15.2 Thermal Conductivity and Specific Heat Capacities

15.2 Thermal Conductivity and Specific Heat Capacities

Thermal transport properties—specifically thermal conductivity ($k$) and specific heat capacity ($C_p$)—govern energy conservation equations, heat exchanger surface area sizing, and transient heating/cooling operations in process plants. These properties appear in key dimensionless transport groups on the NCEES PE Chemical Exam, notably the Prandtl number ($\text{Pr} = C_p \mu / k$) and the thermal diffusivity ($\alpha = k / (\rho C_p)$).

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1. Thermal Conductivity Fundamentals & Fourier's Law

Fourier's Law of Heat Conduction

Thermal conduction is the transfer of internal thermal energy via microscopic molecular collisions and electron/phonon transport. In one dimension, Fourier's Law states:

$q_x = -k \frac{dT}{dx}$

Where:

• $q_x$ = conductive heat flux ($\text{W/m}^2$ or $\text{Btu}/(\text{hr}\cdot\text{ft}^2)$).
• $k$ = thermal conductivity.
• $dT/dx$ = temperature gradient ($\text{K/m}$ or $^\circ\text{F/ft}$).

Units and Conversions

• SI Unit: $\text{W}/(\text{m}\cdot\text{K}) = \text{J}/(\text{s}\cdot\text{m}\cdot\text{K})$.
• US Customary Unit: $\text{Btu}/(\text{hr}\cdot\text{ft}\cdot^\circ\text{F})$.
• Conversion Factor: $1.0\text{ Btu}/(\text{hr}\cdot\text{ft}\cdot^\circ\text{F}) = 1.7307\text{ W}/(\text{m}\cdot\text{K}) \iff 1.0\text{ W}/(\text{m}\cdot\text{K}) = 0.5778\text{ Btu}/(\text{hr}\cdot\text{ft}\cdot^\circ\text{F})$ $1.0\text{ cal}/(\text{s}\cdot\text{cm}\cdot^\circ\text{C}) = 418.4\text{ W}/(\text{m}\cdot\text{K})$

The Prandtl Number ($\text{Pr}$)

The dimensionless Prandtl number measures the relative rate of momentum diffusion to thermal diffusion in the velocity and thermal boundary layers:

$\text{Pr} \equiv \frac{\nu}{\alpha} = \frac{\mu / \rho}{k / (\rho C_p)} = \frac{C_p \mu}{k}$

Representative magnitudes:

• Liquid metals (mercury, sodium): $\text{Pr} \sim 0.004 - 0.03$ (conduction dominates over convection).
• Air and light gases: $\text{Pr} \approx 0.69 - 0.72$ (momentum and thermal boundary layers grow at identical rates).
• Liquid water ($20^\circ\text{C}$): $\text{Pr} \approx 7.0$.
• Viscous lubricating oils, glycols: $\text{Pr} \sim 50 - 10,000$ (momentum boundary layer is vastly thicker than the thermal boundary layer).

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2. Gas Thermal Conductivity Estimation

Monatomic Gases (Kinetic Theory)

For dilute monatomic gases (helium, argon), molecules possess only 3 translational degrees of freedom ($C_v = \frac{3}{2} R$). Elementary kinetic theory relates thermal conductivity directly to dynamic viscosity:

$k = \frac{5}{2} \mu c_v = \frac{5}{2} \frac{\mu}{M} C_v = \frac{15}{4} \frac{R}{M} \mu$

Where $c_v$ is specific heat capacity ($\text{J}/(\text{kg}\cdot\text{K})$), $C_v$ is molar heat capacity ($\text{J}/(\text{mol}\cdot\text{K})$), and $M$ is molecular weight ($\text{kg/mol}$).

Polyatomic Gases: The Modified Eucken Correlation

Polyatomic molecules possess rotational and vibrational energy storage modes in addition to translation. Translational energy transfers at the collision rate (factor $\approx 2.5$), while internal rotational/vibrational energy diffuses more slowly (factor $\approx 1.0$). The modified Eucken correlation partitions these modes:

$k = \frac{\mu}{M} \left( C_v + \frac{9}{4} R \right) = \frac{\mu}{M} \left( C_p + \frac{5}{4} R \right) = \frac{\mu}{M} \left( C_p + 1.25 R \right)$

Where:

• $k$ = gas thermal conductivity ($\text{W}/(\text{m}\cdot\text{K})$).
• $\mu$ = gas dynamic viscosity ($\text{Pa}\cdot\text{s} = \text{kg}/(\text{m}\cdot\text{s})$).
• $M$ = molecular weight ($\text{kg/mol}$; e.g., for methane $M = 0.01604\text{ kg/mol}$).
• $C_p, C_v$ = ideal gas molar heat capacities ($\text{J}/(\text{mol}\cdot\text{K})$).
• $R$ = universal gas constant ($8.3145\text{ J}/(\text{mol}\cdot\text{K})$).

