13.1 Steady & Unsteady Conduction, Extended Surfaces (Fins)

Steady & Unsteady Conduction, Extended Surfaces (Fins)

Conduction is the mode of heat transfer in which energy is transferred from more energetic particles of a substance to adjacent, less energetic particles as a result of microscopic interactions: atomic lattice vibrations (phonons) and molecular collisions, alongside free electron transport in conducting media. In mining engineering operations—ranging from the thermal management of high-power continuous miner electric motors, dragline brake assemblies, and underground transformer vaults to steam pipeline insulation in thermal power co-generation units—mastery of steady and transient conduction is vital for equipment reliability and safety.

---

1. Fourier's Law of Heat Conduction & Thermal Conductivity Mechanisms

Fourier's law is the phenomenological governing law of conductive heat transfer. It states that the time rate of heat transfer through a material is proportional to the negative gradient in temperature and to the area through which the heat flows at right angles to that gradient:

$q = -k A \frac{dT}{dx} \quad \implies \quad q'' = \frac{q}{A} = -k \frac{dT}{dx}$

Where:

• $q = \text{Heat transfer rate } (\text{W})$
• $q'' = \text{Heat flux } (\text{W/m}^2)$
• $k = \text{Thermal conductivity of the material } (\text{W/m}\cdot\text{K})$
• $A = \text{Cross-sectional area normal to heat flow } (\text{m}^2)$
• $\frac{dT}{dx} = \text{Temperature gradient in the direction of heat flow } (\text{K/m})$

[!NOTE] The Negative Sign: The negative sign in Fourier's law is a physical requirement of the Second Law of Thermodynamics, ensuring that heat flows spontaneously down the temperature gradient from higher temperature to lower temperature (where $dT/dx < 0$, making $q > 0$).

+-----------------------------------------------------------------------------------------+
|                       FOURIER'S LAW & THERMAL RESISTANCE CONCEPT                        |
|                                                                                         |
|             Higher Temp (T1)                      Lower Temp (T2)                       |
|             +---------------------------------------------------+                       |
|             | ---> Heat Flux q'' = -k*(dT/dx)                   |                       |
|             |                                                   |                       |
|   Area A    |        Material Thermal Conductivity = k          |   Thickness L         |
|             |                                                   |                       |
|             +---------------------------------------------------+                       |
|             x = 0                                               x = L                   |
|                                                                                         |
|   Thermal Resistance Analogy:                                                           |
|   [T1] o--------------------/\/\/\/\/\/--------------------o [T2]                       |
|                              R_th = L / (k*A)                                           |
|   Heat Flow:   q = (T1 - T2) / R_th = Delta_T / R_th                                    |
+-----------------------------------------------------------------------------------------+

Microscopic Mechanisms of Thermal Conductivity ($k$)

Thermal conductivity varies across states of matter based on microscopic energy transport carriers:

$k = k_e + k_{\text{ph}}$

Where $k_e$ is the electronic contribution (free electron migration) and $k_{\text{ph}}$ is the lattice vibration contribution (phonons).

1. Pure Metals: Dominated by free electrons ($k_e \gg k_{\text{ph}}$). Pure copper ($k \approx 386\text{ W/m}\cdot\text{K}$) and aluminum ($k \approx 205\text{ W/m}\cdot\text{K}$) exhibit extremely high conductivity. Under the Wiedemann-Franz Law, the ratio of thermal conductivity to electrical conductivity ($\sigma_e$) is proportional to absolute temperature:

$\frac{k}{\sigma_e T} = L_0 = 2.44 \times 10^{-8} \ \text{W}\cdot\Omega/\text{K}^2 \quad (\text{Lorentz Number})$

