4.4 Steady & Transient Conduction, Forced & Natural Convection

Steady & Transient Conduction, Forced & Natural Convection

Thermal energy transport is governed by three fundamental mechanisms: conduction (molecular diffusion across stationary matter), convection (combined conduction and bulk fluid advection), and thermal radiation (electromagnetic wave emission). Mastery of 1D steady-state resistance networks, composite radial insulation sizing, extended surface fin mechanics, transient lumped capacitance response, and forced/natural convection correlations is essential for the NCEES PE Mechanical exam.

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|                         HEAT TRANSFER MECHANISMS & GOVERNING LAWS                                 |
|                                                                                                   |
|   [CONDUCTION: FOURIER'S LAW]       [CONVECTION: NEWTON'S COOLING]     [OVERALL RESISTANCE]       |
|   - q = -k * A * (dT/dx)            - q = h * A * (T_s - T_\infty)      - q = \Delta T / R_total   |
|   - R_th = L / (k*A) (Plane)        - R_conv = 1 / (h*A)               - q = U * A * \Delta T     |
|   - R_th = ln(r2/r1)/(2*\pi*k*L)     - Nu = h*L / k = f(Re, Pr, Ra)     - Critical Radius: k/h    |
|                                                |                                                  |
|                                                v                                                  |
|                                     [TRANSIENT & FINS]                                            |
|                                     - Lumped Cap: Bi = h*Lc / k < 0.1                             |
|                                     - \theta(t) / \theta_i = exp(-t / \tau)                       |
|                                     - Fin Sizing: m = \sqrt{h*P / (k*Ac)}                         |
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1. 1D Steady-State Conduction & Thermal Resistance Networks

Fourier's Law of Heat Conduction

In differential form, Fourier's law states that local heat flux $q''$ is proportional to the negative temperature gradient:

$$q'' = \frac{q}{A} = -k \nabla T = -k \frac{dT}{dx}$$

Where:

• $q = \text{Heat transfer rate } (\text{W or } \text{BTU/hr})$
• $k = \text{Thermal conductivity } (\text{W/(m}\cdot\text{K)} \text{ or } \text{BTU/(hr}\cdot\text{ft}\cdot^\circ\text{F)})$
• $A = \text{Cross-sectional heat transfer area perpendicular to heat flow}$

Thermal Resistance Analogies for Common Geometries

Analogous to Ohm's electrical law ($I = \frac{\Delta V}{R}$), thermal heat rate is $q = \frac{\Delta T}{R_{\text{th}}}$.

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|                             THERMAL RESISTANCE FORMULA SUMMARY                                    |
|                                                                                                   |
|   1. Plane Wall (Thickness L, Area A):               R_{\text{plane}} = \frac{L}{k A}              |
|                                                                                                   |
|   2. Cylindrical Layer (Radii r_1 to r_2, Length L):  R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2 \pi k L} |
|                                                                                                   |
|   3. Spherical Shell (Radii r_1 to r_2):             R_{\text{sph}} = \frac{1/r_1 - 1/r_2}{4 \pi k} |
|                                                                                                   |
|   4. Convection Boundary (Surface Area A):           R_{\text{conv}} = \frac{1}{h A}              |
|                                                                                                   |
|   5. Radiation Boundary (h_r Linearized):            R_{\text{rad}} = \frac{1}{h_r A}             |
|      where h_r = \epsilon \sigma (T_s + T_{\text{surr}})(T_s^2 + T_{\text{surr}}^2)                |
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Composite Cylindrical Pipe with Internal and External Convection

For an insulated industrial pipe carrying hot fluid $T_{\infty,i}$ at convection coefficient $h_i$, with inner radius $r_1$, outer pipe radius $r_2$, outer insulation radius $r_3$, and ambient air $T_{\infty,o}$ at $h_o$:

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|                         COMPOSITE CYLINDRICAL RESISTANCE CIRCUIT                                  |
|                                                                                                   |
|   T_{\infty,i} ---[ R_conv,i ]--- T_1 ---[ R_pipe ]--- T_2 ---[ R_ins ]--- T_3 ---[ R_conv,o ]--- T_{\infty,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}         |
|                                                                                                   |
|   Heat Rate: q = \frac{T_{\infty,i} - T_{\infty,o}}{R_{\text{total}}} = U_o A_o (T_{\infty,i} - T_{\infty,o}) |
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Critical Radius of Insulation

