13.2 Forced/Natural Convection, Radiation & Heat Exchanger LMTD/NTU

Forced/Natural Convection, Radiation & Heat Exchanger LMTD/NTU

Thermal energy exchange in industrial mining complexes encompasses convective cooling of heavy mining haul truck engines, radiation inside coal-fired power boilers, and heat recovery via shell-and-tube heat exchangers. This section establishes the theoretical foundations and empirical correlations for convective transport, radiative enclosures, and heat exchanger thermal design.

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1. Convective Heat Transfer Fundamentals & Non-Dimensional Numbers

Convection heat transfer occurs between a solid surface and a moving fluid governed by Newton's Law of Cooling:

$$q = h A_s (T_s - T_{\infty}) \quad \implies \quad q'' = h (T_s - T_{\infty})$$

Where $h\ (\text{W/m}^2\cdot\text{K})$ is the local or average convective heat transfer coefficient. Because $h$ is a complex function of fluid properties ($\rho, \mu, c_p, k$), flow velocity ($V$), and surface geometry ($L$), dimensional analysis groups these variables into foundational dimensionless numbers.

+-----------------------------------------------------------------------------------------+
|                         FOUNDATIONAL NON-DIMENSIONAL NUMBERS                            |
|                                                                                         |
|   Dimensionless Group     Definition                 Physical Significance              |
|   --------------------    -----------------------    ---------------------------------  |
|   Nusselt Number (Nu)     Nu = h * L / k_fluid       Ratio of convective to conductive  |
|                                                      heat transfer across fluid layer.  |
|   Reynolds Number (Re)    Re = rho * V * L / mu      Ratio of dynamic inertia forces    |
|                              = V * L / nu            to viscous shear forces.           |
|   Prandtl Number (Pr)     Pr = nu / alpha            Ratio of momentum diffusivity to   |
|                              = mu * cp / k_fluid     thermal diffusivity.               |
|   Grashof Number (Gr)     Gr = g*beta*DeltaT*L^3/nu^2 Ratio of natural buoyancy forces   |
|                                                      to viscous hydrodynamic forces.    |
|   Rayleigh Number (Ra)    Ra = Gr * Pr               Governs laminar-to-turbulent       |
|                                                      transition in natural convection.  |
+-----------------------------------------------------------------------------------------+

Boundary Layer Mechanics & The Prandtl Number ($\text{Pr}$)

When a fluid flows over a surface, two distinct boundary layers develop:

1. Hydrodynamic Boundary Layer ($\delta$): Region where velocity transitions from $u=0$ at the wall (no-slip condition) to $0.99 V_{\infty}$.
2. Thermal Boundary Layer ($\delta_t$): Region where fluid temperature transitions from $T_s$ at the wall to $0.99 T_{\infty}$.

The relative thickness of these two boundary layers is governed directly by the Prandtl number:

$$\frac{\delta}{\delta_t} \approx \text{Pr}^{1/3}$$

• Liquid Metals ($\text{Pr} \sim 0.005\text{--}0.03$): $\delta_t \gg \delta$. Thermal conduction outpaces momentum diffusion; thermal boundary layer extends far into free stream.
• Gases / Air ($\text{Pr} \sim 0.70\text{--}0.80$): $\delta_t \approx \delta$. Velocity and temperature profiles develop at nearly identical rates.
• Water ($\text{Pr} \sim 2\text{--}10$): $\delta > \delta_t$. Viscous hydrodynamic boundary layer is thicker than the thermal layer.
• Heavy Oils ($\text{Pr} \sim 100\text{--}100,000$): $\delta \gg \delta_t$. High viscosity suppresses momentum while thermal energy diffuses slowly.

Regime Selection: Forced vs. Natural vs. Mixed Convection

The relative dominance of natural versus forced convection is determined by the ratio $\frac{\text{Gr}}{\text{Re}^2}$:

• $\frac{\text{Gr}}{\text{Re}^2} \ll 1$: Forced convection dominates; buoyancy effects are entirely negligible.
• $\frac{\text{Gr}}{\text{Re}^2} \gg 1$: Natural (free) convection dominates; forced velocity effects are negligible.
• $\frac{\text{Gr}}{\text{Re}^2} \approx 1$: Mixed convection; both buoyancy and forced inertia forces must be accounted for ($Nu_{\text{combined}} = [Nu_{\text{forced}}^n \pm Nu_{\text{natural}}^n]^{1/n}$).