1. Alloys: Impurity atoms and structural dislocations scatter free electrons, significantly depressing thermal conductivity (e.g., carbon steel $k \approx 45\text{ W/m}\cdot\text{K}$, austenitic stainless steel $k \approx 15\text{ W/m}\cdot\text{K}$).
2. Non-Metallic Solids (Dielectrics): Heat conducts exclusively through lattice phonons ($k_e \approx 0$). In disordered structures (glass, refractory bricks), phonon scattering by amorphous boundaries yields low conductivity ($k \approx 0.8\text{--}1.4\text{ W/m}\cdot\text{K}$).
3. Liquids: Energy is transported by longitudinal acoustic molecular vibrations. In general, thermal conductivity in liquids decreases with increasing temperature, with water being a notable anomaly where $k$ rises to a maximum near $130^{\circ}\text{C}$ before decreasing.
4. Gases: Governed by kinetic theory of gases: $k = \frac{1}{3} \rho \bar{v} \lambda c_v \propto \frac{\sqrt{T}}{\sqrt{M}}$. Thermal conductivity in gases is independent of pressure across standard industrial operating ranges ($0.01\text{--}10\text{ bar}$) and increases with $\sqrt{T}$.

---

2. Three-Dimensional General Heat Conduction Equations

Applying the First Law of Thermodynamics to a differential control volume yields the general three-dimensional transient heat conduction equation with internal heat generation.

Cartesian Coordinates $(x, y, z)$

$\frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x}\right) + \frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y}\right) + \frac{\partial}{\partial z}\left(k \frac{\partial T}{\partial z}\right) + \dot{q}_g = \rho c_p \frac{\partial T}{\partial t}$

For a homogeneous, isotropic material with constant thermal conductivity ($k = \text{const}$):

$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}_g}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t} \quad \implies \quad \nabla^2 T + \frac{\dot{q}_g}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$

Where $\alpha = \frac{k}{\rho c_p}\ (\text{m}^2/\text{s})$ is the thermal diffusivity, quantifying a medium's ability to conduct thermal energy relative to its capacity to store thermal energy.

+-----------------------------------------------------------------------------------------+
|                       SPECIAL FORMS OF CONDUCTION EQUATION                              |
|                                                                                         |
|   1. Fourier Equation:    nabla^2 T = (1/alpha)*(dT/dt)          (No heat gen, unsteady)|
|   2. Poisson Equation:    nabla^2 T + (q_g_dot / k) = 0          (Steady, heat gen)     |
|   3. Laplace Equation:    nabla^2 T = 0                          (Steady, no heat gen)  |
+-----------------------------------------------------------------------------------------+

Cylindrical Coordinates $(r, \phi, z)$

$\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 T}{\partial \phi^2} + \frac{\partial^2 T}{\partial z^2} + \frac{\dot{q}_g}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$

Spherical Coordinates $(r, \theta, \phi)$

$\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial T}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 T}{\partial \phi^2} + \frac{\dot{q}_g}{k} = \frac{1}{\alpha} \frac{\partial T}{\partial t}$

---

3. One-Dimensional Steady Conduction & Thermal Resistance Networks

Under steady-state conditions without internal heat generation, heat flow is governed by direct thermal circuit analogies ($q = \Delta T / R_{\text{th}}$).

+-----------------------------------------------------------------------------------------+
|                       THERMAL RESISTANCE FORMULAS BY GEOMETRY                           |
|                                                                                         |
|   Geometry               Conduction Resistance R_th         Heat Rate q                 |
|   --------------------   --------------------------------   -------------------------   |
|   Plane Slab (Area A)    R_th = L / (k*A)                   q = k*A*(T1 - T2) / L       |
|   Cylinder (Length L)    R_th = ln(r2/r1) / (2*pi*k*L)      q = 2*pi*k*L*Delta_T/ln(r2/r1|
|   Sphere (Radii r1, r2)  R_th = (r2 - r1) / (4*pi*k*r1*r2)  q = 4*pi*k*r1*r2*Delta_T/(r2-r1|
|   Convective Boundary    R_conv = 1 / (h*A)                 q = h*A*(T_s - T_inf)       |
+-----------------------------------------------------------------------------------------+

