14.3 Heat Transfer: Conduction, Convection, and Radiation

14.3 Heat Transfer: Conduction, Convection, and Radiation

Core FE Exam Principle: Heat transfer occurs whenever a temperature gradient exists. Heat is transferred via three primary modes: conduction (molecular diffusion in solids/fluids), convection (bulk fluid motion plus diffusion), and radiation (electromagnetic wave emission). On the FE exam, solving heat transfer problems requires constructing thermal resistance networks, applying non-dimensional numbers, and computing radiation and heat exchanger performance.

Conduction Heat Transfer & Fourier's Law

Conduction is energy transfer from more energetic particles of a substance to adjacent less energetic particles as a result of interactions between particles.

Fourier's Law of 1D Heat Conduction

where:

• $\dot{Q}_{cond}$ = Heat transfer rate ($\text{W}$ or $\text{Btu/h}$).
• (k) = Thermal conductivity of the material ($\text{W/(m}\cdot\text{K)}$ or $\text{Btu/(h}\cdot\text{ft}\cdot{}^\circ\text{F)}$).
• (A) = Cross-sectional area perpendicular to heat flow ($\text{m}^2$ or $\text{ft}^2$).
• $\frac{dT}{dx}$ = Temperature gradient in the direction of heat flow ($\text{K/m}$).

For 1D steady-state heat conduction through a flat plane wall of thickness (L) with uniform thermal conductivity (k) and surface temperatures $T_1$ and $T_2$:

where $R_{cond} = \frac{L}{k A}$ is the thermal resistance to conduction ($\text{K/W}$).

Thermal Resistance Networks & Composite Geometry

Thermal circuits allow complex multi-layer heat transfer systems to be solved using an analogy to Electrical Ohm's Law ($I = \frac{\Delta V}{R} \iff \dot{Q} = \frac{\Delta T}{R_{total}}$).

Composite Plane Walls (Series & Parallel Networks)

For a series composite wall exposed to fluids on both sides:

Cylindrical Conduction (Pipes and Tubes)

For 1D radial heat conduction through a hollow cylinder of inner radius $r_1$, outer radius $r_2$, length (L), and thermal conductivity (k):

Convection Heat Transfer & Dimensionless Correlations

Convection involves energy transfer between a solid surface and an adjacent moving liquid or gas.

Newton's Law of Cooling

where (h) is the convective heat transfer coefficient ($\text{W/(m}^2\cdot\text{K)}$), $T_s$ is solid surface temperature, and $T_\infty$ is fluid bulk temperature.

Non-Dimensional Numbers in Convection

To evaluate (h) for external or internal flows, NCEES problems utilize empirical dimensionless correlations:

1. Nusselt Number ((Nu)): Ratio of convective to conductive heat transfer across fluid layer: $$Nu = \frac{h L_c}{k_f}$$
2. Reynolds Number ((Re)): Ratio of inertial forces to viscous forces (determines laminar vs. turbulent flow): $$Re = \frac{\rho V L_c}{\mu} = \frac{V L_c}{\nu}$$
3. Prandtl Number ((Pr)): Ratio of momentum diffusivity to thermal diffusivity: $$Pr = \frac{\nu}{\alpha} = \frac{\mu c_p}{k_f}$$
4. General Forced Convection Correlation: $Nu = C Re^m Pr^n$.
   • Flat plate laminar flow ($Re_x < 5 \times 10^5$): $Nu_x = 0.332 Re_x^{1/2} Pr^{1/3}$.
   • Fully developed turbulent pipe flow (Dittus-Boelter): $Nu_D = 0.023 Re_D^{0.8} Pr^n$ (where (n=0.4) for heating, (n=0.3) for cooling).

Thermal Radiation Heat Transfer

Radiation is energy emitted by matter in the form of electromagnetic waves (photons) as a result of changes in the electronic configurations of atoms or molecules. Unlike conduction and convection, radiation requires no intervening medium and occurs most efficiently in a vacuum.

Stefan-Boltzmann Law for Ideal Blackbodies

An ideal blackbody absorbs all incident radiation and emits maximum possible thermal energy at absolute temperature (T):

where $\sigma = 5.670 \times 10^{-8} \text{ W/(m}^2\cdot\text{K}^4)$ (or $0.1714 \times 10^{-8} \text{ Btu/(h}\cdot\text{ft}^2\cdot{}^\circ\text{R}^4)$) is the Stefan-Boltzmann constant.