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2. Forced Convection Empirical Correlations

1. Fully Developed Laminar Flow in Circular Tubes ($\text{Re}_D < 2300$)

For fully developed internal laminar pipe flow, the Nusselt number is a constant value independent of Reynolds and Prandtl numbers:

$$\text{Constant Surface Heat Flux } (q_s'' = \text{const}): \quad \text{Nu}_D = \frac{h D}{k} = \frac{48}{11} \approx 4.36$$ $$\text{Constant Surface Temperature } (T_s = \text{const}): \quad \text{Nu}_D = \frac{h D}{k} = 3.66$$

2. Fully Developed Turbulent Flow in Circular Tubes ($\text{Re}_D > 10,000$)

The classical Dittus-Boelter Equation applies for smooth circular tubes under moderate temperature differences ($0.6 \le \text{Pr} \le 160$, $L/D > 10$):

$$\text{Nu}_D = 0.023 \, \text{Re}_D^{0.8} \, \text{Pr}^n$$ $$\text{Where: } \quad n = 0.4 \quad (\text{Heating of fluid: } T_s > T_b), \qquad n = 0.3 \quad (\text{Cooling of fluid: } T_s < T_b)$$

When large temperature differences cause significant radial variation in fluid dynamic viscosity, the Sieder-Tate Correlation is applied:

$$\text{Nu}_D = 0.027 \, \text{Re}_D^{0.8} \, \text{Pr}^{1/3} \left(\frac{\mu_b}{\mu_w}\right)^{0.14}$$

Where $\mu_b$ is evaluated at bulk temperature and $\mu_w$ is evaluated at the tube wall temperature.

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3. Natural (Free) Convection Correlations

Natural convection is driven by density gradients in the presence of a body force (gravity $g$). The volumetric coefficient of thermal expansion $\beta$ is defined as:

$$\beta = -\frac{1}{\rho}\left(\frac{\partial \rho}{\partial T}\right)_P \approx \frac{1}{T_f} \quad (\text{for ideal gases, where } T_f = (T_s + T_{\infty})/2 \text{ in Kelvin})$$ $$\text{Rayleigh Number } \text{Ra}_L = \text{Gr}_L \cdot \text{Pr} = \frac{g \beta (T_s - T_{\infty}) L^3}{\nu \alpha}$$

Vertical Plates and Cylinders of Height $L$

• Laminar Flow ($\text{Ra}_L < 10^9$): $\overline{\text{Nu}}_L = 0.59 \, \text{Ra}_L^{1/4}$
• Turbulent Flow ($\text{Ra}_L > 10^9$): $\overline{\text{Nu}}_L = 0.10 \, \text{Ra}_L^{1/3}$

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4. Radiation Fundamentals & Governing Laws

Thermal radiation is electromagnetic radiation emitted by matter as a result of its temperature, propagating at the speed of light ($c_0 = 3 \times 10^8\text{ m/s}$) across wavelengths $\lambda = 0.1\text{--}100\ \mu\text{m}$ (spanning ultraviolet, visible, and infrared spectra).

+-----------------------------------------------------------------------------------------+
|                            FUNDAMENTAL RADIATION LAWS                                   |
|                                                                                         |
|   Law                     Mathematical Formula                  Physical Meaning        |
|   --------------------    ---------------------------------     ----------------------- |
|   Stefan-Boltzmann Law    E_b = sigma * T^4                     Total emissive power of |
|                           sigma = 5.67e-8 W/(m^2*K^4)           an ideal black body.    |
|   Planck's Law            E_b_lambda(lambda, T)                 Spectral distribution   |
|                           = C1 / [lambda^5 * (exp(C2/lambda*T) - 1)] of black body radiation.|
|   Wien's Displacement     lambda_max * T = 2898 micron*K        Peak emission wavelength|
|                           = 2.898e-3 m*K                        shifts inversely with T.|
|   Kirchhoff's Law         alpha_lambda = epsilon_lambda         Monochromatic absorptivity|
|                           alpha = epsilon (Grey Body)           equals emissivity at Teq|
+-----------------------------------------------------------------------------------------+

Surface Radiation Properties

For any incident radiative irradiation $G\ (\text{W/m}^2)$ striking a surface:

$$\alpha + \rho_r + \tau_r = 1$$

Where $\alpha$ is absorptivity, $\rho_r$ is reflectivity, and $\tau_r$ is transmissivity.