Composite Plane Walls in Series & Overall Heat Transfer Coefficient ($U$)

For a composite wall consisting of multiple material layers $1, 2, \dots, n$ with convective fluid boundaries on both sides:

+-----------------------------------------------------------------------------------------+
|                        COMPOSITE WALL THERMAL NETWORK CIRCUIT                           |
|                                                                                         |
|   Fluid (T_inf1, h1) | Layer A (L_A, k_A) | Layer B (L_B, k_B) | Fluid (T_inf2, h2)     |
|                      |                    |                    |                        |
|   [T_inf1] o---/\/\/\/---[T1]---/\/\/\/---[T2]---/\/\/\/---[T3]---/\/\/\/---o [T_inf2] |
|                R_conv1          R_cond,A         R_cond,B         R_conv2               |
|               1/(h1*A)          L_A/(k_A*A)      L_B/(k_B*A)      1/(h2*A)              |
+-----------------------------------------------------------------------------------------+

$R_{\text{total}} = \frac{1}{h_1 A} + \sum_{i=1}^n \frac{L_i}{k_i A} + \frac{1}{h_2 A}$

$\dot{q} = \frac{T_{\infty,1} - T_{\infty,2}}{R_{\text{total}}} = U A (T_{\infty,1} - T_{\infty,2})$

$\frac{1}{U} = \frac{1}{h_1} + \sum_{i=1}^n \frac{L_i}{k_i} + \frac{1}{h_2} \quad \implies \quad \frac{1}{UA} = \sum R_{\text{th}}$

Radial Systems: Composite Cylindrical Pipes

For an insulated industrial pipe of inner radius $r_1$, outer metal radius $r_2$, and insulation outer radius $r_3$, carrying hot fluid at $T_{\infty,i}$ ($h_i$) surrounded by ambient air at $T_{\infty,o}$ ($h_o$):

$R_{\text{total}} = \frac{1}{2\pi r_1 L h_i} + \frac{\ln(r_2/r_1)}{2\pi k_{\text{pipe}} L} + \frac{\ln(r_3/r_2)}{2\pi k_{\text{ins}} L} + \frac{1}{2\pi r_3 L h_o}$

$\dot{q} = \frac{T_{\infty,i} - T_{\infty,o}}{R_{\text{total}}} = U_i A_i \Delta T = U_o A_o \Delta T$

---

4. Critical Radius of Insulation

Adding insulation to a flat plane wall always increases conductive resistance ($L/kA$), thereby reducing heat transfer. However, adding insulation to a curved surface (cylinder or sphere) introduces two competing thermal effects:

1. Conduction Resistance: Increases logarithmically with insulation outer radius $r$ ($R_{\text{cond}} = \frac{\ln(r/r_1)}{2\pi k L}$).
2. Convection Resistance: Decreases linearly with external surface area ($R_{\text{conv}} = \frac{1}{2\pi r L h_o}$).

+-----------------------------------------------------------------------------------------+
|                        CRITICAL RADIUS OF INSULATION BEHAVIOR                           |
|                                                                                         |
|   Thermal Resistance (R)                                                                |
|     ^                                                                                   |
|     |           / R_total = R_cond + R_conv                                             |
|     |          /                                                                        |
|     |         /                 \                                                       |
|     |        /     R_cond        \                                                      |
|     |       /                     \                                                     |
|     |      /                       \_ R_conv                                            |
|     |     /                          \_                                                 |
|     |    /    R_total MINIMUM          \_                                               |
|     |   +------------*--------------------> (Max Heat Dissipation Rate)                 |
|     |   |            |                                                                  |
|     +---+------------+--------------------------------------------> Outer Radius (r)     |
|        r_1          r_cr = k / h                                                        |
|                                                                                         |
|   - If r_initial < r_cr: Adding insulation INCREASES heat transfer (ideal for cables).  |
|   - If r_initial > r_cr: Adding insulation DECREASES heat transfer (steam pipes).       |
+-----------------------------------------------------------------------------------------+

