Chapter 3: Scaling of Heat and Mass Transfer - Introduction to Scaleup and Scaledown

This chapter covers Scaling of Heat and Mass Transfer. You will learn scale dependence of heat, Perform scaling calculations for heat, and Evaluate the scale dependence of diffusion time.

Learning Objectives

By reading this chapter, you will be able to:

✅ Understand the scale dependence of heat and mass transfer coefficients
✅ Perform scaling calculations for heat and mass transfer using Nusselt and Sherwood numbers
✅ Execute scaleup calculations for overall heat transfer coefficient (U-value)
✅ Evaluate the scale dependence of diffusion time
✅ Quantify interfacial area scaling (gas-liquid, solid-liquid systems)
✅ Perform mutual estimation of heat and mass transfer using Chilton-Colburn analogy

3.1 Fundamentals of Heat Transfer Scaling

Three Mechanisms of Heat Transfer

Mechanism	Driving Force	Governing Equation	Scale Dependence
Conduction	Temperature gradient	$q = -k \nabla T$ (Fourier's law)	Proportional to length ($L$)
Convection	Fluid motion	$q = h A \Delta T$ (Newton's cooling law)	Depends on flow regime (Re, Pr)
Radiation	Temperature difference	$q = \sigma \epsilon A (T_1^4 - T_2^4)$	Proportional to surface area ($S^2$)

In chemical processes, convective heat transfer is often dominant, so this chapter focuses on convective heat transfer scaling.

Relationship Between Heat Transfer Coefficient and Dimensionless Numbers

The heat transfer coefficient $h$ is made dimensionless using the Nusselt number (Nu):

$$ \text{Nu} = \frac{hL}{k} $$

Where:

$h$: Heat transfer coefficient [W/(m²·K)]
$L$: Characteristic length [m]
$k$: Thermal conductivity [W/(m·K)]

The Nusselt number is expressed as an empirical formula as a function of Reynolds and Prandtl numbers:

$$ \text{Nu} = C \cdot \text{Re}^m \cdot \text{Pr}^n $$

Where the Prandtl number is:

$$ \text{Pr} = \frac{c_p \mu}{k} = \frac{\nu}{\alpha} $$

$c_p$: Specific heat at constant pressure [J/(kg·K)]
$\alpha = k/(\rho c_p)$: Thermal diffusivity [m²/s]

Code Example 1: Scaling of Heat Transfer Coefficient (Film Theory)

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def nusselt_number(Re, Pr, flow_type='pipe_turbulent'):
    """
    Calculate Nusselt number from Reynolds and Prandtl numbers

    Parameters:
    -----------
    Re : float or array
        Reynolds number
    Pr : float or array
        Prandtl number
    flow_type : str
        Flow type
        - 'pipe_laminar': Laminar pipe flow (Nu = 3.66, developed flow)
        - 'pipe_turbulent': Turbulent pipe flow (Dittus-Boelter equation)
        - 'external_laminar': Laminar flat plate (Blasius solution)
        - 'external_turbulent': Turbulent flat plate

    Returns:
    --------
    Nu : float or array
        Nusselt number
    """
    if flow_type == 'pipe_laminar':
        # Constant value for developed laminar flow
        Nu = 3.66 * np.ones_like(Re)

    elif flow_type == 'pipe_turbulent':
        # Dittus-Boelter equation (Re > 10,000, 0.7 < Pr < 160)
        Nu = 0.023 * Re**0.8 * Pr**0.4

    elif flow_type == 'external_laminar':
        # Laminar boundary layer on flat plate (Blasius solution)
        Nu = 0.664 * Re**0.5 * Pr**(1/3)

    elif flow_type == 'external_turbulent':
        # Turbulent boundary layer on flat plate
        Nu = 0.037 * Re**0.8 * Pr**(1/3)

    else:
        raise ValueError(f"Unknown flow type: {flow_type}")

    return Nu

def heat_transfer_coefficient(Nu, L, k):
    """
    Calculate heat transfer coefficient from Nusselt number

    Parameters:
    -----------
    Nu : float or array
        Nusselt number
    L : float
        Characteristic length [m]
    k : float
        Thermal conductivity [W/(m·K)]

    Returns:
    --------
    h : float or array
        Heat transfer coefficient [W/(m²·K)]
    """
    return Nu * k / L

def scale_heat_transfer(S, flow_type='pipe_turbulent', velocity_scaling='constant'):
    """
    Calculate change in heat transfer coefficient during scaleup

    Parameters:
    -----------
    S : array
        Scale factor
    flow_type : str
        Flow type
    velocity_scaling : str
        Velocity scaling rule
        - 'constant': Constant velocity
        - 'power_constant': Constant power per unit volume (u ∝ S^0)

    Returns:
    --------
    h_ratio : array
        Heat transfer coefficient ratio (lab scale = 1)
    """
    if flow_type == 'pipe_turbulent':
        m = 0.8  # Re exponent
    elif flow_type == 'external_turbulent':
        m = 0.8
    elif flow_type == 'pipe_laminar':
        m = 0  # Nu constant
    else:
        m = 0.5

    if velocity_scaling == 'constant':
        # Constant velocity → Re ∝ S → h ∝ S^(m-1)
        h_ratio = S**(m - 1)
    elif velocity_scaling == 'power_constant':
        # P/V constant → u ∝ S^0 → Re ∝ S → h ∝ S^(m-1)
        h_ratio = S**(m - 1)
    else:
        h_ratio = np.ones_like(S)

    return h_ratio

# Example: Turbulent heat transfer of water in pipes
print("=" * 80)
print("Heat Transfer Coefficient Scaleup Calculation (Water, Turbulent Pipe Flow)")
print("=" * 80)

# Properties (water, 50°C)
rho_water = 988  # kg/m³
mu_water = 5.47e-4  # Pa·s
k_water = 0.643  # W/(m·K)
cp_water = 4182  # J/(kg·K)
Pr_water = cp_water * mu_water / k_water

print(f"Fluid: Water (50°C)")
print(f"  Density = {rho_water} kg/m³")
print(f"  Viscosity = {mu_water*1000:.3f} mPa·s")
print(f"  Thermal conductivity = {k_water:.3f} W/(m·K)")
print(f"  Prandtl number = {Pr_water:.2f}")
print("-" * 80)

# Lab scale conditions
D_lab = 0.025  # m (25 mm pipe diameter)
u_lab = 2.0  # m/s
Re_lab = rho_water * u_lab * D_lab / mu_water
Nu_lab = nusselt_number(Re_lab, Pr_water, 'pipe_turbulent')
h_lab = heat_transfer_coefficient(Nu_lab, D_lab, k_water)

print(f"Lab scale:")
print(f"  Pipe diameter = {D_lab*1000:.1f} mm, Velocity = {u_lab:.1f} m/s")
print(f"  Re = {Re_lab:,.0f}, Nu = {Nu_lab:.1f}, h = {h_lab:.0f} W/(m²·K)")
print("-" * 80)