• Opaque Body ($\tau_r = 0$): $\alpha + \rho_r = 1$.
• Black Body ($\alpha = 1, \rho_r = 0, \tau_r = 0$): Perfect absorber and perfect emitter ($\epsilon = 1$).
• Grey Body: A surface whose monochromatic emissivity $\epsilon_{\lambda}$ and absorptivity $\alpha_{\lambda}$ are independent of wavelength across the spectral band: $\epsilon = \alpha = \text{const} < 1$.

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5. View Factor (Shape Factor) Algebra & Enclosure Theory

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

$$F_{ij} = \frac{1}{A_i} \int_{A_i} \int_{A_j} \frac{\cos\theta_i \cos\theta_j}{\pi R^2} dA_j dA_i$$

+-----------------------------------------------------------------------------------------+
|                             VIEW FACTOR ALGEBRA RULES                                   |
|                                                                                         |
|   1. Reciprocity Relation:     A_i * F_ij = A_j * F_ji                                  |
|   2. Summation Rule (Enclosure): sum_{j=1}^N F_ij = 1                                   |
|   3. Flat / Convex Surface:    F_ii = 0 (A surface cannot see itself)                   |
|   4. Concave Surface:          F_ii > 0 (A concave cavity radiates to itself)           |
|   5. Superposition Rule:       F_1(2+3) = F_12 + F_13                                   |
|                                F_(2+3)1 = (A_2*F_21 + A_3*F_31) / (A_2 + A_3)           |
+-----------------------------------------------------------------------------------------+

Standard 2-Surface View Factor Geometries

1. Hemisphere Cavity (Surface 1) closed by Flat Disc Base (Surface 2):
   • Base 2 is flat: $F_{22} = 0 \implies F_{21} = 1$.
   • By Reciprocity: $A_1 F_{12} = A_2 F_{21} \implies (2\pi R^2) F_{12} = (\pi R^2)(1) \implies F_{12} = 0.5$.
   • By Summation: $F_{11} + F_{12} = 1 \implies F_{11} = 1 - 0.5 = 0.5$.
2. Infinite Concentric Cylinders (Inner 1, Outer 2):
   • Inner cylinder is convex: $F_{11} = 0 \implies F_{12} = 1$.
   • By Reciprocity: $A_1 F_{12} = A_2 F_{21} \implies F_{21} = \frac{A_1}{A_2} = \frac{r_1}{r_2}$.
   • By Summation: $F_{22} = 1 - F_{21} = 1 - \frac{r_1}{r_2}$.

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6. Radiative Heat Exchange Between Grey Surfaces & Radiation Shields

Using the radiation electrical network analogy, each real surface has a surface resistance and each pair of surfaces shares a space resistance:

+-----------------------------------------------------------------------------------------+
|                        TWO-SURFACE RADIATION NETWORK CIRCUIT                            |
|                                                                                         |
|   [E_b1 = sigma*T1^4]                                       [E_b2 = sigma*T2^4]         |
|            o---/\/\/\/\/\/---[J1]---/\/\/\/\/\/---[J2]---/\/\/\/\/\/---o                |
|                 R_surface,1           R_space            R_surface,2                    |
|               (1-eps1)/(eps1*A1)     1/(A1*F12)        (1-eps2)/(eps2*A2)               |
|                                                                                         |
|   Net Heat Exchange Rate:                                                               |
|   q_12 = sigma * (T1^4 - T2^4) / [ (1-eps1)/(eps1*A1) + 1/(A1*F12) + (1-eps2)/(eps2*A2) ]|
+-----------------------------------------------------------------------------------------+

1. Two Infinite Parallel Grey Plates ($A_1 = A_2 = A, F_{12} = 1$)

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

2. Radiation Shields

A radiation shield is a thin, highly reflective sheet placed between two radiating surfaces to introduce additional surface and space resistances without generating or absorbing work.

