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Chapter 4: Problem 26

As we move toward a hydrogen-based energy economy for use in fuel cells. The use of fuel cells to operate appliances ranging from computers to automobiles is rapidly becoming a reality. In the immediate future, fuel cells will use hydrogen to produce electricity, which some have said will lead to a hydrogen- based economy instead of a petroleum-based economy. A large component in the processing train for fuel cells is the water gas shift membrane reactor. (M. Gummala, N. Gupla, B. Olsomer, and Z. Dardas. Paper \(103 c, 2003,\) AIChE National Meeting, New Orleans. LA.) $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{CO}_{2}+\mathrm{H}_{2}$$ Here \(\mathrm{CO}\) and water are fed to the membrane reactor containing the catalyst. Hydrogen can diffuse out the sides of the membrane while \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{CO}_{2}\) cannot. Based on the following information, plot the concentrations and molar flow rates of each of the reacting species down the length of the membrane reactor. Assume the following. The volumetric feed is \(10 \mathrm{dm}^{3} / \mathrm{min}\) at 10 atm, and the equil molar feed of \(\mathrm{CO}\) and water vapor with \(C_{\mathrm{T} 0}=0.4 \mathrm{mol} / \mathrm{dm}^{3}\) The equilibrium constant is \(K_{e}=1.44 .\) The \(k\) specific reaction rate constant is \(1.37 \mathrm{dm}^{6} / \mathrm{mol} \mathrm{kg}\) cat \(\cdot \mathrm{min},\) and the mass transfer coefficient for hydrogen. \(k_{\mathrm{CH}_{2}}=0.1 \mathrm{dm}^{3} / \mathrm{kg}\) cat \(\cdot\) min. What is the reactor volume necessary to achieve \(85 \%\) conversion of \(\mathrm{CO} ?\) Compare with a PFR. For that same reactor volume. what would be the conversion if the feed rate were doubled?

Short Answer

Expert verified
The reactor volume necessary to achieve 85% conversion of CO in a membrane reactor can be found by setting up and solving a numerical integration involving molar flow rate and concentration equations. The resulting reactor volume can be compared to that of a plug flow reactor (PFR). If the feed rate is doubled, new conversions can be calculated and compared to the initial 85% conversion.

Step by step solution

01

Evaluate given data and formulate equations

Given data: Volumetric feed rate = 10 dm鲁/min Initial mole concentration (C_t0) = 0.4 mol/dm鲁 Equilibrium constant (K_e) = 1.44 Specific reaction rate constant (k) = 1.37 dm鈦�/(mol kg路cat路min) Mass transfer coefficient for H鈧� (k_CH鈧�) = 0.1 dm鲁/(kg路cat路min) The reaction equation is: CO + H鈧侽 鈬� CO鈧� + H鈧� We will be using the following equations: 1. For the rate of the reaction: \(r = kC_{CO}C_{H_2O} - \frac{C_{CO_2}C_{H_2}}{K_e}\) 2. The differential equation for the molar flow rates in the reactor: \(dF_i = \frac{dV}{W}(r_i)\), where i represents the reacting species (CO, H鈧侽, CO鈧�, H鈧�) 3. For the mass transfer of hydrogen: \(r_{H_2} = k_{CH_2}(C_{H_2} - C_{H_2,start})\)
02

Set up the numerical integration for the molar flow rates down the reactor length

We need to solve the differential equation for molar flow rates (dF_i) throughout the reactor length, so we will use numerical integration. Set up the integration using the equations from Step 1 with given data. Before we can solve for the molar flow rates, we need to convert the volumetric feed rate to molar feed rate. We know that: Molar flow rate = Volumetric flow rate 脳 Concentration
03

Convert the volumetric feed rate to molar feed rate

Initial molar flow rate of CO (F_CO0) = 10 dm鲁/min 脳 0.4 mol/dm鲁 = 4 mol/min Initial molar flow rate of H鈧侽 (F_H2O0) = 10 dm鲁/min 脳 0.4 mol/dm鲁 = 4 mol/min Initial molar flow rate of CO鈧� (F_CO20) = 0 mol/min (no CO鈧� at the start) Initial molar flow rate of H鈧� (F_H20) = 0 mol/min (no H鈧� at the start)
04

Solve the integration for molar flow rates

Set up and solve the numerical integration using the initial molar flow rates (F_CO0, F_H2O0, F_CO20, F_H20), rate equations from Step 1, and the equations for mass transfer. Results will give us: - Molar flow rates of CO (F_CO), H鈧侽 (F_H2O), CO鈧� (F_CO2), and H鈧� (F_H2) along the reactor length - Concentrations of CO (C_CO), H鈧侽 (C_H2O), CO鈧� (C_CO2), and H鈧� (C_H2) along the reactor length
05

