IDEAL REACTORS (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471: LECTURE 4
Fall 2003
IDEAL REACTORS
One of the key goals of chemical reaction engineering is to quantify the relationship between production rate, reactor size, reaction kinetics and selected operating conditions. This requires a mathematical model of the system, which in turn rests on application of conservation laws to a well-defined control volume of the reaction system and on use of appropriate constitutive expressions for the reaction rates. The concepts of ideal reactors allow us to quantify reactor reactor performance as a function of its size and selected operating conditions.
To illustrate this useful concept we deal here with a single, homogeneous phase, single reaction at constant temperature. We introduce then the the ideal batch reactor, reactor, and two ideal ideal continuous flow reactors. In each case we apply the conservation of species mass principle which states (Rate of Accumulation) = (Rate of Input) – (Rate of Output) + (Rate of Generation)
(2-1)
Equation (2-1) is applied to an appropriately selected control volume, the largest arbitrarily selected volume of the system in which there are no gradients in composition.
2.1
Batch Reactor
The ideal batch reactor is assumed to be perfectly perfectly mixed. This implies that at a given moment in time the concentration is uniform throughout the vessel. The volume, V in the development below is assumed equal to the volume of o f the reaction mixture. This is then equal to the reactor volume VR in case of gas phase reaction but not in case of liquids (V< VR , then). The batch reactor can be an autoclave of V = const (Figure 2.1-a) and a constant pressure, P = const) (Figure 2.1-b) vessel. The former is almost almost always encountered in practice. Our goal is: a)
To find a relationship between species concentration (reactant conversion) and time on stream.
b)
To relate reactor size and production rate.
1
ChE 471: LECTURE 4
Fall 2003
IDEAL REACTORS
One of the key goals of chemical reaction engineering is to quantify the relationship between production rate, reactor size, reaction kinetics and selected operating conditions. This requires a mathematical model of the system, which in turn rests on application of conservation laws to a well-defined control volume of the reaction system and on use of appropriate constitutive expressions for the reaction rates. The concepts of ideal reactors allow us to quantify reactor reactor performance as a function of its size and selected operating conditions.
To illustrate this useful concept we deal here with a single, homogeneous phase, single reaction at constant temperature. We introduce then the the ideal batch reactor, reactor, and two ideal ideal continuous flow reactors. In each case we apply the conservation of species mass principle which states (Rate of Accumulation) = (Rate of Input) – (Rate of Output) + (Rate of Generation)
(2-1)
Equation (2-1) is applied to an appropriately selected control volume, the largest arbitrarily selected volume of the system in which there are no gradients in composition.
2.1
Batch Reactor
The ideal batch reactor is assumed to be perfectly perfectly mixed. This implies that at a given moment in time the concentration is uniform throughout the vessel. The volume, V in the development below is assumed equal to the volume of o f the reaction mixture. This is then equal to the reactor volume VR in case of gas phase reaction but not in case of liquids (V< VR , then). The batch reactor can be an autoclave of V = const (Figure 2.1-a) and a constant pressure, P = const) (Figure 2.1-b) vessel. The former is almost almost always encountered in practice. Our goal is: a)
To find a relationship between species concentration (reactant conversion) and time on stream.
b)
To relate reactor size and production rate.
1
Let us consider a single irreversible reaction A → P P with an n-th order irreversible rate of reaction
− R = kC n
(2-2)
At t = 0 a batch of volume V is filled with fluid of concentration CAo. Reaction is started (nAo= CAoVo). Find how reactant conversion depends on reaction reaction time? Also determine determine the production production rate as a function of reaction time.
We apply (eq 2-1) to reactant A: 0 − 0 + ( R A V ) =
dn A dt
=
d ( d (VC A ) dt
(2-3)
a) V = const
b) P = const
FIGURE 2-1: Schematic of Batch Reactors 2
In our case due to the fact that
∑ j=1
υ
j
= 0 , V= const irrespective of the batch reactor type, so that
eq (2-3) becomes dC A dt
−
= R A
dC A dt
(2-4)
= (− R A ) = kC An ;
t = 0
C A = C Ao
(2-5)
Separation of variables and integration yields: t
C A
dC A
∫ dt = − ∫ kC o
C A A0
n A
=
1
C Ao
dC ∫ k C C A A
A n A
(2-6)
2
t
t o =
1 C A1−n k 1 − n
t − 0 =
C Ao
C A
1
[C k (1 − n)
1−n Ao
− C A1 −n ]
1−n
t =
(2-7)
[1 − (1 − x k (1 − n) C Ao
1−n
A
)
[(1 − x
A
]
(2-8a)
or t =
1 n −1 Ao
k (n −1)C
1− n
)
−1]
(2-8b)
Once order of reaction, n, is specified (as shown below for n=0, 1, 2, 1.5), the relation between t and xA is readily found n=0 C Ao x A
t =
t =
k
n = 1.5;
t =
n =1 1 1 ln k 1 − x A 1
0.5k C Ao
n=2 1 1
t =
kC Ao 1 − x A
− 1
(2-9)
1 − 1 0.5 (1− x A )
Production Rate of Product P can be related by stoichiometry to he consumption rate of A as F P
mol F Ao S = 1
A
The production rate of P is given by: F P = F P =
(moles of P processed per batch) (reaction time + shut down time per batch) C AoV t + t s
A
=
NOTA BENE:
C AoV x A 1 n−1 k (n − 1)C Ao
[(1 − x
A
1 −n
)
] + t
(2-10)
(2-11)
s
Equation (2-11) is valid only for systems of constant density. Thus, it is valid for all systems, gas or liquid, conducted in an autoclave at V = const (see Figure 2-1a). It is also valid for gaseous systems with no change in the
3
number of moles
(∑
ν j
= 0) conducted in P = const. system at T = const
(Figure 2-1b). The first equality in equation (2-11) gives the general result, the second equality presents the result for an n-th order irreversible reaction with resect to reactant A.
To use this equation the shut down time, i.e. the time needed between batches, ts, must be known. Consider now the following second order reaction with stoichiometry A = P. mol − R A = 0.1C A2 L min
a)
Find the batch reactor volume needed to produce FP = 38 kmol/min if reactor shut down time is 60 minutes and the desired conversion is 0.95. Initial reactant concentration is CAo = 1 (mol L).
Using the right form of equation (2-9) for n = 2 we get the reaction time. t =
1 1 − 1 = = 190.0(min) 0.1∗ 1 1− 0.95 1 − 0.95 1
Then, solving equation (2-11) for the volume we get V =
b)
F P (t + t s ) C Ao x A
=
1.38(190 + 60) 1 x0.95
= 10,000 L = 10m 3
What is the maximum production rate, FP, achievable in the above batch reactor of volume V=10m3 if ts, T, CAo all are fixed at previous values.
Consider eq (2-11) for production rate as a function of conversion F P =
4
C AoV x A
x A + t s kC Ao 1 − x A 1
F P = 10
3
x A (1 − x A ) x A
+ 6(1 − x A
= 10
10 x A x A 1 − x A
+ 60
3
=
10 x A x A 1 − x A
+6
x A − x A 2
= 10
3
6 − 5 x A
4
This expression has a maximum which we can locate by differentiation dF P dx A
= 0 ⇒ (1 − 2 x A )(6 − 5 x A ) + 5 ( x A − x A2 ) = 0
6 − 5 x A − 12 x A
+ 10 x A2 + 5 x A − 5 x A2 = 0 2 6 − 12 x A + 5 x A = 0 6 ± 36 − 30 6 − 6 x A = = = 0.710 1, 2
5
5
Clearly, the positive sign is not permissible as conversion cannot exceed unity. We need to check whether the answer is a maximum or a minimum. dF P dx A dF P dx A
> 0 for x A < 0.710 < 0 for x A > 0.710
Maximum at xA = 0.710. F P max
= 10
3
0.710 − 0.710 2 6 − 5 x0.710
= 84.0
An increase in productivity of
mol min 84 − 38 x100 = 121% can be achieved at the expense of more 38
unreacted A to be recycled.
One must include the cost of separation into the real economic optimization.
2.2
Continuous Flow Reactors (Steady State)
2.2.1 Continuous Flow Stirred Tank Reactor (CFSTR or CSTR or STR) The CSTR is assumed perfectly mixed, which implies that there are no spatial gradients of composition throughout the reactor. Since the reactor operates at steady state, this implies that a single value of species concentration is found in each point of the reactor at all times and this is
5
equal to the value in the outflow. The outflow stream is a true representative of the reaction mixture in the reactor.
(
F A = F Ao 1 − x A
F AO
)
C A
C AO
FIGURE 2-2: Schematic of a Continuous Flow Stirred Tank Reactor (CSTR)
What does the above idealization of the mixing pattern in a CSTR imply? It postulates that the rate of mixing is “instantaneous” so that the feed looses its identify instantly and all the reaction mixture is at the composition of the outlet. Practically this implies that the rate of mixing from macroscopic level down to a molecular scale is orders of magnitude faster than the reaction rate and is so fast in every point of the vessel.
Then the mass balance of eq (2-1) can be applied to the whole volume of the reactor recognizing that at steady state the accumulation term is identically zero. Again, taking a simple example of an irreversible reaction A
→ P application of eq (2-1) to reactant A yields:
FAo – FA + ((R A)V=0
(2-12)
Molar flow rate of unreacted A in the outflow by definition is given by FA = F Ao (1-xA) = Q CAo (1-xA). The production rate of P is given by F P = (− R )V = ( R P )V
(2-13)
Reactor volume is given by eq (2-12) V =
F Ao x A (− R A )
=
Q oC Ao (− R A )
A
(2-14)
6
Reactor space time is defined by τ
=
V Qo
=
C Ao x A (− R A )
(2-15)
Using stoichiometry we readily develop the relation between production rate, FP, and reactor volume, V . nd
Let us consider again the example of our 2 order reaction, A = P, with the rate below:
− R A = kC An = 0.1C A2
mol
L min
Find CSTR volume needed to process FP = 38 mol/min. Suppose we choose again xA = 0.95 for our exit conversion.
From eq (2-13) we get 2
2
FP = 0.1 C Ao (1-xA) V
And solving for volume V V =
F P 0.1C (1 − x A ) 2 Ao
2
=
38 0.1 x1(1 − 0.95)
2
= 152,000 L = 152m 3
If we consider eq (2-13) it is clear that now the maximum production rate is obtained when the reaction rate is the highest. That for n-th order reactions is at zero conversion. So the maximum FP from VCSTR = 12,000 L is obtainable at xA = 0.
F P max =0.1 x 1 x 152,000 = 15,200 mol/min. The penalty or this enormous production rate is that the product is at “zero” purity. Hence, the separation costs would be enormous. The average rate in a CSTR is equal to the rate at exit conditions. (− R A ) = (− R A ) exit
2 = 0.1C Ao (1 − x A )2exit = 0.1 x1(1 − 0.95) 2 = 2.5 x10 −4 (mol / L min)
7
2.2.2 Plug Flow Reactor (PFR) The main assumptions of the plug flow reactor are:
i) perfect instantaneous mixing
perpendicular to flow, ii) no mixing in direction of flow
This implies piston like flow with the reaction rate an d concentration that vary along reactor C AO
C AO
F AO
F Ao dx A =
(
F A = F Ao 1 − x A
(− R )d A
FIGURE 2-3: Schematic of a Plug Flow Reactor (PFR)
Since there are now composition gradients in the direction of flow, the control volume is a differential volume
∆ V to
which eq (2-1) is applied. Let us again use the mass balance on
reactant A
F A
V
− F A
V +∆ V
+ R A ∆V = 0
(2-16)
−∆ F + R ∆ V = 0
∆ F A = (lim )(− R A ) ∆V →0 ∆ V ∆V →0
− lim −
dF A dV
= (− R A )
Since F
(2-17)
= F o (1 − x A ) then dF = − F odx A
so that F Ao
dx A dV
= (− R A )
(2-18)
With initial conditions: V = 0
x A
=0
(2-19)
8
Upon separation of variables in (eq 2-18) and integration: V
∫
x A
dV = F Ao
o
dx A
∫ ( − R
A
o
)
(2-20)
For an n-th order reaction (with ε A V =
F Ao
x A
dx A
∫ (1− x
n Ao o
kC
A )
n
=
= 0) we get
[(1 − x
Qo
A
n−1 Ao
(n − 1)
kC
]
)1−n − 1
(2-21)
The expression for the PFR space time τ
=
V Qo
=
1 n −1 Ao
kC
[(1− x (n − 1)
A
1− n
)
− 1]
(2-22)
is now identical to the expression for reaction time t in the batch reactor. For the example of the second order reaction used earlier we get
x A
1 V = F Ao ∫ 2 2 = 2 kC Ao (1 − x A ) kC Ao 1 − x A o o x A
V =
F Ao 2 Ao
kC
dx A
F Ao
1 − 1 = 1 − x A
F Ao 2 Ao
kC
x A 1 − x A
FAo = Qo CAo τ
=
V Qo
=
1
x A
(Same as the expression for reaction time t in the batch reactor)
kC Ao 1 − x A
Let us consider our example of the second order reaction and find the PFR volume needed to produce FP = 38 mol/min (− R A ) = 0.1C A 2
mol
L min
when C Ao = 1
mol L
and desired conversion xA = 0.95.
From stoichiometry it follows that F Ao x A = F P
F Ao
=
F P x A
9
Substitution in the expression for reactor volume (eq (2-21)) we get: V = V =
x A = 2 kC Ao x A 1 − x A F P
38 0.1 x1(1 − 0.95)
F P kC Ao (1 − x A ) 2
= 7,600 L = 7.6m3
The maximum production rate from that volume can be obtained at zero conversion F P = kC Ao (1 − x A )V 2
F P max
= 0.1 x1 x7600 = 760mol /min
Average rate in PFR − R A (− R A ) entrance
=
F Ao
A
V
2 = 0.1C Ao = 0.1 = 10 −1
(− R A ) exit = 0.1C Ao (1 − 0.95) 2
2
=
F P V
=
38 7,600
= 5.0 x10 −3
mol min
mol
L min
mol = 2.5 x10 −4 L min)
Clearly there is a big variation in the reaction rate between the entrance and exit of the plug flow reactor (PFR).
2.3 STY – Space Time Yield Volumetric Reactor Productivity - RVP
Reactor volumetric productivity (RVP) is defined by: R P
=
F P V
(2-23)
nd
For our 2 order reaction example of stoichiometry A=P, RVP for the two continuous flow reactors is:
CSTR R P PFR R P
2 2 = ( R P )exit = (− R A ) exit = kC Ao (1 − x A )
=
F P V
= kC Ao2 (1 − x A )
(2-24a) (2-24b)
10
For the same exit conversion kC Ao (1 − x A ) 2
( R P ) PFR
=
( R P ) CSTR
2
kC Ao
=
1 (1 − x A )
At xA = 0.95 ( R P ) PFR
= 20
( R P ) CSTR
Indeed 20 x 7,600 L = 152,000 L This is why higher CSTR volume is needed. At xA = 0
( R P ) PFR
= (R P )CSTR
There is no difference! Let us consider another example to illustrate some important points.
Ex:
2A + 3B = P + S – stoichiometry
mol 2 - rate of reaction r = 0.1C A C B L min C Ao = 2
mol
L
and
x A
= 0.95 - are the feed reactant concentration and desired conversion
FP = 10 mol/min is the desired production rate
Assume first that we will operate at stoichiometric ratio so that C Bo = 3 (mol/L). The reaction occurs in the liquid phase so that a)
ε A
= 0. Find the needed reactor volume.