Temperature and Pressure Dependencies in Gases

• Temperature: At low to moderate pressures, gas thermal conductivity increases with temperature ($k \propto T^n$, where $n \approx 0.65 - 1.0$) because molecular velocities increase and internal vibrational modes become thermally active.
• Pressure: For dilute gases ($P < 10\text{ bar}$), thermal conductivity is virtually independent of pressure. At high pressures ($P_r > 1$), dense gas clustering increases $k$, especially near the critical point.

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3. Liquid Thermal Conductivity & The Water Anomaly

General Liquid Behavior

In liquids, heat is conducted via acoustic waves and lattice-like molecular vibrational collisions. As temperature increases, liquid thermal expansion increases average intermolecular separation, dampening collisional energy transmission between adjacent molecular cages. Therefore, for almost all organic and inorganic liquids, thermal conductivity decreases with increasing temperature:

$k_L(T) \approx k_0 [1 - \beta (T - T_0)]$

Typical organic liquid values range between $0.10$ and $0.20\text{ W}/(\text{m}\cdot\text{K})$ at ambient conditions (e.g., benzene $\approx 0.144\text{ W}/(\text{m}\cdot\text{K})$, ethanol $\approx 0.167\text{ W}/(\text{m}\cdot\text{K})$).

The Thermal Anomaly of Liquid Water

Liquid water is the famous, critical exception to the rule! Due to its extensive three-dimensional hydrogen bonding network, liquid water at ambient conditions contains open, ice-like cage structures. As water is heated from $0^\circ\text{C}$ to $\approx 130^\circ\text{C}$ ($403\text{ K}$), thermal energy travels through the hydrogen-bond network by a fast, quasi-lattice (phonon-like) mechanism rather than by slow molecular translation alone. Rising temperature raises molecular vibration amplitude and collision frequency, which strengthens that vibrational transfer faster than the progressive rupture of hydrogen bonds weakens the network, so $k$ increases.

Do not attribute the rise to densification: liquid water becomes steadily less dense above $4^\circ\text{C}$, falling from $999.97\text{ kg/m}^3$ at $4^\circ\text{C}$ to roughly $935\text{ kg/m}^3$ at $130^\circ\text{C}$. Only above $\approx 130^\circ\text{C}$ does the combination of continued thermal expansion and extensive hydrogen-bond breakdown finally dominate, causing $k$ to decline and then collapse toward the critical point. Furthermore, water's thermal conductivity ($0.60 - 0.68\text{ W}/(\text{m}\cdot\text{K})$) is 3 to 5 times higher than almost all organic solvents:

   Thermal Conductivity k [W/(m*K)]
     ^
 0.7 |                      * * * (Peak at ~130°C: k ≈ 0.687)
     |                   *         *
 0.6 | Saturated Water *             *           
     |-----------------------------------------------------
 0.2 |                               Organic Liquids (Benzene, Toluene)
 0.1 |   \                                (k decreases monotonically)
     |    \---------------------
 0.0 +----------------------------------------------------> Temperature (°C)
     0        50       100      130      150      200

Liquid Estimation Methods

• Latini Correlation: $k_L = \frac{A (1 - T_r)^{0.38}}{T_r^{1/6}}$ Where $A$ is a characteristic dimensional constant tabulated by chemical family.
• Robbins-Kingrea Correlation: Employs normal boiling point, heat of vaporization $\Delta H_{vb}$, and liquid density $\rho_L$ to estimate $k_L$.

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4. Heat Capacity Formulations

Fundamental Thermodynamic Relations

Molar heat capacities at constant volume ($C_v$) and constant pressure ($C_p$) are defined as:

$C_v \equiv \left( \frac{\partial U}{\partial T} \right)_V, \quad C_p \equiv \left( \frac{\partial H}{\partial T} \right)_P$

For any homogeneous fluid phase, the general thermodynamic relation is:

$C_p - C_v = -T \frac{(\partial V / \partial T)_P^2}{(\partial V / \partial P)_T} = \frac{T V \beta^2}{\kappa_T}$

Where $\beta = \frac{1}{V} (\partial V / \partial T)_P$ is the volumetric thermal expansivity and $\kappa_T = -\frac{1}{V} (\partial V / \partial P)_T$ is the isothermal compressibility. For an ideal gas ($P V = R T$), this reduces to Mayer's relation: $C_p^\circ - C_v^\circ = R$.