Determine the reaction volume needed for 85% conversion of CO

To find the reactor volume for 85% conversion, we look at the point where the molar flow rate or concentration of CO decreases to 15% of its initial value while maintaining the mass balance with other species. Let F_CO_final be the molar flow rate of CO after achieving 85% conversion. F_CO_final = 0.15 脳 F_CO0 = 0.15 脳 4 = 0.6 mol/min Now, find the reactor volume (V_reactor) that corresponds to this 85% conversion.
06

Compare the reactor volume with a PFR

Compare the reactor volume necessary to achieve 85% conversion using a membrane reactor with that of a PFR (plug flow reactor).
07

Determine the conversion if feed rate is doubled

If the feed rate is doubled while keeping the same reactor volume, we need to determine the new conversion of CO. We can perform Steps 3, 4, 5, using the new feed rate (20 dm鲁/min) or (F_CO0 = 8 mol/min) and F_H2O0 = 8 mol/min) to find the new conversion. Calculate the new conversion considering the given reactor volume and doubled feed rate, and compare it with the initial 85% conversion. By following these steps, we have determined the necessary reaction volume, compared it with a PFR, and evaluated the conversion if the feed rate is doubled in a membrane reactor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Membrane Reactor
Membrane reactors play a crucial role in modern reaction engineering by combining reaction and separation processes in a single unit. This integration enhances reaction efficiency and selectivity. In a membrane reactor, a semipermeable membrane separates different chemical species while allowing specific ones to pass through. For instance, in the water gas shift reaction, hydrogen is continuously removed by the membrane as it is produced.
This selective separation not only shifts the equilibrium in favor of further reaction to produce more hydrogen but also purifies the hydrogen for subsequent applications. Membrane reactors are especially beneficial for processes requiring high purity and yield, which are essential for the hydrogen economy.
  • Combines reaction and separation.
  • Improves selectivity and efficiency.
  • Essential for reactions needing high purity like water gas shift.
Water Gas Shift Reaction
The water gas shift reaction is a chemical reaction that plays a vital role in the large-scale production of hydrogen, which is essential for the hydrogen economy. The reaction involves carbon monoxide reacting with water vapor to produce carbon dioxide and hydrogen:
\[ \text{CO} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + \text{H}_2 \]
In industrial settings, this reaction is carried out in reactors like the membrane reactor to increase the yield of hydrogen. The equilibrium constant, denoted as \( K_e \), helps in determining the extent of the reaction under given conditions. Optimizing the conditions such as temperature and pressure can shift the equilibrium to produce more hydrogen.
  • Produces hydrogen from CO and water vapor.
  • Influenced by equilibrium and reaction condition optimization.
  • Key step in hydrogen production for fuel cells.
Hydrogen Economy
The hydrogen economy represents an energy system where hydrogen is the primary energy carrier, supplementing or replacing traditional carbon-based fuels. This shift is driven by the need to reduce carbon emissions and reliance on fossil fuels. Hydrogen can be produced from various sources including water and biomass, but the water gas shift reaction is central to its efficient production.
Fuel cells, which convert hydrogen into electricity, are pivotal components of this economy. They can power numerous applications from vehicles to residential power systems. The success of a hydrogen economy depends on the development and optimization of processes for hydrogen generation, like ensuring efficient membrane reactor operations.
  • Focuses on hydrogen as an energy carrier.
  • Aims to reduce carbon emissions.
  • Relies on efficient hydrogen production processes.
Numerical Integration
Numerical integration is a mathematical tool used to solve differential equations that cannot be solved analytically. In reaction engineering, it is particularly useful for calculating parameters and performance across systems like membrane reactors.
For instance, in analyzing molar flow rates and concentrations along the length of a reactor, numerical integration helps to sum up incremental changes. This allows engineers to predict behavior over time and optimize reactor designs for desired outputs. Techniques like the Euler method or Runge-Kutta methods are often employed to handle the complexity of these calculations.
  • Used to solve differential equations in reactor models.
  • Helps calculate changes over a system, such as a reactor.
  • Essential for reactor optimization and performance prediction.

