Question

The following reaction is taking place in a batch reactor, A → R → S, where R is the desired product and A is the reactant, with a initial concentration CA₀ = 10 moles/volume). The rate equations for this reaction are provided below. Draw the graph of concentrations of A, R, and S (CA, C_R, and CS) with respect to time t, when time changes in increments of 0.04 hours. Solve the problem to obtain (1) maximum concentration of R, (2) maximum yield of R, (3) maximum profit, where profit = 100 × concentration of product R - 10 × concentration of raw material, (4) maximum profit, where profit = value of product less cost of removing impurities A and S. Product value is the same as given for (3) but the cost of removing A is 1 unit, whereas the cost of removing S is 25 units (5) maximum profit, where profit = product value - raw material cost (same as (3)) - labor cost (25 units per unit time).

CA_t = CA₀ exp (−k₁t)

CR_t = CA₀k₁[ exp (−k₁t) k₂ − k₁ − exp (−k₂t) k₂ − k₁]

CS_t = CA₀[1 − exp (−k₁t) k₂ − k₁ − exp (−k₂t) k₂ − k₁ ]

where k₁= 10 per hour, k₂ = 0.1 per hour

Now include the temperature (T) effects on the reaction in terms of the reaction constants k₁ and k₂given below and resolve the above five optimization problems.

k₁(T) = 19531.2 exp (−2258.0/T)

k₂(T) = 382700. exp (−4517.0/T)

(Note: This is a numerical from applied optimization course)

Answer 1

Graph of change of concentration of each substance that participates in the reaction:

(1) CR = 927.94 mol / volume

(2) R = CR * 100 / (CR + CS + CA) = 89.53%

(3) Maximum Gain = 100 * CR - 10 * CA = 92791.61

(4) It is necessary to know the value of R, which you do not have, but in the following way you could have the result:

MG = Value of R - 1 - 25

(5) Same situation as previous question:

Maximum Benefit = Value of R - Cost of Raw Material - 25 * (0.36 h)

NOTE: You must know the reaction temperature profile or temperature equation as a function of time for the last part.