CH5010
CRT.
Total Marks: 25
Quiz I
01.Sep.2014
1. (3 marks) The following data are available for a reaction. Determine the activation
energy and the pre exponent for the rate constant, assuming Arrhenius temperature
dependency.
Temp (K)
Rate Const (s-1)
500
8 x 10-3
400
5 x 10-5
2. (5 marks) A liquid phase first order reaction A B is carried out in a batch reactor
of 200 lit size, at 75˚C. The rate constant at the operating conditions is 0.003 min-1.
Initially, the reactor is charged with 100 moles of pure A. Determine the time
necessary to obtain a conversion of 80%.
3. (5 marks) A first order liquid phase reaction A B is to be conducted in an ideal
CSTR under isothermal conditions. At the operating temperature, the reaction rate
constant is 0.1 min-1. If pure A is fed to the reactor at a concentration of 1 mol/lit, and
at a flow rate of 10 lit/min, and if the conversion desired is 80%, determine the
volume of reactor needed.
4. (5 marks) A gas phase reaction A B + C is carried out isothermally in a PFR,
with negligible pressure drop. The following data are available. Incoming total molar
flow rate is 100 mol/min, and the volumetric flow rate = 25 lit/min. The reaction is
first order with respect to A. The rate constant is k = 0.05 min-1. Mole percentage of A
in the feed is 50% and the rest are inert gases. The desired conversion is 50%. Under
steady state conditions, determine (i) the parameter epsilon (ε) and (ii) the volumetric
flow rate at the outlet. (You don’t need to calculate the volume of the reactor).
5. (7 marks) Consider the liquid phase reaction A B occurring at constant
temperature in a recycle reactor. The fresh feed contains pure A at a concentration of
5 mol/lit, and is coming at a flow rate of 10 lit/min. The reaction is first order with a
rate constant of 0.05 min-1. The conversion required is 50%. Determine the volume of
the reactor needed, when the recycle ratio R= 1.
0
1
2
R
3
Answers:
−E
1. Using Arrhenius expression k = k0 e RT , with R = 8.314 J mole-1 K-1, we get E =
84,390 J mole-1 and k0 = 5.24 × 106 s-1.
2. V = 200 lit, k = 0.003 min-1, NA0 = 100 mol, x = 0.8 and T = 75 C. For a batch
reactor, with first order kinetics, N A = N A0 (1 − x ) = N A0e − kt . Therefore, 1 − x = e− kt or
t=
− ln (1 − x ) − ln (1 − 0.8 )
=
≈536.5 min ≈ 9 h.
k
0.003
3. k = 0.1 min-1, CA0 = 1 mol/lit, Q = 10 lit/min, FA0 = 10 mol/min, x = 0.8, V = ?. The
design equation gives FA 0 + V ( rA out ) = FA out = FA0 (1 − x ) . i.e.
V=
FA0 x
F x
FA0 x
10 × 0.8
= A0 =
=
= 400 lit
− rA out k C A out kC A0 (1 − x ) 0.1×1× 0.2
4. FT-in = 100 mol/min, Qin = 25 lit/min, k = 0.05 min-1, x = 0.5
Stoichiometric table
Species
A (FA)
B (FB)
C (FC)
Inerts (FI)
Total (FT)
Inlet (x=0)
0.5
0
0
0.5
1
Outlet (x=1)
0
0.5
0.5
0.5
1.5
1.5 − 1
= 0.5
1
Volumetric flow rate at the outlet = Qin (1+ε x) = 25 × (1+0.5 ×0.5) =31.25 lit/min
Parameter ε =
5. CA,0 = 5 mol/lit, Q = 10 lit/min, k = 0.05 min-1, Xo = 0.5, R = 1.
R
1
X o = × 0.5 = 0.25 and FA,0 = 5 × 10 = 50 mol/min
R +1
2
X0
V
dX
The design equation is
= ∫
− rA
( R + 1) FA0
R
Therefore,
R +1
0.5
0.5
X0
V
dX
1
dX
−1  1 − 0.5 
1
 0.75 
= ∫
=
=
ln 
ln 
=

∫
2 × 50 0.25 kC A0 (1 − x ) 0.05 × 5 0.25 (1 − x ) 0.25  1 − 0.25  0.25  0.5 
 0.75 
V = 400 × ln 
 ≈ 162 lit
 0.5 

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