Derivation for a Cylindrical Geometry

Expressing total thermal resistance per unit length $R'$ as a function of insulation outer radius $r$:

$R' = \frac{\ln(r/r_1)}{2\pi k} + \frac{1}{2\pi r h}$

Differentiating with respect to $r$ and setting the derivative to zero to find the extremum:

$\frac{dR'}{dr} = \frac{1}{2\pi k r} - \frac{1}{2\pi h r^2} = 0 \quad \implies \quad \frac{1}{k} = \frac{1}{h r}$

$r_{\text{cr, cylinder}} = \frac{k}{h}$

Evaluating the second derivative reveals $\left.\frac{d^2 R'}{dr^2}\right|{r=r{\text{cr}}} > 0$, confirming that $R'$ is at a mathematical minimum, and heat dissipation $q$ is at a maximum.

Derivation for a Spherical Geometry

$R_{\text{total}} = \frac{r - r_1}{4\pi k r r_1} + \frac{1}{4\pi r^2 h} \quad \implies \quad \frac{dR_{\text{total}}}{dr} = \frac{1}{4\pi k r^2} - \frac{2}{4\pi h r^3} = 0$

$r_{\text{cr, sphere}} = \frac{2k}{h}$

[!IMPORTANT] Practical Design Rules:

• Electrical Conductors & Cables: Keep outer radius $r_o \le r_{\text{cr}}$ so that adding dielectric insulation maximizes heat dissipation, preventing copper core thermal runaway.
• Steam & Process Fluid Pipelines: Ensure the bare pipe outer radius $r_1 > r_{\text{cr}}$, so that every millimeter of insulation added immediately reduces costly heat losses.

---

5. One-Dimensional Steady Conduction with Uniform Heat Generation

When thermal energy is generated uniformly throughout a solid volume at rate $\dot{q}_g\ (\text{W/m}^3)$ (e.g., Joule heating $I^2 R/V$ in electrical busbars, core nuclear reactions, or concrete curing exotherm), internal temperature distributions become parabolic.

+-----------------------------------------------------------------------------------------+
|                        HEAT GENERATION: PLANE SLAB VS SOLID CYLINDER                    |
|                                                                                         |
|   Plane Slab (-L <= x <= +L, Cooled to Tw):    Solid Cylinder (0 <= r <= R, Cooled Tw): |
|                                                                                         |
|               Centerline (x = 0)                           Centerline (r = 0)           |
|                    T_max = T0                                   T_max = T0              |
|                       /\                                           /\                   |
|                      /  \                                         /  \                  |
|                     /    \                                       /    \                 |
|           Tw ______/      \______ Tw                   Tw ______/      \______ Tw       |
|                   -L      +L                                   0       R                |
|                                                                                         |
|   T(x) - Tw = (q_g*L^2 / 2k) * [1 - (x/L)^2]   T(r) - Tw = (q_g*R^2 / 4k) * [1 - (r/R)^2]|
|   T_max - Tw = q_g * L^2 / (2*k)               T_max - Tw = q_g * R^2 / (4*k)           |
+-----------------------------------------------------------------------------------------+

1. Plane Slab of Thickness $2L$ Cooled Symmetrically ($T(x = \pm L) = T_w$)

Starting from $\frac{d^2 T}{dx^2} + \frac{\dot{q}g}{k} = 0$, integrating twice and applying symmetry $\left.\frac{dT}{dx}\right|{x=0} = 0$:

$T(x) = T_w + \frac{\dot{q}_g L^2}{2k}\left(1 - \frac{x^2}{L^2}\right)$

$T_{\text{max}} = T_0 = T_w + \frac{\dot{q}_g L^2}{2k}$

If one surface ($x=0$) is perfectly insulated and the opposite face ($x=L$) is maintained at $T_w$, the boundary conditions are identical, and maximum temperature occurs at the insulated wall with $\Delta T_{\text{max}} = \frac{\dot{q}_g L^2}{2k}$.

