4.3 Radiation Heat Transfer, Surface Emissivity & View Factors

4.3 Radiation Heat Transfer, Surface Emissivity & View Factors

Thermal radiation is the emission of energy in the form of electromagnetic waves (photons) emitted by all matter having a temperature above absolute zero. Unlike conduction and convection, thermal radiation requires no intervening physical medium and transmits most efficiently through a vacuum. In HVAC engineering, radiation governs cooling and heating load calculations through solar heat gains, interior surface radiative exchange between walls and occupants, uninsulated steam/chilled pipe heat losses, attic radiant barriers, and low-emissivity (Low-e) window coatings.

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1. Fundamental Laws of Blackbody Thermal Radiation

Thermal radiation occupies the electromagnetic spectrum between wavelengths of $\lambda = 0.1\text{ }\mu\text{m}$ and $100\text{ }\mu\text{m}$, encompassing ultraviolet ($0.1-0.4\text{ }\mu\text{m}$), visible light ($0.4-0.7\text{ }\mu\text{m}$), and infrared ($0.7-100\text{ }\mu\text{m}$).

+-----------------------------------------------------------------------------------------+
| THERMAL RADIATION IN THE ELECTROMAGNETIC SPECTRUM                                       |
+-----------------------------------------------------------------------------------------+
|   Ultraviolet    |   Visible Spectrum   |              Infrared (Thermal)               |
|   0.1 - 0.4 um   |     0.4 - 0.7 um     |                 0.7 - 100 um                  |
|                  |                      |   Near-IR   |   Mid-IR    |   Far-IR (Longwave) |
|                  |                      |  0.7-3.0 um |  3.0-50 um  |      50-100 um      |
+-----------------------------------------------------------------------------------------+

1. Stefan-Boltzmann Law

A blackbody is an idealized thermodynamic body that absorbs all incident radiation (absorptivity $\alpha = 1.0$) and emits the maximum possible energy at every wavelength for a given temperature. The total hemispherical emissive power of a blackbody ($E_b$) is proportional to the fourth power of its absolute temperature:

$E_b = \sigma T^4$

Where:

• $E_b = \text{blackbody emissive power } (\text{Btu/(hr}\cdot\text{ft}^2) \text{ or W/m}^2)$
• $T = \text{absolute thermodynamic temperature in degrees Rankine } ({}^\circ\text{R} = {}^\circ\text{F} + 459.67) \text{ or Kelvin } (\text{K} = {}^\circ\text{C} + 273.15)$
• $\sigma = \text{Stefan-Boltzmann constant}$: $\sigma = 0.1714 \times 10^{-8} \quad \left[\frac{\text{Btu}}{\text{hr}\cdot\text{ft}^2\cdot{}^\circ\text{R}^4}\right] = 5.670 \times 10^{-8} \quad \left[\frac{\text{W}}{\text{m}^2\cdot\text{K}^4}\right]$

Exam Trap — Temperature Scale: Always convert temperatures to absolute units (${}^\circ\text{R}$ or $\text{K}$) before evaluating fourth-power radiation formulas. $(T_1 - T_2)^4 \neq T_1^4 - T_2^4$. Evaluating $(200 - 70)^4$ instead of $(659.67^4 - 529.67^4)$ results in a catastrophic error.

2. Wien's Displacement Law

The wavelength $\lambda_{max}$ at which blackbody monochromatic spectral emissive power ($E_{b\lambda}$) reaches its peak is inversely proportional to absolute temperature:

$\lambda_{max} T = 5193\text{ }\mu\text{m}\cdot{}^\circ\text{R} = 2898\text{ }\mu\text{m}\cdot\text{K}$

• Solar Radiation: The sun's surface temperature is $T_{sun} \approx 10,400^\circ\text{R}$ ($5778\text{ K}$), resulting in $\lambda_{max} = 2898 / 5778 \approx 0.50\text{ }\mu\text{m}$ (peak in the visible light band).
• Terrestrial & Building Surfaces: Building walls and piping at $70^\circ\text{F}$ ($529.67^\circ\text{R} = 294.3\text{ K}$) have $\lambda_{max} = 2898 / 294.3 \approx 9.85\text{ }\mu\text{m}$ (far-infrared, longwave thermal radiation).