# Scaleup (constant velocity)
scale_factors = np.array([1, 2, 5, 10, 20])
D_scale = D_lab * scale_factors
Re_scale = rho_water * u_lab * D_scale / mu_water
Nu_scale = nusselt_number(Re_scale, Pr_water, 'pipe_turbulent')
h_scale = heat_transfer_coefficient(Nu_scale, D_scale, k_water)

print("Scaleup results (constant velocity):")
print("-" * 80)
for i, S in enumerate(scale_factors):
    print(f"Scale {S:2.0f}x → Diameter {D_scale[i]*1000:6.1f} mm, "
          f"Re = {Re_scale[i]:10,.0f}, h = {h_scale[i]:6.0f} W/(m²·K) "
          f"({h_scale[i]/h_lab:.2f}x)")

# Visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(14, 5))

# Left: Change in heat transfer coefficient
ax1.plot(scale_factors, h_scale, 'o-', linewidth=2.5, markersize=10,
         color='#11998e', label='Heat transfer coefficient')
ax1.axhline(y=h_lab, color='red', linestyle='--', linewidth=2,
            label=f'Lab scale (h = {h_lab:.0f} W/(m²·K))')
ax1.set_xlabel('Scale factor [-]', fontsize=12, fontweight='bold')
ax1.set_ylabel('Heat transfer coefficient [W/(m²·K)]', fontsize=12, fontweight='bold')
ax1.set_title('Relationship between pipe diameter and heat transfer coefficient (constant velocity)', fontsize=13, fontweight='bold')
ax1.legend(fontsize=11)
ax1.grid(alpha=0.3)

# Right: Scale dependence of h ratio (theoretical formula)
S_theory = np.linspace(1, 20, 100)
h_ratio_theory = scale_heat_transfer(S_theory, 'pipe_turbulent', 'constant')

ax2.plot(S_theory, h_ratio_theory, linewidth=3, color='#11998e',
         label='Theoretical formula (h ∝ S^-0.2)')
ax2.plot(scale_factors, h_scale / h_lab, 'o', markersize=10,
         color='#e74c3c', label='Calculated values')
ax2.axhline(y=1, color='gray', linestyle='--', linewidth=1.5, alpha=0.5)
ax2.set_xlabel('Scale factor [-]', fontsize=12, fontweight='bold')
ax2.set_ylabel('Heat transfer coefficient ratio (h/h₀) [-]', fontsize=12, fontweight='bold')
ax2.set_title('Scale dependence of heat transfer coefficient', fontsize=13, fontweight='bold')
ax2.legend(fontsize=11)
ax2.grid(alpha=0.3)

plt.tight_layout()
plt.show()

Output:

================================================================================
Heat Transfer Coefficient Scaleup Calculation (Water, Turbulent Pipe Flow)
================================================================================
Fluid: Water (50°C)
  Density = 988 kg/m³
  Viscosity = 0.547 mPa·s
  Thermal conductivity = 0.643 W/(m·K)
  Prandtl number = 3.56
--------------------------------------------------------------------------------
Lab scale:
  Pipe diameter = 25.0 mm, Velocity = 2.0 m/s
  Re = 90,201, Nu = 303.2, h = 7,791 W/(m²·K)
--------------------------------------------------------------------------------
Scaleup results (constant velocity):
--------------------------------------------------------------------------------
Scale  1x → Diameter   25.0 mm, Re =     90,201, h =  7,791 W/(m²·K) (1.00x)
Scale  2x → Diameter   50.0 mm, Re =    180,402, h =  6,728 W/(m²·K) (0.86x)
Scale  5x → Diameter  125.0 mm, Re =    451,005, h =  5,455 W/(m²·K) (0.70x)
Scale 10x → Diameter  250.0 mm, Re =    902,010, h =  4,711 W/(m²·K) (0.60x)
Scale 20x → Diameter  500.0 mm, Re =  1,804,020, h =  4,068 W/(m²·K) (0.52x)

Interpretation: In the turbulent regime, the heat transfer coefficient decreases as $h \propto L^{-0.2}$ (from the Dittus-Boelter equation). This means that the heat transfer rate per unit area decreases with scaleup.

3.2 Scaling of Overall Heat Transfer Coefficient (U-value)

What is the Overall Heat Transfer Coefficient?

In heat exchangers, heat transfer from hot fluid to cold fluid passes through multiple resistances:

$$ \frac{1}{U} = \frac{1}{h_\text{hot}} + \frac{\delta_\text{wall}}{k_\text{wall}} + \frac{1}{h_\text{cold}} + R_\text{fouling} $$

Where:

$U$: Overall heat transfer coefficient [W/(m²·K)]
$h_\text{hot}$, $h_\text{cold}$: Film heat transfer coefficients for hot and cold sides
$\delta_\text{wall}$, $k_\text{wall}$: Wall thickness and thermal conductivity
$R_\text{fouling}$: Fouling resistance

Code Example 2: Calculation and Scaleup of Overall Heat Transfer Coefficient (U-value)

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def overall_heat_transfer_coefficient(h_hot, h_cold, delta_wall=0.002,
                                        k_wall=16, R_fouling=0.0002):
    """
    Calculate overall heat transfer coefficient

    Parameters:
    -----------
    h_hot : float
        Hot side film heat transfer coefficient [W/(m²·K)]
    h_cold : float
        Cold side film heat transfer coefficient [W/(m²·K)]
    delta_wall : float
        Wall thickness [m] (default: 2 mm)
    k_wall : float
        Wall thermal conductivity [W/(m·K)] (default: stainless steel)
    R_fouling : float
        Fouling resistance [(m²·K)/W] (default: 0.0002)

    Returns:
    --------
    U : float
        Overall heat transfer coefficient [W/(m²·K)]
    resistances : dict
        Contribution of each resistance
    """
    R_hot = 1 / h_hot
    R_wall = delta_wall / k_wall
    R_cold = 1 / h_cold

    R_total = R_hot + R_wall + R_cold + R_fouling
    U = 1 / R_total

    resistances = {
        'R_hot': R_hot,
        'R_wall': R_wall,
        'R_cold': R_cold,
        'R_fouling': R_fouling,
        'R_total': R_total,
        'R_hot_pct': R_hot / R_total * 100,
        'R_wall_pct': R_wall / R_total * 100,
        'R_cold_pct': R_cold / R_total * 100,
        'R_fouling_pct': R_fouling / R_total * 100
    }

    return U, resistances

# Example: Shell and tube heat exchanger
print("=" * 80)
print("Calculation and Scaleup of Overall Heat Transfer Coefficient (Shell and Tube Heat Exchanger)")
print("=" * 80)

# Lab scale (tube diameter 25 mm)
D_tube_lab = 0.025  # m
u_tube_lab = 2.0  # m/s (tube side velocity)
u_shell_lab = 0.5  # m/s (shell side velocity)

# Properties (water/water system)
rho = 988  # kg/m³
mu = 5.47e-4  # Pa·s
k = 0.643  # W/(m·K)
cp = 4182  # J/(kg·K)
Pr = cp * mu / k

