CP33 Set # July-October 3 Design of Ideal Batch Reactors operated under Isothermal Conditions It is important to have this note set with you during all lecture classes. A batch reactor is such that a batch of reactants is fed to the reactor at the beginning, the reactants are allowed to react for a certain period of time, and the products are removed from the reactor at the end. While the reaction is progressing, nothing is fed to the batch reactor and nothing is taen away from the batch reactor. In an ideal batch reactor, the reacting mixture is assumed to be well mixed and therefore the properties of the reacting mixture are uniformly distributed through out the reactor. C A C A is the uniform concentration of the reactant A within the reacting mixture at time t Design equation for reactant A in the batch reactor is obtained by writing the mass balance for reactant A over the volume of the reacting mixture during time as follows: mass of A entering the reactor volume during time = mass of A leaving the reactor volume during time + mass of A accumulated within the reactor volume during time + mass of A disappearing by the reaction during time which becomes = + dn A M A + r A M A. where N A is the number of moles of A within the reactor at time t, M A is the molar mass of A, dm A N A is the mass of A accumulated within the reactor during time, and r A is the molar rate at which A is disappearing because of the progression of the reaction. Note that r A is a negative quantity and therefore we have used the positive quantity r A in the above mass balance. Also, note that the unit of r A is in general moles per volume per time, and therefore r A is multiplied by the molar mass of A to get the reaction rate in compatible unit for the mass balance given by..
Removing M A from. and rearranging it, we get the design equation for reactant A in an ideal batch reactor as follows: = r A. Woring out in terms of the number of moles of A, N A : The time t f required to reduce the number of moles of A present in the batch reactor from N Ao moles at the commencement of the reaction to N Af moles at time t f can be evaluated by integrating. as follows: t f = NAf N Ao r A = NAo N Af r A.3 where r A should be expressed as a function of N A in order to be able to perform the above integration. Woring out in terms of the concentration of A, C A : The concentration of A in the batch reactor is defined as Number of moles of A in the reacting mixture at time t C A Total volume of the reacting mixture at time t Equation.4 gives N A = C A. Substituting this in., we get = N A.4 dc A = r A.5 When the volume of the reacting mixture remains a constant,.5 yields CAf CAo t f = dc A = C Ao r A C Af r A dc A.6 where C Ao and C Af are the respective concentrations of A at the beginning and at the end of the reaction, and r A should be expressed as a function of C A in order to perform the above integration. If volume does not remain constant during the reaction, the solution procedure gets slightly more complicated which will be discussed latter in this note set. Woring out in terms of the conversion of A, x A : Conversion of A is defined by x A N Ao N A N Ao.7 which gives N A = N Ao x A. Substituting it in., we get Upon integration of.8, we get N Ao = r A.8 t f = xaf N Ao r A.9 where x Af is the final conversion of A, and r A should be expressed as a function of x A in order to perform the above integration. If the volume of the reacting mixture does not remain a constant during the reaction then should also be expressed as a function of x A, which will be wored out in Example.3 of this note set.