Batch (ts = 60 min)
− R A = 0.2C Ao (1 − x A )(C Bo − 3 3 (1 − x A ) − R A = 0.2C Ao − R A = 0.2C
(1 − x A )
3
a
C Ao x A )2
2
2
3 Ao
b
3 2
(1 − x A )
2
2
11
Reaction time is: t =
2 Ao
0.2C t = t =
x A
1
∫ (1−dx x
A
3 2
1
g 2
0.2 x2 a
2
o
x A
dx
∫ (1 − x)
3
A
=
o
)3
1 1
x A
1
1.8 2 (1 − x )
2 o
1 1 1 1 1 − = − 3.6 (1− x A ) 2 36 (1− 0.95) 3 1
t = 110.83 min 1 F P = V= b)
2
C Ao x A V t + t s
170.83 10 0.95
=
0.95 xV 110.83 + 60
= 10
= 1,798 L = 1.8m3
CSTR F Ao 2 V F Ao V = V =
V =
= =
F P
- from stoichiometry
1
A
− R A
F Ao
A
− R A
- basic design equation (2-14)
=
2 F P ( − R A )
=
2 F P 2
0.2C A C B
F P
C 3 3 0.1C Ao (1 − x A ) Bo − x A C Ao 2 10 0.1 × 2 × 9(1 − 0.95)
3 =
10
2
F P
= 3 Ao
0.1C 1
1.8 0.05
3
3 2
(1 − x A )
3
= 4,444( L ) = 44.4m 3
3 2 3 (− R A ) = (− R A ) exit = 0.2 xC (1 − x A ) exit 2 3 Ao
3
2
-4
= 0.2x2x9(1-0.95) = 4.5 x10 (mol/L min)
12
c)
PFR F Ao
2 F P
=
- from stoichiometry
x A
Basic design equation (2-21) V = F Ao
x A
dx A
= F Ao ∫
o
A
o
2 F P
x A
∫ − R
V = 0.2C
=
3 x 2 A
3 Ao
0.2C
3 (1 − x) 2
dx
=
3
o
1
100
dx
∫ (1 − x )
2
3 Ao
V =
x A
1
18 x0.95 2 (1 − x A )
2
100
2
(1− x )2 x A
dx ∫ 18 x 0.95 (1 − x )
3
o
−1
1 3 1 1,167( L ) 1.17 m − = = 2 10 × 0.95 (1− 0.95) 50
Now the rate, at stoichiometric feed ratio, along the PFR as a function of conversion is
− R A = 0.2C
3 Ao
3 2
2
(1− x A )
3
= 3.6(1− x A ) 3
PFR reactor volume as function of conversion at stoichiometric feed ratio is V =
2 F P 3.6 x A
x A
dx
∫ (1 − x)
3
o
=
x A
F P 1.8 x A
dx
∫ (1 − x)
3
o
Hence, the production rate from a given PFR volume as a function of conversion (at stoichiometric feed rate) is F P stoich =
1.8
A
V
x A
∫ (1 −dx x) o
3
=
3.6 1
A
V
(1− x A ) 2
−1
How much can we increase the production rate by doubling CBo to CBo = 6 (mol/L), i.e. by using B in excess?
13
Now the rate as a function of conversion is: 2
2 C Bo 3 3 3 3 2 − x A = 0.2 x2 x − R A = 0.2C Ao(1− x A ) (1 − x )(2 − x A ) 2 C Ao 2
− R A = 3.6(1− x A )(2 − x A )2 a)
Batch x A
t = C Ao
dx A
2
x A
dx
∫ (− R ) = 3.6 ∫ (1 − x )(2 − x )
2
A
o
o
To integrate use partial fractions: A 1 − x
+
B + Cx
=
(2 − x)
2
4 A − 4 Ax + Ax
2
A(2 − x) 2
+ ( B + Cx)(1 − x) 2 (1 − x)(2 − x )
+ B − Bx + Cx − Cx 2 = 1
4A+B=1
4A-3A=1
A=1
-4A-B+C=0
-3A-B-0
B=-3A
A-C=0
C=A = 1 dx
∫ (1 − x )(2 − x )
2
=∫
dx 1 − x
x A
dx 2 − x = ∫ − 1 − x (2 − x )2 o
dx −
+
x−3 (2 − x ) 1
(2 − x )
2
2
dx
dx
x 1 = −ln(1 − x ) + ln(2 − x ) − 2 − x o
A
=− l n(1 − x A ) + 0 + ln(2 − x A ) − ln2 −
= ln t =
1 2 − x A
+
1 2
2 − x A 1 2 x A ln + 1 − = − 2(1 − x A 2(2 − x A ) 2(1 − x A ) 2 2 − x A 2 − x A
2 − x A x A − ln 1.8 2(1 − x A 2(2 − x A ) x 1
A
t =
B=-3
= 0.95
ln 1.05 − 0.95 = 1 ln10.5 − 0.95 1.8 2 x0.05 2 x1.05 1.8 2.1 1
t = 1.055 min
Batch ill advised at these conditions since ts >> t!
14
1 F P new
∆ F P F P old
=
2
C Ao x A V t + t s
× 100 =
=
0.95 x1,798 1.06 + 60
27.97 − 10 10
= 27.97
mol min
× 100 = 179.7%
By operating at double the stoichiometric requirement of B we increase, at same xA, the production rate of the batch reactor by 180%.
b)
CSTR 2
− R = 3.6(1 − x )(2 − x A ) − 3.6(1 − 0.95)(2 − 0.95) 2
mol − R A = 3.6 x0.05 x1.052 = 0.19845 L min F P new = R P V =
− R A 0.19845 mol V = × 44,494 = 4,410 2 2 min
∆ F P
4,410 − 10
F P old
× 100 =
10
x100 = 44,000%
In a CSTR we increase the production rate by 44,000%!
c)
PFR x A
V = F Ao
dx
∫ 3.6(1 − x )(2 − x )
2
o
F P new
Vx A 2
x A
∫
= x
dx
3.6 o (1 − x )(− x) F P new
=
F P new
=
F P new
1.8 x AV
A
dx
∫ (1 − x )(2 − x )
3
3
o
1.8 A V
2 − x A − x A ln 2(1 − x A ) 2(2 − x A ) ln
1.8 0.95 x1,167 2 − 0.95 0.95 2(1 − 0.95)
−
2(2 − 0.95)
=
1.8 x0.95 1,167 1.05 0.95 ln − 0.1 2.10
= 1,051 mol / min
15
∆ F P F P old
1,051 − 10
x100 =
10
x100
= 10,419%
In a PFR over 10,000% increase in FP is obtained.
We present below these ratios of production rate obtainable at
nonstoichiometric ratio of
C Bo C Ao = 2 (C Bo C Ao) stoich and at stoichiometric ratio of CBo/CAo = 3/2 for our example reaction.
This ratio is: For a PFR: 1.8 x A V F P (non − stoich ) F P ( stoich)
2 − x A x A − 2(1 − x A 2(2 − x A V
=
3.6 x A 1 (1 − x A )
Specifically for F P (non − stoich) F P ( stoich)
1
ln
A
2
2
= 2
−1
−1
2 − x A − x A ln 2(1 − x A 2(2 − x A )
= 0.95 we get 1
=
(1 − x A )
0.05 2 0.95
−1
1.05 0.75 2 l n 2 − l n10.5 − 2.10 2 x 0.05 2 x1.05
=
199.5 ln10.5
− 0.45228045
= 105.0
For a CSTR F P (non − stoich) F P ( stoich)
=
3.6(1 − x A )(2 − x A ) 3.6(1− x A )
2 − 0.98 2 F P − 1 − 0.98 F P
( non stoich)
( stoich)
3
2
=
(2 − x A )
2
(1 − x A )
2
=
1.05 2 = = 212 = 441 0.05
Let us examine the situation when the reaction just considered occurs at P = const, T = const in the gaseous phase. Then due to stoichiometry we have
2B + 3B = P + S
∑
υ
j
= 1 + 1 − 2 − 3 = −3 16
Consider stoichiometric feed of reactants at CBo/CAo =3/2. y Ao
ε A
2
= = 0.4
(−υ A ) = 2
5
= y Ao
∑
υ j
(−υ A )
(−3)
0.4
=
=−0.6
2
C Bo C A = C Ao
1 − x A
C B
1 − 0.6 x A
− R A = 0.2 xC
3 Ao
3 2
2
= C Ao
(1 − x A )3 (1 − 0.6 x A )
3
C Ao
−
3 2
x A
1 − 0.6 x A
= 3.6
(1 − x A )3 (1 − 0.6 x A )
3
CSTR F P = V = V =
( − R A ) 2 2 F P
− R A
=
V 2 x10(1 − 0.6 x0.95)3 3.6(1 − 0.95)
3
10 (1 − 0.57)3 3
1.8
0.05
= 44,444 x0.433 = 3,534( L)
Tremendous reduction in required volume compared to the ε A = 0 case occurs! PFR x A
V = F Ao
∫ − R o
V =
dx A
F P 1.8 x A
=
A
x A = 0.95
∫ 0
2 F p x A (1 − 0.6 x)3 dx x A
∫
3.6(1 − x )
3
o
(1 − 0.6 x)3 (1 − x)
3
dx
=
10 1.8 x0.95
0.95
∫ 0
1 − 06 x) 1 − x
3
dx
V = 125.6( L)
Again a significant reduction in PFR reactor volume requirement is observed. Why?
17
2.4
Graphic Comparison of PFR and CSTR
x A V = F Ao CSTR (− R A ) exit x V = ∫ dx A F Ao PFR o − R A Ao
The graphic representation of the above two design equations is represented below for an n-th order reaction. Clearly, for fixed feed conditions and feed rate and for chosen desired conversion the volume of the CSTR will always be larger than or equal to the PFR volume.
1
V = area of box F A CSTR
− R A
o
V = area under the curve F A PFR o
x
FIGURE 2-4: Graphical Comparison of CSTR and PFR
18
Ideal Reactors and Multiple Reactions Isothermal Operation (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471
Lecture 6
October 2005
Ideal Reactors and Multiple Reactions Isothermal Operation Selection of a proper flow pattern is the key factor in achieving desired selectivities and yields in multiple reactions. For every multiple reaction system of known stoichiometry it is possible to determine “a priori” which limiting flow pattern – complete backmixing (CSTR) or no mixing (PFR) will yield superior yields or selectivities. The consideration of yields often is more important than reactor size in choosing the preferred reactor flow pattern. From Lecture 1 we know that all multiple reaction systems can be represented by a set of R independent reactions among the S chemical species present in the system: S
∑υ A = 0 for ij
i = 1,2,3... R
j
(1*)
j=1
These stoichiometric relationships allow one to relate moles produced (or depleted) of each species to the molar extents of the R reactions: F j = F jo +
R
∑υ X & ij
(1)
i
i =1
The rate of reaction of each species is given through the rates of the R independent reactions, r i, i = 1,2,…R. R
R j =
∑ υ r i =1
(2)
ij i
CSTR – Ideal Stirred Tank Continuous Flow Reactor F jo
F j
Qo
Q V
The design equation (i.e., the mass balance for species j) can be written for R species, j = 1,2,3…R: R
F jo − F j +
∑ υ r V = 0 i =1
(3)
ij i
for j = 1, 2,...R
1
ChE 471
Lecture 6
October 2005
If the reaction rate r i for each independent reaction i can be represented by an n-th order form, of eq (4a) r i = k i
S
∏ C
α ij
j
(4a)
j =1
then at P = const, T = const, the rate of the i-th reaction, r i, can be represented in terms of molar & of the reactions by: extents X i α ij
R ⎛ ⎞ & ⎜ ⎟ + F υ X ∑ jo ij i S ⎟ i =1 ∑ α ⎜ r i = k i ∏ C tot ,o ⎜ S R ⎟ j =1 & ⎜⎜ F tot ,o + ∑∑υ ij X i ⎟⎟ j =1 i =1 ⎝ ⎠ ij
(4b)
where α ij is the reaction order of reaction i with respect to species j, Ftot,o is the total initial molar flow rate. & . Substitution of equations (2) and (4b) into (3) results in set of R nonlinear equations in X i
Three types of problems described below arise: a)
Given the feed flow rates, reactor size V and rate forms for all reactions one can calculate & ' s and from equation (1) get the composition of the outlet all the reaction extents X i stream.
In addition, from Lecture 1, at P = const, T = const:
⎛ R S ⎞ & F ⎟ Q = Qo ⎜⎜1 + ∑∑υ ij X i tot , o ⎟ ⎝ i =1 j =1 ⎠
(5)
The exit volumetric flow rate can be computed and effluent concentrations calculated C j =
b)
c)
F j Q
(6)
Given the feed molar flow rates and composition, and the desired partial composition of the outflow, as well as the reaction rates, one can calculate the reactor size from equation (3) and the composition of other species in the outflow. Given molar feed rates and outflow molar flow rates for a given reactor size the rate of reaction for each species can be found from equation (3).
2
ChE 471
Lecture 6
October 2005
PFR – Plug Flow Reactor F jo
F j
Qo
Q V
The design equation (i.e. the differential mass balance for species j) j) can be written for R species dF j
R
∑
= υ ij r i dV i =1 j = 1, 2, ...R
(7)
The initial conditions are
= 0 V =
F j = F jo
(7a)
Using equations (1) and (4b) the above set of R first order differential equations can be & ' s as functions of V. integrated simultaneously and solved for X i a) b)
Given the feed flow rate and composition, and the form of the reaction rates, one could determine what volume V is required to attain the desired product distribution. Given the feed and reactor reactor volume and reaction reaction rate forms, forms, one can determine the exit product distribution.
Batch Reactor – Autoclave of Constant Volume
The R species (for j=1, 2, 3..R) mass balances yield: dn j dt
R
= ∑υ ij r i
(8)
i =1
Initial conditions are: t=0
n j = n jo
(8a)
Moles and extents are related by:
3
ChE 471
Lecture 6
n j = n jo +
October 2005
R
∑υ X ij
(9)
i
i =1
For j = 1, 2, 3…S The rate form as a function of extents is given by α ij
⎛ R ⎞ s ⎜ ∑υ ij X i ⎟ α ⎟ r i = k i Π C jo ⎜1 + i =1 ⎜ ⎟ n jo j = 1 ⎜ ⎟ ⎝ ⎠
(10)
ij
s
α ij
R ⎛ ⎞ r i = k i Π ⎜ C jo + ∑υ ij ξ i ⎟ i =1 ⎝ ⎠ j = 1
where ξ i =
X i V
(10a)
(10b)
One can solve the set of R first order differential equations to calculate the product distribution in time, or the desired time needed for a prescribed product distribution. The above approach, while well suited for the computer, does not provide us with the insight as to which flow pattern is better in a given process until we actually compute the answers for both limiting cases. In order to get better insight in the role of the flow pattern in product distribution in multiple reactions we will consider some simple systems and use the notions of yields and selectivity. Classification of Multiple Reactions A + B = R ⎫ ⎬ parallel C + D = P ⎭ A + B = R ⎫
⎬ competitive
A + 2 B = S ⎭ A + B = R ⎫
⎬ consecutive (series reactions)
R = S ⎭ A + B = R ⎫
⎬ mixed reactions
R + B = S ⎭
4
ChE 471
Lecture 6
October 2005
In Lecture 1 we have defined the various yields R
∑ ip i ⎛ P ⎞ R p i =1 = R y ⎜ ⎟ = { point (relative) yield ⎝ A ⎠ − R A − υ r ∑ iA i υ r
i =1
Point (relative) yield measures the ratio of the production rate of a desired product P and the rate of disappearance of the key reactant A. Point yield is a function of composition composition and this varies along a PFR reactor, varies in time time in a batch reactor, and is a constant number in a CSTR.
⎛ P ⎞ F P − F Po {overall (relative) yield ⎟= − A F F ⎝ ⎠ Ao A
Y ⎜
Overall (relative) yield gives the ratio of the overall product product P produced consumption of reactant A.
and the total
In a CSTR the overall and point yield are identical. P ⎛ P ⎞ Y ( ) = y ⎜ ⎟ A ⎝ A ⎠
In a PFR the overall yield is the integral average of the point yield: 1 ⎛ P ⎞ Y ⎜ ⎟ = ⎝ A ⎠ F Ao − F A
F Ao
∫
F A
⎛ P ⎞ ⎟dF A A ⎝ ⎠
y⎜
Overall operational yield is also often used, defined as the number of moles of the desired product produced per mole of key reactant fed to the system.
⎛ P ⎞ F P − F Po ⎟= A F Ao ⎝ ⎠
⎜
The relationship to overall relative yield is obvious
⎛ P ⎞ ⎛ P ⎞ ⎟ = Y ⎜ ⎟ x A ⎝ A ⎠ ⎝ A ⎠
⎜
where xA is the overall conversion of A
5
ChE 471
Lecture 6
x A =
October 2005
F Ao − F A F AO
None of the above yields has been normalized, i.e., their maximum theoretical value may be more or less than one as dictated by stoichiometric coefficients. A normalized yield can be introduced by ⎛ P ⎞ y ⎜ ⎟ ⎛ P ⎞ ⎝ A ⎠ y ⎜ ⎟ = ⎝ A ⎠ y ⎛ P ⎞ ⎟ max ⎜ ⎝ A ⎠
⎛ P ⎞ ⎟ is obtained by assuming that only the reactions leading from A to R occur. ⎝ A ⎠
where y max ⎜
Point selectivity and overall selectivity measure the ratio of formation of the desired product and one or more of the unwanted products, e.g.