+-----------------------------------------------------------------------------------------+
|                             RADIATION SHIELD ATTENUATION                                |
|                                                                                         |
|   For n identical thin shields (emissivity eps_s = eps_1 = eps_2) inserted between      |
|   two large parallel plates:                                                            |
|                                                                                         |
|   (q_12)_with_n_shields = [ 1 / (n + 1) ] * (q_12)_without_shield                       |
|                                                                                         |
|   - 1 Shield  (n=1): Reduces radiative heat transfer by 50%  (Factor = 1/2)             |
|   - 2 Shields (n=2): Reduces radiative heat transfer by 66.7% (Factor = 1/3)            |
|   - 3 Shields (n=3): Reduces radiative heat transfer by 75%  (Factor = 1/4)             |
+-----------------------------------------------------------------------------------------+

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7. Heat Exchanger Analysis: Log Mean Temperature Difference (LMTD)

Heat exchangers facilitate thermal transfer between two fluids separated by a solid wall. The fundamental rating equation is:

$$\dot{Q} = U A \Delta T_m$$

Where $U$ is the overall heat transfer coefficient, $A$ is the heat transfer area, and $\Delta T_m$ is the Log Mean Temperature Difference (LMTD).

+-----------------------------------------------------------------------------------------+
|                        PARALLEL FLOW VS COUNTER FLOW PROFILES                           |
|                                                                                         |
|   PARALLEL FLOW:                              COUNTER FLOW:                             |
|   Temp                                        Temp                                      |
|    ^  T_h,in                                   ^  T_h,in                                |
|    |  *                                        |  *                                     |
|    |   \  Hot Fluid                             |   \  Hot Fluid       T_c,out          |
|    |    \---------> T_h,out                    |    \-------------> *                  |
|    |                 *                         |                     \  Cold Fluid      |
|    |    /---------> T_c,out                    |    /-----------------\                 |
|    |   /  Cold Fluid                           |   /                   * T_c,in         |
|    |  * T_c,in                                 |  * T_h,out                             |
|    +----------------------> Area               +----------------------> Area            |
|       DeltaT1       DeltaT2                       DeltaT1       DeltaT2                 |
|       (Inlet)       (Outlet)                      (Left End)    (Right End)             |
+-----------------------------------------------------------------------------------------+

$$\text{LMTD} = \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)}$$

• Parallel Flow: $\Delta T_1 = T_{h,\text{in}} - T_{c,\text{in}}$, $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{out}}$.
• Counter Flow: $\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}}$, $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}}$.

Special Mathematical Cases of LMTD

1. Balanced Counter-Flow ($C_h = C_c \implies \Delta T_1 = \Delta T_2 = \Delta T$): The temperature difference is constant throughout the exchanger. Direct evaluation of LMTD yields $0/0$; applying L'Hôpital's rule gives:

$$\text{LMTD} = \Delta T_1 = \Delta T_2$$

1. Phase Change Components (Condensers and Boilers/Evaporators): One fluid changes phase at constant saturation temperature ($C_{\text{max}} \to \infty$). The temperature profile is identical regardless of flow direction:

$$\text{LMTD}_{\text{counter}} = \text{LMTD}_{\text{parallel}}$$

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8. Heat Exchanger Analysis: The Effectiveness-NTU ($\epsilon\text{-NTU}$) Method

When fluid outlet temperatures are unknown, the LMTD method requires tedious iterative trials. The Effectiveness-NTU method solves performance directly from known inlet conditions.

