The rate of disappearance of A, -rA, is almost always a function of the con- centrations of the various species present. When a single reaction is occurring,
Sec, 2.3 Applications of the Design Equations for. Continuous-Flow Reactors 41 each of the concentrations can be expressed as a function of the conversion X (see Chapter 3); consequently, - r A can be expressed as a function of X .
Ii particularly simple functional dependence, yet one that occurs on many occasions, is - r A = kCAo(1 - X ) . For this dependence, a plot of the reciprocal rate of reaction ( - l / r A ) as a function of conversion yields a curve
similar to the one shown in Figure 2- 1 I where
To illustrate the design of a series of reactors, we consider the isothermal gas-phase decomposition reaction
A _ _ j B + C
The laboratory measurements given in Table 2-1 show the chemical reaction rate as a function of conversion. The temperature was 300°F (422.2 K), the total pressure 10 atm (1013 kPa), and the initial charge an equimolar mixture of A and inerts.
‘TABLE 2-1 RAW DATA
X - r A (mol/dm3
.
s)If we know -rA as a function of X, we can size any isothermal reaction system.
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85
0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.00100
‘The rate data in Table 2- 1 have been converted to reciprocal rates, 1 I - rA in Table 2-2, which are now used to arrive at the desired plot of l l - r A as a function of X , shown in Figure 2-1. We will use this figure to illustrate how one can size each of the reactors in a number of different reactor sequences.
The volumetric feed to each reactor sequence will be 6.0 dm3/s. First, though, some initial conditions should be evaluated. If a reaction is carried out isother- mally, the rate is usually greatest at the start of the reaction when the concen- tration of reactant is greatest [i.e., when there is negligible conversion (X zz
O)].
Hence (1 / - r A ) will be small. Near the end of the reaction, when the reactant concentration is small (i.e., the conversion is large), the reaction rate will be small. Consequently, (l/-rA) is large. For irreversible reactions of greater than zero-order,42 Conversion and Reactor Sizing Chap. 2
X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85
-rA l / - r A
0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001
189 192 200 222 250 303 400 556 800 1000
0 0 . 2 0.4 0.6 0.8 1.0 Conversion, X
Figure 2-1 Processed Rata.
1000
eo0
600
400
200
A + B + C - _ ' - - + m as X - + I
rA
-
-
-
-
For rmersible reactions in which the equilibrium conversion is X,, 1
rA
A
e
B + C - - - + m as X + X ,These characteristics are illustrated in Figure 2-1. The majority of reactions exhibit qualitatively similar curves for isothermal operation.
Example 2-2 Sizing a CSTR
(a) Using the data in either Table 2-2 or Figure 2- 1 , calculate the volume necessary to achieve 80% conversion in a CSTR. (b) Shade the area in Figure 2-1 which when multiplied by F,,,,, would give the volume of a CSTR necessary to achieve 80% con- version (Le., X = 0.8).
Solution
From Example 2-1, knowing the entering conditions u,, = 6 dm3/s, Po = 10 atm, yAO = 0.5, To = 422.2 K, we can use the ideal gas law to calculate the entering molar flow rate of A, Le.,
Design
&
equationPlots of ll-rA vs. X are sometimes
referred to as Levenspiel plots (after Octave L.evenspie1)
43
Sec. 2.3 Applications of the Design Equations for Continuous-Flow Reactors
(a) Equation (2-13) gives the volume of a CSTR as a function of FAo, X, and - - r A : (2.-13) In a CSTR, the composition, temperature, and conversion of the effluent stream are identical to that of the fluid within the reactor, since perfect mixing i’s assumed.
Therefore, we need to find the value of -rA (or reciprocal thereof) at X = 0.8. From either Table 2-2 or Figure 2-1 we see that when X = 0.8, then
dm3. s I / - r A = 800 -
mol Substitution into Equation (2- 13) gives
V = 0.867 -
S (E2-.2.1)
= 554.9 dm3 = 554.9 L
(b) Shade the area in Figure 2-1 which when multiplied by FA, yields the CSTR volume. Rearranging Equation (2- 13) gives
(2-13)
(E2-:2.2)
Conversion, X Figure E2-2.1 Levenspiel CSTR plot.
In Figure E2-2.1 the value of VIF,, is equal to the area of a rectangle with a heLght l l - r A = 800 d1n3.s/mol and a base X = 0.8. This rectangle is shaded in the figure.
To calculate the reactor volume, we multiply the area of the rectangle by FAO.
44 Conversion and Reactor Sizing Chap. 2
dm.s
V = 0.867 S
[
800mol
(OX)] = 554.9 dm3The CSTR volume necessary to achieve 80% conversion at the specified temper&- ture and pressure is 555 dm3.
