Holooly Plus Logo
Holooly Plus Logo

Question : Reagent A undergoes an essentially irreversible isomerizatio...

Reagent A undergoes an essentially irreversible isomerization reaction that obeys first-order kinetics:

A\rightarrow B

Both A and B are liquids at room temperature and both have very high boiling points. A 1000-gallon glass-lined kettle is available for carrying out the reaction. The kettle may be maintained at essentially isothermal conditions by a heat transfer fluid that circulates through a jacket on its external surface. The heat transfer fluid may be cooled or heated as required by circulation through appropriate heat exchangers. Since the reaction is exothermic, Kladko (25) wished to consider the possibility of using cold reactant feed to provide a heat sink for some of the energy liberated by reaction. By controlling the rate of addition of feed, they could also obtain a measure of control over the rate of energy release by reaction. Hence, a semibatch mode of operation appeared to be an attractive alternative. Since cold incoming reactant would crack the hot glass liner, they considered the possibility of starting with 1500 lb of product B in the reactor so as to provide a thermal and material sump. The sump not only acts as a thermal sink for the cold incoming reactant, but also dilutes it, thereby reducing the reaction rate and the rate of energy release by reaction.

If the temperature of the reactor contents is maintained constant at 163^{\circ}C, determine the total amounts of species A and B in the reactor as functions of time when it is loaded according to the following schedule:

Time, t Feed rate of A
(h) (lb/h)
0_3 175
3_6 225
6_7 275
7_8 325
8_11 400
11_12 325
12_13 275
13_14 225
14_15 175
15_16 100
16_17 50
17+ 0

As we shall see in Illustration 10.7, this type of filling schedule is necessary to avoid dramatic exotherms that would result from sudden termination of the feed and to ensure that the heat transfer capability of the system is not exceeded.

Data and permissible assumptions are as follows:

1. The reactor contents are perfectly mixed.

2. The rate expression is first-order in species A.

3. At 163^{\circ}C the reaction rate constant is 0.8

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

A material balance involving the amount of species A contained within the reactor at any time can be written as

 

input = accumulation + disappearance by reaction

 

F_{A0}=\frac{dn_{A}}{dt}+kC_{A}{V}'_{R} (A)

 

where n_{A} is the instantaneous number of moles of species A contained within the reactor and {V}'_{R} is the instantaneous volume occupied by the liquid solution. This equation is similar to equation (8.6.1), but it lacks the term corresponding to the effluent stream and has been written in terms of moles.

 

Now, at any time,

 

C_{A}=\frac{n_{A}}{{V}'_{R}} (B)

 

Thus,

 

F_{A0}=\frac{dn_{A}}{dt}+kn_{A} (C)

 

Equation (A) can be rewritten in terms of the mass of species A present in the reactor as

 

\Phi _{A0}=\frac{dm_{A}}{dt}+km_{A} (D)

 

where \Phi _{A0} is the mass rate of flow of species A into the reactor.

 

Equation (C) may be solved in piecewise fashion to determine the mass of species A present in the reactor as a function of time. The solution can be written as

 

\int_{m_{A_{i}}}^{m_{A(t)}}\frac{dm_{A}}{\Phi _{A0}-km_{A}}=t-t_{i}

 

where one uses a constant value of \Phi _{A0} appropriate to the time interval in question and m_{A_{i}} is the mass of A in the reactor at the start of the time interval (time t_{i}). Thus,

 

\frac{1}{k}\ln \left [ \frac{\Phi _{A0}-km_{Ai}}{\Phi _{A0}-km_{A}} \right ]=t-t_{i}

 

Rearranging yields

 

m_{A}=m_{Ai}e^{-k(t-t_{i})}+\frac{\Phi _{A0}}{k}[1-e^{-k(t-t_{i})}] (E)

 

Using 0.8 h^{-1} for k and values of \Phi _{A0} given in the filling schedule, equation (E) can be solved in piecewise fashion.