⎛ P ⎞ R p ⎟= U ⎝ ⎠ Ru
S =
s⎜
F p − F po F u − F uo
A general rule:
⎛ P ⎞ ⎟ ⎝ A ⎠ > 0 PFR produces more P.
dy⎜
If
dC A
⎛ P ⎞ ⎟ ⎝ A ⎠ < 0 CSTR produces more P
dy⎜
If
dC A
⎛ P ⎞ ⎟ is not a monotonic function of CA either reactor type may produce more P depending A ⎝ ⎠
If y⎜
on operating conditions. The case of monotonic point yield is illustrated below for the case with ε A = 0 . ⎛ P ⎞ ⎟ ⎝ A ⎠
⎛ P ⎞ ⎟ ⎝ A ⎠
y⎜
y ⎜
\ \ \ Area = Cp in PFR Area = Cp in CSTR C A
C Ao
C A C A
6
C Ao
C A
ChE 471
Lecture 6
R
I.
Liquid Systems or Gases With
S
∑∑υ
ij
October 2005
=0
i =1 j =1
Competitive Reactions a1 A + b1 B= p1P a2 A + b2 B = s2S r 1 = k 1C A1 A C B 1 B α
Given the rate
α
r 2 = k 2 C A 2 A C B 2 B α
α
Point yield then is: p1 p1 r 1 a1 ⎛ P ⎞ R p = = ⎟= ⎝ A ⎠ − R A a1 r 1 + a2 r 2 1 + a 2 r 2
y ⎜
a1 r 1
p1 a1 ⎛ P ⎞ ⎟= ⎝ A ⎠ 1 + a 2 k 2 C (α −α ) C (α A B
y ⎜
2 A
1 A
2 B −α 1 B )
a1 k 1
⎛ P ⎞ p1 ⎟= ⎝ A ⎠ a1
y max ⎜
1 ⎛ P ⎞ ⎟= ⎝ A ⎠ 1 + a 2 k 2 C (α −α ) C (α A B
y ⎜
2 A
1 A
2 B −α 1 B )
a1 k 1
⎛ P ⎞ ⎟ to be as high as possible. This implies: A ⎝ ⎠
We want y ⎜
i) ii) iii)
if α 2 A < α 1 A , α 2 B < α 1 B , keep CA and CB as high as possible. PFR is better than CSTR. if α 2 A = α 1 A ,α 2 B < α 1 B , keep CB as high as possible. PFR is better than CSTR. if α 2 A = α 1 A ,α 2 B > α 1 B , keep CB as low as possible. CSTR is better than PFR. Try for yourself other combinations.
7
ChE 471
Lecture 6
October 2005
Example 1 3
A = P 2A = S
RP = 1.0 C A (kmol/m s) 2 3 RS = 0.5 C A (kmol/m s)
Determine C p in a) CSTR, b) PFR. The feed contains C Ao = 1 (kmol/ms), C po = 0. Conversion of 98% is desired. R p 1 ⎛ P ⎞ R p = = ⎟= ⎝ A ⎠ − R A R p + 2 Rs 1 + C A
y ⎜
⎛ P ⎞ ⎟ =1 A ⎝ ⎠
⎛ P ⎞ ⎛ P ⎞ ⎟ = y⎜ ⎟ ⎝ A ⎠ ⎝ A ⎠
y max ⎜
y ⎜
To keep point yield as high as possible, it is necessary to keep C A low everywhere. will be better than PFR. Let us show this quantitatively.
CSTR
a) CSTR By setting the overall yield equal to the point yield we can solve for the exit concentration of product P. C p 1 ⎛ P ⎞ ⎛ P ⎞ Y ⎜ ⎟ = = y⎜ ⎟ = ⎝ A ⎠ C Ao − C A ⎝ A ⎠ 1 + C A C p =
C Ao − C A
1 + C A
=
C Ao x A
1 + C Ao (1 − x A )
=
1 x 0.98 1 + 1 − 0.98
= 0.961(kmol / m 3 )
C p ⎛ P ⎞ = 0.980 ⎟= ⎝ A ⎠ C Ao − C A
Y ⎜
Overall yield
⎛ P ⎞ ⎟ = 0. 961 ⎝ A ⎠
Overall operational yield ⎜
Required reactor size (space time) τ =
C Ao − C A
− R A
=
C Ao x A 2 C Ao (1 − x A ) + C Ao (1 − x A ) 2
8
= 48.0( s)
ChE 471
Lecture 6
October 2005
b) PFR Product concentration is obtained by integration of the point yield C Ao
C Ao
A
C A
⎛ P ⎞ C p = ∫ y ⎜ ⎟dC A = ⎝ A ⎠ C
dC A
∫ 1 + C
A
⎛ 1 + C Ao ⎞ 1 + 1 ⎞ 3 ⎟⎟ = ln⎛ ⎜ ⎟ = 0.673(kmol / m ) ⎝ 1 + 1 − 0.98 ⎠ ⎝ 1 + C A ⎠
C p = ln⎜⎜
⎛ P ⎞ ⎟ = 0.687 ⎝ A ⎠
Overall yield Y ⎜
⎛ P ⎞ ⎟ = 0. 673 ⎝ A ⎠
Overall operational yield ⎜
C Ao
Required reactor size
τ =
∫
C A
dC A
− R A
C Ao
=
∫
C A
dC A C A (1 + C A )
⎛ C Ao (1 + C A ) ⎞ ⎡ 1 + 1 − 0.98 ⎤ ⎟⎟ = ln ⎢ ⎥ = 3.2( s) + − + C ( 1 C ) ( 1 0 . 98 )( 1 1 ) ⎣ ⎦ A ⎠ ⎝ A
τ = ln⎜⎜
Plug flow reactor is considerably smaller but CSTR gives a better yield and higher concentration of the desired product.
Example 2 A + B = P A + A = S
3
R p = 1.0 C AC B (kmol/m s) 2 3 Rs = 0.5 C A (kmol/m s) 3
Given F Ao = F Bo = 1 (kmol/s;
C Ao = C Bo = 1 (kmol/m ) C Po = C So = 0 and desired ⎛ P ⎞ ⎛ P ⎞ ⎛ P ⎞ conversion x A = 0.98, determine C p , Y ⎜ ⎟, ⎜ ⎟, S ⎜ ⎟ and required reactor space time for ⎝ A ⎠ ⎝ A ⎠ ⎝ S ⎠ a) CSTR, b) PFR.
R p C B 1 ⎛ P ⎞ R p = = = ⎟= ⎝ A ⎠ − R A R p + 2 Rs C B + C A 1 + C A
y ⎜
C B
9
ChE 471
Lecture 6
October 2005
To maximize point yield one should keep the reactant concentration ratio CA/CB as low as possible everywhere. R
Eliminate CB in terms of C p and CA using C j = C jo +
∑υ ξ ij
i
i =1
C A = C Ao − ξ 1 − 2ξ 2 C P = C Po + ξ 1
1
ξ 1 = C P − C Po ξ 2 =
2
[C Ao − C A − C P ]
Now: C B = C Ao − ξ 1 = C Bo − C p C S = C So + ξ 2 = 0 +
1 2
[C
Ao
− C A − C p ]
1
= [C Ao − C A − C Bo + C B ] 2
a) CSTR C p C Ao −C p ⎛ P ⎞ ⎛ P ⎞ = y⎜ ⎟ = ⎟= ⎝ A ⎠ C Ao − C A ⎝ A ⎠ C Ao − C p + C A
Y ⎜
Solve for C p C p − (C Ao _ C Bo )C p + C Bo (C Ao − C A ) = 0 2
C Ao = C Bo = 1; C A = C Ao (1 − x A ) = 1 − 0.98 C p2 − 2C p + 0.98 = 0 3 C p = 1 − 1 − 0.98= 0.859( kmol / m ⎫ 3 C A = 0.02kmol / m
C B = 0.141kmol / m
3
C S = 0.061kmol / m
3
⎪ ⎪ ⎬exit stream composition ⎪ ⎪ ⎭
⎛ P ⎞ ⎟ = 0.877 ⎝ A ⎠
Overall yield Y ⎜
10
i = 1, R
ChE 471
Lecture 6
⎛ P ⎞ ⎟ = 0. 859 ⎝ A ⎠
Overall operational yield ⎜
⎛ P ⎞ C p = 14.2 ⎟= ⎝ S ⎠ C s
Overall selectivity S ⎜
Required Reactor Size τ =
C Bo − C B
− R B
=
C Ao − C B C A C B
=
1 − 0.141 0.02 x 0.141
= 306( s )
b) PFR dF p dF A
=
⎛ P ⎞ = − y⎜ ⎟ R A ⎝ A ⎠ R p
Since F j = QC j and Q = const dC p dC A
=−
C B o − C p C Bo − C p + C A
at C A = C Ao , C p = 0
Rearrange: dC A dC p
+
C A C Bo − C p
d ⎛ ⎜
= −1
⎞ −1 ⎟= dC p ⎜⎝ C Bo − C p ⎠⎟ C Bo − C p at C p = 0
C A
CA = CAo
Integrate from the indicated initial condition: C A C Ao − C p C A
1 − C p
−
⎛ C Bo − C p ⎞ ⎟⎟ = ln⎜⎜ C Bo C Bo ⎝ ⎠
C Ao
= 1 + ln(1 − C p )(*)
Substitute known quantities: 3
C A = C Ao (1-x A) = 1-0.98 = 0.02 (kmol/m )
11
October 2005
ChE 471
Lecture 6
October 2005
Solve for C p by trial and error: φ (C p ) = C p − 0.98 − (1 − C p )ln(1 − C p ) = 0 Dφ (C p ) = 2 + ln(1 − C p ) n
n +1 p
C
= C − n p
φ (C p )
Newton-Raphson Algorithm
n
Dφ (C p )
This yields: C p = 0.613 kmol / m 3 ⎫
⎪
C A = 0.02 kmol / m ⎪ Exit stream composition 3⎬ C B = 0.387 kmol / m ⎪ 3
3 C s = 0.184 kmol / m ⎪ ⎭
The last two concentrations above are evaluated using the stoichiometric relationship.
⎛ P ⎞ ⎟ = 0.626 ⎝ A ⎠ ⎛ P ⎞ Overall operational yield ⎜ ⎟ = 0. 613 ⎝ A ⎠ Y ⎜
Overall yield
⎛ P ⎞ ⎟ = 3.33 ⎝ S ⎠
Overall selectivity S ⎜
Required reactor space time: C Bo
τ =
∫
C B
dC B
− RB
C Bo =1
=
∫
C B = 0.387
dC B C A C B
From (*) C A = (1 − C p ) 1 + ln(1 − C p )
From stoichiometry C B = C Bo − C p = 1 − C p
Thus C A = C B (1 + lnC B )
12
ChE 471
Lecture 6
C Bo =1
τ =
∫
1+ lnC Bo
dC B C B (1 + lnC B ) 2
C B = 0.387
=e
∫
1+ lnC B
e
October 2005
−u
u
du
τ = e{ E 1 (1 + lnC B ) − E 1 (1 + lnC Bo )} = e{ E 1 (0.05) − E 1 (1) }
= 2.71{2.4679 − 0.2194} = 6.1( s) where ∞
E 1 ( z ) =
∫
e
z
−u
u
du
exponential integral
E1(z) values tabulated in M. Abramowitz & A. Stegun “Handbook of Mathematical Functions”, Dover Press, N.Y. 1964. Comparison of CSTR & PFR
Reaction System:
A + B = P (desired) A+A=S
Decision variables:
xA = 0.98 F M B / A = Bo = 1 F Ao
Rate Form:
r 1 = 10 CACB (kmol/m s) 2 3 r 2 = 0.5 CA (kmol/m s)
3
Operational yield Overall selectivity Reactor space time
CSTR 0.859 14.2 306 (s)
PFR 0.613 3.3 6.1 (s)
Optimal Ideal Reactor 0.950 63.0 150 (s)
The last column of the above Table was computed based on an ideal reactor model shown below. We have B entering a plug flow reactor while FAo is distributed from the side stream into the 3 reactor in such a manner that CA = 0.02 kmol/m everywhere in the reactor. F Ao
PFR
F Bo
13
ChE 471
Lecture 6
October 2005
From the expression for the point yield C B 1 ⎛ P ⎞ = ⎟= ⎝ A ⎠ C B + C A 1 + C A
y ⎜
C B
it is clear that one needs to keep CA low and CB high. With the constraint of FAo = FBo the above ideal reactor accomplishes that requirement in a n optimal manner. Could such a “porous wall” reactor with plug flow be constructed? It depends on the nature of the reaction mixture. However, we learn from the above that with our choice of decision variables maximum selectivity is 63, we can never do better than that! We also learn that a good reactor set up is a cascade of CSTR’s. F Ao
F Bo
The total number of reactors used will depend on economics With 2 reactors we get selectivity of over 20, with five we are close to optimum.
Examining the effect of decision variables we see that C p increases with increased conversion of A. For conversions larger than 0.98 the reactor volume becomes excessive. If we took MB/A > 1 that would improve the yield and selectivity but at the expense of having to recycle more unreacted B. Let us ask the following question. How much excess B would we have to use in a PFR in order C p = 14. 2 at CAo = 1 to bring its overall selectivity to the level of a single CSTR i.e., S = C s 3
(kmol/m ) and xA = 0.98. So the goal is to choose C Bo in order to get at the exit of plug flow: C p C s
= 14.2
From stoichiometry: C s =
1 2
[C
Ao
− C A − C p ] =
1 2
[C x
Ao A
− C p ] =
14
1 2
[0.98 − C ] p
ChE 471
Lecture 6
October 2005
The prescribed desired selectivity is: 2C p 0.98 − C p
= 14.2⇒C p = 0.859(kmol / m 3 )
C s = 0.061(kmol / m ) 3
The integrated equation for PFR is: ⎛ C p ⎞ C C A ⎟⎟ − Ao = ln⎜⎜1 − C Bo − C p C Bo ⎝ C Ao ⎠ The initial concentration of B is now the only unknown. Evaluate it by trial & error. 0.02 C Bo − 0.859 C Bo = 3.79
−
⎛ 0.859 ⎞ ⎟⎟ gives = ln⎜⎜1 − C Bo C Bo ⎠ ⎝ 1
kmol m
3
3
Since CAo = l (kmol/m ), MB/A = 3.79 – almost four times more B than A should be introduced in the feed to get the selectivity in a PFR to the level of a CSTR. A great excess of unreacted B has to be separated in the effluent: 3
CB = CBo – C p = 3.790-0.859 = 2.931 (kmol/m ) Consecutive Reactions aA = p1P p2P = s S
Two basic problems arise: a) conduct the reaction to completion, b) promote production of the intermediate. The first problem is trivial and can be reduced to a single reaction problem. Use the slowest reaction in the sequence to design the reactor. In order to maximize the production of intermediates PFR flow pattern is always superior to a CSTR flow pattern.
