(25%) 1) The following reversible, elementary, liquid phase reaction occurs in a CSTR:
→ B
A←
The entering flow rate is 10 dm3 /s with an entering concentration of 2 M of A and the
feed temperature is 300 K. What is the reactor volume necessary to achieve 90% of the
equilibrium conversion in a CSTR operated adiabatically?
Additional information:
CpA = CpB = 60 cal/mol/K
∆H°rxn = -10,000 cal/mol A
Solution:
To begin, you need to use the conversion obtained from the energy balance
cal
C pA (T − To ) 60 mol K ( T − 300 K)
X EB = = = X EB
−∆H˚rxn 10000
cal
mol A
Now, you can plot this line on the X eq versus T curve (see below)
Where the lines intersect is the operating temperature, T, (~ 398 K)
and equilibrium conversion, X eq , (~0.59)
You want 90% of this value (as stated in the problem)
X = 0.90*X eq = 0.90*0.59 = 0.53 = X
Now you can use this value in the design equation for a
first-order, reversible reaction using a CSTR
kτ
X= 1
1+kτ 1+
KC
To get values of k and KC at 398 K, use the plots given (see below)
k ~ 2.4 s-1
KC ~ 2
Plug in values and solving for space time, τ
2.4 s-1τ V dm 3
0.53 = ⇒ τ =1.1 s = ⇒ V= 1.1 s*10 = 11 dm 3
-1 1 v s
1+2.4 s τ 1+ o
2
344/W’00 ExamII Open Book
, 20
18
16
14
k 12
(s
10
^-
1) 8
6
4
2
0
300 320 340 360 380 400 420 440 460 480 500
Temperature
Answer: V = ___11___________dm3
344/W’00 ExamII Open Book
→ B
A←
The entering flow rate is 10 dm3 /s with an entering concentration of 2 M of A and the
feed temperature is 300 K. What is the reactor volume necessary to achieve 90% of the
equilibrium conversion in a CSTR operated adiabatically?
Additional information:
CpA = CpB = 60 cal/mol/K
∆H°rxn = -10,000 cal/mol A
Solution:
To begin, you need to use the conversion obtained from the energy balance
cal
C pA (T − To ) 60 mol K ( T − 300 K)
X EB = = = X EB
−∆H˚rxn 10000
cal
mol A
Now, you can plot this line on the X eq versus T curve (see below)
Where the lines intersect is the operating temperature, T, (~ 398 K)
and equilibrium conversion, X eq , (~0.59)
You want 90% of this value (as stated in the problem)
X = 0.90*X eq = 0.90*0.59 = 0.53 = X
Now you can use this value in the design equation for a
first-order, reversible reaction using a CSTR
kτ
X= 1
1+kτ 1+
KC
To get values of k and KC at 398 K, use the plots given (see below)
k ~ 2.4 s-1
KC ~ 2
Plug in values and solving for space time, τ
2.4 s-1τ V dm 3
0.53 = ⇒ τ =1.1 s = ⇒ V= 1.1 s*10 = 11 dm 3
-1 1 v s
1+2.4 s τ 1+ o
2
344/W’00 ExamII Open Book
, 20
18
16
14
k 12
(s
10
^-
1) 8
6
4
2
0
300 320 340 360 380 400 420 440 460 480 500
Temperature
Answer: V = ___11___________dm3
344/W’00 ExamII Open Book
