Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 5 (Q.No. 1)
1.
An isothermal aqueous phase reversible reaction, P 
R, is to be carried out in a mixed flow reactor. The reaction rate in k.mole/m3 .h is given by, r = 0.5CP - 0.125CR. A stream containing only P enters the reactor. The residence time required (in hours) for 40% conversion of P is
R, is to be carried out in a mixed flow reactor. The reaction rate in k.mole/m3 .h is given by, r = 0.5CP - 0.125CR. A stream containing only P enters the reactor. The residence time required (in hours) for 40% conversion of P isDiscussion:
5 comments Page 1 of 1.
Merlin said:
5 years ago
t/cp0 = xa/-ra.
t/cp0 = xa/(0.5CP - 0.125CR),
t/cp0 = xa/(0.5(cp0(1-xa)-0.125cp0*xa),
t = xa/(0.5(1-xa)-(0.125xa)).
t = 0.4/(0.5 * (1-0.4)-0.125*0.4).
t = 0.4/0.25 = 1.6.
t/cp0 = xa/(0.5CP - 0.125CR),
t/cp0 = xa/(0.5(cp0(1-xa)-0.125cp0*xa),
t = xa/(0.5(1-xa)-(0.125xa)).
t = 0.4/(0.5 * (1-0.4)-0.125*0.4).
t = 0.4/0.25 = 1.6.
(1)
Ashish Jaiswal said:
5 years ago
-rA=0.5Cp - 0.125Cr.
-rA=0.5Cpo(1-Xa) - 0.125(Cro-CroXa),
-rA = 0.5 Cpo(1-Xa) - 0.125Cpo(M+Xa),
For MFR:
(t/Cpo) = Xa/-rA,
t = 0.4/((0.5(1-0.4))-0.125(0+0.4)),
t = 1.6.
-rA=0.5Cpo(1-Xa) - 0.125(Cro-CroXa),
-rA = 0.5 Cpo(1-Xa) - 0.125Cpo(M+Xa),
For MFR:
(t/Cpo) = Xa/-rA,
t = 0.4/((0.5(1-0.4))-0.125(0+0.4)),
t = 1.6.
(1)
M. Manoj kumar said:
5 years ago
Cp=Cpo(1-x).
Cq=Cqo + 1/1Cpo(x).
Cqo=0.
(T/Cpo ) =x/rp,
X=0.4.
Solving and eliminating Cpo, we get T=8/5 =1.6.
Cq=Cqo + 1/1Cpo(x).
Cqo=0.
(T/Cpo ) =x/rp,
X=0.4.
Solving and eliminating Cpo, we get T=8/5 =1.6.
Moazzam said:
5 years ago
Please give an explanation of the answer.
ROHITH said:
6 years ago
The correct Answer is B.
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