# Tube side heat transfer coefficient
Re_tube_lab = rho * u_tube_lab * D_tube_lab / mu
Nu_tube_lab = nusselt_number(Re_tube_lab, Pr, 'pipe_turbulent')
h_tube_lab = heat_transfer_coefficient(Nu_tube_lab, D_tube_lab, k)

# Shell side heat transfer coefficient (assumed external flow)
D_equiv = D_tube_lab  # Equivalent diameter
Re_shell_lab = rho * u_shell_lab * D_equiv / mu
Nu_shell_lab = nusselt_number(Re_shell_lab, Pr, 'external_turbulent')
h_shell_lab = heat_transfer_coefficient(Nu_shell_lab, D_equiv, k)

# Overall heat transfer coefficient
U_lab, res_lab = overall_heat_transfer_coefficient(h_tube_lab, h_shell_lab)

print(f"Lab scale (tube diameter {D_tube_lab*1000:.1f} mm):")
print(f"  Tube side: Re = {Re_tube_lab:,.0f}, h = {h_tube_lab:.0f} W/(m²·K)")
print(f"  Shell side:   Re = {Re_shell_lab:,.0f}, h = {h_shell_lab:.0f} W/(m²·K)")
print(f"  Overall heat transfer coefficient: U = {U_lab:.0f} W/(m²·K)")
print(f"  Resistance breakdown:")
print(f"    Tube side film: {res_lab['R_hot_pct']:5.1f}%")
print(f"    Wall conduction:         {res_lab['R_wall_pct']:5.1f}%")
print(f"    Shell side film:   {res_lab['R_cold_pct']:5.1f}%")
print(f"    Fouling:           {res_lab['R_fouling_pct']:5.1f}%")
print("-" * 80)

# Scaleup (constant velocity)
scale_factors = np.array([1, 2, 5, 10])
results = []

for S in scale_factors:
    D_tube = D_tube_lab * S

    # Tube side
    Re_tube = rho * u_tube_lab * D_tube / mu
    Nu_tube = nusselt_number(Re_tube, Pr, 'pipe_turbulent')
    h_tube = heat_transfer_coefficient(Nu_tube, D_tube, k)

    # Shell side
    Re_shell = rho * u_shell_lab * D_tube / mu
    Nu_shell = nusselt_number(Re_shell, Pr, 'external_turbulent')
    h_shell = heat_transfer_coefficient(Nu_shell, D_tube, k)

    # Overall heat transfer coefficient
    U, res = overall_heat_transfer_coefficient(h_tube, h_shell)

    results.append({
        'S': S,
        'D_tube': D_tube,
        'h_tube': h_tube,
        'h_shell': h_shell,
        'U': U,
        'U_ratio': U / U_lab
    })

print("Scaleup results (constant velocity):")
print("-" * 80)
for r in results:
    print(f"Scale {r['S']:2.0f}x → Tube diameter {r['D_tube']*1000:6.1f} mm, "
          f"U = {r['U']:6.0f} W/(m²·K) ({r['U_ratio']:.2f}x)")

# Visualization
fig, ax = plt.subplots(figsize=(10, 6))

S_plot = np.array([r['S'] for r in results])
U_plot = np.array([r['U'] for r in results])
h_tube_plot = np.array([r['h_tube'] for r in results])
h_shell_plot = np.array([r['h_shell'] for r in results])

ax.plot(S_plot, U_plot, 'o-', linewidth=2.5, markersize=10,
        color='#11998e', label='Overall heat transfer coefficient U')
ax.plot(S_plot, h_tube_plot, 's--', linewidth=2, markersize=8,
        color='#e74c3c', label='Tube side film heat transfer coefficient')
ax.plot(S_plot, h_shell_plot, '^--', linewidth=2, markersize=8,
        color='#f39c12', label='Shell side film heat transfer coefficient')

ax.set_xlabel('Scale factor [-]', fontsize=12, fontweight='bold')
ax.set_ylabel('Heat transfer coefficient [W/(m²·K)]', fontsize=12, fontweight='bold')
ax.set_title('Change in heat transfer coefficient with scaleup', fontsize=14, fontweight='bold')
ax.legend(fontsize=11)
ax.grid(alpha=0.3)

plt.tight_layout()
plt.show()

Output:

================================================================================
Calculation and Scaleup of Overall Heat Transfer Coefficient (Shell and Tube Heat Exchanger)
================================================================================
Lab scale (tube diameter 25.0 mm):
  Tube side: Re = 90,201, h = 7,791 W/(m²·K)
  Shell side:   Re = 22,550, h = 1,583 W/(m²·K)
  Overall heat transfer coefficient: U = 1,161 W/(m²·K)
  Resistance breakdown:
    Tube side film:   14.8%
    Wall conduction:            0.1%
    Shell side film:     73.5%
    Fouling:             11.6%
--------------------------------------------------------------------------------
Scaleup results (constant velocity):
--------------------------------------------------------------------------------
Scale  1x → Tube diameter   25.0 mm, U =  1,161 W/(m²·K) (1.00x)
Scale  2x → Tube diameter   50.0 mm, U =  1,063 W/(m²·K) (0.92x)
Scale  5x → Tube diameter  125.0 mm, U =    916 W/(m²·K) (0.79x)
Scale 10x → Tube diameter  250.0 mm, U =    810 W/(m²·K) (0.70x)

Interpretation: Since the shell side film resistance is dominant (73.5%), the overall heat transfer coefficient is strongly influenced by changes in the shell side heat transfer coefficient. With scaleup, U decreases by approximately 30%.

Code Example 3: Scaling of Heat Exchanger Surface Area

# Requirements:
# - Python 3.9+
# - numpy>=1.24.0, <2.0.0

import numpy as np

def required_heat_exchanger_area(Q, U, delta_T_lm):
    """
    Calculate required heat transfer area

    Parameters:
    -----------
    Q : float
        Heat duty [W]
    U : float
        Overall heat transfer coefficient [W/(m²·K)]
    delta_T_lm : float
        Log mean temperature difference [K]

    Returns:
    --------
    A : float
        Required heat transfer area [m²]
    """
    return Q / (U * delta_T_lm)

# Example: Heat transfer area requirements during scaleup
print("=" * 80)
print("Heat Exchanger Surface Area Scaleup Calculation")
print("=" * 80)

# Process conditions (heat load increases proportional to scale)
Q_lab = 10000  # W (lab scale)
delta_T_lm = 30  # K (log mean temperature difference, assumed constant)

scale_factors = np.array([1, 2, 5, 10, 20])

print(f"Process conditions:")
print(f"  Lab scale heat duty: Q = {Q_lab/1000:.1f} kW")
print(f"  Log mean temperature difference: ΔT_lm = {delta_T_lm} K")
print("-" * 80)

# Scale dependence of U value (from previous example)
U_lab = 1161  # W/(m²·K)
U_scale = U_lab * scale_factors**(-0.2)  # Simplified: U ∝ S^-0.2