15
ChE 471
Lecture 6
October 2005
p r − p 2 r 2 p1 p 2 r 2 ⎛ P ⎞ R p = 11 = − ⎟= − A R a r a a r 1 ⎝ ⎠ 1 A ⎛ P ⎞ p y max ⎜ ⎟ = 1 ⎝ A ⎠ a y ⎜
β p 2 r 2 p 2 k 2 C P ⎛ P ⎞ y ⎜ ⎟ = 1 − = 1− α p1 r 1 p1 k 1C A ⎝ A ⎠
One needs to keep CA high and C p/CA low which is best accomplished in a PFR. Example 1 A=P P=S
r 1 = 1.0 CA (kmol/m s) α = 1 3 r 2 = 0.5 CP (kmol/m s) β = 1 3
3
Starting with CAo = 1 (kmol/m ) and CPo = Cso = 0 find the maximum attainable CP in a) CSTR, b) PFR. We could continue to use the point yield approach. CSTR Stirred Tank Reactor C − 0.5C P ⎛ P ⎞ (C Ao − C A ) ⎟(C Ao − C A ) = A C A ⎝ A ⎠
C P = y ⎜
Solve for CP C P =
C A (C Ao − C A )
0.5(C Ao + C A )
Find optimal CA at which the CSTR should operate. dC P dC A
= 0 ⇒ (C Ao − 2C A )(C Ao + C A ) − C A (C Ao − C A ) = 0
C A + 2C Ao C A − C Ao = 0 2
2
C Aopt = C Ao C Pmax =
2 − 1 = 0.414( kmol / m 3 )
0.414(1 − 0.414) 0.5(1 + 0.414)
= 0.343(kmol / m 3 )
C S = C Ao − C A − C P = 0.243kmol / m
Overall yield
3
⎛ P ⎞ ⎟ = 0.585 ⎝ A ⎠
Y ⎜
16
ChE 471
Lecture 6
⎛ P ⎞ ⎟ = 0. 343 ⎝ A ⎠ ⎛ P ⎞ S ⎜ ⎟ = 1.4 ⎝ S ⎠
⎜
Operational yield Overall selectivity Required reactor space time: τ =
b)
C Ao − C Aopt
− R A
PFR dC P
C Aopt
= − y⎜ ⎟ = ⎝ A ⎠
at CA = CAo,
dC A
C Ao − C Aopt
−
1 − 0.414 0.414
= 1.4( s)
Plug Flow Reactor ⎛ P ⎞ 0.5C P − C A
dC A
dC P
=
C A
CP = 0
−
0.5 C p C A
= −1
d ⎡ C P ⎤
1
⎢ ⎥ == − dC A ⎢⎣ C A ⎥⎦ C A C P C A
= 2[ C Ao − C A ]
C P = 2 C A
C Ao − C A
Find CA at the reactor exit by dC P dC A
= 0 ⇒ C Ao − 2 C A = 0
C Aopt =
C Ao
4
C Aopt = 0.25( kmol / m ) 3
C Pmax = 2 0.25 1 − 0.25 = 0.5(kmol / m 3 ) 3 C s = 1 − 0.5 − 0.25 = 0.25( kmol / m )
Overall yield Operational yield
⎛ P ⎞ 2 ⎟ = = 0.667 A ⎝ ⎠ 3 ⎛ P ⎞ ⎜ ⎟ = 0. 5 ⎝ A ⎠ Y ⎜
17
October 2005
ChE 471
Lecture 6
October 2005
⎛ P ⎞ ⎟ = 2.0 ⎝ A ⎠
S ⎜
Overall selectivity
Required space time: C Ao
dC A
∫
τ =
C Aopt
C A
= ln
C Ao C Aopt
= ln4 = 2ln2 = 1.39( s)
The same results can be obtained by using the design equations (i.e. mass balance)for P & A. CSTR C Ao − C A
τ =
C A =
C A C P
τ =
C A − 0.5C P
dC p d τ
C P =
C Ao
1 + τ C Aτ
1 + 0.5τ
=
C Aoτ
(1 + τ )(1 + 0.5τ )
= 0 ⇒ gives τ opt = 1.4( s ), etc.
PFR dC A
= −C A C A = C Ao e −τ
d τ τ = 0, dC P d τ
τ = 0 d d τ
C A = C Ao = 1
= C A − 0.5C P C P = 0
(e 0.5τ C P ) = C Ao e −τ e 0.5τ = C Ao e −0.5τ
C P = 2(e dC P d τ
− 0.5τ
− e −τ )
= 0 ⇒ τ opt = 2ln 2, etc.
18
ChE 471
Lecture 6
October 2005
Mixed Reactions
This is the most frequently encountered type of multiple reactions which can be viewed as a combination of competitive and consecutive reactions. We can solve the problems involving these reactions either by setting R design equations for R components or by utilizing the concept of the point yield in simpler reaction schemes. Example 1 3
2A + B = R
-R A = 2k 1 CA CB (kmol/m min)
2B + R = S
R s = k 2 CB CR (kmol/m min)
3
3
k 1 = 10 k 2 = 1 (m /kmol min). R is the desired product. Find CR in a) CSTR, b) PFR, when CRo 3 = Cso = 0. Decision variable CAo = CBo = 1 (kmol/m ). We can write two point yields: k C C − k 2 C B C R k C ⎤ 1⎡ ⎛ R ⎞ R y ⎜ ⎟ = R = 1 A B = ⎢1 − 2 R ⎥ 2k 1C A C B 2⎣ k 1 C A ⎦ ⎝ A ⎠ − R A C A −
k 2
C R
k C C − k 2 C B C R k 1 ⎛ R ⎞ R = y ⎜ ⎟ = R = 1 A B ⎝ B ⎠ − R B k 1C A C B + 2k 2 C B C R C + 2k 2 C A R k 1
The point yield y
R
⎝ A depends only on CR and CA and is simpler to use.
a)
CSTR
Stirred Tank Reactor k C A − 2 C R k 1 ⎛ R ⎞ C R = y ⎜ ⎟(C Ao − C A ) = (C Ao − C A ) 2C A ⎝ A ⎠ C R =
C A (C Ao − C A )
⎛ k 2 ⎞ k ⎜⎜ 2 − ⎟⎟C A + 2 C Ao k 1 ⎝ k 1 ⎠
=
C A (C Ao − C A )
1.9C A + 0.1C Ao
Find optimum CA in a CSTR.
⎛ k ⎞ k k 2 = 0 ⇒ ⎜⎜ 2 − 2 ⎟⎟C A2 + 2 2 C Ao C A − 2 C Ao =0 dC A k k k 1 ⎠ 1 1 ⎝ dC R
19
ChE 471
Lecture 6
C Ao
C Aopt =
2 C Aopt =
C Rmax =
k 1 k 2
+1
1 2 x10 + 1
= 0.183(kmol / m 3 )
0.183(1 − 0.183) 1.9 x 0.183 + 0.1
= 0.334 (kmol / m 3 )
From stoichiometry C B − C Bo − C S =
1
3 2
(C Ao − C A ) + 2C R = 1 −
(C Ao − C A ) − C R =
3 2
(1 − 0.183) + 2 x0.334
1
(1 − 0183) − 0.334 2 2 C B = 0.443( kmol / m 3 ); C S = 0.0745(kmol / m 3 )
⎛ R ⎞ ⎛ R ⎞ Y ⎜ ⎟ = 0.409; Y ⎜ ⎟ = 0.600 ⎝ A ⎠ ⎝ B ⎠
Overall yield Operational yield Overall selectivity
⎛ R ⎞ ⎛ R ⎞ ⎜ ⎟ = 0.334; ⎜ ⎟ = 0. 334 ⎝ A ⎠ ⎝ B ⎠ ⎛ R ⎞ S ⎜ ⎟ = 4.48 = 4.5 ⎝ S ⎠
Required reactor space time: τ =
b)
C Ao − C A
2k 1C A C B
PFR
=
1 − 0.183 2 x1 x0.183 x0.443
= 5.0(min)
Plug Flow Reactor
1 k C ⎛ R ⎞ = − y⎜ ⎟ = − + 2 R dC A 2 2k 1 C A ⎝ A ⎠ C A = C Ao ,C R = 0 dC R
d ⎛ − k 2 ⎞ = − 1 C − 2k k 2 k 2 C C 1 1 ⎜ A R ⎟ A dC A ⎝ 2 ⎠ C R =
⎛ k 2 ⎞ ⎜⎜ 1− ⎟⎟ ⎝ 2 k 1 ⎠
C Ao
k 2
0.95 C A0.05 − C A − C A C Ao = 1.9 ⎛ k 2 ⎞ ⎜⎜ 2 − ⎟⎟ ⎝ k 1 ⎠ 2 k 1 A
C
20
October 2005
ChE 471
Lecture 6
For optimal CA at reactor exit:
⎛ k ⎞ = 0 ⇒ C A = ⎜⎜ 2 ⎟⎟ dC A ⎝ 2k 1 ⎠ dC R
opt
⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ ⎜ k 2 ⎟ ⎜⎜ 1− ⎟⎟ ⎝ 2 k 1 ⎠
1
C Aopt = (0.05) C Rmax =
1 2
= 0.0427(kmol / m 3 )
10.95 x0.0427 0.05 − 0.0427 1.9 3
C B = 1 −
C S =
0.95
2
= 0.427(kmol / m 3 )
(1 − 0.0427) + 2 x 0.427 = 0.418( kmol / m 3 )
(1 − 0.0427) − 0.427 = 0.0517(kmol / m 3 )
⎛ R ⎞ ⎛ R ⎞ Y ⎜ ⎟ = 0.446; Y ⎜ ⎟ = 0.734 ⎝ A ⎠ ⎝ B ⎠
Overall yield
⎛ R ⎞ ⎛ R ⎞ ⎜ ⎟ = 0.427; ⎜ ⎟ = 0. 427 ⎝ A ⎠ ⎝ B ⎠ ⎛ R ⎞ S ⎜ ⎟ = 8.27 = 8.3 ⎝ S ⎠
Operational yield Overall selectivity
Reactor space time: C Ao
τ =
∫
C A
dC A
2C A C B .
From stoichiometry C B = C Bo −
3 2
(C Ao − C A ) + 2C R
but all along the PFR 0.05
C R =
C A
− C A
1.9
21
October 2005
ChE 471
Lecture 6
C B = C Bo −
3
2
(C Ao − C A ) +
(C
0.05 A
2 1.9 2 ⎞ 2 0.05 ⎛ C B = ⎜1.5 − C A − 0.5 ⎟C A + 1 . 9 1 . 9 ⎝ ⎠ C B = 0.447C A + 1.05C A
0.05
1
− C A )
− 0.5
dC A
∫
τ = 0.5
October 2005
0.447C A2 + 1.05C A1.05 − 0.5C A 0.0427
integrate numerically (2.759 min )
Caution must be exercised when using the point yield concept and finding maximum concentrations in mixed reactions. Sometimes formal answers will lie outside the physically permissible range if the other reactant is rate limiting.
For example in Example 1 if we take C Ao = 2
kmol ⎞
; C Bo = 1
⎝ m ⎠ 3
kmol ⎞
⎝ m3 ⎠
We are feeding the reactants in stoichiometric ratio for reaction 1. Following the above described procedure in a CSTR we would find C A opt = C Rmax =
C Ao
= 0.366(kmol / m 3 )
2 x10 + 1 0.366( 2 − 0.366) 1.9 x 0.366 + 0.2
= 0.668(kmol / m 3 )
However these values are not attainable since from stoichiometry it follows that: C B = 1 −
3 2
( 2 − 0.366) + 2 x 0.668 = −0.115 < 0
This indicates that B is not introduced in sufficient amount to allow the reactions to proceed to that point. 3
If CBo = 1.115 (kmol/m ) then the above C R max can be obtained (theoretically) at CB = 0 and that would require an infinitely large reactor. Thus the maximum reactor size that is allowed would determine C Bmin . 3
(kmol/m ) (99% conversion of B). Calculate the resulting CA and CR from
22
Say C Bmin =0.01
ChE 471
Lecture 6
C A = C Ao − C R =
4
C R −
2
3 3 C A (C Ao − C A )
October 2005
(C Bo − C Bmin )
1.9C A + 0.1C Ao C A = 0.461( kmol / m 3 ) 3 C R = 0.660( kmol / m )
That yields:
C S = 0.110( kmol / m ) 3
⎛ R ⎞ ⎛ R ⎞ Y ⎜ ⎟ = 0.429; Y ⎜ ⎟ = 0.667 ⎝ A ⎠ ⎝ B ⎠
Overall yield Operational yield Overall selectivity
⎛ R ⎞ ⎛ R ⎞ ⎜ ⎟ = 0.330; ⎜ ⎟ = 0. 660 ⎝ A ⎠ ⎝ B ⎠ ⎛ R ⎞ S ⎜ ⎟ = 6.0 ⎝ S ⎠
Reactor space time: τ = II.
2 − 0.461 2 x1 x 0.461 x0.01
= 170min
Systems with change in total volumetric flow rate (ε A ≠ 0) gases
& Same approach may be used but one needs to deal with X extents rather than concentrations. i&
Use relationships from Lecture 1.
Summary
PFR promotes more reactions of higher order with respect to reactions of lower order. CSTR favors reactions of lower order with respect to those of higher order. In consecutive reactions better yields are achieved always in PFR than in a CSTR for an intermediate product. Select judiciously the objective function to be optimized. Remember: Optimizing overall yield does not necessarily lead to the same result as maximizing the production rate (or concentration) of the desired product or as maximizing selectivity. Be aware of the relationship of the design equations and reaction stoichiometry.
23
NONISOTHERMAL OPERATION OF IDEAL REACTORS Continuous Flow Stirred Tank Reactor (CSTR) (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471
Fall 2005 LECTURE 7
NONISOTHERMAL OPERATION OF IDEAL REACTORS Continuous Flow Stirred Tank Reactor (CSTR)
T o
T
F jo , Qo
F j
V
T mo
V
T o
or
T F j
F jo , Qo
T m
T mo
T m
Figure 1: Schematic of CSTR with jacket and coil Assumptions: Homogeneous system s
a) Single Reaction
∑ υ A = 0 j = 1
j
j
b) Steady state A CSTR is always assumed perfectly mixed so that the concentration of every species is uniform throughout the reactor and equal to the concentration in the outflow. Due to assumption of perfect mixing the temperature, T, throughout the reactor is uniform and equal to the temperature of the outflow. The only difference between an “isothermal” CSTR treated previously and the general case treated now, is that now we do not necessarily assume that reactor temperature, T, and feed temperature, To, are equal. A CSTR can be jacketed or equipped with a cooling (heating) coil. Two basic types of problems arise: 1. Given feed composition and temperature, rate form and desired exit conversion and temperature find the necessary reactor size to get the desired production rate and find the necessary heat duty for the reactor. 2. Given feed conditions and flow rate and reactor size together with cooling or heating rates, determine the composition and temperature of the effluent stream. To solve either of the above two problems we need to use both a species mass balance on the system and the energy balance. Consider a single reaction s
∑ υ A = 0
(1)
aA + bB = pP
(1a)
j = 1
j
j
or
1
ChE 471
Fall 2005 LECTURE 7
Suppose the reaction is practically irreversible and the rate of reaction, which is a function of composition and temperature is given by:
⎛ kmol ⎞ ⎟ 1 424 3 123⎝ m 3 s ⎠
r = k o e
− E / RT
C A C B ⎜ α
β
Arrhenius Temperature Dependence of the rate
(2)
concentration dependence n-th order reaction n = α + β
For a single reaction we can always eliminate all concentrations in terms of conversion of the limiting reactant A (Lecture 1). For liquids
C j
υ C ⎛ ⎞ = C Ao ⎜⎜ M j / A − j x A ⎟⎟; M j / A = jo υ A ⎠ C Ao ⎝
For gases
C j
υ ⎛ ⎞ ⎜ M j / A − j x A ⎟ υ A ⎟ T o P = C Ao ⎜⎜ ⎟ TP 1 + ε A x A ⎜⎜ ⎟⎟ o ⎝ ⎠
ε A
= y Ao
where
∑υ
j
(−υ A )
In a CSTR we assume Po = P = const i.e constant pressure. The rate now becomes: For liquids β
r = k o e
− E / RT
(α + β ) Ao
C
b ⎞ (1 − x A ) ⎛ ⎜ M B / A − x A ⎟ a ⎠ ⎝ α
(3a)
For gases β
b ⎞ ⎛ (1 − x A ) ⎜ M B / A − x A ⎟ a ⎠ α + β ⎝ E RT ⎛ T ⎞ r = k o e − / ⎜ o ⎟ C Ao (α + β ) α + β (1 + ε A x A ) ⎝ T ⎠ α
(3b)
Recall that at steady state the basic conservation equation is:
(Rate of input)− (Rate of output)+ (Rate of generation by reaction)= 0 2
(4)
ChE 471
Fall 2005 LECTURE 7
Apply equation (4) to mass of species A: (or to any species j) F Ao − F A
+ υ A rV = 0;
F Ao x A
= arV = 0
C Ao x A
= ar τ
F jo − F j + υ j rV = 0
(5)
& = rV X
(5a)
where τ =
V Qo
The energy balance of course cannot contain a generation term (in absence of nuclear reactions) and hence can be written as: s
∑ j =1
~ F jo H jo
s
~
− ∑ F j H j + q& = 0
(6)
j =1
⎛ kmol j ⎞ - molar flow rate of species j in the feed ⎟ ⎝ s ⎠ ⎛ kmol j ⎞ F j ⎜ ⎟ - molar flow rate of species j in the outflow ⎝ s ⎠ ~ ⎛ J ⎞ ⎟⎟ - virtual partial molal enthalpy of species j in the outflow mixture H j ⎜⎜ kmol j ⎝ ⎠ ~ ⎛ J ⎞ ⎟⎟ - virtual partial molal enthalpy of species j in the feed H jo ⎜⎜ ⎝ kmol j ⎠ ⎛ J ⎞ q&⎜ ⎟ - rate of heat addition from the surroundings i.e from the jacket or coil to the ⎝ s ⎠ F jo ⎜
reaction mixture in the reactor The energy balance given by equation (6) is not general in the sense that the following assumptions have already been made in order to present it in that form:
3
ChE 471
Fall 2005 LECTURE 7
Assumptions involved in deriving eq (6):
1. Potential energy changes are negligible with respect to internal energy changes. 2. Kinetic energy changes are negligible with respect to internal energy changes. 3. There is no shaft work involved i.e the only work term is the expansion (flow) work. J The virtual partial molal enthalpy, H÷ j ⎜ ⎟ , is defined by: ⎝ mol j & tot H = m
s
∑
~ F j H j
j =1
⎛ kg ⎞ - mass flow rate of the mixture ⎟ ⎝ s ⎠ ⎛ J ⎞ ⎟⎟ - enthalpy per unit mass of the mixture H ⎜⎜ kg ⎝ ⎠ & tot ⎜ m
In principle the virtual partial molal enthalpy can be evaluated by the following procedure (for gases): ~ o H j (T , P, y j ) = Δ H fj
T
+ ∫ C T *o
dT + [ H (T , P ) − H T , P ] + [ H (T , P, y j ) − H (T , P )] *
~ j *j
~ j
~ j
~ j
~ j
Δ H - standard enthalpy of formation for species j at the pressure Po of standard state and temperature ~ f jo
o
o
o
T (enthalpy of 1 kmole of pure j at T , P ) T
∫ C o
T
~ f j*
dT
- change in enthalpy of species j, if it behaved as an ideal gas, due to change in temperature o
from standard state temperature T to the temperature of interest T . C * is the specific heat of j in ideal ~ p j
gas state. In reality C pjo data for real gases obtained at atmospheric or lower pressures can be used. T
Δ H + o
~ f j
∫ C dT = H (T , P ) enthalpy per mol of j for pure j, if it behaved like an ideal gas, at T and P . *
*
f j
o
o
~ j
T o
Since enthalpy of an ideal gas does not depend on pressure this is also the enthalpy per kmol of j, if it *
were an ideal gas, at T and P, H ( T , P ). ~ j
H (T , P) − H (T , P ) = pressure correction factor i.e the difference between the enthalpy of j being a *
~ j
~ j
*
real gas at T, P, ( H (T , P ) and enthalpy of j being an ideal gas at T, P ( H (T , P ) ). ~ j
~ j
4
ChE 471
Fall 2005 LECTURE 7
Frequently the pressure correction is read off appropriate charts and is given by:
⎛ H * − H j ⎞ ⎜ ~ j ~ ⎟ ⎜ ⎟ * H j (T , P) − H j (T , P ) = −T cj ⎜ ⎟ T cj ~ ~ ⎜⎜ ⎟⎟ ⎝ ⎠T ,P r
r
Where T cj – critical temperature of species j.
