5.3 Thermal Radiation & Industrial Heat Exchanger Design

5.3 Thermal Radiation & Industrial Heat Exchanger Design

Thermal radiation is the electromagnetic energy emitted by matter as a result of its finite temperature. Unlike conduction and convection, radiation requires no intervening medium and travels at the speed of light in a vacuum ($c_0 \approx 3 \times 10^8\text{ m/s}$).

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1. Fundamentals of Thermal Radiation & Blackbody Laws

Blackbody Radiation

A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of wavelength or direction, and emits the maximum possible thermal radiation at a given temperature.

• Stefan-Boltzmann Law: Total emissive power of a blackbody $E_b$ is proportional to the fourth power of its absolute temperature:

$E_b = \sigma T^4 \quad [\text{W/m}^2]$

Where:

• $\sigma = 5.670374 \times 10^{-8} \text{ W/m}^2\cdot\text{K}^4$ = Stefan-Boltzmann Constant

• $T$ = Absolute temperature of the surface in Kelvin ($\text{K} = ^\circ\text{C} + 273.15$)

• Wien's Displacement Law: The wavelength $\lambda_{max}$ at which blackbody spectral emissive power is maximum is inversely proportional to absolute temperature:

$\lambda_{max} T = 2897.8 \quad [\mu\text{m}\cdot\text{K}]$

Example: The sun ($T \approx 5800\text{ K}$) has peak emission at $\lambda_{max} \approx 0.5\mu\text{m}$ (visible green-yellow light), while a surface at $300\text{ K}$ peaks at $\lambda_{max} \approx 9.66\mu\text{m}$ (infrared spectrum).

Real Surface Properties & Kirchhoff's Law

Real surfaces emit and reflect less radiation than a blackbody:

• Emissivity ($\epsilon$): $\epsilon = \frac{E(T)}{E_b(T)} \le 1.0$
• Absorptivity ($\alpha$): Fraction of incident radiation absorbed.
• Reflectivity ($\rho$): Fraction reflected.
• Transmissivity ($\tau$): Fraction transmitted through medium.

For any surface, energy conservation dictates:

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

For an opaque body ($\tau = 0$), $\alpha + \rho = 1.0$.

Kirchhoff's Law of Radiation: For any surface in thermal equilibrium with its surroundings, monochromatic directional emissivity equals monochromatic directional absorptivity ($\epsilon_\lambda = \alpha_\lambda$). For a diffuse-gray surface, total emissivity equals total absorptivity ($\epsilon = \alpha$).

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2. View Factor Algebra & Radiative Heat Exchange

The View Factor (or shape factor) $F_{ij}$ is defined as the fraction of radiation leaving diffuse surface $i$ that strikes diffuse surface $j$ directly.

Fundamental View Factor Relations

1. Reciprocity Relation:

   $A_i F_{ij} = A_j F_{ji}$

2. Summation Rule (for an enclosure of $N$ surfaces forming a complete interior space):

   $\sum_{j=1}^{N} F_{ij} = 1.0 \quad \text{for } i = 1, 2, \dots, N$

3. Superposition Rule: If surface $j$ is subdivided into parts $k$ and $l$ ($A_j = A_k + A_l$):

   $F_{i(k+l)} = F_{ik} + F_{il}$

4. Flat or Convex Surfaces: Cannot see themselves, so $F_{ii} = 0$. Concave surfaces have $F_{ii} > 0$.

Net Radiative Exchange Between Diffuse-Gray Surfaces

• Two Infinite Parallel Plates ($ A_1 = A_2 = A$):

  $q_{12}'' = \frac{q_{12}}{A} = \frac{\sigma (T_1^4 - T_2^4)}{\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} - 1}$

• Small Object 1 Enclosed by Large Surface 2 ($A_1 \ll A_2, F_{12} = 1.0$):