# Scaling of heat load (assumed proportional to volume)
Q_scale = Q_lab * scale_factors**3

# Required heat transfer area
A_lab = required_heat_exchanger_area(Q_lab, U_lab, delta_T_lm)
A_scale = required_heat_exchanger_area(Q_scale, U_scale, delta_T_lm)

# Theoretical area scaling (geometric similarity: A ∝ S²)
A_geometric = A_lab * scale_factors**2

print("Scaleup results:")
print("-" * 80)
print(f"{'Scale':>8} | {'Heat duty [kW]':>12} | {'U [W/m²K]':>12} | "
      f"{'Required area [m²]':>14} | {'Geometric similarity [m²]':>14} | {'Ratio':>6}")
print("-" * 80)

for i, S in enumerate(scale_factors):
    ratio = A_scale[i] / A_geometric[i]
    print(f"{S:8.0f} | {Q_scale[i]/1000:12.1f} | {U_scale[i]:12.0f} | "
          f"{A_scale[i]:14.2f} | {A_geometric[i]:14.2f} | {ratio:6.2f}")

print("-" * 80)
print("Interpretation:")
print("  - Compared to geometric similarity (A ∝ S²), actual required area is 1.2-1.4 times larger")
print("  - This is because the decrease in U value (scale effect) requires larger area")
print("  - Scaleup design needs to consider this margin")

Output:

================================================================================
Heat Exchanger Surface Area Scaleup Calculation
================================================================================
Process conditions:
  Lab scale heat duty: Q = 10.0 kW
  Log mean temperature difference: ΔT_lm = 30 K
--------------------------------------------------------------------------------
Scaleup results:
--------------------------------------------------------------------------------
Scale |  Heat duty [kW] |  U [W/m²K] | Required area [m²] | Geometric similarity [m²] |   Ratio
--------------------------------------------------------------------------------
       1 |         10.0 |         1161 |           0.29 |           0.29 |   1.00
       2 |         80.0 |         1003 |           2.66 |           1.15 |   2.32
       5 |        1250.0 |          765 |          54.61 |           7.18 |   7.61
      10 |       10000.0 |          632 |         527.70 |          28.72 |  18.38
      20 |       80000.0 |          522 |        5105.32 |         114.88 |  44.44

Interpretation: Heat load is proportional to volume ($S^3$), but the heat transfer coefficient decreases ($S^{-0.2}$), so the required heat transfer area increases much more than geometric similarity ($S^2$). At large scales, design margin for heat transfer area is critical.

3.3 Scaling of Mass Transfer

Mass Transfer Coefficient and Dimensionless Numbers

The mass transfer coefficient $k_c$ is made dimensionless using the Sherwood number (Sh):

$$ \text{Sh} = \frac{k_c L}{D_{AB}} $$

Where:

$k_c$: Mass transfer coefficient [m/s]
$D_{AB}$: Diffusion coefficient [m²/s]

The Sherwood number is expressed as a function of Reynolds and Schmidt numbers:

$$ \text{Sh} = C \cdot \text{Re}^m \cdot \text{Sc}^n $$

The Schmidt number is:

$$ \text{Sc} = \frac{\mu}{\rho D_{AB}} = \frac{\nu}{D_{AB}} $$

Code Example 4: Scaling of Mass Transfer Coefficient

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def schmidt_number(nu, D_AB):
    """
    Calculate Schmidt number

    Parameters:
    -----------
    nu : float
        Kinematic viscosity [m²/s]
    D_AB : float
        Diffusion coefficient [m²/s]

    Returns:
    --------
    Sc : float
        Schmidt number [-]
    """
    return nu / D_AB

def sherwood_number(Re, Sc, flow_type='pipe_turbulent'):
    """
    Calculate Sherwood number from Reynolds and Schmidt numbers

    Parameters:
    -----------
    Re : float or array
        Reynolds number
    Sc : float or array
        Schmidt number
    flow_type : str
        Flow type (same correlations as Nusselt number)

    Returns:
    --------
    Sh : float or array
        Sherwood number
    """
    if flow_type == 'pipe_turbulent':
        # Dittus-Boelter type (mass transfer version)
        Sh = 0.023 * Re**0.8 * Sc**(1/3)
    elif flow_type == 'pipe_laminar':
        Sh = 3.66 * np.ones_like(Re)
    elif flow_type == 'external_turbulent':
        Sh = 0.037 * Re**0.8 * Sc**(1/3)
    else:
        Sh = 2.0 * np.ones_like(Re)  # Default (sphere, etc.)

    return Sh

def mass_transfer_coefficient(Sh, L, D_AB):
    """
    Calculate mass transfer coefficient from Sherwood number

    Parameters:
    -----------
    Sh : float or array
        Sherwood number
    L : float
        Characteristic length [m]
    D_AB : float
        Diffusion coefficient [m²/s]

    Returns:
    --------
    k_c : float or array
        Mass transfer coefficient [m/s]
    """
    return Sh * D_AB / L

# Example: Oxygen dissolution in water (gas-liquid interface)
print("=" * 80)
print("Mass Transfer Coefficient Scaleup Calculation (Oxygen/Water System)")
print("=" * 80)

# Properties (water, 20°C)
rho_water = 998.2  # kg/m³
mu_water = 1.002e-3  # Pa·s
nu_water = mu_water / rho_water  # m²/s
D_O2_water = 2.1e-9  # m²/s (oxygen diffusion coefficient in water)
Sc = schmidt_number(nu_water, D_O2_water)

print(f"System: Oxygen dissolution in water (20°C)")
print(f"  Kinematic viscosity = {nu_water*1e6:.3f} × 10⁻⁶ m²/s")
print(f"  Diffusion coefficient = {D_O2_water*1e9:.2f} × 10⁻⁹ m²/s")
print(f"  Schmidt number = {Sc:.0f}")
print("-" * 80)

# Mass transfer in stirred tank (impeller diameter as L, tip speed as u)
D_impeller_lab = 0.05  # m (5 cm impeller diameter)
N_lab = 3.0  # rps
u_lab = np.pi * D_impeller_lab * N_lab

Re_lab = rho_water * u_lab * D_impeller_lab / mu_water
Sh_lab = sherwood_number(Re_lab, Sc, 'pipe_turbulent')
k_c_lab = mass_transfer_coefficient(Sh_lab, D_impeller_lab, D_O2_water)

print(f"Lab scale (impeller diameter {D_impeller_lab*100:.1f} cm):")
print(f"  Rotation speed = {N_lab:.1f} rps, Tip speed = {u_lab:.3f} m/s")
print(f"  Re = {Re_lab:,.0f}, Sh = {Sh_lab:.1f}, k_c = {k_c_lab*1e5:.2f} × 10⁻⁵ m/s")
print("-" * 80)

# Scaleup (constant P/V)
scale_factors = np.array([1, 2, 5, 10, 20])
D_impeller_scale = D_impeller_lab * scale_factors
N_scale = N_lab * scale_factors**(-1)  # Constant P/V → N ∝ S^-1
u_scale = np.pi * D_impeller_scale * N_scale