Real Gray Surfaces & Radiation Exchange

Real surfaces emit less radiation than a blackbody, quantified by emissivity $\epsilon$ ($0 \le \epsilon \le 1$). The net radiation heat exchange between two diffuse, gray surfaces is:

where:

• $F_{12}$ = View factor (fraction of radiation leaving surface 1 that strikes surface 2 directly).
• Reciprocity Relation: $A_1 F_{12} = A_2 F_{21}$.
• Small object in a large enclosure ($A_1 \ll A_2$, $F_{12} = 1$): $$\dot{Q}_{rad} = \epsilon_1 \sigma A_1 \left( T_1^4 - T_{surr}^4 \right)$$

Heat Exchangers: Log Mean Temperature Difference (LMTD)

Heat exchangers facilitate thermal energy transfer between two fluids at different temperatures separated by a solid wall.

Overall Heat Transfer Coefficient ((U))

Log Mean Temperature Difference ($\Delta T_{lm}$)

For a counter-flow heat exchanger:

• $\Delta T_1 = T_{h,in} - T_{c,out}$
• $\Delta T_2 = T_{h,out} - T_{c,in}$

For a parallel-flow heat exchanger:

• $\Delta T_1 = T_{h,in} - T_{c,in}$
• $\Delta T_2 = T_{h,out} - T_{c,out}$

Comprehensive Worked Engineering Example

Problem Statement

A steel steam pipe ($k_{pipe} = 45 \text{ W/(m}\cdot\text{K)}$) with inner radius $r_1 = 0.05 \text{ m}$ and outer radius $r_2 = 0.06 \text{ m}$ carries saturated steam at $T_{\infty,1} = 200^\circ\text{C}$. The inner convective heat transfer coefficient is $h_1 = 1200 \text{ W/(m}^2\cdot\text{K)}$. To reduce heat losses, the pipe is covered with a layer of glass wool insulation ($k_{ins} = 0.04 \text{ W/(m}\cdot\text{K)}$) of thickness $t = 0.04 \text{ m}$, yielding an outer insulation radius of $r_3 = 0.10 \text{ m}$. The insulated pipe is exposed to ambient air at $T_{\infty,2} = 20^\circ\text{C}$ with outer convection coefficient $h_2 = 15 \text{ W/(m}^2\cdot\text{K)}$.

For a pipe length of $L = 10.0 \text{ m}$, calculate:

1. The total thermal resistance $R_{total}$ of the network in K/W.
2. The steady rate of heat loss $\dot{Q}$ from the steam to ambient air.
3. The outer surface temperature $T_{s,3}$ of the insulation.

Step-by-Step Solution

Step 1: Compute Individual Thermal Resistances

• Inner convection resistance $R_{conv,1}$:

  $$A_1 = 2 \pi r_1 L = 2 \pi (0.05) (10.0) = 3.1416 \text{ m}^2$$ $$R_{conv,1} = \frac{1}{h_1 A_1} = \frac{1}{1200 \times 3.1416} = 0.000265 \text{ K/W}$$

• Pipe wall conduction resistance $R_{pipe}$:

  $$R_{pipe} = \frac{\ln(r_2 / r_1)}{2 \pi k_{pipe} L} = \frac{\ln(0.06 / 0.05)}{2 \pi \times 45 \times 10.0} = \frac{0.18232}{2827.43} = 0.0000645 \text{ K/W}$$

• Insulation conduction resistance $R_{ins}$:

  $$R_{ins} = \frac{\ln(r_3 / r_2)}{2 \pi k_{ins} L} = \frac{\ln(0.10 / 0.06)}{2 \pi \times 0.04 \times 10.0} = \frac{0.51083}{2.51327} = 0.20325 \text{ K/W}$$

• Outer convection resistance $R_{conv,2}$:

  $$A_3 = 2 \pi r_3 L = 2 \pi (0.10) (10.0) = 6.2832 \text{ m}^2$$ $$R_{conv,2} = \frac{1}{h_2 A_3} = \frac{1}{15 \times 6.2832} = 0.01061 \text{ K/W}$$

Step 2: Sum Resistances and Find Rate of Heat Loss $\dot{Q}$

Step 3: Compute Outer Insulation Surface Temperature $T_{s,3}$

Using the outer convection resistance link:

Final Answer: Total resistance $R_{total} = 0.214 \text{ K/W}$, heat loss rate $\dot{Q} = 840.4 \text{ W}$, and outer surface temperature $T_{s,3} = 28.9^\circ\text{C}$.