1. Design equation
4.3 Tubular Reactors
Gas-phase reactions are carried out primarily in tubular reactors where the flow is generally turbulent. By assuming that there is no dispersion and there are no radial gradients in either temperature, velocity, or concentration, we can model the flow in the reactor as plug-flow. Laminar reactors are discussed in Chapter 13 artd dispersion effects in Chapter 14. The differential form of the dlesign equation
PFR mole
dX
-must be used when there is a pressure drop in the reactor or heat exchange between the PFR and the surroundings. In the absence of pressure drop or heat exchange the integral form of the plug JEow design equation is used,
balance FA0
-
- - r AdW
PFR design equation
Substituting the rate law for the special case of a second-order reaction gives us
Rate law dX
For constant-temperature and constant-pressure gas-phase reactions, the con- centration is expressed as a function of conversion:
Stoichiometry (gas-phase)
Combine
and then substituted into the design equation:
The entering concentration CAO can be taken outside the integral sign since it is not a function of conversion. Since the reaction is carried out isothermally, the specific reaction rate constant, k, can also be taken outside the integral sign.
For an isothermal reaction, k i r
constant
From the integral equations in Appendix A. 1, we find that
Reactor volume for a second-order gas-phase reaction
1 48 Isothermal Reactor Design Chap. 4
If we &vide both sides of Equation (4-17) by the cross-sectional area of the reactor, A , , we obtain the following equation relating reactor length to con- version:
A plot of conversion along the length of the reactor is shown for four dif- ferent reactions and values of E in Figure 4-7 to illustrate the effects of volume
The importance of changes in volumetric flow rate (i.e., E ;ti 0 ) with reaction
L . 1
I I I f ’ 1 1 ” ’ ” ’o 1.0 2.0 3.0 4.0 5.0 6.0 7.0 a0 90 10.0 11.0 12.0 13.0 14.0
L(m)
Figure 4-7 Conversion as a function of distance down the reactor.
change on reaction parameters. The following typical parameter values were chosen to arrive at these curves:
k = 5.0 dm3/11101* s A, = 1 .O dm2 C,, = 0.2 moUdm3 u0 =‘1 dm3/s
1
( 1
+
&)2X2 ~ ( 1 + c ) l n ( l - X ) + ~ ~ X + 1 - x meters
We observe from this figure that for identical rate-law parameters, the reac- tion that has a decrease in the total number of moles (i.e., c = -0.5) pill have the highest conversion for a fixed reactor length. This relationship should be expected for fixed temperature and pressure because the volumetric flow rate,
u = (1 - 0 . 5 X ) ~ ~
Look at the poor design that could result
Sec. 4.3 Tubular Reactors 1 49
decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles (e.g., .A
+
B and E = 0). Similarly, reactants that produce an increase in the total rmmber of moles upon reaction (e.g., E = 2 ) will spend less time iin the reactor than reactants of reactions for which E is zero or negative.Example 4-3 Neglecting Volume Change with Reaction
The gas-phase cracking reaction
A __$ 2B-I-C
is to be carried out in a tubular reactor. The reaction is second-order and the param- eter values are the same as those used to construct Figure 4-7. If 60% conversilon is desired, what error will result if volume change is neglected (E = 0 ) in sizing the reactor?
Solution
In Fiigure E4-3.1 (taken from Figure 4-3, we see that a reactor length of 1.5 m is requiired to achieve 60% conversion for E = 0. However, by correctly accounting for voluine change [ E
=
(1)(2+
1 - 1) = 23, we see that a reactor length of 5.0 m would be required. If we had used the 1.5-m-long reactor, we would have achieved only 40% conversion.X
L(m) Figure E4-3.1
Example 4-4 Producing 300 Million Pounds per Year of Ethylene in a Plug-Flow Reactor: Design of a Full-scale Tubular Reactor
Ethylene ranks fourth in the United States in total pounds of chemicals produced each year and it is the number one organic chemical produced each year. Over 35 billion pounds were produced in 1997 and sold for $0.25 per pound. Sixty-five per- cent of the ethylene produced i s used in the manufacture of fabricated plastics, 20%
for ethylene oxide and ethylene glycol, 5% for fibers, and 5% for solvents.
.
150 Isothermal Reactor Design Chap. 4 The uses Determine the plug-flow reactor volume necessary to produce 300 million pounds of ethylene a year from cracking a feed stream of pure ethane. The reac- tion is irreversible and follows an elementary rate law. We want to achieve 80%
conversion of ethane, operating the reactor isothermally at 1100 K at a pressure of 6 atm.
Solution
CZH, C,Hq+H,
Let A = C2Hs, B = C,H,, and C = H2. In symbols, A
--+
B + C The molar flow rate of ethylene exiting the reactor is6 lb 1 year 1 day 1 h Ibmol year 365 days 24h 3600s
x' ?%%
FB=30OX10 -X-X-X-
lb mol
= 0.340
-
S
Next calculate the molar feed rate of ethane. To produce 0.34 lb mol/s of eth- ylene when 80% conversion is achieved,
lb mol
0'34 - 0.425 -
FAO =
0.x
- S1. Plug-flow design equation:
Rearranging and integrating for the case of no pressure drop and isothermal operation yields
(E4-4.1) 2. Ratelaw:'
- r A = kCA with k = 0.072 s-l at 1000 K (E4-4.2) The activation energy is 82 kcaVg mol.
concentration of ethane is calculated as follows:
3. Stoichiometry. For isothermal operation and negligible pressure drop, the
'Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975); I d . Eng. Chem., 59(5), 70 (1967).
Combining the design equation, rate law, and stoichiometry
Analytical solution
Sec. 4.3 Tubular Reactors
Gas phase, constant T and P:
u = u 0 - = F T u o ( l + E X ) :
FTO
1-x c x
c -AO
- (1 +EX)
4. We now combine Equations (E4-4.1) through (E4-4.3) to obtain
dX =
X
0 kCAO(1-X)
=
kCA0 (1 -
x)/
(1+
EX) (1 +EX)dX151
(EI-4.3)
(Et-4.4)
(E41-4.5)
Since the reaction is carried out isothermally, we can take k outside the inte- gral sign and use Appendix A.l to cany out our integration.