⎛ H * − H ⎞ ⎜ ~ j ~ j ⎟ T = is the correction read off the charts at appropriate reduced temperature T and r ⎜ T ⎟ T cj ⎜ ⎟ cj ⎝ ⎠ reduced pressure Pr =
P pcj
Pcj – critical pressure of species j.
Finally
(
)− H (T , P ) = π is the correction factor which accounts for the nonideality of the
H j T , P, y j ~
~
j
j
mixture. _________________________________________________________________________ For liquids (in a first approximation) ~ H j
T
= Δ H f + ~
o
j
∫ C dT + π pj
j
T o
Here we will assume: Gases:
a. ideal mixtures π j
=0
b. ideal gas behavior H = H j
*
~
~
Liquids:
a.
ideal mixture π j =0
Now the energy balance of eq (6) based on the above assumption can be written as
5
ChE 471
Fall 2005 LECTURE 7 T
s
∑ F ∫ C jo
j =1
T o
& = q& dT + Δ H r T X
(7a)
& = q& dT + Δ H r T X
(7b)
p j
~
T
s
∑ F ∫ C j
j =1
~
T o
p j
o
where s
s
T
Δ H r = ∑υ j Δ H + ∑υ j ∫ C p ~ f jo
T
j = 1
j = 1
j
dT
T o
is the heat of reaction at temperature T . Finally, in preliminary reactor design we assume that the heat of reaction does not vary much with temperature
Δ H r ≈ const ≈ Δ H r ≈ ΔH r T o
T
and that some mean value of the specific heat can be used T
∫ C
p j
T o
dT = C p j (T − T o ) ~
~
Equations (7) can then be written as: s
∑ F
jo
j =1
& = q& C p j (T − T o ) + Δ H r X ~
or ρ C p Q(T o
& + q& = 0 − T ) + (− Δ H r ) X
(8)
where & = X
F Ao x A
( −υ A )
Δ H r = A
=
C Ao Qo x A
( −υ A )
Δ H r ( −υ A )
Δ H r - heat of reaction for the stoichiometry as written
Δ H r
A
- heat of reaction per mole of A. ρ C p Q (T o
− T ) + − Δ H r
A
)C
Ao
Qo x A
+ q& = 0
(9)
or ρ C p Q( T o − T ) + ( −Δ H r ) rV + qś = 0
(9a)
6
ChE 471
Fall 2005 LECTURE 7
This final form of the energy balance resulting from all of the above assumptions can be interpreted as a “heat balance” i.e as rate of input of sensible heat by the flowing stream minus rate of output of sensible heat by exit stream plus heat generated by reaction plus heat added from the surroundings must add to zero. In addition we now need an energy balance on the jacket or coil and a constitutive relationship for heat transfer rate, q& . For a jacket at steady state (assuming that the jacket is well mixed too) ρ m Qm C p m ( T m o
− T m ) − qś= 0
(10)
ρ m , Qm C pm - are density, volumetric flow rate and mean specific heat of the fluid flowing through the jacket. T mo - inlet jacket temperature.
Tm – exit jacket temperature. Let q&
= UA(T m − T )
(11)
⎛ J ⎞ ⎟ - overall heat transfer coefficient. ⎜ m 3 o C s ⎟ ⎝ ⎠
U ⎜
2
A (m ) – area for heat transfer between reactor and jacket. From eq (10) T m
=
κ m
=
+ κ mT 1 + κ m
T mo
(12)
where UA
ρ m C pm Qm
For a coil with plug flow of heating/cooling medium: dT m = κ m (T − T m ) dz z
= 0T m = T m
(13) (13a)
o
z = fractional length of the coil.
7
ChE 471
Fall 2005 LECTURE 7
T m ( z ) = T + e
−κ m z
(T mo
− T )
(14)
For an n-th order irreversible reaction G(T) always has a sigmoidal shape and at high temperatures tends to a horizontal asymptote CAo. Finally, in dimensionalized form the two equations that have to be solved simultaneously are:
= ar τ ~ T o − T + β A a r τ − ω (T − T mo ) = 0
(15)
C Ao x A
(16)
where r is given by equation (3) and β˜ A
−Δ H r ⎟ =⎜ ⎝ ρ C p A
ω =
ω =
κ m
κ 1 + κ m κ κ m
=
=
ρ m C p m Qm
UA
ρ m C p m Qm
+ UA
for a jacketed reactor
ρ C pQ
(1 − e ) −κ m
UA
for a reactor with coil
κ =
ρ m C p m Qm
UA
ρ C p Q
When we deal with problems of type 1, to find reactor size for given feed and product stream conditions we use directly eq (15). τ =
V C Ao x A Qo
=
ar
=
C Ao x A
− R A
Then calculate the desired heating or cooling rate from eq (16)
− q& = ρ C p Q(T o − T ) + − Δ H r )ar V A
When we deal with problems of type 2 and try to find the operating conditions for a given reactor, then we must solve eqs (15) and (16) simultaneously by trial and error for xA and T. Sometimes we can solve explicitly for xA from eq (15) in terms of temperature x A = xA(T). Substituting this relationship into eq (16) we get
8
ChE 471
Fall 2005 LECTURE 7
T o − T + β Aar T τ −
(T − T mo ) = 0
r T indicates that the rate is now a function of temperature only since r (x A,T) = r (xA(T),T) T o + T mo 1 + ω − T ~ ~ β A β A
= ar T τ = C Ao x A
(15)
{
1 2 3
G (T )
L (T )
Reactor operating temperature is given by the intersection of the straight line L(T) and curvilinear function G(T).
For an n-th order irreversible reaction G(T) always has a sigmoidal shape and at high temperatures tends to a horizontal asymptote C Ao . For an endothermic reaction Δ H r > 0 ⇒ β˜ A
<0
and line L has a negative slope. L
C Ao
G
G
L
C Ao x A
− slope =
C Ao x A
T
1+ β A
T o + T mo
T
1 + ω Figure 2:
Operating point for an endothermic reaction in a CSTR.
Several conclusions can be reached for endothermic reactions: i) ii)
L & G can intersect at most once and for a given set of parameters only one steady state exists. , For a given reactor and flow rate, fixed τ , given feed, To and heating medium feed temperature, T , and given heat transfer properties, , the more endothermic the reaction, the larger β˜ and mo
iii)
A
the smaller the slope of the L line, therefore the lower the operating temperature and conversion. For a fixed reaction, feed flow rate and composition and given reactor, fixed G curve, an increase in feed temperature moves the L-line to the right while its slope remains unchanged. Hence, the operating T and xA are increased.
For an exothermic reaction Δ H r < 0 and β˜ A
> 0. 9
The L-line has a positive slope.
ChE 471
Fall 2005 LECTURE 7
3
C Ao x A G L
slope =
2
C Ao x A
1+ β A
1
T o + T mo
T
T
1 + ω Figure 3:
Operating point(s) for an exothermic reaction in a CSTR.
Several conclusions can be reached for exothermic reactions: i) ii) iii)
L & G can intersect sometimes at more than one place and for a given set of conditions, more than one steady state may be possible. For fixed τ and reaction, fixed G, an increase in the feed temperature moves the L line to the right increasing the operating T and xA. An increase in coolant T mo has the same effect. For a fixed τ and reaction, fixed G, β˜ , an increase in heat removal, increase in , rotates the A
L-line in the counterclockwise direction and moves the intercept at the abscissa to the left if T o > T mo or to the right if T mo > T o . For exothermic reactions it is important to calculate the adiabatic temperature rise and the maximum adiabatic Δ T . From eq (16) with
≡0
(T − T o )ad = β˜ AC Ao x A
(18)
Δ T ad . max = β˜ A C Ao
(19)
Maximum fractional temperature rise is also called the Prater number, β . β =
ΔT ad . max β˜ A C Ao =
T o T o Now let us consider a number of simple illustrative examples.
Example 1.
The following information is given. Irreversible reaction A
→R 10
(20)
ChE 471
Fall 2005 LECTURE 7
⎛ mol − R A = e 25e −20,000 / RT C A ⎜ ⎝ L min ⎛ mol ⎞T = 350 o K = 77 o C C Ao = 1⎜ ⎟ o ⎝ L ⎠ Δ H R = −100,000 A
ρ C p
cal mol A
cal
= 500(
) Lo C
CSTR
a.
Isothermal. Find reactor space time needed for xA = 0.9 at 350 K and heat to be removed. ˚
τ =
C Ao
− C A
=
− R A
C Ao x A
− R A
1 x0.9
= 25
e e
−
20 , 000
=
1.987 x 300
0.9 0.0233
τ = 38.59 = 36 min
= Q ρ C p
&Δ H (T − T o ) + X R
& = F x X Ao A
= Q ρ C Ao x A (Δ H R
q&
A
)
⎛ cal ⎞ = Q ρ C (T − T ) + QC x (ΔH ⎟ p o Ao A R ⎝ min ⎠
q& ⎜
A
)
⎛ cal ⎞ = ρ C (T − T ) + C x (Δ H ) = 1 x0.9(− 100,00)0 = −90,000cal / L to be removed. ⎜ ⎟ p o Ao A R Q ⎝ L ⎠ ⎛ mol ⎞ ⎟⎟ Note that the reaction rate at 350K is only 0.0233 ⎜⎜ L min ⎝ ⎠ q&
A
b.
Adiabatic
Find τ for xA = 0.9. Q ρ C p (T − T o ) = ( Δ H R A )QC Ao x A T = T o
+
(− Δ H )C
T = 350 +
R A
Ao
ρ C p 100,000 x1 500
x A
0.9 = 530 K (257 o C )
11
ChE 471
Fall 2005 LECTURE 7
τ =
C Ao
− C A
=
C Ao x A
1 x 0.9
=
=
− R A e 25 e − 20, 000 / 1.987 x 530 τ = 2.21 x10 −3 min = 0.133s − R A
0.9 407
Note that at 530K the reaction rate has increased by orders of magnitude to
⎛ mol ⎞ ⎟⎟ − R A = 407⎜⎜ L min ⎝ ⎠ Nonisothermal
Find τ , q& / Q given desired x A τ =
q& Q
C Ao x A
− R A
= 0.9 and desired T = 400 K .
0.9
= 25
e e
−
=
20 , 000 1.987 x 400
0.9 0.849
= ρ C p (T − T o ) + C Ao x A (−Δ H R
A
= 1.06 (min )
)
= 500(400 − 350) + (−90,000) = 25,000 − 90,000 = −65,000cal / L Note the value of the rate which is kmol ⎞ − R A = 0.849⎛ ⎜ ⎟ ⎝ c min ⎠
Example 2
A well stirred bench scale reactor (CSTR) is used for a first order exothermic reaction A → R (practically irreversible) under the following conditions: τ = 1.0(min) C Ao = 1.0( mol / liter ) ; T o T mo
= 350K ; ω = 1.0 =
= 350K ρ mC pmQm
UA
ρ m C pm Qm
+ UA
•
ρ C p Q
⎛ K lit ⎞ (−Δ H R ) = 200⎜ ⎟= ⎝ mol ⎠ ρ C p 20,000 ⎛ mol ⎞ − 25 r = e e RT C A ⎜ ⎟ ⎝ lit min ⎠
β÷ A
A
12
ChE 471
Fall 2005 LECTURE 7
a) b) c) d)
Find the operating temperature and exit conversion. How many steady states are possible under these conditions? What is the maximum adiabatic temperature rise? How would you change some of the operating conditions (Tmo, To, ) in order to operate at a unique steady state of high conversion. e) What start up program should one use in order to have the reactor settle in the steady state of high conversion? Solution Solve simultaneously eqs (15) & (16)
= ar τ = k 0 e − E / RT C Ao (1 − x A )τ 25 k 0 = e E = 20,000 cal / mol
C Ao x A
T o + ω T mo 1 + ω ~ T − ~ β A β A
1
(15)
= ar τ = k 0 e − E / RT C Ao (1 − x A )τ
(16)
Eliminate xA from (15). x A
=
k 0 e − E / RT τ
1 + k 0 e − E / RT τ Substitute into eq (16) T o + ω T mo 1 + ω − T ~ ~ β A β A
k o e −
E / RT
=
τ C Ao
1 + k 0 e − E / RT τ
144 4 244 4 3
14 4 244 3
L
G
Substitute in the values of parameters:
⎧ 20,000 ⎫ ⎬ 1+1 350 + 1 x350 1.9877T ⎭ ⎩ T − = x1 x1 20,000 ⎫ 200 200 ⎧ 1 + exp⎨25 − ⎬ 1.987T ⎭ ⎩ 20,000 ⎫ ⎧ exp⎨25 − ⎬ 1.987T ⎭ ⎩ 0.01T − 3.5 = (*) 20,000 ⎫ ⎧ 1 + exp⎨25 − ⎬ 1.987T ⎭ ⎩ exp⎨−
Solve * by trial and error for T. Then obtain the corresponding conversion by:
⎧ ⎩
exp⎨25 − x A
=
20,000 ⎫
⎬
1.987T ⎭ 20,000 ⎫
(**)
⎧ 1 + exp⎨25 − ⎬ 1.987T ⎭ ⎩ 13
ChE 471
Fall 2005 LECTURE 7
We can also represent equation (*) graphically as shown below:
Example 1 1
0.9
0.8
0.7
0.6
A
X 0.5
0.4
0.3
0.2
0.1
0 320
340
360
380
400
420
440
460
480
500
Temperature, K
Figure 4:
Operating points for the CSTR of Example 2.
Figure 5 on next page shows the dramatic temperature excursions that the reactor can experience during start-up before it settles to a steady state.
14
ChE 471
Fall 2005 LECTURE 7
Error!
Figure 5:
Transient CSTR operation for Example 2
15
ChE 471
Fall 2005 LECTURE 7
a)
There are three intersections of the L and G line indicating 3 possible steady states. (1) (2) (3)
T = 353K ⎨ ⎩ x A = 0.03 T = 408 K ⎨ ⎩ x A = 0.576 T = 439 K ⎨ ⎩ x A = 0.886
b)
Three steady states are possible.
c)
Δ T ad max = β˜ A C Ao = 200 x1 = 200 K T ad max = T o + ΔT ad max = 550 K !
d)
Increase To or Tmo to bypass the lower bump in the curve. Increase
if possible.