In radial systems, adding insulation increases conduction resistance ($R_{\text{cond}} \propto \ln(r_o/r_i)$) but decreases external convection resistance ($R_{\text{conv}} = \frac{1}{2\pi r_o L h}$) due to the expanding surface area. Differentiating total resistance with respect to outer radius $r_o$ yields the critical radius of insulation ($r_{\text{cr}}$):

$$r_{\text{cr, cylinder}} = \frac{k_{\text{ins}}}{h_o}, \quad r_{\text{cr, sphere}} = \frac{2 k_{\text{ins}}}{h_o}$$

   Total Thermal Resistance R_th
       ^
       |        R_total = R_cond + R_conv
       |          \                      /
       |           \                    /
       |            \                  /
       |             \___          ___/
       |                 \________/
       |                      |
       |                      v  Minimum Resistance (Maximum Heat Loss q_max)
       +----------------------(r_cr)---------------------------> Outer Radius r_o
                             r_cr = k / h

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2. Extended Surfaces: Fin Heat Transfer

Fins increase convective surface area to enhance heat dissipation from components where the convection heat transfer coefficient $h$ is low (e.g., air-cooled engine cylinders, electronics heat sinks).

+---------------------------------------------------------------------------------------------------+
|                             FIN DIFFERENTIAL EQUATION & PARAMETERS                                |
|                                                                                                   |
|   Governing ODE:        \frac{d^2 \theta}{dx^2} - m^2 \theta = 0                                  |
|                         where \theta(x) = T(x) - T_\infty                                         |
|                                                                                                   |
|   Fin Parameter:        m = \sqrt{\frac{h P}{k A_c}}   [\text{m}^{-1} \text{ or } \text{ft}^{-1}]   |
|                                                                                                   |
|   Fin Characteristic:   M = \sqrt{h P k A_c} \theta_b = \sqrt{h P k A_c} (T_b - T_\infty)         |
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Fin Tip Boundary Condition Formulations

Where corrected fin length $L_c$ accounts for convective tip area: $L_c = L + \frac{A_c}{P}$ (for rectangular fin: $L_c = L + t/2$; for round pin fin: $L_c = L + D/4$).

Fin Efficiency ($\eta_f$) and Fin Effectiveness ($\epsilon_f$)

• Fin Efficiency ($\eta_f$): Ratio of actual fin heat transfer to the ideal heat transfer if the entire fin were maintained at base temperature $T_b$: $$\eta_f = \frac{q_f}{h A_{\text{fin}} \theta_b} = \frac{\tanh(m L_c)}{m L_c}$$
• Fin Effectiveness ($\epsilon_f$): Ratio of fin heat transfer to heat transfer from the bare base area without the fin: $$\epsilon_f = \frac{q_f}{h A_c \theta_b} = \frac{\eta_f A_{\text{fin}}}{A_c}$$ Design Guideline: Fins are economically justified only if $\epsilon_f \ge 2.0$.

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3. Transient Conduction & Lumped Capacitance Method

When a solid body at initial temperature $T_i$ is suddenly exposed to a convective environment at $T_\infty$, spatial temperature gradients inside the body may be negligible if conduction resistance is much smaller than external convection resistance.

+---------------------------------------------------------------------------------------------------+
|                         LUMPED CAPACITANCE CRITERIA & TRANSIENT RESPONSE                          |
|                                                                                                   |
|   Biot Number Criterion:    Bi = \frac{h L_c}{k_{\text{solid}}} < 0.1                              |
|                             where Characteristic Length L_c = \frac{V_{\text{solid}}}{A_s}       |
|                                                                                                   |
|   Temperature Response:     \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{t}{\tau}\right) |
|                             = \exp(-Bi \cdot Fo)                                                  |
|                                                                                                   |
|   Thermal Time Constant:    \tau = \frac{\rho V c_p}{h A_s} = R_{\text{conv}} C_{\text{thermal}}   |
|                                                                                                   |
|   Fourier Number:           Fo = \frac{\alpha t}{L_c^2} = \frac{k t}{\rho c_p L_c^2}             |
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Characteristic Lengths ($L_c = V / A_s$):