  $q_{12} = \epsilon_1 A_1 \sigma (T_1^4 - T_2^4)$

Radiation Shields

Radiation shields are thin, highly reflective sheets placed between radiating surfaces to reduce net radiative heat transfer. Inserting $N$ thin shields of identical emissivity $\epsilon_s$ equal to plate emissivities ($\epsilon_1 = \epsilon_2 = \epsilon_s$) reduces net heat transfer rate to:

$q_{shielded} = \frac{1}{N + 1} q_{unshielded}$

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3. Industrial Heat Exchangers & Overall Heat Transfer Coefficient ($U$)

Heat exchangers are devices that facilitate heat transfer between two fluid streams at different temperatures without mixing them.

Primary Heat Exchanger Configurations

1. Double-Pipe (Concentric Tube):
   • Parallel-Flow: Both hot and cold fluids enter at the same end and flow in the same direction.
   • Counter-Flow: Hot and cold fluids enter at opposite ends and flow in opposite directions. (Produces higher outlet cold fluid temperature and larger average temperature difference).
2. Shell-and-Tube: Most common industrial type; contains a bundle of tubes inside a outer shell equipped with baffles (1-shell pass, 2-tube pass; 2-shell pass, 4-tube pass).
3. Cross-Flow: Fluids flow perpendicular to each other (e.g., automobile radiator); streams can be unmixed or mixed.

Overall Heat Transfer Coefficient ($U$)

The thermal resistance across a clean tubular heat exchanger wall is:

$\frac{1}{U A} = \frac{1}{U_i A_i} = \frac{1}{U_o A_o} = \frac{1}{h_i A_i} + R_{f,i}'' \frac{1}{A_i} + \frac{\ln(r_o/r_i)}{2 \pi k L} + R_{f,o}'' \frac{1}{A_o} + \frac{1}{h_o A_o}$

Where $R_f''$ represents fouling factors ($\text{m}^2\cdot\text{K/W}$) resulting from chemical deposits, rust, or biological scaling.

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4. Logarithmic Mean Temperature Difference (LMTD) Method

The LMTD method is preferred for heat exchanger sizing when fluid inlet and outlet temperatures are known or specified.

$q = U A F \Delta T_{lm}$

Where:

• $\Delta T_{lm}$ = Logarithmic Mean Temperature Difference for pure counter-flow
• $F$ = LMTD Correction Factor ($F \le 1.0$; $F = 1.0$ for pure counter-flow or parallel-flow)

Counter-Flow LMTD Equation

$\Delta T_{lm,CF} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$

Where:

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

(Note: If $\Delta T_1 = \Delta T_2$, then $\Delta T_{lm} = \Delta T_1$.)

Multipass LMTD Correction Factor ($F$)

For shell-and-tube or cross-flow heat exchangers, $F$ is obtained from standard charts using two dimensionless ratios:

$P = \frac{t_{out} - t_{in}}{T_{in} - t_{in}} \quad \text{(Thermal Effectiveness Ratio)}$

$R = \frac{T_{in} - T_{out}}{t_{out} - t_{in}} = \frac{\dot{m}_c c_{p,c}}{\dot{m}_h c_{p,h}} = \frac{C_c}{C_h} \quad \text{(Heat Capacity Ratio)}$

Where $T$ refers to shell-side fluid and $t$ refers to tube-side fluid.

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5. Effectiveness-NTU ($\epsilon$-NTU) Method

When fluid outlet temperatures are unknown, the LMTD method requires tedious iteration. The $\epsilon$-NTU Method directly solves performance without iteration.