Example.: Consider the elementary liquid-phase reaction A products taing place in an ideal batch reactor at constant temperature. Determine the time taen for the concentration of A to become half of its initial concentration. Solution: Since the given reaction is elementary, the reaction rate equation can be written as r A = C A. where, which is a function of temperature, remains constant since the reaction is said to tae place at constant temperature. Substituting r A given by. in the design equation for A in an ideal batch reactor, given by., we get = C A. Since the concentration C A in the batch reactor is defined as C A N A /, see.4, we could rewrite. in terms of C A as dc A = C A The given reaction is a liquid phase-reaction, and therefore it is acceptable to assume that the density of the reacting mixture remains a constant. As far as there is no nuclear reaction taing place, the total mass of the reacting mixture remains a constant. Since the mass and the density of the reacting mixture are constants, the volume of the reacting mixture is also a constant, which helps to simplify the above equation to dc A = C A. Integrating. with the initial condition C A = C Ao at t =, we get which gives C A = C Ao exp t t f = ln C Af/C Ao.3 where t f is the time taen to reach the final concentration of A and C Af is the final concentration of A. Therefore, the time taen to halve the initial concentration could be calculated as follows: t f = ln.5c Ao/C Ao = ln.5.4 Note: Since r A taes the unit mole per volume per time and C A taes the unit mole per volume,. gives the unit of as per time. For example, if r A is in moles/litre.sec and C A is in moles/litre then is in sec. Therefore, t f in.4 would tae the unit of time, as it should be. But, the unit of is not always per time, and it depends on the form of the rate equation. Suppose the rate equation were given by r A = C A then the unit of would be volume per mole per time so that r A could tae the unit mole per volume per time since C A taes the unit mole/volume. 3
Example.: Determine the time taen to reach 9% conversion of A in the reaction of Example.. Solution: Using C A = N A /, we could rewrite. in terms of N A as = N A From the definition of conversion, see.7, we get N A = N Ao x A. Substituting this in the above equation, we get N Ao = N Ao x A which gets simplified to = x A.5 Equation.5, when integrated with the conditions x A = at t = and x A = x Af at t = t f, gives t f = xaf x A = [ ln xa ] x Af = ln x Af.6 Since x Af =.9 we get t f = ln..7 Example.3: Consider the gas-phase reaction A B + C, for which the rate equation is given by r A = C A.8 Conversion of A required is expected to be 9%. Determine the time required to carry out the above reaction in an ideal batch reactor operated at constant temperature and constant pressure. Solution: Substituting r A given by.8 in the design equation for reactant A in an ideal batch reactor, given by., we get = CA.9 Since the reaction considered in this example is a gas-phase reaction, it is convenient to wor it out in terms of conversion of A. In order to do that, we must first write the above differential equation in terms of N A, which is done as follows: Substituting C A = N A / in.9, we get a differential equation in terms of N A as = NA = 4 N A.
From the definition of x A given by.7, we get Combining. and., we get N A = N Ao x A. = N Ao which is the differential equation in terms of conversion of A. x A. If the volume remains a constant then. shall be integrated to obtain the time required to achieve 9% conversion of A as t f = C Ao.9 where C Ao = N Ao / for a constant-volume batch reactor. x A.3 But, the given reaction is a gas-phase reaction in which moles of reactant give 3 moles of product. Besides, the temperature and the pressure remain constants. Therefore, the volume of the reacting mixture would change with time. Let us use the ideal gas equation of state to describe the behaviour of the given gas mixture. We then get P = N T R T.4 where N T denotes the total number of moles of the reacting mixture at time t, and is given by N T = N A + N B + N C + N I.5 where N A, N B, N C and N I are the respective number of moles of A, B, C, and inert gas I present in the reactor at time t. Stoichiometry gives N Ao N A = N B N Bo Equation.6 combined with. gives = N C N Co.6 N B = N Bo + N Ao x A / and N C = N Co + N Ao x A.7 The inert gas does not react, and therefore N I = N Io.8 Using.7 and.8 in.5, we get N T = + N Ao x A.9 where = N Ao + N Bo + N Co + N Io, denotes the total number of moles of the reacting mixture at the commencement of the reaction. 5
Substituting N T from.9 in.4, we get = R T P + N Ao x A..3 Since the initial volume of the reacting mixture o also satisfies the ideal gas equation of state, given by P o o = R T o,.3 can be rewritten as = P o P T o + N Ao x A T o..3 Since the reactor is operated at constant pressure and constant temperature,.3 becomes = o + N Ao x A..3 Substituting from.3 in., we get = N Ao x A x A o + N Ao x A = C Ao + N Ao x A.33 where C Ao is the initial concentration of A given by N Ao / o. Equation.33 shall be integrated to obtain the time required to achieve 9% conversion of A as t f = C Ao.9 x A + N Ao dx x A A.34 If only pure A is present at the commencement of the reaction, then.34 reduces to t f = C Ao.9 +.5 x A x A.35 For a general gas-phase reaction, is related to x A by where a A + b B + c C + p P + q Q + s S +,.36 ɛ A = N Ao = P o P T T o o + ɛ A x A..37 [ ] p + q + s + a + b + c + a.38 and = N Ao + N Bo + N Co + + N P o + N Qo + N Ro + + N Io is the total number of moles of the reacting mixture at the commencement of the reaction. Self-study 3: Determine the numerical values of the integrations in.3 and.35. 6