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2. Real Surface Properties: Emissivity, Absorptivity & Low-e Technology

Real surfaces emit and absorb less radiation than an ideal blackbody at the same temperature.

+-----------------------------------------------------------------------------------------+
| RADIATION ENERGY BALANCE AT A SURFACE BOUNDARY                                          |
+-----------------------------------------------------------------------------------------+
| Incident Radiation (Irradiation G) --------------------->                               |
|                                   |                                                     |
|                                   +---> Reflected:   rho * G  (Reflectivity)            |
|                                   |                                                     |
|                                   +---> Absorbed:    alpha * G (Absorptivity)           |
|                                   |                                                     |
|                                   +---> Transmitted: tau * G  (Transmissivity)          |
|                                                                                         |
| Conservation of Energy: alpha + rho + tau = 1.0  (For Opaque Surfaces: alpha + rho = 1) |
+-----------------------------------------------------------------------------------------+

Definitions of Surface Radiative Properties

1. Emissivity ($\varepsilon$): Ratio of total radiation emitted by a real surface to that emitted by a blackbody at the same temperature ($0 \le \varepsilon \le 1.0$): $E = \varepsilon E_b = \varepsilon \sigma T^4$
2. Absorptivity ($\alpha$): Fraction of incident irradiation $G$ absorbed by the surface ($0 \le \alpha \le 1.0$).
3. Reflectivity ($\rho$): Fraction of incident irradiation reflected by the surface ($0 \le \rho \le 1.0$).
4. Transmissivity ($\tau$): Fraction of incident irradiation transmitted through the medium ($0 \le \tau \le 1.0$).

For any surface, energy conservation dictates:

$\alpha + \rho + \tau = 1.0$

For opaque bodies (building walls, metals, unglazed surfaces, $\tau = 0$):

$\alpha + \rho = 1.0 \implies \rho = 1.0 - \alpha$

Kirchhoff's Law & Gray-Diffuse Surface Approximation

Under thermodynamic equilibrium, directional spectral emissivity equals directional spectral absorptivity. For a gray and diffuse surface (where radiative properties are independent of wavelength $\lambda$ and emission angle $\theta$ over the thermal spectrum):

$\varepsilon = \alpha$

Low-Emissivity (Low-e) Coatings & Selective Surfaces

Building window glazings utilize microscopically thin silver or metal oxide coatings designed as spectrally selective surfaces:

• High Shortwave Transmittance: Transmits visible solar radiation ($\tau_{sol} \approx 0.70-0.80$) to provide daylighting and winter passive solar gain.
• Low Longwave Emissivity: Reflects longwave infrared radiation emitted by indoor furnishings and heating equipment ($,\varepsilon_{glass} \approx 0.04$ vs standard uncoated glass $\varepsilon = 0.84$), reducing radiant heat loss by over $90%$ through the glazing cavity.

3. Radiation View Factors (Shape / Configuration Factors)

The view factor $F_{ij}$ (also called shape factor or configuration factor) is defined as the fraction of diffuse radiation leaving surface $i$ that directly intercepts surface $j$:

$F_{ij} = \frac{\dot{q}_{i \to j}}{A_i E_{b,i}}$

+-----------------------------------------------------------------------------------------+
| RADIATION VIEW FACTOR FUNDAMENTAL RULES                                                 |
+-----------------------------------------------------------------------------------------+
| 1. Reciprocity Theorem:       A_1 * F_12 = A_2 * F_21                                   |
| 2. Enclosure Summation Rule:  Sum_{j=1}^N F_ij = 1.0  (for an N-surface enclosure)      |
| 3. Flat / Convex Surface:     F_ii = 0 (A flat or convex surface cannot see itself)     |
| 4. Concave Surface:           F_ii > 0 (A concave cavity sees part of its own surface)  |
| 5. Superposition Rule:        F_1(2+3) = F_12 + F_13                                    |
+-----------------------------------------------------------------------------------------+

Fundamental View Factor Relationships

1. Reciprocity Relation: Connects view factors between two surfaces of arbitrary areas $A_i$ and $A_j$: $A_i F_{ij} = A_j F_{ji}$
2. Summation Rule for an Enclosure: For an enclosure consisting of $N$ surfaces completely enclosing a volume: $\sum_{j=1}^N F_{ij} = F_{i1} + F_{i2} + \cdots + F_{iN} = 1.0$
3. Self-Viewing Factor:
   • For a flat or convex surface: $F_{ii} = 0$
   • For a concave surface: $F_{ii} > 0$

