Second fractional conversions

It has been shown that coke yield as a fraction of feed does give a linear relationship with second-order conversion (13) indicating a positive coke yield at 2ero conversion. This coke yield at 2ero conversion is the additive coke contribution to the total coke yield and is related to feed properties, particularly Conradson carbon content. The amount of this additive coke is significantly less than the Conradson carbon value of the feed (14), probably in the range of 50% of the Conradson carbon. 

Nc = 0.0 gmol, Nq = 0.0 gmol, respectively. A mixture of A and B is charged into a 1-liter reactor. Determine the holding time required to achieve 90% fractional conversion of A (X = 0.9). The rate constant is k = 1.0 X 10 [(liter) /(gmoP min)] and the reaction is first order in A, second order in B and third order overall. 

Solution The first-order rate constant is 0.693/2.1=0.33 so that the fractional conversion for a first-order reaction will be 1 — exp(—0.227) where f is in seconds. The inlet and outlet pressures are known so Equation (3.27) can be used to And t given that [L/Mom ] = 57/9.96 = 5.72s. The result is f = 5.91 s, which is 3.4% higher than what would be expected if the entire reaction was at Pout- The conversion of the organic compound is 86 percent. 

Figure 5.46. A burst of warm feed at 310 K for the first 1400 seconds causes the system to shift into a higher temperature and higher conversion steady state. The fractional conversion curves A and C are for normal startup and B and D are for the startup with a warm feed period.

Each of these reactions is first-order in A and first-order in S. The inlet concentration of A is equal to 5 moles/liter. The reactor combination is to be operated under conditions such that the fraction conversion of A based on the inlet concentration is 0.4 leaving the first reactor and 0.6 leaving the second reactor. 

The F(t) curve for a system consisting of a plug flow reactor followed by a continuous stirred tank reactor is identical to that of a system in which the CSTR precedes the PFR. Show that the overall fraction conversions obtained in these two combinations are different when the reactions are other than first-order. Derive appropriate expressions for the case of second-order irreversible reactions and indicate how the reactors should be ordered so as to maximize the conversion achieved. 

D. Use fA = 0.008 as a first estimate of the average fraction conversion for the second increment in... 

The first column lists all the species involved (including inert species, if present). The second column lists the basis amount of each substance (in the feed, say) this is an arbitrary choice. The third column lists the change in the amount of each species from the basis or initial state to some final state in which the fractional conversion is fA. Each change is in terms of fA, based on the definition in equation 2.2-3, and takes the stoichiometry into account. The last column lists the amounts in the final state as the sum of the second and third columns. The total amount is given at the bottom of each column. 

A second-order reaction may typically involve one reactant (A -> products, ( -rA) = kAc ) or two reactants ( pa A + vb B - products, ( rA) = kAcAcB). For one reactant, the integrated form for constant density, applicable to a BR or a PFR, is contained in equation 3.4-9, with n = 2. In contrast to a first-order reaction, the half-life of a reactant, f1/2 from equation 3.4-16, is proportional to cA (if there are two reactants, both ty2 and fractional conversion refer to the limiting reactant). For two reactants, the integrated form for constant density, applicable to a BR and a PFR, is given by equation 3.4-13 (see Example 3-5). In this case, the reaction stoichiometry must be taken into account in relating concentrations, or in switching rate or rate constant from one reactant to the other. 

The estimated time required to achieve a fractional conversion of 0.99 is 1.80 h, and the temperature at this time is 333.5 K, if the reactor operates adiabatically. The fA(t) profile is given by the values listed in the second and last columns the T(t) profile is given by the third and last columns. 

At elevated temperatures, acetaldehyde (CH3CHO, A) undergoes gas-phase decomposition into methane and carbon monoxide. The reaction is second-order with respect to acetaldehyde, with kA = 22.2 L mol-1 min-1 at a certain T. Determine the fractional conversion of acetaldehyde that can be achieved in a 1500-L CSTR, given that the feed rate of acetaldehyde is 8.8 kg min-1, and the inlet volumetric flow rate is 2.5 m3 min-1. Assime T and P arc unchanged... 

Calculate the ratio of the volumes of a CSTR and a PFR ( Vst pf) required to achieve, for a given feed rate in each reactor, a fractional conversion (/A) of (i) 0.5 and (ii) 0.99 for the reactant A, if the liquid-phase reaction A - products is (a) first-order, and (b) second-order with respect to A. What conclusions can be drawn Assume the PFR operates isothermally at the same T as that in the CSTR. 

