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

$\dot{Q}_{cond} = -k A \frac{dT}{dx}$

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):

$\dot{Q}_{cond} = \frac{k A (T_1 - T_2)}{L} = \frac{T_1 - T_2}{R_{cond}}$

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:

$\dot{Q} = \frac{T_{\infty,1} - T_{\infty,2}}{\sum R_{th}} = \frac{T_{\infty,1} - T_{\infty,2}}{\frac{1}{h_1 A} + \frac{L_A}{k_A A} + \frac{L_B}{k_B A} + \frac{1}{h_2 A}}$

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):

$\dot{Q}_{cyl} = \frac{2 \pi k L (T_1 - T_2)}{\ln(r_2 / r_1)} = \frac{T_1 - T_2}{R_{cyl}}$

$\text{Cylindrical Conduction Resistance: } R_{cyl} = \frac{\ln(r_2 / r_1)}{2 \pi k L} \quad [\text{K/W}]$

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

$\dot{Q}_{conv} = h A (T_s - T_\infty) = \frac{T_s - T_\infty}{R_{conv}}$

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):

$\dot{E}_b = \sigma A T^4$

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:

$\dot{Q}_{12} = \epsilon_{eff} \sigma A_1 F_{12} \left( T_1^4 - T_2^4 \right)$

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))

$\dot{Q} = U A \Delta T_{lm}$

$\frac{1}{U A} = R_{total} = \frac{1}{h_i A_i} + \frac{\ln(r_o/r_i)}{2 \pi k L} + \frac{1}{h_o A_o}$

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

$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln\left( \frac{\Delta T_1}{\Delta T_2} \right)}$

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})

$R_{total} = R_{conv,1} + R_{pipe} + R_{ins} + R_{conv,2}$ $R_{total} = 0.000265 + 0.0000645 + 0.20325 + 0.01061 = 0.21419 \text{ K/W}$ $\dot{Q} = \frac{T_{\infty,1} - T_{\infty,2}}{R_{total}} = \frac{200^\circ\text{C} - 20^\circ\text{C}}{0.21419 \text{ K/W}} = \frac{180}{0.21419} = 840.38 \text{ W}$

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

Using the outer convection resistance link: $\dot{Q} = \frac{T_{s,3} - T_{\infty,2}}{R_{conv,2}} \implies T_{s,3} = T_{\infty,2} + \dot{Q} R_{conv,2}$ $T_{s,3} = 20^\circ\text{C} + (840.38 \text{ W} \times 0.01061 \text{ K/W}) = 20 + 8.92 = 28.92^\circ\text{C}$

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}).