4.5 Radiation Heat Transfer, View Factors & Heat Exchanger Sizing (LMTD/NTU)

Radiation Heat Transfer, View Factors & Heat Exchanger Sizing (LMTD/NTU)

Thermal radiation and heat exchanger design encompass the final foundational pillars of thermal-fluid engineering. Unlike conduction and convection, thermal radiation requires no intervening physical medium, propagating via electromagnetic waves at the speed of light. Meanwhile, heat exchangers represent the practical mechanical hardware—ranging from compact plate-fin units to multi-pass shell-and-tube condensers—that transfers thermal energy between fluid streams. This section develops blackbody radiation physics, gray enclosure network algebra, and the dual LMTD and $\epsilon$-NTU heat exchanger design frameworks.

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|                         RADIATION & HEAT EXCHANGER ARCHITECTURE                                   |
|                                                                                                   |
|   [RADIATION LAWS]             [ENCLOSURE NETWORKS]           [HEAT EXCHANGER DESIGN]             |
|   - E_b = \sigma * T^4         - Surface: (1-\epsilon)/(\epsilon*A)   - Duty: q = U*A*F*\Delta T_lm        |
|   - Wien: \lambda_max*T = 2898 - Space: 1 / (A_i * F_ij)      - LMTD: Counter vs Parallel Flow    |
|   - Real: \alpha+\rho+\tau = 1 - Shields: q_N = q_0 / (N+1)   - \epsilon-NTU: q = \epsilon*C_min*\Delta T_max |
+---------------------------------------------------------------------------------------------------+

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1. Fundamentals of Thermal Radiation & Blackbody Laws

Thermal radiation spans the electromagnetic spectrum from $\lambda = 0.1\;\mu\text{m}$ to $100\;\mu\text{m}$ (ultraviolet, visible, and infrared). A blackbody is an ideal surface that absorbs 100% of all incident radiation regardless of wavelength or direction, and emits the maximum possible energy at any given temperature.

Stefan-Boltzmann Law

The total hemispherical emissive power $E_b$ of a blackbody is proportional to the fourth power of its absolute temperature:

$$E_b = \sigma T^4$$

Where:

• $\sigma = \text{Stefan-Boltzmann Constant} = 5.670 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4) = 0.1714 \times 10^{-8}\text{ BTU/(hr}\cdot\text{ft}^2\cdot^\circ\text{R}^4)$
• $T = \text{Absolute Temperature (MUST be in Kelvin or Rankine)}$

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|                         SPECTRAL RADIATION DISTRIBUTION & WIEN'S LAW                              |
|                                                                                                   |
|   Planck's Spectral Law:   E_{b\lambda}(\lambda, T) = \frac{C_1}{\lambda^5 [\exp(C_2 / \lambda T) - 1]} |
|                            where C_1 = 3.742 \times 10^8 W·\mu m^4/m^2, C_2 = 1.439 \times 10^4 \mu m·K   |
|                                                                                                   |
|   Wien's Displacement Law: \lambda_{\max} T = 2897.8\;\mu\text{m}\cdot\text{K}                     |
|                            = 5216.4\;\mu\text{m}\cdot^\circ\text{R}                               |
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   Spectral Emissive Power E_b\lambda
       ^
       |          T_1 = 5800 K (Sun: Peak in visible ~0.5 \mu m)
       |           /\
       |          /  \
       |         /    \      T_2 = 1500 K (Industrial Furnace)
       |        /      \       /\
       |       /        \_____/  \       T_3 = 300 K (Ambient: Peak in IR ~10 \mu m)
       |      /                   \_______/\
       +-----+------------------------------+---------------------------> Wavelength \lambda (\mu m)
            0.4   0.7 (Visible Spectrum)

Real Surface Properties & Kirchhoff's Law

For real surfaces interacting with irradiation $G$ (incident radiation power per unit area):

$$\alpha + \rho + \tau = 1$$

Where $\alpha = \text{absorptivity}$, $\rho = \text{reflectivity}$, $\tau = \text{transmissivity}$. For opaque solids, $\tau = 0 \implies \alpha + \rho = 1$.

• Emissivity ($\epsilon$): Ratio of radiation emitted by a real surface to that emitted by a blackbody at the same temperature: $\epsilon = \frac{E(T)}{E_b(T)} \le 1.0$.
• Kirchhoff's Law: Under conditions of thermal equilibrium ($T_{\text{surface}} = T_{\text{source}}$), total hemispherical emissivity equals absorptivity: $\epsilon = \alpha$.
• Gray and Diffuse Surface Approximation: A gray surface has spectral properties independent of wavelength ($\epsilon_\lambda = \text{const}$), while a diffuse surface emits and reflects uniformly in all directions. Under this approximation, $\epsilon = \alpha$ applies universally.

