The companion Web site offers extensive enrichment opportunities and additional content, including. Professional Reference Shelf, containing advanced content on reactors, weighted least squares, experimental planning, laboratory reactors, pharmacokinetics, wire gauze reactors, trickle bed reactors, fluidized bed reactors, CVD boat reactors, detailed explanations of key derivations, and more. Register your product at informit. Learn algorithms for solving classic computer science problems with this concise guide covering everything from fundamental ….
Fuller, John N. A Comprehensive Reference for Electrochemical Engineering Theory and Application From chemical and electronics manufacturing, to hybrid …. Cutlip - University of Connecticut. Skip to main content. Start your free trial. Scott Fogler. To promote the transfer of key skills to real-life settings, Fogler presents three styles of problems: Straightforward problems that reinforce the principles of chemical reaction engineering Living Example Problems LEPs that allow students to rapidly explore the issues and look for optimal solutions Open-ended problems that encourage students to use inquiry-based learning to practice creative problem-solving skills About the Web Site umich.
Show and hide more. Table of contents Product information. Mole Balances 1. Conversion and Reactor Sizing 2. Rate Laws 3. Stoichiometry 4. Isothermal Reactor Design: Conversion 5. Collection and Analysis of Rate Data 7. Multiple Reactions 8. Reaction Mechanisms, Pathways, Bioreactions, and Bioreactors 9. Catalysis and Catalytic Reactors Unsteady-State Nonisothermal Reactor Design For now, we will assume that conversion X will be less that 0. CDP2-A g Critique the answers to this problem.
The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactor of geological proportions a CSTR or PFR approximately the size of the Los Angeles Basin , or as we saw in the case of the batch reactor, a very long time. P a Note: This problem can have many solutions as data fitting can be done in many ways.
See Polymath program Pfireflies. See Polymath program Pcrickets. See Polymath program Pants. So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same. Insect Activation Energy Cricket Firefly Ant Honeybee P d There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would be helpful is the maximum and the minimum temperature that these insects can endure before death.
Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless. The temperature increases as we go from top to bottom of the column and consequently the rate of corrosion should increase. There is virtually no HCN in the bottom of the column.
These two opposing factors results in the maximum of the corrosion rate somewhere around the middle of the column. Therefore doubling the temperature will not necessarily double the reaction rate, and therefore halve the cooking time.
When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than double that of boiling water. Therefore, So, option 4 is correct. For any reaction, the rate law cannot be written on the basis of the stoichiometric equation. It can only be found out using experimental data. In the evaluation of the specific reaction rate constant at o C the gas constant that should have been used was 8.
In the same equation, the temperatures used should have been in K rather than oC. The units for calculated k at oC are incorrect. The dimension of the reaction rate obtained is incorrect. This is due to the fact that the rate law that has been taken is wrong. P c Example For the concentration of N2 to be constant, the volume of reactor must be constant. Therefore the reverse reaction decreases. The units of the rate constant, k, will differ depending on whether partial pressure or concentration units are used.
See below for an example. This can be done with mass balances on each element involved in the reaction. Once all the coefficients are found, you can then calculate the yield coefficients by simply assuming the reaction proceeds to completion and calculating the ending mass of the cells. The rate law is to be obtained from the experimental data.
It has been mentioned as an elementary reaction in the problem statement but in the proposed solution the rate law is based on the reaction equation that has been divided by stoichiometric coefficient of A. P a Example There would be no error!
The initial liquid phase concentration remains the same. P h Individualized solution. P i Individualized solution. So, we get 0. But actually it is a second order reaction. Also when we increase the particle size from position A, we reach at point B, again there is a decrease in the conversion. Assume the reaction temperature is K. P f The points of the problem are: 1 To note the significant differences in processing times at different temperature i.
One minute to react and to fill and empty. It does not matter if the reactor is red or black. In case of a , For a batch reactor. PFR with pressure drop: Alter the Polymath equations from part c. See Polymath program Pf-pressure. CB 0 L 10 ft lb mol P psig Assume that the reactions are irreversible and first order. Case 1: gal v0 X 0. Also, the reverse reaction begins to overtake the forward reaction near the exit of the reactor. See Polymath program Pc.
The cost of this storage could prove to be the more expensive alternative. A cost analysis needs to be done to determine which situation would be optimal. What is the maximum number of moles of ethylene glycol CH2OH 2 you can make in one 24 hour period? The feed rate of ethylene cholorhydrin will be adjusted so that the volume of fluid at the end of the reaction time will be dm3. Now suppose CO2 leaves the reactor as fast as it is formed.
First try equal number of moles of A and B added to react. See Polymath program Pb. This results in a conversion of. CDGA d We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole balance on species A will change slightly. Because species B is added to two different reactors we will also need a mole balance for species B. Since for every mole of CH3 2O consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.
This would result in the pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is true for colder temperatures. See Polymath program Pi.
P j For equal molar feed in hydrogen and mesitylene. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again. Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions. P m Individualized solution P Solution is in the decoding algorithm given with the modules ICM problem P a Assume that all the bites will deliver the standard volume of venom.