2. Solid Cylinder of Radius $R$ Cooled at Surface ($T(r=R) = T_w$)

Integrating $\frac{1}{r}\frac{d}{dr}\left(r \frac{dT}{dr}\right) + \frac{\dot{q}g}{k} = 0$ with $\left.\frac{dT}{dr}\right|{r=0} = 0$:

$T(r) = T_w + \frac{\dot{q}_g R^2}{4k}\left(1 - \frac{r^2}{R^2}\right)$

$T_{\text{max}} = T_0 = T_w + \frac{\dot{q}_g R^2}{4k}$

---

6. Extended Surfaces (Fins): Governing Equations & Performance

Extended surfaces (fins) are engineered protrusions designed to enhance heat dissipation from a primary surface into a surrounding fluid by increasing effective convective surface area. They are ubiquitous on mining compressor cylinder heads, electronic power converters, and air-cooled heat exchangers.

+-----------------------------------------------------------------------------------------+
|                             FIN DIFFERENTIAL CONTROL VOLUME                             |
|                                                                                         |
|   Base (x=0)                  dx                                Tip (x=L)               |
|   +----------+          +------------+                         +--------+               |
|   | T = Tb   | ======>  | q_x        | ======> q_(x+dx)        |        |               |
|   |          |          +------------+                         +--------+               |
|   +----------+                |                                                         |
|                               v dQ_conv = h * (P*dx) * (T - T_inf)                      |
|                                                                                         |
|   Governing Fin Equation:   d^2(theta)/dx^2 - m^2 * theta = 0                           |
|   Where:  theta(x) = T(x) - T_inf,     m = sqrt( h*P / (k*Ac) )                         |
+-----------------------------------------------------------------------------------------+

Derivation of Governing Fin Equation

Applying energy balance on a 1D differential element of length $dx$, perimeter $P$, and cross-sectional area $A_c$:

$q_x - q_{x+dx} - dQ_{\text{conv}} = 0 \quad \implies \quad -\left(-k A_c \frac{dT}{dx}\right) + \left[-k A_c \frac{d}{dx}\left(T + \frac{dT}{dx} dx\right)\right] - h (P dx)(T - T_{\infty}) = 0$

$\frac{d^2 T}{dx^2} - \frac{h P}{k A_c}(T - T_{\infty}) = 0$

Defining excess temperature $\theta(x) = T(x) - T_{\infty}$ and fin parameter $m = \sqrt{\frac{h P}{k A_c}}\ (\text{m}^{-1})$:

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

General solution: $\theta(x) = C_1 e^{mx} + C_2 e^{-mx} = C_3 \cosh(mx) + C_4 \sinh(mx)$.

The Three Canonical Fin Boundary Conditions

[!TIP] Corrected Fin Length ($L_c$): Instead of evaluating cumbersome convective tip hyperbolic series, apply the insulated-tip formula using a corrected length: $L_c = L + \frac{A_c}{P}$. For a rectangular fin of thickness $t$, $L_c = L + \frac{t}{2}$; for a pin fin of diameter $d$, $L_c = L + \frac{d}{4}$.

Fin Efficiency ($\eta_{\text{fin}}$) vs. Fin Effectiveness ($\epsilon_{\text{fin}}$)

1. Fin Efficiency ($\eta_{\text{fin}}$): The ratio of actual heat transferred by the fin to the maximum hypothetical heat transfer if the entire fin surface were maintained at base temperature $\theta_0$:

$\eta_{\text{fin}} = \frac{Q_{\text{actual}}}{Q_{\text{max}}} = \frac{Q_{\text{fin}}}{h A_{\text{fin}} \theta_0} = \frac{\sqrt{h P k A_c} \theta_0 \tanh(mL)}{h (P L) \theta_0} = \frac{\tanh(mL)}{mL} \quad (\text{for insulated tip})$