Example 2-3 Sizing a PFR
The reaction described by the data in Tables 2-1 and 2-2 is to be carried out in a PFR. The entering molar flow rate of A is 0.867 mol/s. Calculate the reactor volume necessary to achieve 80% conversion in a PFR (a) First, use one of the integration formulas given in Appendix A.4 to determine the PFR reactor volume. (b) Next, shade the area in Figure 2-1 which when multiplied by FA0 would give the PFR vol- ume. (c) Make a qualitative sketch of the conversion, X , and the rate of reaction, -rA, down the length (volume) of the reactor.
Solution
(a) For the PFR, the differential form of the mole balance is
Rearranging and integrating gives
(2-15)
(2-16) For 80% conversion, we will use the five-point quadratic formula with AX = 0.2.
f-+- 4
+-+-
-rA(X = 0 ) -rA(0.2) -rA(0.4) -rA(0.6) -rA(0.8)
4 2
1
Using values of 1 / - r, in Table 2-2 yields
s.dm3 V = (0.867 mol/s)(0.2/3)[189
+
4(200)+
Z(250)+
4(400)+
(800)Jmol
= (0.867 mol/s)(259.3 s dm3/m01)
= 225 dm3
(b) The integral in Equation (2-16) can be evaluated for the area under the curve of a plot of ( l / - r A ) versus X .
-!!-
=F*O
%
= area under the curve between X = 0 and X = 0.8 (see appropriate shaded area in Figure E2-3.1)Sec. 2 3 Applications of the Design Equations for Continuous-Flow Reactors 45
0.2 0.4 0.6 0.8
Conversion, X Figure E2-3.1 Levenspiel PFR plot.
The: product of this area and FAO will give the tubular reactor volume necessary to achieve the specified conversion of A. For 80% conversion, the shaded area is roughly equal to 260 dm3*(s/mol). The tubular reactor volume can be deterimined by multiplying this area [in dm3.(s/mol)] by FA,, (mol/s). Consequently, for an entering molar flow rate of 0.867 mol/s the PFR volume necessary to achieve 80%
conversion is 225 dm3.
(c) Sketch -rA and X down the length of the reactor. We know that as we proceed down the reactor and more and more of the reactant i s consumed, the concentration of reactant decreases, as does the rate of disappearance of A. However, the conver- sion increases as more and more reactant is converted to product. For X = 0.2 we cak!ulate the corresponding reactor volume using Simpson’s rule with AX = 0.1.
mol * s
1
dm3[I89
+
4(192)+
2001 ~L
= 33.4 d m 3
For X = 0.4, we can again use Simpson’s rule with AX = 0.2:
v
= FAO -“I
3 L -rA(x = 0)+
- T A ( X 4= 0.2) +-rA(X = 0.4)0.2
{ 189+
4(200)+
2.501= 71.6 d m 3
46 Conversion and Reactor Sizing Chap. 2
V(dm3) X
We can continue in this manner to arrive at Table E2-3.1.
TABLE E2-3.1, CONVERSION PROFILE
0 33.4 71.6 126
0 0.2 0.4 0.6
- r A ( s ) 0.0053 0.005 0.004 0.0025
which is shown in Figure E2-3.2.
1
X
0
0 V (d;) 250
Figure E2-3.2 Conversion profile.
225 0.8 0.00125
Rather than using Simpson’s rule we could have used the data in Table 2-2 to fit - r A ( X ) to a polynomial and then used POLYMATH to integrate the design equa- tion to obtain the conversion profile.
Example 2-4 Comparing CSTR and PFR Sizes
It is interesting to compare the volumes of a CSTR and a plug-flow reactor (PFR) required for the same job. To do this we shall use the data in Figure 2-1 to learn which reactor would require the smaller volume to achieve a conversion of 60%: a CSTR or a PFR. The feed conditions are the same in both cases. The entering molar flow rate is 5 mol/s.
Solution For the CSTR:
dm3. s X = (400) ( 0 . 6 ) ~ 240 -
V
=
[+)
molGenerally, the isothermal tubular reactor volume is smaller than the CSTR for the same conversion
Sec. 2.3 Applications of the Design Equations for Contfnuous-Flow Reactors 47
Tlhis is also the area of the rectangle with vertices (X, l/-rA) of (0, 0), (0, 400), (0.6, 400), and (0.6,O). The CSTR volume necessary to achieve 60% conversion is
5 mol 240 dm3.s = 1200 dm3
v=(-)(
mol1
For the plug-flow (tubular) reactor:
F - = - r dX
A o d V A (2-13
Integrating and rearranging Equation (2-1 5) yields
0.3
.
= X [ 189
+
4(222)+
4001 dm3.s= 148
-
molThe PFR volume necessary to achieve 60% conversion is V =
pF1) [
148 dm3 s = 740 dm3For the same flow rate FA,o the plug-flow reactor requires a smaller volume than the CSTR to achieve a conversion of 60%. This comparison can be seen in Figure E2-4.1. For isothermal reactions of greater than zero order, the PFR will always rquire a srrtaller volume than the CSTR to achieve the same conversion.
6oa 1 -r A
-
i
" 0 0.2 a4 a6 0.8