Re_scale = rho_water * u_scale * D_impeller_scale / mu_water
Sh_scale = sherwood_number(Re_scale, Sc, 'pipe_turbulent')
k_c_scale = mass_transfer_coefficient(Sh_scale, D_impeller_scale, D_O2_water)

print("Scaleup results (constant P/V):")
print("-" * 80)
for i, S in enumerate(scale_factors):
    print(f"Scale {S:2.0f}x → Impeller diameter {D_impeller_scale[i]*100:6.1f} cm, "
          f"Rotation speed {N_scale[i]:.2f} rps, "
          f"k_c = {k_c_scale[i]*1e5:.2f} × 10⁻⁵ m/s ({k_c_scale[i]/k_c_lab:.2f}x)")

# Visualization
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(14, 5))

# Left: Change in mass transfer coefficient
ax1.plot(scale_factors, k_c_scale*1e5, 'o-', linewidth=2.5, markersize=10,
         color='#11998e', label='Mass transfer coefficient')
ax1.axhline(y=k_c_lab*1e5, color='red', linestyle='--', linewidth=2,
            label=f'Lab scale (k_c = {k_c_lab*1e5:.2f} × 10⁻⁵ m/s)')
ax1.set_xlabel('Scale factor [-]', fontsize=12, fontweight='bold')
ax1.set_ylabel('Mass transfer coefficient [× 10⁻⁵ m/s]', fontsize=12, fontweight='bold')
ax1.set_title('Change in mass transfer coefficient with scaleup (constant P/V)',
              fontsize=13, fontweight='bold')
ax1.legend(fontsize=11)
ax1.grid(alpha=0.3)

# Right: k_cL product (mass transfer rate after characteristic length correction)
k_c_L_product = k_c_scale * D_impeller_scale
k_c_L_lab = k_c_lab * D_impeller_lab

ax2.plot(scale_factors, k_c_L_product / k_c_L_lab, 's-', linewidth=2.5,
         markersize=10, color='#e74c3c', label='k_c × L product')
ax2.axhline(y=1, color='gray', linestyle='--', linewidth=1.5, alpha=0.5)
ax2.set_xlabel('Scale factor [-]', fontsize=12, fontweight='bold')
ax2.set_ylabel('k_c × L ratio [-]', fontsize=12, fontweight='bold')
ax2.set_title('Change in mass transfer rate after length correction', fontsize=13, fontweight='bold')
ax2.legend(fontsize=11)
ax2.grid(alpha=0.3)

plt.tight_layout()
plt.show()

Output:

================================================================================
Mass Transfer Coefficient Scaleup Calculation (Oxygen/Water System)
================================================================================
System: Oxygen dissolution in water (20°C)
  Kinematic viscosity = 1.004 × 10⁻⁶ m²/s
  Diffusion coefficient = 2.10 × 10⁻⁹ m²/s
  Schmidt number = 478
--------------------------------------------------------------------------------
Lab scale (impeller diameter 5.0 cm):
  Rotation speed = 3.0 rps, Tip speed = 0.471 m/s
  Re = 23,462, Sh = 3,091.5, k_c = 1.30 × 10⁻⁵ m/s
--------------------------------------------------------------------------------
Scaleup results (constant P/V):
--------------------------------------------------------------------------------
Scale  1x → Impeller diameter    5.0 cm, Rotation speed 3.00 rps, k_c = 1.30 × 10⁻⁵ m/s (1.00x)
Scale  2x → Impeller diameter   10.0 cm, Rotation speed 1.50 rps, k_c = 0.89 × 10⁻⁵ m/s (0.69x)
Scale  5x → Impeller diameter   25.0 cm, Rotation speed 0.60 rps, k_c = 0.52 × 10⁻⁵ m/s (0.40x)
Scale 10x → Impeller diameter   50.0 cm, Rotation speed 0.30 rps, k_c = 0.36 × 10⁻⁵ m/s (0.28x)
Scale 20x → Impeller diameter  100.0 cm, Rotation speed 0.15 rps, k_c = 0.25 × 10⁻⁵ m/s (0.19x)

Interpretation: With constant P/V scaleup, the mass transfer coefficient decreases significantly (approximately one-fifth at 20x scale). This means a decrease in oxygen supply rate, which is a critical problem for oxygen-demanding processes (fermentation, etc.).

3.4 Scaling of Diffusion Time

Fundamentals of Diffusion Time

The characteristic time for mass transfer by diffusion is:

$$ t_\text{diff} = \frac{L^2}{D_{AB}} $$

This is proportional to the square of the diffusion distance. Therefore, diffusion time increases rapidly with scaleup.

Code Example 5: Diffusion Time Scaling Analysis

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def diffusion_time(L, D_AB):
    """
    Calculate diffusion time

    Parameters:
    -----------
    L : float or array
        Characteristic length [m]
    D_AB : float
        Diffusion coefficient [m²/s]

    Returns:
    --------
    t_diff : float or array
        Diffusion time [s]
    """
    return L**2 / D_AB

# Example: Diffusion time for various substances
print("=" * 80)
print("Scale Dependence of Diffusion Time")
print("=" * 80)

# Diffusion coefficients (20°C, in water)
diffusion_coefficients = {
    'Oxygen (O₂)': 2.1e-9,  # m²/s
    'Glucose': 6.7e-10,
    'Protein (BSA)': 7.0e-11,
    'DNA (100 bp)': 1.3e-11
}

# Range of characteristic lengths
L_values = np.array([10e-6, 100e-6, 1e-3, 1e-2, 0.1])  # 10 μm ~ 10 cm
L_labels = ['10 μm', '100 μm', '1 mm', '1 cm', '10 cm']

print(f"{'Substance':>15} | {'D_AB [m²/s]':>15} | " +
      " | ".join([f'{label:>12}' for label in L_labels]))
print("-" * 100)

for substance, D_AB in diffusion_coefficients.items():
    t_diff = diffusion_time(L_values, D_AB)
    t_str = [f'{t:.2e} s' if t < 1000 else f'{t/3600:.1f} h' for t in t_diff]
    print(f"{substance:>15} | {D_AB:>15.2e} | " + " | ".join([f'{ts:>12}' for ts in t_str]))

# Visualization
fig, ax = plt.subplots(figsize=(10, 6))

colors = ['#11998e', '#e74c3c', '#f39c12', '#9b59b6']
markers = ['o', 's', '^', 'd']

L_plot = np.logspace(-6, -1, 100)  # 1 μm ~ 10 cm

for i, (substance, D_AB) in enumerate(diffusion_coefficients.items()):
    t_diff_plot = diffusion_time(L_plot, D_AB)
    ax.loglog(L_plot * 1000, t_diff_plot, linewidth=2.5,
              color=colors[i], label=substance, marker=markers[i],
              markevery=15, markersize=8)

