A high value of the Thiele modulus during a biochar particle conversion process corresponds to a Regime I conversion. Right or wrong?

Short Answer

Expert verified
In conclusion, the given statement is correct. A high value of the Thiele modulus during a biochar particle conversion process corresponds to a Regime I conversion, where the reaction rate is limited by diffusion, and the reaction takes place near the surface of the catalyst particle.

Step by step solution

01

Understanding Thiele Modulus

Thiele modulus (φ) is a dimensionless number that helps us evaluate the relative rates of diffusion and reaction in porous catalyst particles. In a nutshell, it is used to describe how fast the diffusion of reactants/products is occurring compared to the reaction rate in the catalyst particle. A high Thiele modulus corresponds to slow diffusion rates, whereas a low Thiele modulus corresponds to fast diffusion rates.
02

Understanding Different Regimes

There are three different diffusion regimes during a biochar particle conversion process: 1. Regime I: When the Thiele modulus is high, diffusion is slow, and the reaction rate is limited by diffusion. This means that the reaction is happening faster than the reactants can diffuse into the catalyst particle or the products can diffuse out of the particle. As a result, the reaction takes place near the surface of the catalyst particle. 2. Regime II: When the Thiele modulus is around 1, the reaction rate and the diffusion rate are comparable, and neither controls the overall process. 3. Regime III: When the Thiele modulus is low, diffusion is fast, and the reaction rate is limited by the reaction itself, rather than diffusion of the reactants/products. Since the diffusion rate is much higher, the reaction takes place uniformly throughout the catalyst particle.
03

Checking the Given Statement

The given statement is: "A high value of the Thiele modulus during a biochar particle conversion process corresponds to a Regime I conversion." As we can see from our understanding of the Thiele modulus and the different diffusion regimes, this statement is correct. Regime I occurs when the Thiele modulus is high, which means diffusion is slow and the reaction rate is limited by diffusion.

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Most popular questions from this chapter

Consider a perfectly stirred reactor to be used for dry wood gasification. The composition of wood is given as overall molecular formula: \(\mathrm{CH}_{1.4} \mathrm{O}_{0.6}\). The reactor volume \((\mathrm{V})\) is \(4 \mathrm{~m}^{3}\), and the mass flow rate of wood is \(2 \mathrm{~kg} \cdot \mathrm{h}^{-1}\). Primary air consists of \(23 \mathrm{wt} \% \mathrm{O}_{2}\) and is fed to the gasifier with a mass flow rate \(\phi_{\mathrm{m}, \text { air. }}\) Oxygen in the air reacts with biomass in an idealized way so as to form only \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). The reaction is given as \(\mathrm{CH}_{1.4} \mathrm{O}_{0.6}+\left|\mathrm{v}_{1}\right| \mathrm{O}_{2} \rightarrow \mathrm{v}_{2} \mathrm{CO}+\mathrm{v}_{3} \mathrm{H}_{2}\) with \(v_{i}\) being the stoichiometric coefficients. The rate of consumption of \(\mathrm{O}_{2}\), \(R_{\mathrm{O} 2,1}\), in \(\left[\mathrm{kmol} \cdot \mathrm{m}^{-3} \cdot \mathrm{s}^{-1}\right]\) is given by $$ R_{\mathrm{O} 2,1}=\mathrm{k}_{1} \mathrm{Y}_{\mathrm{O} 2} \exp \left(-\mathrm{T}_{\mathrm{a} 1} / \mathrm{T}\right) $$ with \(\mathrm{Y}_{\mathrm{O} 2}\) being the \(\mathrm{O}_{2}\) mass fraction, \(\mathrm{k}_{1}=10^{7} \mathrm{kmol} \cdot \mathrm{m}^{-3} \cdot \mathrm{s}^{-1}\), and \(\mathrm{T}_{\mathrm{a} 1}=2.5 \times\) \(10^{4} \mathrm{~K}\). The reactor is operated at steady state and at isothermal conditions with \(\mathrm{T}=1000 \mathrm{~K}\). a. Calculate \(\lambda\). b. Suppose that just enough air is fed into the reactor for complete wood conversion into \(\mathrm{CO}\) and \(\mathrm{H}_{2}\). Compute \(\phi_{\mathrm{m}, \text { air }}\) c. Write down the conservation equations for total mass and \(\mathrm{O}_{2}\) (mass fraction), respectively. d. Determine \(\mathrm{Y}_{\mathrm{O} 2}\) in the reactor by solving the equations. N. B. \(\mathrm{Y}_{\mathrm{O}_{2}}>0\), though just enough air is introduced in the reactor for complete conversion.

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