Ideal Gas Heat Capacity Polynomials

At low pressures, ideal gas heat capacity $C_p^\circ(T)$ is a function of temperature alone, representing translational, rotational, and vibrational molecular modes:

$C_p^\circ(T) = a + b T + c T^2 + d T^3$ $\frac{C_p^\circ(T)}{R} = a_1 + a_2 T + a_3 T^2 + a_4 T^3 + a_5 T^4$

NIST Shomate Equation: Widely used in modern thermochemical software and tables:

$C_p^\circ = A + B t + C t^2 + D t^3 + \frac{E}{t^2}$

Where $t = T / 1000$ ($T$ in $\text{K}$) and $C_p^\circ$ is in $\text{J}/(\text{mol}\cdot\text{K})$.

Liquid Heat Capacity and Residual Property ($C_p^L > C_p^\circ$)

Liquid heat capacity is always significantly greater than ideal gas heat capacity at the same temperature ($C_p^L > C_p^\circ$). In liquids, close intermolecular potential wells hinder free rotation and create low-frequency intermolecular vibrational modes (librations and cage oscillations) that store additional sensible energy. The difference is the residual heat capacity:

$C_p^L = C_p^\circ + C_p^R$

Rowlinson-Bondi Correlation:

$\frac{C_p^L - C_p^\circ}{R} = 1.45 + \frac{0.45}{1 - T_r} + 0.25 \omega \left[ 17.11 + \frac{25.2 (1 - T_r)^{1/3}}{T_r} + \frac{1.74}{1 - T_r} \right]$

Where $\omega$ is Pitzer's acentric factor.

Kopp's Rule (Neumann-Kopp Law) for Condensed Phases

When experimental calorimetric data is unavailable for solid salts, minerals, or liquid compounds, Kopp's rule approximates molar heat capacity at $20-25^\circ\text{C}$ as the sum of atomic heat capacities of constituent elements:

$C_p = \sum_{i} n_i c_{p,i}$

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5. Comprehensive Worked Numerical Example

Problem Statement

An engineer is sizing a gas cooler and a catalyst bed. Solve the following three engineering evaluations:

1. Modified Eucken Gas Thermal Conductivity: A gas reactor effluent contains pure ethane ($M = 30.07\text{ kg/kmol} = 0.03007\text{ kg/mol}$) at $400.0\text{ K}$ and $1.0\text{ bar}$. At this condition, dynamic viscosity is $\mu = 1.250 \times 10^{-5}\text{ Pa}\cdot\text{s}$ and molar heat capacity is $C_p = 65.80\text{ J}/(\text{mol}\cdot\text{K})$. Estimate the thermal conductivity $k$ in $\text{W}/(\text{m}\cdot\text{K})$ and evaluate the Prandtl number $\text{Pr}$.
2. Liquid Cooler Comparison: In a shell-and-tube exchanger, compare the heat transfer fluid properties of liquid water at $40^\circ\text{C}$ ($k = 0.630\text{ W}/(\text{m}\cdot\text{K})$, $\mu = 0.653\text{ cP}$, $C_p = 4178\text{ J}/(\text{kg}\cdot\text{K})$) against liquid toluene at $40^\circ\text{C}$ ($k = 0.131\text{ W}/(\text{m}\cdot\text{K})$, $\mu = 0.500\text{ cP}$, $C_p = 1750\text{ J}/(\text{kg}\cdot\text{K})$). Calculate the Prandtl number for each fluid.
3. Kopp's Rule Solid Heat Capacity: Estimate the solid molar heat capacity ($C_p$) and specific heat capacity ($c_p$) in $\text{kJ}/(\text{kg}\cdot\text{K})$ of an alumina catalyst support, aluminum oxide ($\text{Al}_2\text{O}_3$, $M = 101.96\text{ g/mol}$), at $25.0^\circ\text{C}$ using Kopp's rule.

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Step 1: Ethane Gas Thermal Conductivity & Prandtl Number

Using the modified Eucken correlation:

$k = \frac{\mu}{M} \left( C_p + 1.25 R \right)$

Evaluate the energy transport factor:

$1.25 R = 1.25 \times 8.3145\text{ J}/(\text{mol}\cdot\text{K}) = 10.393\text{ J}/(\text{mol}\cdot\text{K})$ $C_p + 1.25 R = 65.80 + 10.393 = 76.193\text{ J}/(\text{mol}\cdot\text{K})$

Compute the viscosity-to-molar-mass ratio:

$\frac{\mu}{M} = \frac{1.250 \times 10^{-5}\text{ kg}/(\text{m}\cdot\text{s})}{0.03007\text{ kg/mol}} = 4.15697 \times 10^{-4}\text{ mol}/(\text{m}\cdot\text{s})$

Calculate thermal conductivity:

$k = (4.15697 \times 10^{-4}) \times 76.193 = \mathbf{0.03167\text{ W}/(\text{m}\cdot\text{K})}$

Compute mass specific heat capacity and Prandtl number:

$c_p = \frac{C_p}{M} = \frac{65.80\text{ J}/(\text{mol}\cdot\text{K})}{0.03007\text{ kg/mol}} = 2188.2\text{ J}/(\text{kg}\cdot\text{K})$ $\text{Pr} = \frac{c_p \mu}{k} = \frac{(2188.2\text{ J}/(\text{kg}\cdot\text{K})) \times (1.250 \times 10^{-5}\text{ Pa}\cdot\text{s})}{0.03167\text{ W}/(\text{m}\cdot\text{K})} = \frac{0.027353}{0.03167} = \mathbf{0.864}$

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Step 2: Liquid Water vs. Toluene Prandtl Numbers

Convert dynamic viscosities from centipoise to SI units ($1\text{ cP} = 10^{-3}\text{ Pa}\cdot\text{s}$):

• Water ($40^\circ\text{C}$): $\mu = 0.653 \times 10^{-3}\text{ Pa}\cdot\text{s}$ $\text{Pr}_{\text{water}} = \frac{C_p \mu}{k} = \frac{(4178\text{ J}/(\text{kg}\cdot\text{K})) \times (0.653 \times 10^{-3}\text{ Pa}\cdot\text{s})}{0.630\text{ W}/(\text{m}\cdot\text{K})} = \frac{2.7282}{0.630} = \mathbf{4.33}$
• Toluene ($40^\circ\text{C}$): $\mu = 0.500 \times 10^{-3}\text{ Pa}\cdot\text{s}$ $\text{Pr}_{\text{toluene}} = \frac{C_p \mu}{k} = \frac{(1750\text{ J}/(\text{kg}\cdot\text{K})) \times (0.500 \times 10^{-3}\text{ Pa}\cdot\text{s})}{0.131\text{ W}/(\text{m}\cdot\text{K})} = \frac{0.875}{0.131} = \mathbf{6.68}$

(Notice: Even though toluene has lower viscosity than water, its thermal conductivity is almost 5 times lower, causing its Prandtl number to be higher than that of water).

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Step 3: Kopp's Rule for Alumina ($\text{Al}_2\text{O}_3$)

Aluminum is a metal (use $c_{p,\text{Al}} = 26.0\text{ J}/(\text{mol}\cdot\text{K})$). Oxygen has $c_{p,\text{O}} = 16.7\text{ J}/(\text{mol}\cdot\text{K})$. Applying Kopp's rule for $\text{Al}_2\text{O}_3$:

$C_p = 2 \times c_{p,\text{Al}} + 3 \times c_{p,\text{O}} = (2 \times 26.0) + (3 \times 16.7) = 52.0 + 50.1 = \mathbf{102.1\text{ J}/(\text{mol}\cdot\text{K})}$

Compute mass specific heat capacity ($M = 101.96\text{ g/mol} = 0.10196\text{ kg/mol}$):

$c_p = \frac{C_p}{M} = \frac{102.1\text{ J}/(\text{mol}\cdot\text{K})}{0.10196\text{ kg/mol}} = 1001.4\text{ J}/(\text{kg}\cdot\text{K}) = \mathbf{1.001\text{ kJ}/(\text{kg}\cdot\text{K})}$

(Experimental calorimetric value for $\alpha-\text{Al}_2\text{O}_3$ at $298\text{ K}$ is $79.0\text{ J}/(\text{mol}\cdot\text{K})$; Kopp's rule provides a rapid, conservative first-order estimate within engineering screening accuracy).

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6. Critical PE Exam Traps & Pitfalls

[!WARNING] Trap 1: The Water Thermal Conductivity Anomaly
Assuming that thermal conductivity decreases with temperature for all liquids will cause you to miss points on the PE exam. Liquid water's thermal conductivity increases with temperature from $0^\circ\text{C}$ to $130^\circ\text{C}$. For all other standard liquid hydrocarbons (benzene, hexane, ethanol), $k$ decreases as temperature rises.

[!WARNING] Trap 2: Molecular Weight Units in the Modified Eucken Formula
In the modified Eucken correlation $k = (\mu / M)(C_p + 1.25 R)$, $R$ is in $\text{J}/(\text{mol}\cdot\text{K})$. Therefore, molecular weight $M$ must be entered in $\text{kg/mol}$ (e.g., $0.02805\text{ kg/mol}$ for ethylene), NOT $\text{g/mol}$. Entering $28.05$ will underestimate thermal conductivity by a factor of 1,000.

[!WARNING] Trap 3: Confusing $C_p$ and $C_v$ in Modified Eucken Formulations
Be careful which heat capacity you are using: $k = (\mu/M)(C_p + 1.25R)$ uses constant-pressure heat capacity $C_p$, whereas $k = (\mu/M)(C_v + 2.25R)$ uses constant-volume heat capacity $C_v$. Because $C_p = C_v + R$, adding $1.25R$ to $C_v$ instead of $C_p$ will introduce significant error.