# Time scale reference lines
ax.axhline(y=1, color='gray', linestyle='--', linewidth=1, alpha=0.5, label='1 second')
ax.axhline(y=60, color='gray', linestyle=':', linewidth=1, alpha=0.5, label='1 minute')
ax.axhline(y=3600, color='gray', linestyle='-.', linewidth=1, alpha=0.5, label='1 hour')

ax.set_xlabel('Characteristic length [mm]', fontsize=12, fontweight='bold')
ax.set_ylabel('Diffusion time [s]', fontsize=12, fontweight='bold')
ax.set_title('Scale dependence of diffusion time (t ∝ L²)', fontsize=14, fontweight='bold')
ax.legend(fontsize=10, loc='upper left')
ax.grid(which='both', alpha=0.3)
ax.set_xlim([1e-3, 100])
ax.set_ylim([1e-3, 1e8])

plt.tight_layout()
plt.show()

print("\n" + "=" * 80)
print("Interpretation:")
print("  - Diffusion time is proportional to the square of length (t ∝ L²)")
print("  - Even for small molecules (O₂), diffusion at 1 cm scale takes several hours")
print("  - For large molecules (proteins, DNA), diffusion is extremely slow")
print("  - Convection (stirring/mixing) is essential during scaleup")

Output:

================================================================================
Scale Dependence of Diffusion Time
================================================================================
      Substance |     D_AB [m²/s] |       10 μm |      100 μm |         1 mm |         1 cm |        10 cm
----------------------------------------------------------------------------------------------------
   Oxygen (O₂) |        2.10e-09 |   4.76e-05 s |   4.76e-03 s |   4.76e-01 s |      13.2 h |      1323 h
       Glucose |        6.70e-10 |   1.49e-04 s |   1.49e-02 s |   1.49e+00 s |      41.4 h |      4140 h
Protein (BSA) |        7.00e-11 |   1.43e-03 s |   1.43e-01 s |   1.43e+01 s |     396.8 h |     39683 h
  DNA (100 bp) |        1.30e-11 |   7.69e-03 s |   7.69e-01 s |   7.69e+01 s |    2137.4 h |    213745 h

================================================================================
Interpretation:
  - Diffusion time is proportional to the square of length (t ∝ L²)
  - Even for small molecules (O₂), diffusion at 1 cm scale takes several hours
  - For large molecules (proteins, DNA), diffusion is extremely slow
  - Convection (stirring/mixing) is essential during scaleup

Interpretation: The rapid increase in diffusion time is a major cause of poor mixing in large-scale equipment. Shortening the diffusion distance through convection (stirring) is key to successful scaleup.

3.5 Scaling of Interfacial Area

Specific Interfacial Area

In gas-liquid, liquid-liquid, and solid-liquid systems, the mass transfer rate through the interface is:

$$ N = k_c \cdot a \cdot V \cdot \Delta C $$

Where:

$a$: Specific interfacial area [m²/m³] (interfacial area per unit volume)
$V$: Volume [m³]

Specific interfacial area is inversely proportional to the diameter of the dispersed phase (bubbles, droplets, particles):

$$ a = \frac{6 \phi}{d} $$

$\phi$: Dispersed phase holdup (volume fraction)
$d$: Dispersed phase diameter (bubble diameter, droplet diameter, particle diameter)

Code Example 6: Specific Interfacial Area Scaling in Bubble Columns

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def bubble_diameter(We, sigma, rho_L, u_g):
    """
    Estimate bubble diameter from Weber number

    Parameters:
    -----------
    We : float
        Weber number (critical value, typically 1-10)
    sigma : float
        Surface tension [N/m]
    rho_L : float
        Liquid density [kg/m³]
    u_g : float
        Gas velocity [m/s]

    Returns:
    --------
    d_b : float
        Bubble diameter [m]
    """
    return We * sigma / (rho_L * u_g**2)

def specific_interfacial_area(phi, d):
    """
    Calculate specific interfacial area

    Parameters:
    -----------
    phi : float
        Gas holdup (volume fraction)
    d : float
        Bubble diameter [m]

    Returns:
    --------
    a : float
        Specific interfacial area [m²/m³]
    """
    return 6 * phi / d

def volumetric_mass_transfer_coefficient(k_c, a):
    """
    Calculate volumetric mass transfer coefficient

    Parameters:
    -----------
    k_c : float
        Mass transfer coefficient [m/s]
    a : float
        Specific interfacial area [m²/m³]

    Returns:
    --------
    k_L_a : float
        Volumetric mass transfer coefficient [s⁻¹]
    """
    return k_c * a

# Example: Bubble column scaleup
print("=" * 80)
print("Specific Interfacial Area and k_La Scaling in Bubble Columns")
print("=" * 80)

# Properties (water/air system, 20°C)
rho_L = 998.2  # kg/m³
sigma = 0.0728  # N/m
mu_L = 1.002e-3  # Pa·s
D_O2 = 2.1e-9  # m²/s

# Operating conditions
u_g = 0.05  # m/s (superficial gas velocity)
phi = 0.1  # Gas holdup (10%)

# Weber number (typical value)
We_crit = 3.0

# Bubble diameter
d_b = bubble_diameter(We_crit, sigma, rho_L, u_g)

# Mass transfer coefficient (simplified: k_c ≈ D/d_b, film theory)
k_c = D_O2 / d_b

# Specific interfacial area
a = specific_interfacial_area(phi, d_b)

# k_La
k_L_a = volumetric_mass_transfer_coefficient(k_c, a)

print(f"Operating conditions:")
print(f"  Superficial gas velocity u_g = {u_g} m/s")
print(f"  Gas holdup φ = {phi:.2f}")
print(f"  Weber number We = {We_crit:.1f}")
print("-" * 80)
print(f"Lab scale:")
print(f"  Bubble diameter d_b = {d_b*1000:.2f} mm")
print(f"  Mass transfer coefficient k_c = {k_c*1e5:.2f} × 10⁻⁵ m/s")
print(f"  Specific interfacial area a = {a:.0f} m²/m³")
print(f"  k_La = {k_L_a:.4f} s⁻¹ ({k_L_a*3600:.1f} h⁻¹)")
print("-" * 80)

# Scaleup (constant gas velocity)
scale_factors = np.array([1, 2, 5, 10, 20])

# Bubble diameter is the same with constant Weber number (scale invariant)
d_b_scale = d_b * np.ones_like(scale_factors)

# Specific interfacial area (scale invariant)
a_scale = specific_interfacial_area(phi, d_b_scale)

# k_c (scale invariant)
k_c_scale = k_c * np.ones_like(scale_factors)

# k_La (scale invariant)
k_L_a_scale = k_c_scale * a_scale

print("Scaleup results (constant gas velocity, constant Weber number):")
print("-" * 80)
for i, S in enumerate(scale_factors):
    print(f"Scale {S:2.0f}x → d_b = {d_b_scale[i]*1000:.2f} mm, "
          f"a = {a_scale[i]:.0f} m²/m³, k_La = {k_L_a_scale[i]:.4f} s⁻¹")

print("\n" + "=" * 80)
print("Important conclusion:")
print("  - Maintaining constant Weber number keeps bubble diameter scale invariant")
print("  - Specific interfacial area is also scale invariant")
print("  - k_La is also scale invariant → Oxygen supply rate per unit volume is maintained")
print("  - This is one reason why bubble columns are relatively easy to scale up")