1. Fin Effectiveness ($\epsilon_{\text{fin}}$): The ratio of heat transferred with the fin to the heat transfer from the bare base area without the fin:

$\epsilon_{\text{fin}} = \frac{Q_{\text{fin}}}{Q_{\text{without fin}}} = \frac{\sqrt{h P k A_c} \theta_0 \tanh(mL)}{h A_c \theta_0} = \sqrt{\frac{k P}{h A_c}} \tanh(mL)$

$\epsilon_{\text{fin}} = \eta_{\text{fin}} \frac{A_{\text{fin}}}{A_c}$

+-----------------------------------------------------------------------------------------+
|                           CRITERIA FOR EFFECTIVE FIN DESIGN                             |
|                                                                                         |
|   - Thermal conductivity k MUST be high (use aluminum k=205 W/m*K or copper k=386).     |
|   - Convective coefficient h MUST be low (fins on air/gas side, NOT liquid/water side). |
|   - Perimeter-to-area ratio (P / Ac) MUST be large (thin, slender, closely spaced fins).|
|   - Justification Threshold: epsilon_fin >= 2.0 (if epsilon_fin < 1, fin acts as        |
|     an insulator and impedes heat dissipation!).                                        |
+-----------------------------------------------------------------------------------------+

---

7. Transient Conduction & The Lumped Capacitance Method

In transient (unsteady-state) conduction, temperature varies with both spatial position and time ($T = T(x, y, z, t)$). When internal conductive resistance within a solid is negligible compared to external convective resistance, internal temperature gradients vanish, and the solid behaves as a single spatially uniform lumped thermal capacitance.

+-----------------------------------------------------------------------------------------+
|                        LUMPED CAPACITANCE ENERGY BALANCE                                |
|                                                                                         |
|   Solid Body: Volume V, Area As, Density rho, Specific Heat cp, Initial Temp T0         |
|                                                                                         |
|               - dE_stored / dt = dQ_conv / dt                                           |
|               - rho * V * cp * (dT/dt) = h * As * (T - T_inf)                           |
|                                                                                         |
|   Temperature Response:                                                                 |
|   (T(t) - T_inf) / (T0 - T_inf) = exp( - (h*As / (rho*V*cp)) * t )                      |
|                                 = exp( - t / tau ) = exp( - Bi * Fo )                   |
+-----------------------------------------------------------------------------------------+

Governing Non-Dimensional Parameters

1. Characteristic Length ($L_c$):

$L_c = \frac{\text{Volume } V}{\text{Surface Area } A_s}$

• Plane slab of thickness $2L$ cooled on both sides: $L_c = \frac{2L A}{2A} = L$
• Long cylinder of radius $R$: $L_c = \frac{\pi R^2 L}{2\pi R L} = \frac{R}{2}$
• Sphere of radius $R$: $L_c = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}$
• Cube of side $L$: $L_c = \frac{L^3}{6L^2} = \frac{L}{6}$

1. Biot Number ($\text{Bi}$): Ratio of internal conductive resistance to external convective boundary resistance:

$\text{Bi} = \frac{R_{\text{cond}}}{R_{\text{conv}}} = \frac{L_c / (k_{\text{solid}} A_s)}{1 / (h A_s)} = \frac{h L_c}{k_{\text{solid}}}$

[!IMPORTANT] Validity Criterion: The lumped capacitance method is strictly valid if and only if $\text{Bi} < 0.1$. When $\text{Bi} < 0.1$, the maximum temperature variation across the solid body is less than 5%, and spatial gradients can be safely neglected.