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2. Radiation View Factors & Enclosure Network Analysis

View Factor Definitions & Fundamental Rules

The view factor (shape factor) $F_{ij}$ is defined as the fraction of diffuse radiation leaving surface $i$ that directly impinges on surface $j$.

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|                             VIEW FACTOR ALGEBRAIC IDENTITIES                                      |
|                                                                                                   |
|   1. Reciprocity Relation:       A_i F_{ij} = A_j F_{ji}                                          |
|                                                                                                   |
|   2. Summation Rule:             \sum_{j=1}^N F_{ij} = 1.0   (for an N-surface closed enclosure)  |
|                                                                                                   |
|   3. Self-Viewing Rule:          F_{ii} = 0 (for planar or convex surfaces)                       |
|                                  F_{ii} > 0 (for concave surfaces that "see" themselves)          |
|                                                                                                   |
|   4. Superposition Rule:         F_{i \to (j+k)} = F_{ij} + F_{ik}                                |
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Radiation Network Circuit Method (Oppenheim Analogy)

Radiosity $J_i$ is the total radiation energy leaving surface $i$ per unit area (emission plus reflected irradiation): $J_i = \epsilon_i E_{b,i} + \rho_i G_i$.

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|                         RADIATION NETWORK RESISTANCE COMPONENTS                                   |
|                                                                                                   |
|   1. Surface Resistance (Potential: E_bi - J_i):      R_{\text{surf},i} = \frac{1-\epsilon_i}{\epsilon_i A_i} |
|                                                                                                   |
|   2. Spatial / Geometric Resistance (Potential: J_i - J_j): R_{\text{space},ij} = \frac{1}{A_i F_{ij}}   |
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   [ Blackbody E_b1 ] ----( R_surf,1 = (1-\epsilon_1)/(\epsilon_1*A_1) )---- [ Radiosity J_1 ]
                                                                           |
                                                  ( R_space = 1 / (A_1 * F_12) )
                                                                           |
   [ Blackbody E_b2 ] ----( R_surf,2 = (1-\epsilon_2)/(\epsilon_2*A_2) )---- [ Radiosity J_2 ]

Net Heat Exchange Between Two Gray Diffuse Surfaces in an Enclosure:

$$q_{12} = \frac{\sigma(T_1^4 - T_2^4)}{\frac{1-\epsilon_1}{\epsilon_1 A_1} + \frac{1}{A_1 F_{12}} + \frac{1-\epsilon_2}{\epsilon_2 A_2}}$$

Special Classic Two-Surface Geometries

Radiation Shields

Inserting $N$ thin, highly reflective radiation shields of emissivity $\epsilon_s$ between two large parallel plates with emissivity $\epsilon_1 = \epsilon_2 = \epsilon$ reduces heat transfer to:

$$q_{N\text{-shields}} = \frac{1}{N + 1} q_{\text{no shield}} \quad (\text{when } \epsilon_s = \epsilon)$$ $$\text{General Case: } q = \frac{A \sigma (T_1^4 - T_2^4)}{\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} - 1\right) + \sum_{i=1}^N \left(\frac{2}{\epsilon_{s,i}} - 1\right)}$$

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3. Heat Exchanger Classifications & Overall Heat Transfer Coefficient ($U$)

Heat exchangers transfer thermal energy between a hot fluid stream (cooling down from $T_{h,\text{in}}$ to $T_{h,\text{out}}$) and a cold fluid stream (heating up from $T_{c,\text{in}}$ to $T_{c,\text{out}}$).

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|                         HEAT EXCHANGER CLASSIFICATION SCHEMES                                     |
|                                                                                                   |
|   1. Double-Pipe (Concentric Tube): Parallel-Flow (Co-current) vs. Counter-Flow (Opposing)        |
|   2. Shell-and-Tube: 1-Shell Pass / 2-Tube Pass (1-2), 2-Shell Pass / 4-Tube Pass (2-4), Baffles  |
|   3. Cross-Flow: Fluids flow perpendicular (Unmixed/Unmixed, Mixed/Unmixed, Mixed/Mixed)          |
|   4. Plate-and-Frame: High surface area per unit volume, corrugated channels, low fouling         |
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Overall Heat Transfer Coefficient ($U$)

The overall heat transfer coefficient accounts for internal convection, tube wall conduction, external convection, and fouling resistance ($R_f$) on both fluid sides:

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

Where $R_{f,i}$ and $R_{f,o}$ are fouling factors (unit thermal resistance due to scale, corrosion, or biological buildup, in $\text{m}^2\cdot\text{K/W}$ or $\text{hr}\cdot\text{ft}^2\cdot^\circ\text{F/BTU}$).

