B1 — Steady-state temperature of a catalyst pellet: 23-Chem-B1 Transport Phenomena, May 2017, Q3

Question 3: B1 — Steady-state temperature of a catalyst pellet (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Pellet diameter	$D_p$	$15\ \text{mm}=0.015\ \text{m}$
Superficial mass velocity	$G$	$3500\ \text{kg/m}^2\text{s}$
Gas temperature	$T_g$	$800\ \text{°C}$
Methane reaction rate	$r''$	$0.05\ \text{mol/m}^2\text{s}$
Heat of reaction (endothermic)	$\Delta H_{rxn}$	$40\ \text{kcal/mol}$
Heat capacity	$c_p$	$3.2\ \text{cal/g}\!\cdot\!\text{K}=13{,}389\ \text{J/kg}\!\cdot\!\text{K}$
Viscosity	$\mu$	$0.03\ \text{cP}=3.0\times10^{-5}\ \text{Pa}\!\cdot\!\text{s}$
Thermal conductivity	$k$	$0.33\ \text{W/m}\!\cdot\!\text{K}$

Find. The steady-state temperature $T_s$ of the catalyst pellet surface.

Fig. B1: The endothermic reaction draws heat from the pellet surface; at steady state the convective heat delivered by the hot gas exactly supplies it, fixing $T_s$ slightly below the gas temperature.

Approach. Write a steady-state energy balance on the pellet surface (convective heat in = heat absorbed by reaction), obtain the convective coefficient from a single-sphere Nusselt correlation, and solve for the small temperature depression.

Heat absorbed by the reaction (per unit external area). Each mole of CH&sub4; reacted absorbs $\Delta H_{rxn}$, so $$q=r''\,\Delta H_{rxn}=(0.05)(40)=2.0\ \text{kcal/m}^2\text{s}=8.37\times10^{3}\ \text{W/m}^2.$$
Reynolds and Prandtl numbers. $$Re=\frac{D_pG}{\mu}=\frac{(0.015)(3500)}{3.0\times10^{-5}}=1.75\times10^{6},\qquad Pr=\frac{c_p\mu}{k}=\frac{(13{,}389)(3.0\times10^{-5})}{0.33}=1.22.$$
Convective coefficient (Ranz–Marshall single sphere). $Nu=2+0.6\,Re^{1/2}Pr^{1/3}=2+0.6(1323)(1.068)=849$, hence $$h=\frac{Nu\,k}{D_p}=\frac{(849)(0.33)}{0.015}\ \Longrightarrow\ \boxed{h\approx1.87\times10^{4}\ \text{W/m}^2\text{K}.}$$
Surface energy balance. The gas supplies the reaction heat by convection: $h(T_g-T_s)=q$, so the pellet sits below the gas temperature by $$\Delta T=\frac{q}{h}=\frac{8.37\times10^{3}}{1.87\times10^{4}}=0.45\ \text{K}\ \Longrightarrow\ \boxed{T_s=800-0.45\approx799.6\ \text{°C}.}$$

Quantity	Result
Heat flux absorbed by reaction	$q=8.37\ \text{kW/m}^2$
Reynolds / Prandtl	$Re=1.75\times10^{6}$, $Pr=1.22$
Convective coefficient	$h\approx1.87\times10^{4}\ \text{W/m}^2\text{K}$
Surface temperature	$T_s\approx799.6\ \text{°C}$ (0.45 K below the gas)

Check — correlation range and the robustness of $T_s$

At $G=3500\ \text{kg/m}^2\text{s}$ the pellet Reynolds number ($1.75\times10^{6}$) exceeds the nominal validity of the Ranz–Marshall sphere correlation; a packed-bed $j_H$ correlation (Gupta–Thodos) gives an even larger $h$ ($j_H=2.06/(\varepsilon Re^{0.575})$ gives $h\approx5$–$6\times10^{4}\ \text{W/m}^2\text{K}$ for bed voidage $\varepsilon=0.35$–$0.45$), hence a smaller $\Delta T$. Either way the intense convection at this mass velocity pins the surface within a fraction of a degree of the gas: $T_s\approx800\ \text{°C}$, marginally cooler because the reaction is endothermic.

Concept & Theory

Topic. Interphase heat transfer to a reacting catalyst pellet; surface energy balance with a reaction heat sink.

Key relations. $q=r''\,\Delta H_{rxn}$; Ranz–Marshall $Nu=2+0.6\,Re^{1/2}Pr^{1/3}$; $h=Nu\,k/D_p$; surface balance $h(T_g-T_s)=q$.

Why this works. At steady state the pellet stores no energy, so whatever heat the endothermic reaction consumes must arrive by convection from the surrounding gas. The reaction rate (not the kinetics) is given, so the heat demand is fixed; only the film coefficient is needed, and a sphere Nusselt correlation supplies it. Because $h$ is enormous at this mass velocity, the required driving force is tiny.

Common pitfall. Getting the sign wrong — an endothermic reaction makes the surface cooler than the gas, not hotter; and forgetting to convert $c_p$ (cal/g·K) and $\Delta H$ (kcal/mol) to SI before combining with $k$ in W/m·K.

Source. Welty et al., Ch. 19–20 (convection to spheres/packed beds); Bird–Stewart–Lightfoot §14 (interphase transport, Ranz–Marshall).

Question 3 of 6: B1 — Steady-state temperature of a catalyst pellet

Question 3: B1 — Steady-state temperature of a catalyst pellet (25 marks)