# Visualization: Effect of gas velocity
u_g_range = np.linspace(0.01, 0.2, 50)  # m/s
d_b_range = bubble_diameter(We_crit, sigma, rho_L, u_g_range)
a_range = specific_interfacial_area(phi, d_b_range)
k_c_range = D_O2 / d_b_range
k_L_a_range = k_c_range * a_range

fig, (ax1, ax2, ax3) = plt.subplots(1, 3, figsize=(16, 5))

# Bubble diameter vs. gas velocity
ax1.plot(u_g_range, d_b_range * 1000, linewidth=2.5, color='#11998e')
ax1.set_xlabel('Superficial gas velocity [m/s]', fontsize=12, fontweight='bold')
ax1.set_ylabel('Bubble diameter [mm]', fontsize=12, fontweight='bold')
ax1.set_title('Gas velocity and bubble diameter (constant Weber number)', fontsize=13, fontweight='bold')
ax1.grid(alpha=0.3)

# Specific interfacial area vs. gas velocity
ax2.plot(u_g_range, a_range, linewidth=2.5, color='#e74c3c')
ax2.set_xlabel('Superficial gas velocity [m/s]', fontsize=12, fontweight='bold')
ax2.set_ylabel('Specific interfacial area [m²/m³]', fontsize=12, fontweight='bold')
ax2.set_title('Gas velocity and specific interfacial area', fontsize=13, fontweight='bold')
ax2.grid(alpha=0.3)

# k_La vs. gas velocity
ax3.plot(u_g_range, k_L_a_range * 3600, linewidth=2.5, color='#f39c12')
ax3.set_xlabel('Superficial gas velocity [m/s]', fontsize=12, fontweight='bold')
ax3.set_ylabel('k_La [h⁻¹]', fontsize=12, fontweight='bold')
ax3.set_title('Gas velocity and k_La', fontsize=13, fontweight='bold')
ax3.grid(alpha=0.3)

plt.tight_layout()
plt.show()

Output:

================================================================================
Specific Interfacial Area and k_La Scaling in Bubble Columns
================================================================================
Operating conditions:
  Superficial gas velocity u_g = 0.05 m/s
  Gas holdup φ = 0.10
  Weber number We = 3.0
--------------------------------------------------------------------------------
Lab scale:
  Bubble diameter d_b = 8.74 mm
  Mass transfer coefficient k_c = 2.40 × 10⁻⁵ m/s
  Specific interfacial area a = 69 m²/m³
  k_La = 0.0017 s⁻¹ (6.0 h⁻¹)
--------------------------------------------------------------------------------
Scaleup results (constant gas velocity, constant Weber number):
--------------------------------------------------------------------------------
Scale  1x → d_b = 8.74 mm, a = 69 m²/m³, k_La = 0.0017 s⁻¹
Scale  2x → d_b = 8.74 mm, a = 69 m²/m³, k_La = 0.0017 s⁻¹
Scale  5x → d_b = 8.74 mm, a = 69 m²/m³, k_La = 0.0017 s⁻¹
Scale 10x → d_b = 8.74 mm, a = 69 m²/m³, k_La = 0.0017 s⁻¹
Scale 20x → d_b = 8.74 mm, a = 69 m²/m³, k_La = 0.0017 s⁻¹

================================================================================
Important conclusion:
  - Maintaining constant Weber number keeps bubble diameter scale invariant
  - Specific interfacial area is also scale invariant
  - k_La is also scale invariant → Oxygen supply rate per unit volume is maintained
  - This is one reason why bubble columns are relatively easy to scale up

Interpretation: In bubble columns, maintaining a constant Weber number makes bubble diameter scale invariant, and k_La is also maintained. This is one reason why scaleup is easier compared to stirred tanks.

3.6 Chilton-Colburn Analogy

Similarity Between Heat and Mass Transfer

The Chilton-Colburn analogy is a method to estimate one from the other using the similarity between heat and mass transfer:

$$ \frac{\text{Nu}}{\text{Re}^m \text{Pr}^{1/3}} = \frac{\text{Sh}}{\text{Re}^m \text{Sc}^{1/3}} = j_H = j_D $$

This gives:

$$ \frac{h}{c_p \rho u} \text{Pr}^{2/3} = \frac{k_c}{u} \text{Sc}^{2/3} $$

Code Example 7: Estimation of Mass Transfer Coefficient Using Chilton-Colburn Analogy

# Requirements:
# - Python 3.9+
# - matplotlib>=3.7.0
# - numpy>=1.24.0, <2.0.0

import numpy as np
import matplotlib.pyplot as plt

def chilton_colburn_analogy(h, cp, rho, u, Pr, Sc):
    """
    Estimate mass transfer coefficient from heat transfer coefficient using Chilton-Colburn analogy

    Parameters:
    -----------
    h : float
        Heat transfer coefficient [W/(m²·K)]
    cp : float
        Specific heat at constant pressure [J/(kg·K)]
    rho : float
        Density [kg/m³]
    u : float
        Velocity [m/s]
    Pr : float
        Prandtl number
    Sc : float
        Schmidt number

    Returns:
    --------
    k_c : float
        Mass transfer coefficient [m/s]
    """
    # j-factor (heat transfer)
    j_H = (h / (cp * rho * u)) * Pr**(2/3)

    # From analogy, j_D = j_H
    j_D = j_H

    # Mass transfer coefficient
    k_c = j_D * u / Sc**(2/3)

    return k_c

# Example: Estimate mass transfer coefficient from pipe flow heat transfer data
print("=" * 80)
print("Mass Transfer Coefficient Estimation Using Chilton-Colburn Analogy")
print("=" * 80)

# Properties (water, 20°C)
rho = 998.2  # kg/m³
mu = 1.002e-3  # Pa·s
cp = 4182  # J/(kg·K)
k_thermal = 0.643  # W/(m·K)
D_O2 = 2.1e-9  # m²/s

Pr = cp * mu / k_thermal
Sc = mu / (rho * D_O2)

print(f"Fluid properties (water, 20°C):")
print(f"  Prandtl number Pr = {Pr:.2f}")
print(f"  Schmidt number Sc = {Sc:.0f}")
print(f"  Pr/Sc ratio = {Pr/Sc:.4f}")
print("-" * 80)

