(a) Starting from the mole balance of feed in a tubular reactor with a first ord
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Question
(a) Starting from the mole balance of feed in a tubular reactor with a first order reaction A rightarrow Product in which the volumetric flow rate, v is constant, show that the volume of the reactor, V, can be written as V = -v_o/k ln C_A/C_A where k is the rate constant and C_A and C_A are the inlet and outlet concentration respectively. (b) The data in Table Q4 were collected for an elementary liquid phase reaction A rightarrow B at room temperature. Using a graphical method calculate the volume of reactor required for 80% conversion of reactant with a flowrate of 2 litre h^-1 and an initial concentration of 7.0 mol litre^-1. Assume the reaction is carried out in a Continuous Stirred Tank Reactor (CSTR). c) From your graph in part (b) or otherwise, calculate the volume required if the reactor is changed to a Plug Flow Reactor (PFR). Comment on your result.Explanation / Answer
The general mass balance equation for a steady state flow can be written as
Rate of moles of A in = Rate of moles of A out+ rate of loss of reactant due to chemical reaction
In a tubular reactor, the concentration of A drops gradullay and hence a differential element has to be considered for analysis. Let FA= molar flow rate of A in, FA+dFA= molar flow rate of A out, K= rate constant, CA= Concentratin of A , dV= differential volume. Now the mass balance equation becomes
FA= FA+dFA+ KCAdV
-dFA= KCAdV, but FA= FAO*(1-XA), CA= CAO*(1-XA), FAO and CAO are inital molar flow rate and concentratios of A, XA= conversion
dFA= FAO*(-dXA)
FAO*dXA)= KCAO*(1-XA)dV
Or dXA/(1-XA)= (KCAO/FAO)*dV
When integrated noting that XA=0 when V=0
-ln(1-XA)= KCAOV/FAO
But Space time T is defined as VCAO/FAO= V/VO, where Vo= Volumetric flow rate
Hecne –ln(1-XA)= K*V/Vo
XA= 1-CA/CAO or 1-XA= CA/CAO
Hence –ln(CA/CAO)= K*V/Vo, V= -(VO/K)* ln (CA/CAO)
rA and CA are releated as
-rA= KCAn, n is the order of reaction.
When taking ln, the equation becomes
Ln(-rA)= lnK+ nlnCA, so a plot of ln(-rA) vs lnCA gives a straight line whose slope is n and intercept is lnK, where K is the rate constant. The plot is generated and shown
from the plot.slope n=2, second order reaction.
for a CSTR, T= space time= CAOXA/KCAO2*(1-XA)2=
or KCAOT= XA/(1-XA)2= 0.8/0.22= 20
lnK from the plot = 0.695, K= e(0.695)= 2/M.hr
2*7*V/2= 20
V= 20*2/14= 2.85 L
for a PFR and for a second order reaction
KCAOT= XA/(1-XA)= 0.8/0.2= 4
2*7*V/2= 4 or V= 4/7 L= 0.57L
for a given conversion, PFR gives a smaller volume of reactor than CSTR.
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