ChE 471
Fall 2005 LECTURE 7
a)
There are three intersections of the L and G line indicating 3 possible steady states. (1) (2) (3)
T = 353K ⎨ ⎩ x A = 0.03 T = 408 K ⎨ ⎩ x A = 0.576 T = 439 K ⎨ ⎩ x A = 0.886
b)
Three steady states are possible.
c)
Δ T ad max = β˜ A C Ao = 200 x1 = 200 K T ad max = T o + ΔT ad max = 550 K !
d) e)
Increase To or Tmo to bypass the lower bump in the curve. Increase See attached Figure 5.
if possible.
Reversible Reactions
For reversible reactions the effect of temperature on equilibrium must be considered. For endothermic reactions, temperature.
Δ H r > 0 , equilibrium conversion increases with increased
x A
x A e
II III I
Figure 6:
T Equilibrium conversion as function of temperature for endothermic reactions.
In region III the reaction can be considered practically irreversible. In region II xAe rises with increase in T. For endothermic reactions the maximum permissible temperature is always the optimal temperature for maximizing conversion or production rate from a given CSTR.
16
ChE 471
Fall 2005 LECTURE 7
The net rate of an endothermic reaction at fixed composition always increases with increased temperature. r = k 10 e
− E 1 / RT
s
∏ C
α j
j
− k 20 e
j =1
since E 1
− E 2 / RT
s
∏ C
β j j
j =1
> E 2
For exothermic reactions, temperature.
Δ H r < 0 , equilibrium conversion decreases with increased
x A
x A e
II III I
T
Figure 7:
Equilibrium conversion as function of temperature for exothermic reactions.
Again in region I the reaction is practically irreversible. In region II it is reversible. x A r = 0.01
r = 0.02 r=0
r = 0.04
0.06
T
Figure 8:
Conversion temperature relation at fixed rates for exothermic reactions.
Now we notice that at fixed composition, fixed xA, the net rate of reaction has a maximum at a certain temperature, Tm. Below that temperature the rate is lower and above it, it is lower. The reason for this is that E 1 < E2. On the above diagram we can pass a line - - - called the locus of maximum rates or a Tm line. For a given conversion the Tm line defines a temperature at which the rate is maximum and vice versa at every T the line defines an xA a which the rate is maximum.
17
ChE 471
Fall 2005 LECTURE 7
We should always select To, T mo, line.
in such a manner as to make sure that we operate on the Tm
The equation for the Tm line is obtained by ∂ r ∂ T
= 0 from
dx A dT
r
=0
For example if we have a reversible reaction aA+bB=pP
and the rate is given by r = k 10e
− E 1 / RT
α β
C A C B
− k 20 e− E / RT C pγ 2
Assuming further that we deal with liquids and that CAo and CBo are given while C po = 0
K c
p ( x Ae ) p a
= C Ao p −a −b
⎛ C Bo
(1 − x Ae )a ⎜⎜
⎝ C Ao
−
b a
⎞
b
(A)
x Ae ⎟⎟
⎠
Assuming an ideal solution −Δ Gr o
K c
= e RT
(B)
Using (A) and (B) we can calculate xAe as a function of temperature. The given rate form, if it is to be viable in the vicinity of equilibrium, must satisfy the following constraint: q
p ⎛ C pγ ⎞ C p ⎜ α β ⎟ = a b ⎝ C A C B ⎠ eq C A C B
The locus of maximum rates i.e the Tm-line can be obtained by
18
∂ r ∂ T
= 0 which results in
ChE 471
Fall 2005 LECTURE 7
T = T m
=
( E 2 − E 1 ) / R
⎡ ⎤ ⎛ p ⎞γ ⎢ ⎥ ⎜ x A ⎟ a ⎢ k 20 E 2 γ −α − β ⎥ ⎝ ⎠ ln⎢ C Ao ( ) β ⎥ k 10 E 1 ⎛ ⎞ C b ⎢ (1 − x A )α ⎜ Bo − x A ⎟ ⎥ ⎢⎣ ⎝ C Ao a ⎠ ⎥⎦
(C)
For a desired conversion eq (C) gives the temperature Tm at which the rate is maximum. Substituting that conversion and temperature into the equation for the rate produces the maximum rate Example 3 - CSTR A ⇔ R
ΔGr o298 = 2,500cal / mol Δ H r o298 = −20,000cal / mol = const ρ C p = 2,000(cal / lit ) = const k 1
8 10 = 51 x2 3e −
12,500
k 10
mol ⎞ = 2⎛ ⎜ ⎟ ⎝ lit ⎠
C Ao
f)
g) h)
RT
(min −1 )
Find he optimal size CSTR necessary to achieve a production rate of FR =100 (mol/min) at xA = 0.9. If the reactor is to be operated adiabatically find the necessary feed temperature. If the feed is available only a T = 298K how should one operate? Can one maintain the desired production rate and conversion? How should 2 CSTR’s in series be operated to minimize the total reactor volume and keep FR and xA at desired levels. Feed is at 298K, T = 350Kis not to be exceeded.
Solution
Find equilibrium constant at 298K K 298
⎛ − ΔGr o298 ⎞ − 2,500 ⎞ ⎟⎟ = exp⎛ = exp⎜⎜ ⎜ ⎟ = 68.2 ⎝ 1.987 x 298 ⎠ ⎝ RT ⎠
Find equilibrium constant as a function of temperature using Van’t Hoff’s equation d ln K Δ H r dT K T
=
2
RT
= K 298e
Δ H r ⎛ 1 1 ⎞ − ⎟ ⎜ R ⎝ 298 T ⎠
10.065 ⎞ = 1.461 x10 −13 exp⎛ ⎜ ⎟ ⎝ T ⎠
19
ChE 471
Fall 2005 LECTURE 7
Find equilibrium conversion variation with temperature:
x Ae
K T
=
1 + K T
298
T
⎛ 10,065 ⎞ ⎟ T ⎠ ⎝ = ⎛ 10,065 ⎞ 1 + 1.461 x10 −13 exp⎜ ⎟ ⎝ T ⎠ 1.461 x10 −13 exp⎜
300
310
0.986 0.982
x Ae
320
330
340
350
360
370
380
0.949 0.870 0.720 0.512 0.311 0.169 0.094 0.045
Find the locus of maximum rates T m
=
( E 2 − E 1 ) R 10,065 = ⎡ k E x A ⎤ ⎛ x A ⎞ ⎜⎜ ⎟⎟ 30 . 51 ln + ln ⎢ 20 2 ⎥ 1 x − − k E 1 x ⎝ A ⎦ A ⎠ ⎣ 10 1 x A
0.94
0.95
0.9
0.85
0.8
0.6
0.4
0.2
0.1
0.01
T m
287
301
308
312
316 326 334 346 356
388
Plot the equilibrium line, xAe, and the locus of maximum rates, Tm, in order to graphically interpret some of the later results. T m 1
0.8
x A e
x A 0.6
0.4
0.2
280
a.
At x A
= 0.9
290
300
T = T m
320
340
360
= 308K
The rate at the operating point is: r = 5 × 10 e 8
−
12,500 1.987 × 308
×2 (1 − 0.9)− 3.422 × 10
21
−
e
32,500 1.987 × 308
⎛ mol ⎞ ⎟ ⎝ lit min ⎠
r = 0.815 ⎜
20
× 2 × 0.9
380
400
T
ChE 471
Fall 2005 LECTURE 7
From the design equation (balance on R) V =
F R R R
=
F R r
V = 1.23 m
=
100 0.815
= 1227 (lit )= 1.23 m3
3
For adiabatic operation T o − T + β÷ A
(−V )r τ = 0
β÷ =
A
T o = T − β÷ A C Ao x A
−Δ H rA ρ C p
= 10
= 308 − 10 × 2 × 0.9 = 290K
For adiabatic operation the feed temperature would have to be 290K. obtained graphically (--- adiabatic line). b.
The same result is
If the feed is available only at 298K we could do one of the following: i)
Cool the feed from 298K to 290K and operate adiabatically 1
T o = 290K
T o = 298K
−qś = Q ρ C p
( − )=
F R
1
T o T o
C A o x A
T = 308K
(
1
ρ C p T o − T o
)
⎛ cal −qś = × 2000 (298 − 290)= 8.889 × 10 ⎜ ⎟ 2 × 0.9 ⎝ min 100
ii)
5
Introduce the feed at 298K and cool the reactor T o = 298K
T = 308K
21
ChE 471
Fall 2005 LECTURE 7
−qś = Q ρ C p (T o − T )+ (−Δ H r )C A A
−qś =
F R C A o x A
(
o
Q x A
)+ (−Δ H ) F
ρ C p T o − T
r
R
⎡ 2000 ⎤ 5 298 308 20,000 −qś = 100⎢ − + ( ) ⎥ = 8.889 × 10 ⎣2 × 0.9 ⎦ ↑
cal
min
The cooling rate requirement is the same, as it should be as there is no work term in the energy balance. But it is easier to remove heat from the higher reactor temperature. iii)
Operate adiabatically with the new feed temperature while maintaining the previous feed 100 rate F A o = ⎜ ⎟ ; F Ao x A = F R 0.9 ⎝ ⎠ Qo
=
F Ao C Ao
=
100 1.8
Now we must find the new operating point by a simultaneous solution of the species and energy balance. C Ao x A T o
= r τ ~
− T + β τ r = 0
Since we need adiabatic operation by combining the above two equations we get the relationship between conversion and temperature T = T o
~
~
+ β r τ = T o + β C A x A T = 298 + 10 × 2 x A = 20 [14.9 + x A ] o
Substituting into the first equation we get: −
12,500
−
[ ] × 2 1 − x − 3.4 ×10 21 e = ⎢5 × 10 8 e ( A ) ⎢ ⎣ 314.5 817.8 ⎫ ⎧⎪ − − ⎪ 14.9 + x 14.9 + x 9 21 − 6.8 ×10 x A e 11.043⎨10 (1− x A )e ⎬ − x A = 0 ⎪⎩ ⎪⎭ 2 x A
1.987 × 2 0 1 4.9 + x A
A
A
By trial and error x A
= 0.88
T = 315.6 = 316 K
22
32,500
(
1.987 × 20 14.9 + x A
)⎥ 1227 ×1.8
⎥ ⎦
100
ChE 471
Fall 2005 LECTURE 7
The purity (conversion) has been reduced somewhat from the required 90%. The production rate now is F R and rate is r . r = 5 ×10 e 8
−
12,500 1.987 × 315.6
(
)− 3.4 ×10
2 1− 0.88
21
−
e
32,500 1.987 × 315.6
2 × 0.88
⎛ mol ⎞ ⎟ ⎝ lit min ⎠
r = 0.0786 ⎜
F R
= F A x A =
mol ⎞ × 0.88 = 97.8 ⎛ ⎜ ⎟ 0.9 ⎝ min ⎠
100
o
and production rate has dropped very little. iii)
We could operate the reactor adiabatically with the new feed temperature and maintain (via a controller) fixed exit composition at x A = 0. 9 which would require adjusting the feed rate.
Now T = T o x A
~
+ β A
C Ao x A
= 298 + 20 × 1.8 = 316 K
= 0.9
The new rate is: r = 5 ×10 e 8
−
12,500 1.987 × 316
× 2 (1 − 0.9)− 3.4
21
10 e
−
32,500 1.987 × 316
× 2 × 0.9
⎛ mol ⎞ = r 0.023 ⎜ ⎟ ⎝ lit min ⎠ Which is much smaller than the maximum rate. The production rate now drops to F R
mol ⎞ = r V = 0.023 × 1227 = 28.2 ⎛ ⎜ ⎟ ⎝ min ⎠
This is too high a penalty to pay for maintaining purity i.e. keeping x A
23
= 0. 9 constant!
ChE 471
Fall 2005 LECTURE 7
C.
For minimum total reactor size we must find the minimum of the following expression τ 1 + τ 2 C A o r m1
With T m
=
x A1 r m1
= 10 9 e
−
0.9 − x A1
+
r m f
12,500
(1 − x )− 6.8 × 10
1.987 T m
A1
−32,500 21
e
1.987 T m1
x A1
10,065
=
⎛ x A ⎞ ⎟ ⎜ 1 − x A ⎟ ⎝ ⎠
30.51 + ln⎜
1
1
r m f
By taking
x A1 T m1
= 0.0818 (mol d d x A1
⎜⎜ ⎝
lit min
)
τ 1 + τ 2 C A o
⎟⎟ = 0 one gets by trial and error ⎠
= 0.72 = 320 K
r m1
mol
= 0.5 ⎜ ⎟ ⎝ lit min
The required volumes are: For the first reactor V 1 =
F A o x A1 r m1
=
100 0.72
⋅
0.9
0.5
= 160 liters
For the second reactor
V 2
=
(
F A o x A f r m f
Total volume V = V 1
− x A
1
)= 100 (0.9 − 0.72) = 244.5 lit 0.9
0.0818
+ V 2 = 405 liters
The graphical representation of the above system is shown below
24
ChE 471
Fall 2005 LECTURE 7
1
T o = 298K
306
320
308
304
0.8
x A
q&1
0.6
− q& 2
V 1 adiabatic
V 2 adiabatic
0.4
0.2
T o 300
320
340
360
380
T
Preheat the feed from 298K to 305.6K ≈ 306K 100 × 2000 (305.6 − 298) = 8.44 × 10 5 q&1 = Q ρ C p (305.6 − 298) = 0.9 × 2
⎛ cal ⎞ ⎜ ⎟ ⎝ min ⎠
Cool the stream from 1 reactor to 304.4K ≈ 304 st
− q& 2 = Q ρ C p ΔT =
cal ⎞ × 2000 (320 − 304.4) = 1.73 × 10 6 ⎛ ⎜ ⎟ 0.9 × 2 ⎝ min ⎠
700
It should be noted that the above calculations can be made easier with a little analytical work to obtain an expression for the maximum rate as a function of conversion. This is done by substituting the T m values T m
=
( E 2 − E 1 ) R ⎡ k E x A ⎤ ln ⎢ 20 2 ⎥ ⎣ k 10 E 1 1 − x A ⎦
for temperature T in the rate expression
(− R )
A m
= r m = k 10e− E
If we recall that e
1
a ln b
R T m
(
C A o 1 − x A
)− k e
− E 2
R T m
20
C A o x A
a
= e ln b = b a
We get: E E ⎛ E − E E − E ⎛ ⎞ ⎛ ⎞ k E k E ⎜ ⎟ r m = ⎜ k 10 ⎜ 10 1 ⎟ C A − k 20 ⎜ 10 1 ⎟ ⎟ k E k E ⎝ 20 2 ⎠ ⎜ ⎝ 20 2 ⎠ ⎟ ⎝ ⎠ 1
2
(1 − x )
2
1
2
1
25
A
o
E 2
E 2 − E 1
E 1
( E − E ) x A 2
1
ChE 471
Fall 2005 LECTURE 7
In our particular example the above expression reduces to: r m
= 3.22 (1 − x A )
1.625
x A
−0.625
Another variation on the above problem is to have two equal size CSTRs in series. maximum production rate then under conditions τ 1 = τ 2 we must find x A1 that satisfies x A1 r m1
=
x A 2
− x A
For
1
r m 2
This happens at x A1
= 0.75 so V 1 = V 2 = 205 liters
In summary the following types of problems may be encountered: I.
~ Given V , Q, T o , T mo , Qm , β , q& , find x A and T . This is equivalent to determining the performance of an existing reactor or of one that is contemplated by design engineers. In this case species mass balance and energy balance are solved simultaneously.
II.
~ Given T o , T mo , β , F R and x A determine the required V and desired operating T .
This is the typical design problem. III.
Production rate is specified. Exit conversion and temperature can be selected together with V in order to optimize some profit function.
This is also a typical design problem. We have not touched here on two very important problems. -
transients in a CSTR control of a CSTR around an unstable steady state.
26
ChE 471
Fall 2005 LECTURE 7
Extension to Multiple Reactions
One should keep in mind that yields and selectivities can be affected dramatically by the choice of operating temperature when activation energies of various reactions are different. The problems consist of a.
Finding the optimum temperature for a desired product distribution irrespective of space time requirements.
b.
Finding an optimal temperature for a given reactor (given τ ) which maximizes the production of the desired product.