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4. Log Mean Temperature Difference (LMTD) Method

The total steady-state heat exchanger duty $q$ is governed by energy conservation across both fluid streams:

$$q = \dot{m}_h c_{p,h} (T_{h,\text{in}} - T_{h,\text{out}}) = \dot{m}_c c_{p,c} (T_{c,\text{out}} - T_{c,\text{in}}) = U A F \Delta T_{\text{lm}}$$

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|                         LMTD TEMPERATURE PROFILE FORMULATIONS                                     |
|                                                                                                   |
|   COUNTER-FLOW PROFILE:                             PARALLEL-FLOW PROFILE:                        |
|   \Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}}   \Delta T_1 = T_{h,\text{in}} - T_{c,\text{in}}|
|   \Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}}   \Delta T_2 = T_{h,\text{out}} - T_{c,\text{out}}|
|                                                                                                   |
|   Log Mean Temperature Difference:   \Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} |
|   (Note: If \Delta T_1 = \Delta T_2, then \Delta T_{\text{lm}} = \Delta T_1)                    |
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   Counter-Flow:                                    Parallel-Flow:
   Hot:   T_h,in ======> T_h,out                    Hot:   T_h,in ======> T_h,out
   Cold:  T_c,out <===== T_c,in                     Cold:  T_c,in ======> T_c,out
   (Enables T_c,out > T_h,out: "Temperature Cross") (T_c,out can NEVER exceed T_h,out)

Multipass Shell-and-Tube Correction Factor ($F$)

For multipass shell-and-tube or cross-flow exchangers, fluid streams do not flow in pure counter-flow. An empirical LMTD correction factor $F \le 1.0$ is applied:

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

Where $\Delta T_{\text{lm, CF}}$ is computed using the counter-flow terminal formulas, and $F$ is evaluated using dimensionless temperature ratio $P$ and heat capacity ratio $R$:

$$P = \frac{t_2 - t_1}{T_1 - t_1} = \frac{T_{\text{tube,out}} - T_{\text{tube,in}}}{T_{\text{shell,in}} - T_{\text{tube,in}}}, \quad R = \frac{T_1 - T_2}{t_2 - t_1} = \frac{T_{\text{shell,in}} - T_{\text{shell,out}}}{T_{\text{tube,out}} - T_{\text{tube,in}}}$$

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

When inlet temperatures ($T_{h,\text{in}}, T_{c,\text{in}}$) and stream heat capacities are known, but fluid outlet temperatures are unknown, the LMTD method requires tedious iterative guessing. The $\epsilon$-NTU method solves for heat duty directly.

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|                         EFFECTIVENESS-NTU (\epsilon-NTU) FORMULATIONS                             |
|                                                                                                   |
|   Heat Capacity Rates:     C_h = \dot{m}_h c_{p,h}, \quad C_c = \dot{m}_c c_{p,c}                 |
|   Capacitance Extremes:    C_{\min} = \min(C_h, C_c), \quad C_{\max} = \max(C_h, C_c)            |
|   Capacity Ratio:          C_r = \frac{C_{\min}}{C_{\max}} \le 1.0                               |
|                                                                                                   |
|   Maximum Possible Heat:   q_{\max} = C_{\min} (T_{h,\text{in}} - T_{c,\text{in}})               |
|                                                                                                   |
|   Effectiveness (\epsilon):   \epsilon = \frac{q}{q_{\max}} \implies q = \epsilon C_{\min} (T_{h,\text{in}} - T_{c,\text{in}}) |
|                                                                                                   |
|   Number of Transfer Units: NTU = \frac{U A}{C_{\min}}                                           |
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Explicit Effectiveness Formulas $\epsilon(NTU, C_r)$:

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|                             EFFECTIVENESS EQUATIONS BY FLOW TYPE                                  |
|                                                                                                   |
|   1. COUNTER-FLOW:          \epsilon = \frac{1 - \exp[-NTU(1-C_r)]}{1 - C_r \exp[-NTU(1-C_r)]}     |
|                             (Special case C_r = 1: \epsilon = \frac{NTU}{1 + NTU})                |
|                                                                                                   |
|   2. PARALLEL-FLOW:         \epsilon = \frac{1 - \exp[-NTU(1+C_r)]}{1 + C_r}                      |
|                                                                                                   |
|   3. PHASE CHANGE (Boilers / Condensers where C_max -> \infty, C_r = 0):                           |
|      \epsilon = 1 - e^{-NTU}   (Valid for ALL flow configurations!)                               |
+---------------------------------------------------------------------------------------------------+

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6. Step-by-Step Worked Engineering Problem

Problem Statement

A counter-flow double-pipe heat exchanger is designed to cool engine lubricating oil ($c_{p,h} = 2.10\text{ kJ/kg}\cdot\text{K}$) from $T_{h,\text{in}} = 100^\circ\text{C}$ to $T_{h,\text{out}} = 60^\circ\text{C}$ using cooling water ($c_{p,c} = 4.18\text{ kJ/kg}\cdot\text{K}$) entering at $T_{c,\text{in}} = 20^\circ\text{C}$ with a flow rate of $\dot{m}_c = 0.80\text{ kg/s}$. The mass flow rate of oil is $\dot{m}_h = 1.20\text{ kg/s}$. The overall heat transfer coefficient is $U = 450\text{ W/m}^2\cdot\text{K}$.

