Batch Reactor
Important Unit Operations of Chemical Engineering
Fundamentals of Heat Transfer
Newtonian and Non-Newtonian Fluids
Hydrostatic Equilibrium
- A batch Reactor is also known as an unsteady state reactor.
- First of all, reactants are charged into the reactor.
- Reactants are well mixed into the reactor and left for a certain period of time to react.
- After the completion of reaction time, the product is withdrawn from the reactor.
- It is an unsteady state type reactor so the composition changes with respect to time.
- But at any point in time composition is the same throughout the reactor.
- A batch reactor is suitable for small-scale production, production of multiple products like in the pharmaceutical industry, in dye industries, etc.
- In Batch, Reactor composition is the same at any instant point of time throughout the reactor.
- So we will consider the reactor as a whole.
- Consider a component "A" as a limiting reactant.
- Write Material Balance Equation for component "A". \[Input = Output + Disapperance + Accumulation ---------- (1) \]
- No reactant is charged or withdrawan during the reaction Therefore, \[Input = Output = 0\]
- So the equation becomes \[Disapperance = - Accumulation -------- (2)\]
- Now, the Disappearance of reactant A due to chemical reaction can be evaluated as follows. Disapprance of A in moles/time \(= (-r_A)V--------(3)\)
- Accumulation of A within the reactor in moles/time can be evaluated as follows We know, \[N_A = N_{A0} (1-X_A)\] Differeantation of above equation with respect to time (t), \[\frac{dN_A}{dt}= \frac{d[ N_{A0} (1-X_A)]}{dt}\] \[\frac{dN_A}{dt}= \frac{dN_{A0}}{dt}×\frac{dX_A}{dt}\] Here, \[\frac {dN_{A0}}{dt}=0\] Therefore, Accumulation is \[\frac{dN_A}{dt}=-N_{A0}\frac{dX_A}{dt}-------(4)\]
- Putting values from equation (3) and (4) in equation (2) \[(-r_A)V=N_{A0}\frac{dX_A}{dt}\]
- Rearranging above equation \[dt=N_{A0}×\frac{dX_A}{(-r_A)V}\]
- Integrating above equation \[t=N_{A0}\int_{0}^{X_A} \frac{dX_A}{(-r_A)V} --------(5)\] It is the general equation, shows the time requied to achieve a conversion \(X_A\) for either isothermal or nonisothermal operation.
- Equation no. (5) can be simplified for many situation.
- Case -I (Denisity of the fluid remain costant) Volume of the fluid is constant, so we can take out it from integral. \[t=\frac {N_{A0}}{V}\int_{0}^{X_A} \frac{dX_A}{(-r_A)}\] \[t=C_{A0}\int_{0}^{X_A} \frac{dX_A}{(-r_A)}--------(6)\] Here, \[X_A=1-\frac{C_A}{C_{A0}}\] \[dX_A=-\frac{dC_A}{C_{A0}}\] Put \(dX_A\) value in equation no. 6. \[t=-\int_{C_{A0}}^{C_A}\frac{dC_A}{(-r_A)}-------(7)\] Above equation is valid for for \(\epsilon_A=0\)
- Case - II (Volume of the reacting mixtue changes proportionaly with conversion) \[V=V_0(1+\epsilon_A X_A)\] Put value of \(V\) in equation no. 5. \[t=N_{A0}\int_{0}^{X_A} \frac{dX_A}{(-r_A)(V_0(1+\epsilon_A X_A))}\] \[t=\frac{N_{A0}}{V_0}\int_{0}^{X_A} \frac{dX_A}{(-r_A)(1+\epsilon_A X_A)}\] \[t=C_{A0}\int_{0}^{X_A} \frac{dX_A}{(-r_A)(1+\epsilon_A X_A)}\]
Important Unit Operations of Chemical Engineering
Fundamentals of Heat Transfer
Newtonian and Non-Newtonian Fluids
Hydrostatic Equilibrium