Common Analytical View Factors

• Small Convex Object 1 Completely Enclosed by Large Enclosure 2: $F_{12} = 1.0 \quad \implies \quad F_{21} = \frac{A_1}{A_2} F_{12} = \frac{A_1}{A_2}, \quad F_{22} = 1 - \frac{A_1}{A_2}$
• Two Concentric Cylinders (Inner 1, Outer 2): $F_{12} = 1.0, \quad F_{21} = \frac{r_1}{r_2}, \quad F_{22} = 1 - \frac{r_1}{r_2}$
• Two Infinite Parallel Plates: $F_{12} = 1.0, \quad F_{21} = 1.0$

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4. Radiative Heat Exchange Between Gray Diffuse Surfaces

1. Small Convex Surface 1 in a Large Enclosure 2

For an uninsulated pipe, duct, or human occupant (surface area $A_1$, emissivity $\varepsilon_1$, temperature $T_1$) located inside a large room (enclosure area $A_2 \gg A_1$, wall temperature $T_2$):

$\dot{q}_{12} = \varepsilon_1 \sigma A_1 (T_1^4 - T_2^4)$

Notice that the emissivity of the large room enclosure ($\varepsilon_2$) has zero effect on the net exchange because the room acts as a blackbody cavity ($A_1/A_2 \to 0$).

2. Two Large Infinite Parallel Gray Plates

For radiant heat transfer across an air gap between two parallel sheets of area $A$ with emissivities $\varepsilon_1$ and $\varepsilon_2$:

$\dot{q}_{12} = \frac{\sigma A (T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1}$

3. Long Concentric Cylinders (Insulated Pipe in Concentric Sleeve)

For inner cylinder 1 (radius $r_1$) and outer cylinder 2 (radius $r_2$):

$\dot{q}_{12} = \frac{\sigma A_1 (T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1} + \left(\frac{1 - \varepsilon_2}{\varepsilon_2}\right)\left(\frac{r_1}{r_2}\right)}$

4. Radiation Shields

Thin, highly reflective sheets (such as polished aluminum foil, $\varepsilon_s \approx 0.05$) placed between two radiating surfaces act as radiation shields to drastically reduce net radiative exchange.

+-----------------------------------------------------------------------------------------+
| EFFECT OF N RADIATION SHIELDS BETWEEN PARALLEL PLATES                                   |
+-----------------------------------------------------------------------------------------+
| Plate 1 (T_1, e_1) | Shield 1 | Shield 2 | ... | Shield N | Plate 2 (T_2, e_2)          |
|                                                                                         |
| When all surfaces have identical emissivity (e_1 = e_2 = e_s):                          |
|                                                                                         |
|   q_with_shields = [ 1 / (N + 1) ] * q_without_shields                                  |
+-----------------------------------------------------------------------------------------+

For $N$ parallel radiation shields placed between two large parallel plates, where all surfaces have identical emissivity $\varepsilon$:

$\dot{q}_{with\text{ }N\text{ }shields} = \frac{1}{N + 1} \dot{q}_{no\text{ }shields}$

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5. Linearized Radiation Heat Transfer Coefficient ($h_r$)

In HVAC cooling and heating load calculations, radiative and convective heat exchanges occur simultaneously from the same surface. Factoring the fourth-power temperature difference $(T_1^4 - T_2^4)$:

$(T_1^4 - T_2^4) = (T_1^2 - T_2^2)(T_1^2 + T_2^2) = (T_1 - T_2)(T_1 + T_2)(T_1^2 + T_2^2)$

Substituting into the net radiation equation for a small body in a large enclosure:

$\dot{q}_{rad} = \varepsilon_1 \sigma A_1 (T_1^4 - T_2^4) = \left[\varepsilon_1 \sigma (T_1 + T_2)(T_1^2 + T_2^2)\right] A_1 (T_1 - T_2) = h_r A_1 (T_1 - T_2)$

Where the linearized radiation coefficient is:

$h_r = \varepsilon_1 \sigma (T_1 + T_2)(T_1^2 + T_2^2) \approx 4 \varepsilon_1 \sigma T_{avg}^3$

Where $T_{avg} = \frac{T_1 + T_2}{2}$ in absolute temperature (${}^\circ\text{R}$ or $\text{K}$).