Determine:

1. The total heat transfer rate (duty) $q$ in kilowatts.
2. The water outlet temperature $T_{c,\text{out}}$.
3. The Log Mean Temperature Difference $\Delta T_{\text{lm}}$ and required heat transfer surface area $A$.
4. Verify the required area using the $\epsilon$-NTU method.

Step-by-Step Solution

Step 1: Compute Heat Transfer Rate ($q$)

• Oil heat capacity rate: $C_h = \dot{m}_h c_{p,h} = (1.20\text{ kg/s}) \times (2.10\text{ kJ/kg}\cdot\text{K}) = 2.520\text{ kW/K} = 2520\text{ W/K}$.
• Heat duty from oil stream: $$q = C_h (T_{h,\text{in}} - T_{h,\text{out}}) = (2.520\text{ kW/K}) \times (100^\circ\text{C} - 60^\circ\text{C}) = 2.520 \times 40 = 100.8\text{ kW}$$

Step 2: Determine Water Outlet Temperature ($T_{c,\text{out}}$)

• Water heat capacity rate: $C_c = \dot{m}_c c_{p,c} = (0.80\text{ kg/s}) \times (4.18\text{ kJ/kg}\cdot\text{K}) = 3.344\text{ kW/K} = 3344\text{ W/K}$.
• Energy balance on cold water stream: $q = C_c (T_{c,\text{out}} - T_{c,\text{in}})$: $$100.8\text{ kW} = 3.344\text{ kW/K} \times (T_{c,\text{out}} - 20^\circ\text{C})$$ $$T_{c,\text{out}} - 20 = \frac{100.8}{3.344} = 30.14^\circ\text{C} \implies T_{c,\text{out}} = 50.14^\circ\text{C}$$

Step 3: Calculate LMTD and Required Surface Area ($A$)

• Terminal temperature differences for counter-flow: $$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 100.0^\circ\text{C} - 50.14^\circ\text{C} = 49.86^\circ\text{C}$$ $$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 60.0^\circ\text{C} - 20.0^\circ\text{C} = 40.00^\circ\text{C}$$
• Log Mean Temperature Difference: $$\Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{49.86 - 40.00}{\ln(49.86 / 40.00)} = \frac{9.86}{\ln(1.2465)} = \frac{9.86}{0.2203} = 44.75^\circ\text{C}$$
• Required area: $$A = \frac{q}{U \Delta T_{\text{lm}}} = \frac{100,800\text{ W}}{(450\text{ W/m}^2\cdot\text{K}) \times (44.75\text{ K})} = \frac{100,800}{20,138} = 5.005\text{ m}^2 \approx 5.01\text{ m}^2$$

Step 4: Verification via $\epsilon$-NTU Method

• Capacitance rates: $C_{\min} = C_h = 2.520\text{ kW/K}$, $C_{\max} = C_c = 3.344\text{ kW/K}$.
• Heat capacity ratio: $C_r = \frac{C_{\min}}{C_{\max}} = \frac{2.520}{3.344} = 0.7536$.
• Maximum possible heat transfer: $$q_{\max} = C_{\min} (T_{h,\text{in}} - T_{c,\text{in}}) = (2.520\text{ kW/K}) \times (100 - 20) = 2.520 \times 80 = 201.6\text{ kW}$$
• Actual effectiveness: $$\epsilon = \frac{q}{q_{\max}} = \frac{100.8\text{ kW}}{201.6\text{ kW}} = 0.5000$$
• Counter-flow NTU inversion formula: $$NTU = \frac{1}{1 - C_r} \ln\left( \frac{1 - \epsilon C_r}{1 - \epsilon} \right) = \frac{1}{1 - 0.7536} \ln\left( \frac{1 - (0.500)(0.7536)}{1 - 0.500} \right)$$ $$NTU = \frac{1}{0.2464} \ln\left( \frac{0.6232}{0.5000} \right) = 4.0584 \times \ln(1.2464) = 4.0584 \times 0.2203 = 0.8941$$
• Calculate area from $NTU = \frac{U A}{C_{\min}}$: $$A = \frac{NTU \times C_{\min}}{U} = \frac{0.8941 \times 2520\text{ W/K}}{450\text{ W/m}^2\cdot\text{K}} = \frac{2253.1}{450} = 5.007\text{ m}^2 \approx 5.01\text{ m}^2$$

(Exact agreement confirms both methods)

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7. Common Exam Traps & PE Pro-Tips