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Most popular questions from this chapter

Dibutyl phthalate (DBP), a plasticizer, has a potential market of 12 million Ib/yr (AIChE Student Contest Problem) and is to be produced by reaction of n-butanol with monobutyl phthalate (MBP). The reaction follows an elementary rate law and is catalyzed by \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (Figure \(\mathrm{P} 4-6\) ). A stream containing MBP and butanol is to be mixed with the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) catalyst immediately before the stream enters the reactor. The concentration of MBP in the stream entering the reactor is \(0.2 \mathrm{tb} \mathrm{mol} / \mathrm{ft}^{3}\), and the molar feed rate of butanol is five times that of MBP. The specific reaction rate at \(100^{\circ} \mathrm{F}\) is \(1.2 \mathrm{ft}^{3} / \mathrm{lb}\) mol \(\cdot \mathrm{h}\) There is a 1000 -gallon CSTR and associated peripheral equipment available for use on this project for 30 days a year (operating 24 h/day). (a) Determine the exit conversion in the available 1000 -gallon reactor if you were to produce \(33 \%\) of the share (i.e., 4 million \(\mathrm{Ib} / \mathrm{yr}\) ) of the predicted market. (Ans.: \(X=0.33\) ) (b) How might you increase the conversion for the same \(F_{\mathrm{AO}} ?\) For example, what conversion would be achieved if a second 1000 -gal CSTR were placed either in series or in parallel with the CSTR? [\(X_{2}=0.55\) (series)] \right. (c) For the same temperature as part (a), what CSTR volume would be necessary to achieve a conversion of \(85 \%\) for a molar feed rate of \(\mathrm{MBP}\) of 1 Ib mol/min? (d) If possible. calculate the tubular reactor volume necessary to achieve \(85 \%\) conversion. when the reactor is oblong rather than cylindrical, with a major-to-minor axis ratio of \(1.3: 1.0 .\) There are no radial gradients in either concentration or velocity. If it is not possible to calculate \(\mathrm{V}_{\mathrm{PRF}}\) explain. (e) How would your results for parts (a) and (b) change if the temperature were raised to \(150^{\circ} \mathrm{F}\) where \(k\) is now \(5.0 \mathrm{ft}^{3} / \mathrm{lb}\) mol \(\cdot \mathrm{h}\) but the reaction is reversible with \(K_{C}=0.3 ?\) (f) Keeping in mind the times given in Table 4-1 for filling, and other operations, how many 1000 -gallon reactors operated in the batch mode would be necessary to meet the required production of 4 million pounds in a 30-day period? Estimate the cost of the reactors in the system. Note: Present in the feed stream may be some trace impurities, which you may lump as hexanol. The activation energy is believed to be somewhere around 25 kcal/mol. Hint: Plot number of reactors as a function of conversion. ( \(A n\) Ans.: 5 reactors) (g) What generalizations can you make about what you learned in this problem that would apply to other problems? (h) Write a question that requires critical thinking and then explain why your question requires critical thinking. [Hint: See Preface. Section B.2]

The reversible isomerization $$\text { m Xylene \(\rightleftarrows\) para-Xylene }$$ follows an elementary rate law. If \(X_{c}\) is the equilibrium conversion, (a) Show for a batch and a PFR: \(t=\tau_{\mathrm{PFR}}=\frac{X_{\mathrm{e}}}{k} \ln \frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\) (b) Show for a CSTR: \(\tau_{\mathrm{PFR}}=\frac{X_{\mathrm{c}}}{k}\left(\frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\right)\) (c) Show that the volume efficiency is $$\frac{V_{\mathrm{PFR}}}{V_{\mathrm{CSTR}}}=\frac{\left(X_{\mathrm{e}}-\mathrm{X}\right) \ln \left(\frac{X_{\mathrm{e}}}{X_{\mathrm{e}}-X}\right)}{X_{\mathrm{c}}}$$ and then plot the volume efficiency as a function of the ratio \(\left(X / X_{\mathrm{e}}\right)\) from 0 to 1 (d) What would be the volume efficiency for two CSTRs in series with the sum of the two CSTR volumes being the same as the PFR volume?

(a) The irreversible elementary reaction \(2 \mathrm{A} \longrightarrow \mathrm{B}\) takes place in the gas phase in an isothermal tubular (plug-flow) reactor. Reactant A and a diluent \(C\) are fed in equimolar ratio, and conversion of \(A\) is \(80 \%\). If the molar feed rate of \(A\) is cut in half, what is the conversion of \(A\) assuming that the feed rate of \(\mathrm{C}\) is left unchanged? Assume ideal behavior and that the reactor temperature remains unchanged. What was the point of this problem? (From California Professional Engineers Exam.) (b) Write a question that requires critical thinking, and explain why it involves critical thinking.