Combined Heat Transfer Coefficient ($h_{total}$)

When ambient air temperature equals surrounding wall temperature ($T_\infty = T_{surr}$):

$\dot{q}_{total} = \dot{q}_{conv} + \dot{q}_{rad} = h_c A (T_s - T_\infty) + h_r A (T_s - T_\infty) = (h_c + h_r) A (T_s - T_\infty) = h_{total} A (T_s - T_\infty)$

$h_{total} = h_c + h_r$

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6. Step-by-Step Worked Example: Uninsulated Steam Pipe Heat Loss

Problem Statement

An uninsulated horizontal steel steam pipe ($4.50\text{-in.}$ outer diameter, $D_o = 0.375\text{ ft}$, length $L = 50.0\text{ ft}$) passes through an open mechanical equipment room. The pipe outer surface is at $T_s = 250.0^\circ\text{F}$ and has an emissivity of $\varepsilon = 0.80$. The mechanical room walls and surrounding air are at $T_\infty = T_{walls} = 70.0^\circ\text{F}$. The natural convection coefficient from the horizontal pipe is $h_c = 1.25\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$.

Calculate:

1. Outer surface area of the pipe $A_s$ ($\text{ft}^2$)
2. Steady-state convective heat loss rate $\dot{q}_{conv}$ ($\text{Btu/hr}$)
3. Steady-state radiative heat loss rate $\dot{q}_{rad}$ ($\text{Btu/hr}$)
4. Linearized radiation coefficient $h_r$ and combined coefficient $h_{total}$
5. Total heat loss rate $\dot{q}_{total}$ ($\text{Btu/hr}$)

Solution Steps

Step 1: Calculate outer pipe surface area $A_s = \pi D_o L = \pi \times 0.375\text{ ft} \times 50.0\text{ ft} = 58.905\text{ ft}^2$

Step 2: Calculate convective heat loss $\Delta T = T_s - T_\infty = 250.0^\circ\text{F} - 70.0^\circ\text{F} = 180.0^\circ\text{F}$ $\dot{q}_{conv} = h_c A_s \Delta T = 1.25\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \times 58.905\text{ ft}^2 \times 180.0^\circ\text{F} = 13,253.6\text{ Btu/hr}$

Step 3: Calculate radiative heat loss using absolute Rankine temperatures $T_s = 250.0 + 459.67 = 709.67^\circ\text{R}$ $T_{walls} = 70.0 + 459.67 = 529.67^\circ\text{R}$ $T_s^4 = (709.67)^4 = 2.5364 \times 10^{11}\text{ }{}^\circ\text{R}^4$ $T_{walls}^4 = (529.67)^4 = 7.8709 \times 10^{10}\text{ }{}^\circ\text{R}^4$ $T_s^4 - T_{walls}^4 = 2.5364 \times 10^{11} - 0.7871 \times 10^{11} = 1.7493 \times 10^{11}\text{ }{}^\circ\text{R}^4$

$\dot{q}_{rad} = \varepsilon \sigma A_s (T_s^4 - T_{walls}^4)$ $\dot{q}_{rad} = 0.80 \times (0.1714 \times 10^{-8}) \times 58.905 \times (1.7493 \times 10^{11})$ $\dot{q}_{rad} = 0.80 \times 0.1714 \times 58.905 \times 174.93 = 14,131.5\text{ Btu/hr}$

Step 4: Calculate linearized radiation coefficient $h_r$ and combined $h_{total}$ $h_r = \frac{\dot{q}_{rad}}{A_s (T_s - T_{walls})} = \frac{14,131.5\text{ Btu/hr}}{58.905\text{ ft}^2 \times 180.0^\circ\text{F}} = 1.333\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$ $h_{total} = h_c + h_r = 1.25 + 1.333 = 2.583\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$

Step 5: Calculate total heat loss rate $\dot{q}_{total} = \dot{q}_{conv} + \dot{q}_{rad} = 13,253.6 + 14,131.5 = 27,385.1\text{ Btu/hr}$

Note that radiation accounts for over $51.6%$ of the total heat dissipation, proving that radiant heat exchange cannot be neglected in uninsulated high-temperature piping.