1. Fluid Heat Capacity Rates

$$C_h = \dot{m}_h c_{p,h}, \qquad C_c = \dot{m}_c c_{p,c}$$ $$C_{\text{min}} = \min(C_h, C_c), \qquad C_{\text{max}} = \max(C_h, C_c), \qquad C_r = \frac{C_{\text{min}}}{C_{\text{max}}} \quad (0 \le C_r \le 1)$$

2. Maximum Possible Heat Transfer Rate ($Q_{\text{max}}$)

$$Q_{\text{max}} = C_{\text{min}} (T_{h,\text{in}} - T_{c,\text{in}})$$

3. Heat Exchanger Effectiveness ($\epsilon$)

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

4. Number of Transfer Units ($\text{NTU}$)

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

+-----------------------------------------------------------------------------------------+
|                        EFFECTIVENESS FORMULAS BY CONFIGURATION                          |
|                                                                                         |
|   Configuration            Capacity Ratio (Cr)    Effectiveness Formula (epsilon)       |
|   ---------------------    -------------------    -----------------------------------   |
|   All Exchangers (Boiler/  C_r = 0                epsilon = 1 - exp( -NTU )             |
|   Condenser phase change)  (C_max -> infinity)                                          |
|   Counter Flow             C_r = 1 (Balanced)     epsilon = NTU / (1 + NTU)             |
|   Parallel Flow            C_r = 1                epsilon = [ 1 - exp(-2*NTU) ] / 2     |
|   Counter Flow (General)   0 < C_r < 1            epsilon = [1 - exp(-NTU*(1-Cr))] /    |
|                                                             [1 - Cr*exp(-NTU*(1-Cr))]   |
+-----------------------------------------------------------------------------------------+

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9. Step-by-Step Worked Problems: Heat Exchanger Design & Radiation Enclosures

Worked Example 1: Mining Compressor Intercooler Sizing via $\epsilon\text{-NTU}$

An air compressor intercooler operates in counter-flow to cool compressed air from $T_{h,\text{in}} = 140^{\circ}\text{C}$ to $T_{h,\text{out}} = 40^{\circ}\text{C}$ using cooling water entering at $T_{c,\text{in}} = 20^{\circ}\text{C}$. The heat capacity rate of air is $C_h = 2.0\text{ kW/K}$ and cooling water is $C_c = 4.0\text{ kW/K}$. The overall heat transfer coefficient is $U = 250\text{ W/m}^2\cdot\text{K}$. Calculate the required heat transfer surface area $A$.

+-----------------------------------------------------------------------------------------+
|                        INTERCOOLER SIZING CALCULATION STEPS                             |
|                                                                                         |
|   STEP 1: Determine Capacity Rates and Ratio                                            |
|           C_min = C_h = 2.0 kW/K = 2000 W/K                                             |
|           C_max = C_c = 4.0 kW/K = 4000 W/K                                             |
|           C_r = C_min / C_max = 2000 / 4000 = 0.50                                      |
|                                                                                         |
|   STEP 2: Calculate Actual Heat Transfer Rate                                           |
|           Q_act = C_h * (T_h,in - T_h,out) = 2.0 kW/K * (140 - 40) = 200 kW             |
|                                                                                         |
|   STEP 3: Calculate Maximum Possible Heat Transfer & Effectiveness                      |
|           Q_max = C_min * (T_h,in - T_c,in) = 2.0 kW/K * (140 - 20) = 240 kW            |
|           epsilon = Q_act / Q_max = 200 / 240 = 0.8333 (83.33%)                         |
|                                                                                         |
|   STEP 4: Solve for NTU for Counter Flow (Cr = 0.50, epsilon = 0.8333)                  |
|           NTU = [ 1 / (1 - C_r) ] * ln[ (1 - epsilon*C_r) / (1 - epsilon) ]             |
|           NTU = [ 1 / 0.50 ] * ln[ (1 - 0.8333*0.50) / (1 - 0.8333) ]                   |
|           NTU = 2.0 * ln[ (1 - 0.41667) / 0.16667 ] = 2.0 * ln[ 0.58333 / 0.16667 ]     |
|           NTU = 2.0 * ln(3.50) = 2.0 * 1.25276 = 2.5055                                |
|                                                                                         |
|   STEP 5: Calculate Area A                                                              |
|           NTU = U*A / C_min  ===>  A = (NTU * C_min) / U                                |
|           A = (2.5055 * 2000 W/K) / (250 W/m^2*K) = 5011 / 250 = 20.04 m^2              |
+-----------------------------------------------------------------------------------------+