = \frac{L_c t}{3}$

Page 36: Efficiency of Annular Fins

• Parameter: $\xi = L_c^{3/2} (h / k A_p)^{1/2}$, evaluated with $r_{2c} / r_1$ curves:

  • $r_{2c} = r_2 + \frac{t}{2}$

  • $L_c = L + \frac{t}{2}$

  • $A_p = L_c t$

Page 37: Fin Effectiveness ($\varepsilon_{\text{fin}}$

• Definition:   $\varepsilon_{\text{fin}} = \frac{\dot{Q}_{\text{fin}}}{\dot{Q}_{\text{no fin}}} = \frac{\dot{Q}_{\text{fin}}}{h A_b (T_b - T_\infty)} = \frac{A_{\text{fin}}}{A_b} \eta_{\text{fin}}$

• For infinitely long fins:   $\varepsilon_{\text{long fin}} = \sqrt{\frac{k p}{h A_c}}$

• Design Rule: Finning is justified only when $\varepsilon_{\text{fin}} > 2$.

Page 38: Total Finned Surface Heat Transfer

• Total heat transfer from finned and unfinned regions combined:   $\dot{Q}_{\text{total, fin}} = \dot{Q}_{\text{unfin}} + \dot{Q}_{\text{fin}} = h (A_{\text{unfin}} + \eta_{\text{fin}} A_{\text{fin}})(T_b - T_\infty)$

• Overall surface effectiveness:   $\varepsilon_{\text{fin, overall}} = \frac{\dot{Q}_{\text{total, fin}}}{\dot{Q}_{\text{total, no fin}}} = \frac{A_{\text{unfin}} + \eta_{\text{fin}} A_{\text{fin}}}{A_{\text{no fin}}}$

Page 39: Proper Fin Length Selection

• Heat dissipation variation relative to an infinite fin ($\tanh(mL)$):

  • $mL = 0.1 \rightarrow 0.100$

  • $mL = 0.5 \rightarrow 0.462$

  • $mL = 1.0 \rightarrow 0.762$

  • $mL = 1.5 \rightarrow 0.905$

  • $mL = 2.0 \rightarrow 0.964$

  • $mL = 5.0 \rightarrow 1.000$

• Criteria: $mL = 5$ represents an infinitely long fin; $mL = 1$ is standard for economic design balance between performance and material cost.

Page 40: Fin Analysis Procedure

• Identify cross-sectional geometry ($k$, $A_c$, $p$).

• Verify boundary condition at the tip (long fin, adiabatic, specified temperature, or convection).

• Select between analytical equations, efficiency charts, or empirical relations.

Page 41: Consolidated Fin Formulas

• Infinite Fin:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty)$

• Adiabatic Tip:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty) \tanh(mL)$

• Specified Tip Temperature:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty) \frac{\cosh(mL) - \left(\frac{T_L - T_\infty}{T_b - T_\infty}\right)}{\sinh(mL)}$

• Convective Tip:   $\dot{Q} = \sqrt{h p k A_c} (T_b - T_\infty) \frac{\sinh(mL) + \left(\frac{h}{mk}\right) \cosh(mL)}{\cosh(mL) + \left(\frac{h}{mk}\right) \sinh(mL)}$

• Chart-Based Method:   $\dot{Q}_{\text{fin}} = \eta_{\text{fin}} h A_{\text{fin}} (T_b - T_\infty)$

Page 42: Solution to Problem 3-121 Gas Turbine Blade

• Parameters: $k = 17\,\text{W/m}\cdot^\circ\text{C}$, $L = 0.053\,\text{m}$, $p = 0.11\,\text{m}$, $A_c = 5.13 \times 10^{-4}\,\text{m}^2$, $T_\infty = 973\,^\circ\text{C}$, $h = 538\,\text{W/m}^2\cdot^\circ\text{C}$, $T_b = 450\,^\circ\text{C}$, adiabatic tip.

• Calculations:   $m L = \left(\frac{h p}{k A_c}\right)^{0.5} L = \left[\frac{(538)(0.11)}{(17)(5.13 \times 10^{-4})}\right]^{0.5} (0.053) = 4.366$   $\sqrt{h p k A_c} = 0.7184\,\text{W/}^\circ\text{C}$   $\dot{Q}_{\text{blade}} = (0.7184)(973 - 450) \tanh(4.366) = 376\,\text{W}$

• Tip temperature:   $T_L = T_\infty + \frac{T_b - T_\infty}{\cosh(mL)} = 973 + \frac{450 - 973}{\cosh(4.366)} = 960\,^\circ\text{C}$

Page 43: Problem 3-128 Annular Finned Tube Setup

• Tube: Outer diameter $D_1 = 5\,\text{cm}$, wall temperature $T_b = 180\,^\circ\text{C}$.