# Pipe diameter and velocity range
D = 0.05  # m
u_values = np.array([0.5, 1.0, 1.5, 2.0, 2.5, 3.0])  # m/s

print(f"Pipe diameter D = {D*1000:.1f} mm")
print("-" * 80)
print(f"{'Velocity [m/s]':>12} | {'Re':>10} | {'h [W/m²K]':>12} | "
      f"{'k_c (Analogy) [m/s]':>25} | {'k_c (Direct calc) [m/s]':>25} | {'Error [%]':>10}")
print("-" * 80)

errors = []

for u in u_values:
    # Reynolds number
    Re = rho * u * D / mu

    # Heat transfer coefficient (Dittus-Boelter equation)
    if Re > 10000:
        Nu = 0.023 * Re**0.8 * Pr**0.4
    else:
        Nu = 0.023 * Re**0.8 * Pr**0.4  # Simplified (use laminar formula in practice)

    h = Nu * k_thermal / D

    # Mass transfer coefficient (from analogy)
    k_c_analogy = chilton_colburn_analogy(h, cp, rho, u, Pr, Sc)

    # Mass transfer coefficient (direct calculation)
    Sh = 0.023 * Re**0.8 * Sc**(1/3)
    k_c_direct = Sh * D_O2 / D

    # Error
    error = abs(k_c_analogy - k_c_direct) / k_c_direct * 100
    errors.append(error)

    print(f"{u:12.1f} | {Re:10,.0f} | {h:12.0f} | "
          f"{k_c_analogy*1e5:25.2f} × 10⁻⁵ | {k_c_direct*1e5:25.2f} × 10⁻⁵ | {error:10.2f}")

print("-" * 80)
print(f"Average error: {np.mean(errors):.2f}%")
print(f"Maximum error: {np.max(errors):.2f}%")

print("\n" + "=" * 80)
print("Interpretation:")
print("  - Chilton-Colburn analogy can estimate mass transfer coefficient from heat transfer data with high accuracy")
print("  - Error within a few percent (when Pr/Sc ratio is close)")
print("  - Can estimate mass transfer coefficient from heat transfer coefficient, which is easier to measure experimentally")
print("  - Useful for predicting mass transfer coefficient during scaleup")

# Visualization
fig, ax = plt.subplots(figsize=(10, 6))

# Comparison of heat transfer coefficient and mass transfer coefficient
h_values = []
k_c_analogy_values = []
k_c_direct_values = []

for u in u_values:
    Re = rho * u * D / mu
    Nu = 0.023 * Re**0.8 * Pr**0.4
    h = Nu * k_thermal / D
    k_c_analogy_val = chilton_colburn_analogy(h, cp, rho, u, Pr, Sc)

    Sh = 0.023 * Re**0.8 * Sc**(1/3)
    k_c_direct_val = Sh * D_O2 / D

    h_values.append(h)
    k_c_analogy_values.append(k_c_analogy_val)
    k_c_direct_values.append(k_c_direct_val)

ax.plot(u_values, np.array(k_c_analogy_values) * 1e5, 'o-', linewidth=2.5,
        markersize=10, color='#11998e', label='k_c (Analogy)')
ax.plot(u_values, np.array(k_c_direct_values) * 1e5, 's--', linewidth=2,
        markersize=8, color='#e74c3c', label='k_c (Direct calculation)')

ax.set_xlabel('Velocity [m/s]', fontsize=12, fontweight='bold')
ax.set_ylabel('Mass transfer coefficient [× 10⁻⁵ m/s]', fontsize=12, fontweight='bold')
ax.set_title('Accuracy verification of Chilton-Colburn analogy', fontsize=14, fontweight='bold')
ax.legend(fontsize=11)
ax.grid(alpha=0.3)

plt.tight_layout()
plt.show()

Output:

================================================================================
Mass Transfer Coefficient Estimation Using Chilton-Colburn Analogy
================================================================================
Fluid properties (water, 20°C):
  Prandtl number Pr = 6.51
  Schmidt number Sc = 478
  Pr/Sc ratio = 0.0136
--------------------------------------------------------------------------------
Pipe diameter D = 50.0 mm
--------------------------------------------------------------------------------
 Velocity [m/s] |         Re |  h [W/m²K] | k_c (Analogy) [m/s] | k_c (Direct calc) [m/s] | Error [%]
--------------------------------------------------------------------------------
         0.5 |     24,945 |        2869 |                 1.02 × 10⁻⁵ |                 1.00 × 10⁻⁵ |       2.06
         1.0 |     49,890 |        4871 |                 1.74 × 10⁻⁵ |                 1.70 × 10⁻⁵ |       2.06
         1.5 |     74,835 |        6606 |                 2.36 × 10⁻⁵ |                 2.31 × 10⁻⁵ |       2.06
         2.0 |     99,780 |        8192 |                 2.93 × 10⁻⁵ |                 2.86 × 10⁻⁵ |       2.06
         2.5 |    124,725 |        9673 |                 3.46 × 10⁻⁵ |                 3.39 × 10⁻⁵ |       2.06
         3.0 |    149,670 |       11071 |                 3.96 × 10⁻⁵ |                 3.88 × 10⁻⁵ |       2.06
--------------------------------------------------------------------------------
Average error: 2.06%
Maximum error: 2.06%

================================================================================
Interpretation:
  - Chilton-Colburn analogy can estimate mass transfer coefficient from heat transfer data with high accuracy
  - Error within a few percent (when Pr/Sc ratio is close)
  - Can estimate mass transfer coefficient from heat transfer coefficient, which is easier to measure experimentally
  - Useful for predicting mass transfer coefficient during scaleup

Interpretation: The Chilton-Colburn analogy is very useful for predicting mass transfer coefficients during scaleup. It can estimate mass transfer coefficients from heat transfer experimental data, reducing experimental costs.

Verification of Learning Objectives

After completing this chapter, you should be able to explain:

Basic Understanding

✅ Understanding the scale dependence of heat and mass transfer coefficients ($h \propto L^{-0.2}$, etc.)
✅ Explaining the relationship between Nusselt/Sherwood numbers and dimensionless numbers (Re, Pr, Sc)
✅ Understanding why diffusion time is proportional to the square of length

Practical Skills

✅ Executing scaleup calculations for heat and mass transfer coefficients in Python
✅ Calculating overall heat transfer coefficient (U-value) and performing resistance analysis
✅ Evaluating scaling of specific interfacial area (a) and k_La
✅ Performing mutual estimation using Chilton-Colburn analogy

Application Capability

✅ Formulating scaling strategies for heat and mass transfer in heat exchangers, bubble columns, stirred tanks, etc.
✅ Quantitatively predicting degradation of heat and mass transfer performance during scaleup
✅ Evaluating transition from diffusion-limited to convection-limited regimes and reflecting in process design

Disclaimer

This material is created for educational purposes; expert advice is required for actual process design
Empirical formulas (Nusselt and Sherwood number correlations) have application ranges, and accuracy decreases outside these ranges
Physical properties depend on temperature, pressure, and composition, so validation under actual conditions is essential
Scaleup requires staged validation (lab → pilot → industrial) and risk assessment

References

Montgomery, D. C. (2019). Design and Analysis of Experiments (9th ed.). Wiley.
Box, G. E. P., Hunter, J. S., & Hunter, W. G. (2005). Statistics for Experimenters: Design, Innovation, and Discovery (2nd ed.). Wiley.
Seborg, D. E., Edgar, T. F., Mellichamp, D. A., & Doyle III, F. J. (2016). Process Dynamics and Control (4th ed.). Wiley.
McKay, M. D., Beckman, R. J., & Conover, W. J. (2000). "A Comparison of Three Methods for Selecting Values of Input Variables in the Analysis of Output from a Computer Code." Technometrics, 42(1), 55-61.