The equations to be solved are given R independent reactions among S species s
∑v
j
ij
=0
A j
i
= 1, 2, ...... R
=1
F jo − F j
j
+ ∑ v ij r i V = 0
(1)
= 1, 2, ...... R
or R
∑ v [ Xś − r V ]= 0 ij
i
i
i=1
Xś1 − r i V = 0 for i = 1, 2,3... R
(1a)
and s
∑ (F H ~ jo
jo
~
− F j H j ) + q& = 0 (2)
j =1
Equation (1) represents independent R species balances and eq (2) is the energy balance based on assumptions of a) b)
no work other than flow work negligible changes in potential and kinetic energy
If we assume further c) ideal gas behavior d) ideal mixture
27
ChE 471
Fall 2005 LECTURE 7 s
∑ F
jo
j = 1
T o
⎜⎜Δ H + ∫ C ⎝ ~ fj T ~ p 0
T o
s
∑ F ∫ C jo
j = 1
T
s
T o
~
pj
∑ F ∫ C jo
j = 1
Where
T
Δ H r
Ti
~
j
dT ⎟⎟ −
⎠
s
T
∑ F ⎜⎜Δ H + ∫ C j
j = 1
⎝
~ fj
T ⎛ dT − ∑ ∑ v ij ⎜⎜Δ H + ∫ C ⎝ ~ fj T ~ p j = 1 i = 1 s
dT −
∑v i=1
pj
T
R
o
R
0
j
~ p j
dT ⎟⎟ + qś = 0
⎠
⎞ dT ⎟⎟ r i V + qś = 0 ⎠
⎛ ⎞ ⎜ ⎟ r V + qś = 0 H Δ ij ⎝ ~ r ⎠ i
(2a)
Ti
is the heat of reaction of reaction i at temperature T .
If we further assume e) constant heats of reaction and f) constant mean specific heats we get
(
R
)+ ∑ (−Δ H ) r V + q = 0
ρ C p Q T o − T
r i
i
ś
i=1
28
(2b)
ChE 471
Fall 2005 LECTURE 7
PROBLEMS
A.
Consider the following simple reaction A ⇔ R
The reaction rate is given by: π = k 1 C A
− k 2
(mol lit s ) ⎛ 83,000 ⎞ × 10 3 (s −1 ) k 1 = exp ⎜ 7 − ⎟ RT ⎠ ⎝ 167,400 ⎞ ⎛ k 2 = exp ⎜18 − ⎟ × 10 3 (s −1 ) RT ⎠ ⎝ C R
Note: the activation energies are given in (J/mol) use the value of the gas constant accordingly.
The heat of reaction is:
Δ H r = − 80,000 ( J mol ) Specific heat of the reaction mixture is: C
~ p
= 40 J (
)
mol K
Assume that these are constant. 1.
The above reaction occurs in the liquid phase. The pressure is high enough so that the liquid will not volatize in the permissible temperature operating range which is: 300 to 900K.
The feed conditions are: Qo
= 100 (lit s );
C Ao
= 1 (mol
lit ) ; T o
= 300K
You have a V = 100 liters CSTR reactor. How would you operate this CSTR in order to maximize the production rate of R?. a)
What is F R (mol s ) -maximum production rate?
b) c) d)
What are x A & T -operating conditions? What is the heat duty for the system? Sketch your system and location and heat duty of all heat exchangers, if any.
29
NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471
Fall 2005 LECTURE 8
NONISOTHERMAL OPERATION OF IDEAL REACTORS Plug Flow Reactor
T o
T
F jo , Qo
F j T m
T m ,Qm
T
T o
T mo
T m
Assumptions:
1. 2. 3.
Homogeneous System Single Reaction Steady State
Two types of problems: 1. 2.
Given desired production rate, conversion and kinetics and other parameters, determine the required reactor size, heat duty and temperature profile. Given reactor size, kinetics, etc., determine the composition of the exit stream.
Let us consider a single reaction s
∑ υ A = 0 j
j =1
(1)
j
with the rate given by s r = k 10e
− E 1 / RT
s
Π C j − k 20 e j = 1 α j
− E 2 / RT
Π C j β j = 1 j
(2)
with M j / A C j
= C Ao
−
υ j υ A
x A
1 + ε A x A
T o P TPo
(3)
The mass balance in the reactor for species j can be written as: dF j dV v
=0
= υ j r
(4)
F j = F jo
(4a) 1
ChE 471
Fall 2005 LECTURE 8
or dx A
F Ao
dV
V =
= ( −v A ) r = − R A
0
x A
(4’)
= 0
(4’a)
The energy balance based on (a) negligible changes in potential and kinetic energy and (b) no work other than flow work is s ˜ ⎜ ⎟ + Ýqv = 0 − F H ∑ j j ⎟ dV ⎜⎝ j =1 ⎠
d
(5)
F j H÷1 = F jo H÷ jo
V = 0
(5a)
Based on further assumptions of (c) ideal mixtures and (d) ideal gases one gets: s
−∑ F j C
dT
~ p j
j =1
dV
s
− ∑ H j =1
~ j
dF j dV
+ qśv = 0
(6a)
Using the idea of (e) mean specific heats which are constant and (f) constant heat of reaction, one gets
−( Q ρ )C p
dT dV
+ (−Δ H r ) r + qśv = 0
(6)
Q ρ = mtot is the mass flow rate which is constant qÝv
J ⎞
⎝ m3s ⎠ is the rate of heat addition per unit reactor volume
The simplest constitutive relationship for the rate of heat exchange is: qśv
= Uav ( T m − T )
(7)
2
m ⎞ av ⎜ 3 - area for heat transfer per unit reactor volume ⎝ m ⎠
The equations to be solved simultaneously are: F ao
dx A dV
+ υ A r = 0
(A)
2
ChE 471
Fall 2005 LECTURE 8
− qv 647 48 − (−Δ H r ) r + U av ( T − T m ) = 0 ś
Q ρ C p m
dT dV
(B)
− qv 647 48 dT m − U av ( T m − T ) = 0 mQm ρ m C p m dV ś
(C)
V = 0; xA = 0; T = To, , (Tm = Tmo for cocurrent flow V = V; (Tm = Tmo for countercurrent flow)
(D)
and G
du dz
+
dp
+ F = 0
dz
(E)
G = ρ u - mass velocity P = pressure z =
V
u=
A Q A
- axial distance - velocity
A – cross sectional reactor area F – frictional losses Equation (E) is the momentum balance. However this equation is usually solved separately and a mean pressure is selected for evaluation of gas concentrations in eq (3). For gases the use of mass fractions, w j, and extent per unit mass, ξ '' is recommended. (See lecture 1). The equations can then be written as: G
G
d ξ ' ' dz
dT dz
z =
0
= r
(8)
= β ' ' r + q''v
(9)
''
ξ
= 0,
T = T o
(10)
3
ChE 471
Fall 2005 LECTURE 8
''
β
=
−Δ H r
; q'' v =
C p
qv C p
(11)
where the rate is expressed by: s
⎡ ⎤∑=1 υ ⎢ ⎥ s «m ⎛ ⎞α ⎢T P ⎥ 1 − E / RT Π C joα ⎜⎜1 + υ j tot ξ © © ⎟⎟ ⎢ o r = k 10e 1 ⎥ s F jo TPo ⎝ ⎠ ⎢ j = 1 1 + ∑ υ j M av ξ ′′⎥ ⎢⎣ ⎥⎦ j = 1
j
j
j
j
o
s
β ⎡ T P ⎛ Ý ⎞ m β − E 2 / RT bot ' ' o ⎜ ⎟ − k 20e Π C jo 1 + υ j ξ ⎢ F jo ⎠ ⎣ TPo 1 + ⎝ j = 1 j j
j
⎤∑ M av ξ "" ⎥⎦ ∑ y M v j
1
( )
(12)
o
M avo - average molecular weight at feed conditions mtot = GA – mass flow rate
«m
tot
F jo
=
M j w jo
w jo – mass faction of j in the feed. For liquids one can write d ξ ξ d τ τ dT d τ τ
= r
(13)
= β˜r + q˜v
(14)
τ = 0
; ξ = 0 ; T = T o
(15)
−Δ H r qv Q''v ˜ ˜ = β = ; qv = ρ C C p
ρ C C p
(16)
ρ
where the rate is given by s r = k 10e
− E 1 /
RT
Π j = 1
(C
jo
)
+ υ jξ
α j
s
− k 20e
− E 2 /
RT
4
Π j = 1
(C
jo
)
β j
+ υ jξ
(17)
ChE 471
Fall 2005 LECTURE 8
τ =
z u
=
V Q
- residence time along the reactor.
From eqs (8) and (9) or (13) & (14) we can always get the following relationship between temperature and extent '' ' '
T = T o + β ξ
+
1
z
q dz G∫ ''
(18a)
v
o
or τ
T = T o + β˜ξ + ∑ q˜v d τ τ
(18b)
o
For adiabatic operation (q'v' = 0, q˜v = 0 ) this yields the equation of the adiabatic line, i.e extent and temperature satisfy the relationship below at any and every point of the reactor '' ' '
T = T o + β ξ
(19a)
T = T o + β˜ξ
(19b)
The maximum fractional adiabatic temperature rise is given by the Prater number just like in the case of a CSTR.
Δ T ad max T o
= β =
(−Δ H )C (−υ )T ρ C r
A
o
Ao
(20)
p
Basic types of problems
1. The temperature in the reactor is prescribed a.
T(z) = To – isothermal reactor. Integrate (8) or (13) and find extent along the the reactor. From eq. (9) or (14) find the heat addition/removal requirement along the reactor and the overall heat duty for the reactor.
b.
T(z) specified. Integrate (8) or (13) find ξ (z). Use ξ (z) and T(z) in eq (9)or (14) to get qv (z)
2. The heat addition (removal) rate is prescribed a)
Adiabatic operation. T = T o + β '' ξ '' integrate
or
5
T = T o + βξ .
Substitute into eq (8) or (13) and
ChE 471
Fall 2005 LECTURE 8
b)
Heat duty is prescribed. q''v ( z) or qv ( z z ) prescribed. Simultaneously integrate integrate (8) or (9) or substitute '' ''
T = T o + β ξ
+
1
z
q dz into (8) and integrate. G∫ ''
v
o
3. Rate of heat addition (removal) controlled by another equation qśv
= Uav ( T − T m )
a)
Tm = const. Integrate eqs (8) and (9) or eqs (13) and (14) simultaneously. This is the case when reactor tubes are immersed in boiling medium or condensing medium.
b)
Tm determined with T and ξ ’’ by equations (A) to (E). Gm
κ m
dT m
= mκ m' (T m − T )
dz Uav m
=
Gm
C p m
=
Qm ρ m Am
Note: With cocurrent cooling coo ling a PFR can be kept isothermal with countercurrent cooling c ooling it cannot in the case of n-th order reactions. Prove that for an exercise. There is always a unique steady state in a PFR. Main problem with PFR’s is: hot spot formation parametric sensitivity and temperature runaway. Classical example of temperature runaway presented by Bilous & Amundson (AIChE J., 2, 117 (1956). PFR cooled from the wall t constant Tm = Twall.
6
ChE 471
Fall 2005 LECTURE 8
440 342.5
420 337.5 400
T 380
360 335 340
330 320
320
310 Tm = 300
300
0
10
20
30
τ
A “hot spot” is formed due to a very small change in wall temperature. The system shows extreme parameter sensitivity. Reaction runaway is the phenomenon when a small change in feed concentration, temperature, flow rate or in coolant temperature triggers a dramatic change in he temperature profile and leads to runaway reactions and explosions. Exact criteria for runaways are difficult to develop. Approximate criteria are given on the enclosed graph.. Example 1 A reversible first order reaction (considered earlier in a CSTR) is now to be per formed in a PFR.
→ (liquid phase) A R ← 8 −12,500 / RT (min−1 ) k 1 = 5 x10 e k 2
= 3.4 x1021 e −32,500 / RT (min−1 )
o Δ H r = −20,000 cal / mol ΔG298 = −2,500 cal / mol ρ C p − 2,000 (cal / lit oC ) C Ao = 2 ( mol / lit )
7
ChE 471
Fall 2005 LECTURE 8
If the feed rate is Q = 100 (lit/min) and the PFR size is V = 1,500 (lit): a)
find final conversion in an isothermal reactor operated at 0, 10, 20, 100 C
b)
determine conversion in an adiabatic reactor if the feed is at i) 0 C, ii) 20 C,
c)
if the maximum permissible temperature is 80 C determine the optimal temperature profile along the reactor necessary to maximize exit conversion.
d)
If the desired conversion is 85% find the minimum reactor volume and the desired heat removal rate along the reactor. Permissible temperature range is 0 to 100 C.
˚
˚
˚
˚
˚
˚
Solution a)
For an isothermal reactor only the mass balance has to be solved τ =
V
x A
= C Ao ∫
dx A
−r A − r A = k 1C A − k 2 C R = C Ao[ k 1(1 − x A ) − k 2 x A ] ⎡ x A( 1 − x Ae ) ⎤ −r A = k 1C Ao⎢1 − x A − ⎥ sin ce k 2 = ⎢⎣ ⎥⎦ x Ae k C K k (− r A ) = 1 Ao ( x Ae − x A ) x Ae = = 1 1 + K k 1 + k 2 x Ae (− r A ) = (k 1 + k 2 )C Ao ( x Ae − x A ) x 1 1 dx A x Ae ⎟⎟ ln⎜⎜ = τ = ∫ k 1 + k 2 o x Ae − x A k 1 + k 2 ⎝ x Ae − x A Qo
o
A
Solve for conversion 1 − ⎡1 − exp(−k 1(1 + )τ )⎤ ⎣ K ⎦ ⎡ ⎛ −k ⎞ ⎤ 1500 = x Ae ⎢1 − exp ⎜ 1 τ ⎟ ⎥ = 15 min τ = 100 ⎝ x ⎠ ⎣ ⎦ Ae
x A
8
k 1 K
=
k 1
(1 − x
Ae
x Ae
ChE 471
Fall 2005 LECTURE 8
We get the following results: T 273 283 293 303 313 323 333 343 353 363 373
K 1494 407 121 40 13.5 5.0 2.0 0.81 0.35 0.16 0.08
k 1 0.0498 0.112 0.239 0.486 0.943 0.755 3.149 5.46 9.17 15.0 23.8
xae 0.999 0.998 0.992 0.975 0.931 0.833 0.662 0.448 0.262 0.139 0.071
xa 0.526 0.813 0.965 0.974 0.931 0.833 0.662 0.448 0.262 0.139 0.071 Same as equilibrium conversion
The reactor space time is so large that above 50 C practically equilibrium conversion is obtained. ˚
a)
The adiabatic operating line is T = T o + β˜ AC Ao x A
−Δ H r 20,000 ⎛⎜ lit oC ˜ β A = = = 10 ⎝ mol 2,000 ρ C p A
= 2⎛ mol ⎞ ⎝ lit ⎠ T = T o + 20 x A C Ao
Substitute this relationship into the mass balance and integrate: dx A
= ( k 1 + k 2) C Ao( x Ae − x A ) d τ τ = 0 x A = 0 E / RT − E / R( T o + 20 x A ) k 1 = k 10e 1 ad = k 10 e 1 C Ao
= k 1C Ao − (
k 1
+ k 2) C Ao x A
− E / R( T +20 x )
o A k 2 = k 20 e 2 K k = 1 x Ae = 1 + K k 1 + k 2
Thus integrate numerically dx A
= k 10 e− E 1 / RT ( T + 20 x ) − ⎝ k 10e d τ τ = 0 x A = 0 o
A
− E 1 / R
(T +20 ) o
x A
+ k 20 e− E 2 / R ( T +20 x ) ⎞ ⎠ x A o
9
A
ChE 471
Fall 2005 LECTURE 8
dx A d τ
= 5 x10
τ = 0
8
−
e
12,500 1.987( T o +20 x A )
; x A = 0
12,500 32,500 − − ⎡ ⎤ 8 1.987( T +20 x ) 21 1.987( T +20 x − ⎢5 x10 e + 3.4 x10 e ⎥ x A ⎣ ⎦ o
A
o
A
Desired result is obtained at τ = 15. Alternatively we could solve by trial and error the following integral: x A
τ = 15 =
dx
∫ o
12,500 − 8 1.987(T o + 20 x )
5 x10 e
12,500 32,500 − − ⎡ ⎤ 8 1.987( T +20 x ) 21 1.987( T +20 x ) x − ⎢5 x10 e + 3.4 x10 e ⎥ ⎣ ⎦ o
o
We find: i)
˚
ii) c)
To = 0 C = 273 K xA = 0.78 Δ T adiabatic = 15.7K = 16K T = 289 K To = 20 C = 293 K xA = 0.94 = xAe Δ T adiabatic = 18.8 = 19 K T = 292 K ˚
In order to maximize conversion at given space time we should follow the line of maximum rates. T m
( E 2 − E 1 / R)
= ln
k 20 E 2 k 10 E 1
+ ln
10, 065
=
x A
1 − x A
⎛
30.51 + ln⎜⎜
x A
⎝1 − x A
⎟⎟
Since maximum permissible temperature is 80 C (353 K) we have to preheat the feed to 33 K, cool the reactor and keep it isothermal a 353 K until the locus of maximum rate is reached and then run along the locus of maximum rates. ˚
The intersection of the isothermal line T = 353 K and the T m line determines up to which point the reactor has to be run isothermally. T = 353 = T m
=
10,065
⎛ x A ⎞ ⎟ ⎝ 1 − x A ⎠
30.51 + ln⎜
10,065 − 353 x 30.51 ⎟ 353 ⎝ ⎠ x A = = 0.119 ⎛10,065 − 353 x 30.1 ⎞ 1 + exp⎜ ⎟ 353 ⎝ ⎠ exp ⎜
10
ChE 471
Fall 2005 LECTURE 8
τ =
0.119
1
∫
k 1 + k 2
o
dx
( x Ae − x A )
=
x Ae ⎞⎟ ln⎜ ( k 1 + k 2 ) ⎝ x Ae − x A ⎠
1
At 80 C (353 K) from the table given earlier ˚
τ =
1 9.17(1 +
1 ) 0.35
0.262 ⎞ ⎝ 0.262 − 0.119 ⎠ = 0.017(min)
ln
The isothermal operation should occur in the very entry section of he reactor. After that the Tm line should be followed. dx A d τ
= 5 x10
− 12,500 8 1.987T m
e
− 32,500 21 1.987T m
(1 − x A ) − 3.42 x10 e
x A
10,065 ⎛ x ⎞ 30.51 + ln ⎜ A ⎟ ⎝ 1 − x A ⎠ τ = 0.017 x A = 0.119 T m
=
Desired result at τ = 15 xA=0.988
Texit = 288K
Really one should preheat only to adiabatic line. Adiabatic line should end at T = 353 K, xA = 0.119. Hence, the fluid must be preheated up to T o = T − β˜ AC Ao x A = 353 − 20 x 0.119 = 350 K The graphical representation of parts (a-c) has the following form:
x A
x A e
T
a.