PF, to model the effect of earliness or lateness of mixing, depending on the sequence, on the performance of a single-vessel reactor. The following two examples explore the consequences of such series arrangements-first, for the RTD of an equivalent single vessel, and second, for the fractional conversion. The results are obtained by methods already described, and are not presented in detail. 

Determine Ihe type of reactor with the smallest volume for the second-order liquid-phase reaction A - products, where (—rA) = k c, and the desired fractional conversion is 0.60. Calculate the volume required. 

A (desired) liquid-phase dimerization 2A -> A2, which is second-order in A0"a2 = for a), is accompanied by an (undesired) isomerization of A to B, which is first-order in A(rB = 1 Ca). Reaction is to take place isothermally in an inert solvent with an initial concentration Ca0 = 5 mol L-1, and a feed rate (q) of 10 L s 1 (assume no density change on reaction). Fractional conversion (/a) is 0.80. 

A batch reactor and a single continuous stirred-tank reactor are compared in relation to their performance in carrying out the simple liquid phase reaction A + B —> products. The reaction is first order with respect to each of the reactants, that is second order overall. If the initial concentrations of the reactants are equal, show that the volume of the continuous reactor must be 1/(1 — a) times the volume of the batch reactor for the same rate of production from each, where a is the fractional conversion. Assume that there is no change in density associated with the reaction and neglect the shutdown period between batches for the batch reactor. 

The last two equations relate the fractional conversion and the meniscus height. For first or second order mechanisms,... 

The fractional conversion of a second order reaction in laminar flow is given by Eq (8) of problem P4.08.01. 

A second order reaction has kC0 = 0.0303. Obtain values of the ratios of fractional conversions along a stream line to the mean values over the cross section as a function of mean residence time over the range of 8 = r/R. At particular values of 0, Eq (3) of problem P4.08.01 can be written... 

The aqueous second order reaction, 2A 2B, has the specific rate k = 1.0 liter/mol-hr and the initial concentration Ca0= 1 mol/liter. Downtime is 1 hr/batch. Cost of fresh reactant is 100/batch and the value of the product is 200xa/batch, where xa is the fractional conversion of A. What is the daily profit for each of these modes of operation ... 

Fig. 10. Fractional conversion versus Damkohler number for half,-, first- and second-order reactions taking place in a single ideal CSTR. Shaded areas represent possible conversion ranges lying between perfectly micromixed flow (M) and completely segregated flow (S). Data taken from reference 32. A = R —...

Determine the fractional conversions in the output stream from the second reactor. 

Three conditions must be fulfilled obtain complete conversion of the reactants, H2 and CI2. The first condition is that thermal equilibrium of the system be favorable. This condition is fulfilled at low and intermediate temperatures, where formation of the product HC1 is thermodynamically favored. At very high temperatures, equilibrium favors the reactants, and thereby serves to limit the fractional conversion. The second requirement is that the overall reaction rate be nonnegligible. There are numerous examples of chemical systems where a reaction does not occur within reasonable time scales, even though it is thermodynamically favored. To initiate reaction, the temperature of the H2-CI2 mixture must be above some critical value. The third condition for full conversion is that the chain terminating reaction steps not become dominant. In a chain reaction system, as opposed to a chain-branching system discussed below, the reaction progress is very sensitive to the competition between chain initiation and chain termination. This competition determines the amount of chain carriers (batons) in the system and thereby the rate of conversion of reactants. 

Ellerstein 201 has derived an expression which utilises the second derivative of the conversion with respect to temperature, and assumes the validity of Eq. (2-9). In terms of fractional conversion, and time derivatives the equation may be written as... 

For second order reactions, graphs showing the fractional conversion for various residence times and reactant feed ratios have been drawn up by Eldridge and PlRET(7). These graphs, which were prepared from numerical calculations based on equation 1.46, provide a convenient method for dealing with sets of equal sized tanks of up to five in number, all at the same temperature. 

The heat removal line is unchanged irrespective of the kinetics, and the fractional conversion XA has a qualitatively similar sigmoidal shape for second-order kinetics. Numerical values as in Example 6-11 have been put into both the mass balance and heat balance equations. Fractional conversion XA from both the mass balance and heat balance equations at effluent temperatures of 300, 325, 350, 375, 400, 425, 450, and 475 K, respectively, were determined using the Microsoft Excel Spreadsheet (Example6-12.xls). Table 6-8 gives the results of... 

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