• Plane Wall of thickness $2L$ (exposed both sides): $L_c = L$
• Long Solid Cylinder of radius $r_o$: $L_c = \frac{\pi r_o^2 L}{2\pi r_o L} = \frac{r_o}{2}$
• Solid Sphere of radius $r_o$: $L_c = \frac{\frac{4}{3}\pi r_o^3}{4\pi r_o^2} = \frac{r_o}{3}$

Total Energy Transfer ($Q_{\text{total}}$) over time $t$:

$$Q(t) = \rho V c_p (T_i - T_\infty) \left[ 1 - \exp\left(-\frac{t}{\tau}\right) \right]$$

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4. Forced Convection Correlations

Convection heat transfer coefficients ($h$) are correlated through dimensionless parameters:

$$Nu = \frac{h L}{k_f}, \quad Re = \frac{V L}{\nu}, \quad Pr = \frac{\nu}{\alpha} = \frac{\mu c_p}{k_f}$$

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|                         INTERNAL FORCED FLOW IN CIRCULAR TUBES                                    |
|                                                                                                   |
|   1. FULLY DEVELOPED LAMINAR FLOW (Re_D < 2300):                                                  |
|      - Uniform Surface Heat Flux (q" = const):   Nu_D = \frac{h D}{k_f} = 4.36                     |
|      - Constant Surface Temperature (T_s = const): Nu_D = \frac{h D}{k_f} = 3.66                  |
|                                                                                                   |
|   2. FULLY DEVELOPED TURBULENT FLOW (Re_D > 10,000, 0.6 <= Pr <= 160):                            |
|      Dittus-Boelter Equation:                                                                     |
|      Nu_D = 0.023 \, Re_D^{0.8} \, Pr^n                                                           |
|      - n = 0.4 for fluid HEATING (T_s > T_b)                                                      |
|      - n = 0.3 for fluid COOLING (T_s < T_b)                                                      |
|      (All fluid properties evaluated at mean bulk temperature T_b = (T_{in} + T_{out})/2)        |
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External Forced Flow Over Flat Plates

• Laminar Boundary Layer ($Re_x < 5 \times 10^5$): $$Nu_x = 0.332 \, Re_x^{1/2} Pr^{1/3}, \quad \overline{Nu}_L = 0.664 \, Re_L^{1/2} Pr^{1/3}$$
• Turbulent Boundary Layer ($Re_x \ge 5 \times 10^5$): $$Nu_x = 0.0296 \, Re_x^{0.8} Pr^{1/3}$$ (Properties evaluated at film temperature $T_f = \frac{T_s + T_\infty}{2}$)

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5. Natural (Free) Convection Correlations

In natural convection, fluid motion is driven entirely by buoyancy forces resulting from temperature-induced density gradients.

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|                             NATURAL CONVECTION DIMENSIONLESS GROUPS                               |
|                                                                                                   |
|   Grashof Number:     Gr_L = \frac{g \beta (T_s - T_\infty) L^3}{\nu^2}                           |
|                                                                                                   |
|   Rayleigh Number:    Ra_L = Gr_L \cdot Pr = \frac{g \beta (T_s - T_\infty) L^3}{\nu \alpha}     |
|                                                                                                   |
|   Ideal Gas Expansivity: \beta = \frac{1}{T_f}   [T_f \text{ in absolute units: Kelvin or Rankine}]|
+---------------------------------------------------------------------------------------------------+

Empirical Correlations for Vertical Plates ($L = \text{Height}$):

• Laminar Regime ($10^4 \le Ra_L \le 10^9$): $$\overline{Nu}_L = 0.59 \, Ra_L^{1/4}$$
• Turbulent Regime ($10^9 \le Ra_L \le 10^{13}$): $$\overline{Nu}_L = 0.10 \, Ra_L^{1/3}$$

Natural vs. Forced Convection Regimes:

• $\frac{Gr}{Re^2} \ll 1$: Forced convection dominates (natural convection neglected).
• $\frac{Gr}{Re^2} \gg 1$: Natural convection dominates (forced convection neglected).
• $\frac{Gr}{Re^2} \approx 1$: Mixed convection (both mechanisms significant).