(a) A liquid-phase isomerization \(A \longrightarrow B\) is carried out in a 1000 -gal CSTR that has a single impeller located halfway down the reactor. The liquid enters at the top of the reactor and exits at the bottom. The reaction is second order. Experimental data taken in a batch reactor predicted the CSTR conversion should be \(50 \%\). However, the conversion measured in the actual CSTR was \(57 \% .\) Suggest reasons for the discrepancy and suggest something that would give closer agreement between the predicted and measured conversions. Back your suggestions with calculations. P.S. It was raining that day. (b) The first-order gas-phase isomerization reaction $$A \stackrel{A}{\longrightarrow} B \text { with } k=5 \min ^{-1}$$ is to be carried out in a tubular reactor. For a feed of pure \(A\) of 5 \(\mathrm{dm}^{3} / \mathrm{min}\), the expected conversion in a PFR is \(63.2 \%\). However. when the reactor was put in operation, the conversion was only \(58.6 \% .\) We should note that the straight tubular reactor would not fit in the available space. One engineer suggested that the reactor be cut in half and the two reactors be put side by side with equal feed to each. However, the chief engineer overrode this suggestion saying the tubular reactor had to be one piece so he bent the reactor in a U shape. The bend was not a good one. Brainstorm and make a list of things that could cause this off-design specification. Choose the most logical explanation/model, and carry out a calculation to show quantitatively that with your model the conversion is 58.6%. (An Ans: 57% of the total) (c) The liquid-phase reaction $$A \longrightarrow B$$ was carried out in a CSTR. For an entering concentration of \(2 \mathrm{mol} / \mathrm{dm}^{3}\) the conversion was \(40 \%\). For the same reactor volume and entering conditions as the CSTR, the expected PFR conversion is 48.6%. However. the PFR conversion was amazingly \(50 \%\) exactly. Brainstorm reasons for the disparity. Quantitatively show how these conversions came about (i.e.., the expected conversion and the actual conversion). (d) The gas-phase reaction $$A+B \longrightarrow C+D$$ is carried out in a packed bed reactor. When the particle size was decreased by \(15 \%\). the conversion remained unchanged. When the particle size was decreased by \(20 \%,\) the conversion decreased. When the original particle size was increased by \(15 \%,\) the conversion also decreased. In all cases, the temperature, the total catalyst weight. and all other conditions remained unchanged. What's going on here?

A microreactor similar to the one shown in Figure P4-19 from the MIT group is used to produce phosgene in the gas phase. $$\begin{array}{c} \mathrm{CO}+\mathrm{Cl}_{2} \rightarrow \mathrm{COCl}_{2} \\ \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \end{array}$$ The microreactor is \(20 \mathrm{mm}\) long. \(500 \mu \mathrm{m}\) in diameter, and packed with catalyst particles \(35 \mu \mathrm{m}\) in diameter. The entering pressure is \(830 \mathrm{kPa}(8.2 \mathrm{atm})\) and the entering flow to each microreactor is equimolar. The molar flow rate of \(\mathrm{CO}\) is \(2 \times 10^{-5} \mathrm{mol} / \mathrm{s}\) and the volumetric flow is \(2.83 \times 10^{-7} \mathrm{m}^{3} / \mathrm{s}\). The weight of catalyst in one microreactor: \(W=3.5 \times 10^{-6} \mathrm{kg}\). The reactor is kept isothermal at \(120^{\circ} \mathrm{C}\). Because the catalyst is also slightly different than the one in Figure \(\mathrm{P} 4-19,\) the rate law is different as well: $$-r_{A}^{\prime}=k_{A} C_{A} C_{B}$$ (a) Plot the molar flow rates \(F_{\mathrm{A}}, F_{\mathrm{B}}\), and \(F_{\mathrm{C}},\) the conversion \(X\), and pressure ratio \(y\) along the length of the reactor. (b) Calculate the number of microreactors in parallel to produce 10.000 kg/year phosgene. (c) Repeat part (a) for the case when the catalyst weight remains the same but the particle diameter is cut in half. If possible compare your answer with part (a) and describe what you find. noting anything unusual. (d) How would your answers to part (a) change if the reaction were reversible with \(K_{\mathrm{C}}=0.4 \mathrm{dm}^{3} / \mathrm{mol} ?\) Describe what you find. (e) What are the advantages and disadvantages of using an array of mi reactors over using one conventional packed bed reactor that provides same yield and conversion? (f) Write a question that involves critical thinking. and explain wh involves critical thinking. (g) Discuss what you learned from this problem and what you believe th the point of the problem. Additional information: \(\alpha=3.55 \times 10^{5} / \mathrm{kg}\) catalyst (based on properties of air and \(\phi=0.4\) ) \(k=0.004 \mathrm{m}^{6} / \mathrm{mol} \cdot \mathrm{s} \cdot \mathrm{kg}\) catalyst at \(120^{\circ} \mathrm{C}\) \(v_{0}=2.83 \cdot 10^{-7} \mathrm{m}^{3} / \mathrm{s}, \rho=7 \mathrm{kg} / \mathrm{m}^{3}, \mu=1.94 \cdot 10^{-5} \mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\) \(A_{c}=1.96 \cdot 10^{-7} \mathrm{m}^{2}, G=10.1 \mathrm{kg} / \mathrm{m}^{2} \cdot \mathrm{s}\)
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