1. Fourier Number ($\text{Fo}$): Dimensionless time variable measuring the rate of heat conduction relative to thermal storage:

$\text{Fo} = \frac{\alpha t}{L_c^2} = \frac{k t}{\rho c_p L_c^2}$

$\text{Bi} \cdot \text{Fo} = \left(\frac{h L_c}{k}\right) \left(\frac{k t}{\rho c_p L_c^2}\right) = \frac{h A_s}{\rho V c_p} t = \frac{t}{\tau}$

Where $\tau = \frac{\rho V c_p}{h A_s} = R_{\text{conv}} C_{\text{th}}$ is the Thermal Time Constant of the system (seconds).

---

8. Step-by-Step Worked Calculation: Industrial Cable Insulation & Pin Fin Heat Sink

Worked Example 1: Critical Insulation Radius for Mining Feeder Cable

A high-voltage electrical cable in an underground mine shaft has a bare copper radius of $r_1 = 5\text{ mm}$ and is coated with rubber insulation ($k = 0.16\text{ W/m}\cdot\text{K}$). The cable is exposed to ventilation air with convective heat transfer coefficient $h = 20\text{ W/m}^2\cdot\text{K}$.

+-----------------------------------------------------------------------------------------+
|                        CABLE INSULATION CALCULATION STEPS                               |
|                                                                                         |
|   1. Calculate Critical Radius:                                                         |
|      r_cr = k / h = (0.16 W/m*K) / (20 W/m^2*K) = 0.008 m = 8.0 mm                      |
|                                                                                         |
|   2. Compare with Bare Cable Radius:                                                    |
|      r_bare = 5.0 mm < r_cr (8.0 mm)                                                    |
|                                                                                         |
|   3. Critical Insulation Thickness:                                                     |
|      t_cr = r_cr - r_1 = 8.0 mm - 5.0 mm = 3.0 mm                                       |
|                                                                                         |
|   4. Physical Interpretation:                                                           |
|      Adding up to 3.0 mm of rubber insulation INCREASES current carrying capacity       |
|      by maximizing heat dissipation rate from the copper core.                          |
+-----------------------------------------------------------------------------------------+

Worked Example 2: Transient Cooling of Steel Bearing Ball

A chrome-steel bearing sphere ($D = 50\text{ mm}$, $\rho = 7800\text{ kg/m}^3$, $c_p = 460\text{ J/kg}\cdot\text{K}$, $k = 40\text{ W/m}\cdot\text{K}$) is quenched from $T_0 = 800^{\circ}\text{C}$ in an agitated oil bath at $T_{\infty} = 50^{\circ}\text{C}$ with $h = 160\text{ W/m}^2\cdot\text{K}$. Determine the time required for the sphere center to cool to $100^{\circ}\text{C}$.

+-----------------------------------------------------------------------------------------+
|                        BEARING QUENCHING CALCULATION STEPS                              |
|                                                                                         |
|   STEP 1: Calculate Characteristic Length (Lc)                                          |
|           Lc = R / 3 = (0.025 m) / 3 = 0.008333 m                                       |
|                                                                                         |
|   STEP 2: Check Biot Number Validity                                                    |
|           Bi = h * Lc / k = (160 * 0.008333) / 40 = 0.03333                             |
|           Since Bi = 0.0333 < 0.1, Lumped Capacitance is VALID!                         |
|                                                                                         |
|   STEP 3: Calculate Thermal Time Constant (tau)                                         |
|           tau = (rho * cp * Lc) / h                                                     |
|               = (7800 kg/m^3 * 460 J/kg*K * 0.008333 m) / (160 W/m^2*K)                |
|               = 29880 / 160 = 186.75 seconds                                            |
|                                                                                         |
|   STEP 4: Calculate Cooling Time t                                                      |
|           (T(t) - T_inf) / (T0 - T_inf) = exp( - t / tau )                              |
|           (100 - 50) / (800 - 50) = 50 / 750 = 0.06667                                 |
|           ln(0.06667) = - t / 186.75                                                    |
|           -2.70805 = - t / 186.75  ===>  t = 505.7 seconds (approx 8.43 minutes)        |
+-----------------------------------------------------------------------------------------+