This means that the initial concentration increases by 5e-9 M for every bite. After 11 bites, no amount of antivenom can keep the number of free sites above This means that the initial concentration of venom would be 5. The best result occurs when a dose of antivenom such that the initial concentration of antivenom in the body is 5.
This means that the initial concentration of venom is 0. From the program below, we see that if an amount of antivenom such that the initial concentration in the blood is 7e-9 M, the patient will die. The concentration of D only increases with time P b Conc. However, the tradeoff is that the reaction rate of species B, and therefore production of B, decrease as temperature drops. So we have to compromise between high selectivity and production. To do this we need expressions for k1, k2, and k3 in terms of temperature.
From the given data we know: Ei ki Ai exp 1. At this temperature the selectivity is only 5. This may result in too much of X and Y, but we know that the optimal temperature is not above K. The optimal temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state for certain that the optimal temperature will be equal to or less than K.
See Polymath program Pf. A moderate pressure would probably be best. Then run the Program for 0. This will give us the concentration of A and B at the time the second martini is ingested. The Polymath code for after the second drink is shown below.
See Polymath program Pd. For the first hour the differential equation for CA becomes: dC A k1C A 2t after that it reverts back to the original equations. So the person has about 4 minutes and 40 seconds to get to their destination. P g A heavy person will have more body fluid and so the initial concentration of CA would be lower.
This means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They will take longer to reach the legal limit, as their initial concentration will be higher.
Doing so will result in build up of the drug in the bloodstream that can cause harmful effects. P d If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also drag the drug to the intestines and may limit its effectiveness. This effect can be seen in the adsorption constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is slow and if k2 increases means that the rate of elimination of Tarzlon increases.
Note that the maximum amount of the drug in the bloodstream is reduced by two. The maximum amount of Tarzlon in the bloodstream is 0. Consequently, the reaction system should be operated at highest possible temperature to maximize SDU. The reaction should also take place in high concentration of A and the concentration of D should be limited by removing through a membrane or reactive distillation. Membrane reactor in which D is diffusing out can be used.
Also low temperatures will help keep the selectivity high. Try to remove D with a membrane reactor or reactive distillation. The selectivity is not dependant on temperature. Also removing D will help keep selectivity high. This is because the first reaction is very fast and the second reaction is slower with no reverse reactions.
This is because after B is formed it will not further get converted to C because the reverse reaction is fast. See Polymath program Pa. K K K Similarly, k2' 3. See Polymath program Pb1. K K K k2' 0. P c Use the Polymath program from part a and change the limits of integration to 0 to P e When the appropriate changes to the Polymath code from part a are made we get the following. P The reactions are 1. The concentration of species are affected by two factors: reaction and pressure drop.
Further down the reactor, the concentration of B drops because the effect of pressure drop outweighs the production of B by reaction. Species E: The reasoning for species E is similar to species B. However, species C and D are also reactants in reaction 3; this contributes to their decline at later times in addition to the pressure drop. But the rate constants for the reactions have been reversed.
If concentration of M increases the slope of line will decrease. P b Example The inhibitor shows competitive inhibition. P e Example See Polymath program Pe.
Part 1 Calculated values of the DEQ variables Variable initial value minimal value maximal value final value t 0 0 24 24 Cc 2. Consequently, the concentration of product is very low compared to the case without uncompetitive inhibition. Part 3 Change the observed reaction rate constant: 0.
Thus CP the plots for this part are approximately the same as the plots in part 1. And at low temperature PSSH results show greatest disparity. Eventually everyone is ill and people start dying. This explains the shape of the figure.
Case2: With drug inhibition Reactions: 1. Thus the stoichiometry equation will be changed. This is not realistic as at some point there will be too many cells to fit into a finite sized reactor. Either a cell death rate must be included or the cells cannot be recycled. PROD max 1 0. Then, Rate of decrease of conc.
Now, since the concentration is very less assuming there is no constraint of sunlight. Since the number of days is coming less than 4. Hence, the initial assumption is verified.
P The following errors are present in this solution- 1. It should be S for Hanes-Woolf form. The expression for intercept is correct but the slope is given wrong. The correct expressions are - 5. As concentration of inhibitor I increases, slope increases. In the given plot, slope for line 1 is more as compared to line 2 in spite of having lower concentrations. This implies that the concentration values are switched.
Also the numerically calculated slope values are wrong. At 40 atm, we had While, at 80 atm we have This is not possible and the model should be discarded. Model f is the worst model of all. Assume H2 in the gas phase reacts with C2H6 adsorbed on the surface and ethane goes directly into the gas phase.
P c Individualized solution. S expression, the constant should be KA2 instead of KDC 2 The overall site balance should include the product, as it too is getting adsorbed. S rC kc CC. Therefore, the inlet temperature we would recommend is K. P d Aspen Problem P e 1 2 At high To, the graph becomes asymptotic to the X-axis, that is the conversion approaches 0.
At low To, the conversion approaches 1. P f. P d The only change to the Polymath code from part b is that the heat of reaction changes sign. The new code is not shown, but the plots are below. This means heat is generated during the reaction and there is no advantage to adding inerts as there was in the endothermic case. See Polymath program Pe.
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