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6. Step-by-Step Worked Engineering Problem

Problem Statement

A Schedule 40 carbon steel steam pipe ($k_p = 45\text{ W/m}\cdot\text{K}$) with outer diameter $D_o = 100\text{ mm}$ ($r_1 = 50\text{ mm} = 0.050\text{ m}$) transports dry steam at $T_{\infty,i} = 220^\circ\text{C}$ with internal convection coefficient $h_i = 1200\text{ W/m}^2\cdot\text{K}$. The pipe is insulated with a layer of fiberglass insulation ($k_{\text{ins}} = 0.040\text{ W/m}\cdot\text{K}$) of thickness $t_{\text{ins}} = 50\text{ mm}$ ($r_2 = 0.100\text{ m}$). The outside ambient air is at $T_{\infty,o} = 20^\circ\text{C}$ with external convection coefficient $h_o = 10\text{ W/m}^2\cdot\text{K}$. Neglect pipe wall thickness (assume $r_{\text{pipe}} = r_1$).

Determine:

1. The critical radius of insulation $r_{\text{cr}}$.
2. The total thermal resistance per unit length $R'_{\text{total}}$ ($\text{K}\cdot\text{m/W}$).
3. The heat loss rate per meter of pipe length $q'$.
4. The outer surface temperature of the insulation $T_{s,o}$.

Step-by-Step Solution

Step 1: Check Critical Radius of Insulation

$$r_{\text{cr}} = \frac{k_{\text{ins}}}{h_o} = \frac{0.040\text{ W/m}\cdot\text{K}}{10\text{ W/m}^2\cdot\text{K}} = 0.004\text{ m} = 4.0\text{ mm}$$

Because the bare pipe outer radius $r_1 = 50\text{ mm} \gg r_{\text{cr}} = 4.0\text{ mm}$, adding insulation is guaranteed to substantially reduce heat loss.

Step 2: Calculate Individual Thermal Resistances per Unit Length ($L = 1.0\text{ m}$)

• Internal convection resistance: $$R'_{\text{conv},i} = \frac{1}{2\pi r_1 h_i} = \frac{1}{2\pi (0.050\text{ m}) (1200\text{ W/m}^2\cdot\text{K})} = \frac{1}{376.99} = 0.00265\text{ K}\cdot\text{m/W}$$
• Insulation conduction resistance: $$R'_{\text{ins}} = \frac{\ln(r_2/r_1)}{2\pi k_{\text{ins}}} = \frac{\ln(0.100 / 0.050)}{2\pi (0.040\text{ W/m}\cdot\text{K})} = \frac{\ln(2.0)}{0.25133} = \frac{0.69315}{0.25133} = 2.7577\text{ K}\cdot\text{m/W}$$
• External convection resistance: $$R'_{\text{conv},o} = \frac{1}{2\pi r_2 h_o} = \frac{1}{2\pi (0.100\text{ m}) (10\text{ W/m}^2\cdot\text{K})} = \frac{1}{6.2832} = 0.15915\text{ K}\cdot\text{m/W}$$
• Total thermal resistance: $$R'_{\text{total}} = R'_{\text{conv},i} + R'_{\text{ins}} + R'_{\text{conv},o} = 0.00265 + 2.7577 + 0.15915 = 2.9195\text{ K}\cdot\text{m/W}$$

Step 3: Calculate Heat Loss per Unit Length ($q'$)

$$q' = \frac{T_{\infty,i} - T_{\infty,o}}{R'_{\text{total}}} = \frac{220^\circ\text{C} - 20^\circ\text{C}}{2.9195\text{ K}\cdot\text{m/W}} = \frac{200\text{ K}}{2.9195} = 68.51\text{ W/m}$$

Step 4: Calculate Outer Surface Temperature of Insulation ($T_{s,o}$) Using the external convection link: $q' = \frac{T_{s,o} - T_{\infty,o}}{R'_{\text{conv},o}}$:

$$T_{s,o} = T_{\infty,o} + q' R'_{\text{conv},o} = 20^\circ\text{C} + (68.51\text{ W/m} \times 0.15915\text{ K}\cdot\text{m/W}) = 20 + 10.90 = 30.9^\circ\text{C}$$

(Safe touch temperature for personnel protection)

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7. Common Exam Traps & PE Pro-Tips