• Fins: Aluminum alloy 2024-T6 ($k = 186\,\text{W/m}\cdot\text{K}$), outer diameter $D_2 = 6\,\text{cm}$, thickness $t = 1\,\text{mm}$.

• Array: 250 fins per meter length (spacing $s = 3\,\text{mm}$).

• Ambient: $T_\infty = 25\,^\circ\text{C}$, $h = 40\,\text{W/m}^2\cdot\text{K}$.

Page 44: Methodology for Problem 3-128

• Determine heat loss from bare tube per meter length.

• Calculate annular fin efficiency $\eta_{\text{fin}}$ via parameter $\xi$.

• Determine combined heat dissipation from 250 fins and 250 unfinned tube spaces.

• Compute net heat transfer increase: $\dot{Q}_{\text{increase}} = \dot{Q}_{\text{total, fin}} - \dot{Q}_{\text{no fin}}$.

Page 45: Annular Fin Efficiency Chart (Figure 3-44)

• Design parameters:   $r_1 = 25\,\text{mm}$, $r_2 = 30\,\text{mm}$, $L = 5\,\text{mm}$   $r_{2c} = r_2 + \frac{t}{2} = 30.5\,\text{mm} \rightarrow \frac{r_{2c}}{r_1} = 1.22$   $\xi = L_c^{3/2} \left(\frac{h}{k A_p}\right)^{1/2} = 0.08$

• Extracted fin efficiency: $\eta_{\text{fin}} = 0.97$.

Page 46: Solution to Problem 3-128

• Heat transfer without fins ($1\,\text{m}$):   $A_{\text{no fin}} = \pi D_1 L = 0.1571\,\text{m}^2$   $\dot{Q}_{\text{no fin}} = h A_{\text{no fin}} (T_b - T_\infty) = (40)(0.1571)(180 - 25) = 974\,\text{W}$

• Single fin heat transfer:   $A_{\text{fin}} = 2 \pi (r_2^2 - r_1^2) + 2 \pi r_2 t = 0.001916\,\text{m}^2$   $\dot{Q}_{\text{fin}} = \eta_{\text{fin}} h A_{\text{fin}} (T_b - T_\infty) = 0.97 (40)(0.001916)(155) = 11.52\,\text{W}$

• Single unfinned section ($s = 3\,\text{mm}$):   $A_{\text{unfin}} = \pi D_1 s = 0.0004712\,\text{m}^2$   $\dot{Q}_{\text{unfin}} = h A_{\text{unfin}} (T_b - T_\infty) = (40)(0.0004712)(155) = 2.92\,\text{W}$

• Total heat transfer and increase:   $\dot{Q}_{\text{total, fin}} = 250 (11.52 + 2.92) = 3610\,\text{W}$   $\dot{Q}_{\text{increase}} = 3610 - 974 = 2636\,\text{W}$

Page 47: Conduction Shape Factor ($S$)

• For steady-state two-dimensional conduction between two surfaces at $T_1$ and $T_2$:   $\dot{Q} = S k (T_1 - T_2)$   $R = \frac{1}{S k}$

• Selected shape factors ($S$):

  1. Horizontal cylinder buried in semi-infinite medium ($L \gg D$, $z > 1.5 D$):      $S = \frac{2 \pi L}{\ln(4z / D)}$

  2. Vertical cylinder in semi-infinite medium ($L \gg D$):      $S = \frac{2 \pi L}{\ln(4L / D)}$

  3. Two parallel cylinders in infinite medium ($L \gg D_1, D_2, z$):      $S = \frac{2 \pi L}{\cosh^{-1}\left(\frac{4z^2 - D_1^2 - D_2^2}{2 D_1 D_2}\right)}$

  4. Row of equally spaced cylinders in semi-infinite medium ($w > 1.5 D$):      $S = \frac{2 \pi L}{\ln\left(\frac{2w}{\pi D} \sinh\left(\frac{2 \pi z}{w}\right)\right)}$

  5. Cylinder centered in infinite wall ($z > 0.5 D$):      $S = \frac{2 \pi L}{\ln(8z / \pi D)}$

  6. Cylinder centered in square bar of same length:      $S = \frac{2 \pi L}{\ln(1.08 w / D)}$