Definitions

• Heat Capacity Rates: $C_h = \dot{m}h c{p,h}$ and $C_c = \dot{m}c c{p,c}$

• Minimum Capacity Rate: $C_{min} = \min(C_h, C_c)$

• Capacity Ratio: $C_r = \frac{C_{min}}{C_{max}} \le 1.0$

• Maximum Possible Heat Transfer Rate ($q_{max}$):

  $q_{max} = C_{min} (T_{h,in} - T_{c,in})$

• Heat Exchanger Effectiveness ($\epsilon$):

  $\epsilon = \frac{q_{actual}}{q_{max}} = \frac{C_h (T_{h,in} - T_{h,out})}{C_{min} (T_{h,in} - T_{c,in})} = \frac{C_c (T_{c,out} - T_{c,in})}{C_{min} (T_{h,in} - T_{c,in})}$

• Number of Transfer Units (NTU):

  $NTU = \frac{U A}{C_{min}}$

Representative $\epsilon$-NTU Formulas

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6. Step-by-Step Worked Engineering Calculation (MELE Board Exam Style)

Problem Statement

A double-pipe counter-flow heat exchanger is used to cool engine lube oil ($c_{p,h} = 2.10\text{ kJ/kg}\cdot\text{K}$) from $T_{h,in} = 110^\circ\text{C}$ to $T_{h,out} = 50^\circ\text{C}$ at a mass flow rate of $\dot{m}h = 1.5\text{ kg/s}$. Cooling water ($c{p,c} = 4.18\text{ kJ/kg}\cdot\text{K}$) enters the heat exchanger at $T_{c,in} = 20^\circ\text{C}$ with a mass flow rate of $\dot{m}_c = 1.2\text{ kg/s}$. The overall heat transfer coefficient is $U = 650\text{ W/m}^2\cdot\text{K}$.

1. Calculate the total heat transfer rate $q$.
2. Calculate the exit temperature of cooling water $T_{c,out}$.
3. Determine the Logarithmic Mean Temperature Difference $\Delta T_{lm}$.
4. Calculate the required heat exchanger heat transfer surface area $A$.

Solution

Step 1: Calculate Heat Transfer Rate $q$

Calculate oil heat capacity rate $C_h$: $C_h = \dot{m}_h c_{p,h} = 1.5\text{ kg/s} \times 2100\text{ J/kg}\cdot\text{K} = 3150\text{ W/K}$

Calculate total heat loss rate from oil $q$: $q = C_h (T_{h,in} - T_{h,out}) = 3150 \times (110 - 50) = 3150 \times 60 = 189,000\text{ W} = 189.0\text{ kW}$

Step 2: Calculate Water Exit Temperature $T_{c,out}$

Calculate water heat capacity rate $C_c$: $C_c = \dot{m}_c c_{p,c} = 1.2\text{ kg/s} \times 4180\text{ J/kg}\cdot\text{K} = 5016\text{ W/K}$

Energy conservation on water side: $q = C_c (T_{c,out} - T_{c,in}) \implies 189,000 = 5016 \times (T_{c,out} - 20)$ $T_{c,out} - 20 = \frac{189,000}{5016} = 37.68^\circ\text{C} \implies T_{c,out} = 57.68^\circ\text{C}$

Step 3: Calculate Logarithmic Mean Temperature Difference (LMTD)

For counter-flow arrangement: $\Delta T_1 = T_{h,in} - T_{c,out} = 110 - 57.68 = 52.32^\circ\text{C}$ $\Delta T_2 = T_{h,out} - T_{c,in} = 50 - 20 = 30.00^\circ\text{C}$

Calculate $\Delta T_{lm}$: $\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{52.32 - 30.00}{\ln(52.32 / 30.00)} = \frac{22.32}{\ln(1.7440)} = \frac{22.32}{0.55618} = 40.13^\circ\text{C}$

Step 4: Calculate Required Heat Exchanger Surface Area $A$

$q = U A \Delta T_{lm} \implies A = \frac{q}{U \Delta T_{lm}}$ $A = \frac{189,000\text{ W}}{650\text{ W/m}^2\cdot\text{K} \times 40.13\text{ K}} = \frac{189,000}{26,084.5} = 7.245\text{ m}^2$

The required surface area for the heat exchanger is $7.25\text{ m}^2$.