Isothermal. Solid lines are operating lines for τ = 15 min
11
ChE 471
Fall 2005 LECTURE 8
x A
x A e T
b.
Adiabatic. Adiabatic line with τ = 15 x A
T m
x A e T max
c.
Operating along the locus of maximum rates
d)
Permissible temperature range is 0 C to 100 C. We want minimum reactor size for xA = 0.85. Preheat to 100 C, run along the locus of maximum rates ˚
˚
˚
x A = 0.85
τ =
∫ o
with
T m
=
τ = 1.8min
dx 12,500 − 8 1.987 T m
5 x10 e
12,500 − 8 1.987T m
⎡ ⎢⎣5 x10
e
+ 3.4 x10
21
−
e
32,500 1.987 T m
⎤ ⎥⎦ x
10,065
⎛ x A ⎞ ⎟ ⎝1 − x A ⎠ Thus with Q = 100 lit/min we need only V = 160 liters
30.51 + ln ⎜
The desired temperature profile along the reactor is presented in the enclosed graph. The heat removal per unit volume is
12
ChE 471
Fall 2005 LECTURE 8
−
qś Q
= ρ C p ( T o − T ) + (−Δ H r )C Ao x A
= 2, 000(100 − T ) + (20,000 x 2) x A This curve is also presented in the figure. The total heat density is: qÝ −⎜ ⎞ = 2,000(100 − 70) + 40,000 x 0.85 ⎝ Q ⎠ tot
= 1.56 x105 (cal / lit ) − qÝtot = 1.56 x107 (cal /min)
With Q = 100 lit/min
For comparison, if cooling failed and reactor ran adiabatically with To = 100 C one would get ˚
x Aad
= 0.068, T exit = 126 o C
The adiabatic temperature profile is shown also on the enclosed figure. Extension to Multiple Reactions s
∑ υ A ij
j = 1
−
dF j dV
j
=0
i
= 1,2,... R
(1)
R
+ ∑ υ ij r i = 0
j = 1,2,... R
(2)
i=1
or R
d X − υ ij i + dV i =1
∑
−
−
d XÝi dV
R
∑ υ r = 0 ij i
i=1
(2a)
+ r i = 0
( )+ qÝ = 0
˜ d F j H j
V =
dV
(3)
v
0 ; F j = F jo ( Xi = Xio ) ; H˜ j = H˜ jo
With the usual assumptions made about the energy balance (see the lecture on CSTR) one gets:
13
ChE 471
Fall 2005 LECTURE 8
s
−∑ F jo C ~ p j
j =1
dT dV
R
(
+ ∑ −Δ H r i =1
Ti
)r + qÝ = 0 i
(4)
v
The equations to be solved for a set of multiple reactions are:
−
d Xi dV
+ r i = 0
− ρ C p Q V =
0
dT
i
= 1,2... R
(
R
+ ∑ −Δ H r
dV ; Xi
i=1
i
= Xio
(A)
)r + qÝ = 0 i
(B)
v
T = T o
ρ Q = const s r i = k i10e
− E 1i / RT
s
Π C j − k i 20e − E 2 RT Π C j j = 1 j = 1 α ij
i
β ij
(C)
with
∑ υ XÝ ⎞⎟ R
ij i ⎜ i =1 1+ ⎟ F jo ρ To ⎜ C j = C jo R s ρ o T ⎜ Ýi ⎟ υ X ij ∑ ∑ ⎜ ⎟ ⎜ 1 + i =1 j=1 ⎟ F bot o ⎠ ⎝
(D)
The constitutive relationship for qv is: qv
= U a (T m − T ) v
a) Tm= const b) Tm is governed by another D.E.
m ρ mQm C p m
V=0 V=V
dT m dV
− qÝv = 0
(E)
Tm = Tmo (cocurrent flow) Tm = Tmo (countercurrent flow)
14
ChE 471
Fall 2005 LECTURE 8
Problems
→ R Consider the reaction introduced in he last lecture ← A
R=k 1CA-k 2CR (mol/lit s) k 1
83,700 ⎞ 3 -1 = exp⎛⎝ 7 − ⎠ x10 (s ) RT
k 2 exp⎜18 −
⎝
167,400 RT
-1 3 ⎟ × 10 (s ) ⎠
Δ H r = − 80,000 ( J / mol) C = 40( J / mol K ) ~ p
Activation energies given in joules. 1.
The above reaction occurs in liquid phase! Permissible temperature range of operation is 300
To = 300 K
;
CAo = 1(mol/lit)
You have a V = 100 liters PFR. How would you operate this reactor if the only objective is to maximize the production rate of R.
2.
a)
What is maximum FR .
b)
What are final x A and ΔT .
c)
What is the profile of heat addition or removal for every 10% of reactor volume.
d)
What is the overall heat duty for the reactor and any heat exchangers preceding it.
e)
Sketch your system.
The above reaction occurs in gas phase. The gas feed ate is Qo
= 100( lit / s) at T o = 300K , Po = 24.6 atm
The feed is 50%A, 50% inerts. Permissible temperature range is 250< T < 900 K. Pressure is constant in the reactor. Gases start to condense below 250 K. Desired conversion is 85%. 15
ChE 471
Fall 2005 LECTURE 8
a)
What reactor volume is needed if you operate along the locus of maximum rates?
b)
What is the distribution of heat duty along the reactor?
c)
What is the production rate of R?
3.
For the above problem what would FR and xA be if you had a reactor (PFR) of V = 100 liters available?
4.
Suppose that the reactor can only be operated adiabatically and the desired conversion is 85%. Minimize the required reactor size. a)
What reactor type do you recommend?
b)
What feed temperature would you use?
c)
What is the heat duty?
16
WALL COOLED TUBULAR REACTORS: AVOIDANCE OF HOT SPOTS (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471
Lecture 9
Fall 2005
WALL COOLED TUBULAR REACTORS: AVOIDANCE OF HOT SPOTS
The pioneering work of Bilous and Amundsen (AIChE J. 2, 117, 1956) showed that a plug flow reactor (PFR) cooled from the wall can exhibit extreme parametric sensitivity to small changes in the wall temperature or heat transfer parameters. Their observation is sketched in Figure 1 below:
Wall Temperatur e T w
T(°C)
1 440
2
420
3
400
4
380
5
360
o
300 C o
320 C o
330 C o
335 C o
337.5 C
5
4
340
− − − − −
3 2
320 1
300 0
10
20
30
τ(s)
FIGURE 1: Temperature Profiles in a Wall Cooled Tubular Reactor as a Function of Wall Temperature
Clearly a small variation in the wall temperature of 2.5 °C (from 335˚C to 337.5 °C) caused an 80˚C local rise in reactor temperature! This 'hot spot' generation can lead to reactor runaway for many reasons. Undesirable reactions of high activation energy can take off at this elevated temperature, vapor formation can cause increased pressure, rate of reaction may reach extreme values, etc. Therefore, it is imperative to operate the wall cooled tubular reactor in such a manner that it does not exhibit such extreme parametric sensitivity and, hence, is not prone to runaway. To identify the parameter space for safe operation one starts with the mass and en ergy balance for the reactor. Assuming for simplicity, constant volumetric flow rate, Q = const., and a single simple irreversible reaction, A → products, the governing equations are: Mass balance on species A: dx A d τ
=
(− R A ) C Ao
(1)
1
ChE 471
Lecture 9
Fall 2005
Energy balance (assuming constant average physical properties and constant heat of reaction): ρ C p
dT d τ
= (−Δ H r )(− R A )− Uav (T − T w )
(2)
A
The initial conditions are: τ = 0 x A
= 0,
T = T o
(2a)
In the above equations: V
τ =
(min ) is the space time measured from the reactor entrance as reactor internal Q volume V is also measured from the entrance. C Ao ⎜
mol
⎟
=
molar feed concentration of A
=
fluid density
=
mean specific heat of the reaction fluid
=
heat of reaction per mole of A
− R A ⎜ 3 ⎟ ⎝ m min
=
rate of reaction of A
T(˚K) xA
= =
temperature in kelvin conversion of A
=
overall heat transfer coefficient
=
area for heat transfer per unit internal reactor volume
= = =
coolant (wall) temperature inlet feed temperature tube (reactor) diameter
⎝ m3
( C (J / kg K
ρ kg / m 3
o
p
J
Δ H r ⎜ ⎟ ⎝ mol A A
mol A
J
U ⎜
⎝ m 2K min
(
2
av m / m
3
⎟
)= d 4
t
Tw (˚K) To (˚K) d t (m)
To simplify the algebra, but without loss of g enerality of the result to be derived, we will take To= Tw thus assuming the inlet and wall temperature to be the same. We can now divide eq (2) by eq (1) and rearrange the result to get: dT dx A
=
−Δ H r
A
C Ao
ρ C p
−
( ) d ρ C (− R )/ C 4U T − T o
t
p
A
(3)
Ao
2
ChE 471
Lecture 9
Fall 2005
Let us define the characteristic reaction time based on the value of the reaction rate at the feed conditions of concentration and temperature: τ R
C Ao
=
=
(− R A )o
C Ao k o e
− E / RT o
C Ao
n
(4)
Here we have assumed for simplicity an n-th order irreversible reaction and
(− R A )o means that
we evaluate the rate at feed conditions of concentrations and temperature. We also have a characteristic heat transfer time: τ h
=
d t ρ C p
4U
(5)
Based on these, two dimensionless groups arise naturally: κ = N = and
τ R τ h
=
4U C Ao
(− R ) d ρ C A o
(−Δ H )C r A
β =
Ao
ρ C p T o
t
(6)
p
(7)
Equation (3) now becomes (recall we assumed T o dT dx A
= β T o − κ
= T w )
T − T o
(− R )/(− R ) A
(8)
A o
This last expression contains the dimensionless rate of reaction,
(− R A ) / (− R A )o , which for an n-
th order irreversible reaction can be written as:
(− R ) = k e (− R ) ke A
− E / RT
o
− E / RT o
A o
o
C Ao
⎛ T o ⎞ ⎜1− ⎟ RT o ⎝ T ⎠ E
C An n
=e
(1 − x ) n
A
(9)
Dimensionless activation energy is: γ
=
E RT o
(10)
and dimensionless temperature is conveniently defined by: θ = γ
T − T o T o
(11)
Upon substitution of eqs (9), (10), (11) into equation (8) we get:
3
ChE 471
Lecture 9
d θ dx A
Fall 2005
κ θ
= β γ −
(
) 1 − x
θ / 1+θ / γ
e
(
A
)
n
(12)
The product βγ in the literature is often denoted by δ or S: S = δ = βγ
(13)
We can rewrite eq (12) as: d θ dx A
−n n κ / 1 /. /1 / = δ e−θ ( +θ γ )(1 − x A ) ⎢eθ ( +θ γ )(1 − x A ) − θ ⎥ δ ⎦ ⎣
(14)
Since the term on the right hand side of eq (14) that multiplies the term in the brackets is always positive the sign of the derivative of the dimensionless temperature with respect to conversion, d θ i.e sgn , depends on the terms in the brackets in eq (14). By taking the coolant temperature dx A Tc to be equal to the feed temperature To, we have assured that
d θ dx A
>0
and temperature rises
with conversion. This is obvious by observing and plotting the terms in the brackets in eq (14). Physically this tells us that if we try to cool an exothermic reaction with the coolant at feed temperature, unless the characteristic heat transfer time is infinitesimal in comparison to characteristic reaction time, i.e κ → ∞ , or stated differently the heat removal rate is infinitely faster than the heat generation rate by reaction at the feed (wall) temperature, then the reactor temperature must rise with conversion. We can allow for this but we want to know what values of parameters δ , κ , γ will keep that temperature rise within safe limits. We can think of the first exponential term in the brackets of eq (14) as the heat generation term by reaction and the second as the heat removal rate. Indeed heat generation term has an Arrhenius dependence on temperature, heat removal rate is linear in temperature. Now the heat generation term is difficult to plot as a function of dimensionless temperature θ because of the presence of the (1-xA)n conversion dependent term. We would need to integrate eq (14), save values of xA vs θ and then we could plot the terms in brackets of eq (14) against θ . However, since we are interested in preventing excessive maximum temperatures in the reactor, we can take a conservative approach by assuming the worst possible scenario. In the n-th order reaction (n > 0) the rate of heat generation is being slowed down as conversion increases with θ as the (1-xA)n term becomes smaller and smaller at increased conversion. The worst possible scenario is then if the heat generation rate (and hence the rate of temperature rise) is not affected by increased conversion. This is the case of zeroth order reaction. For a conservative approach assume n = 0 and now consider the term in the brackets of eq (14)
4
ChE 471
Lecture 9
F (θ ) = e
θ / (1+θ / γ )
−
κ δ
Fall 2005
θ = y1 (θ ) − y 2 (θ )
(15)
γ Clearly, as θ = 0, y1 (0) = 1; and as θ → ∞, y1 → e and y1 is bounded. At the same time, as θ = 0, y 2 (0) = 0; and as θ → ∞, y2 is unbounded. Thus, y1 (θ ) is a sigmoidal (S-shape)
curve and y 2 (θ ) is a straight line through the origin as shown in Figure 2. Depending on the value of the ratio of the two dimensionless groups κ / δ the three scenarios depicted in Figure 2 may occur.
1 y1 (θ )
2
3
γ
y 2 (θ )
e
y1 (θ )
, 2 , 3 are y 2 (0 ) for κ decreasing δ 1
0
θ max1
θ max 2
θ max 3
θ
FIGURE 2: Sketch of y1 (θ ), y 2 (θ ) of Eq. (15)
The three straight lines in Figure 2 represent y 2 (θ ) at three different values of the parameter κ / δ . Please, recall that at κ / δ → ∞ our y 2 (θ ) would be a vertical line guaranteeing θ = 0 , i.e T = To = Tw. For the three scenarios depicted above
⎛ κ ⎞ > ⎛ κ ⎞ > ⎛ κ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ δ ⎠1 ⎝ δ ⎠ 2 ⎝ δ ⎠ 3
(16)
and the rate of cooling decreases as we go from case 1 to case 3. Also remember that the value d θ = 0 and yields the maximum of θ at which y1 (θ ) = y 2 (θ ) signifies F (θ ) = 0 i.e. dx A temperature that is observed in the reactor at such conditions. So, in case of rapid heat removal rate (Case 1) the maximum temperature θ max 1 is relatively small and of no concern. At the
5
ChE 471
Lecture 9
Fall 2005
certain lower value of heat transfer rate than in case 1, y 2 (θ ) not only intersects but is a tangent of y1 (θ ) (Case 2). In this case the maximum temperature θ max 2 is still tolerable but any further, even very small, decrease in κ / δ would lead to an excessive maximum temperature as depicted in Case 3 where clearly θ max 3 is a hot spot to be avoided. This tells us that in order to prevent a hot spot we must have: y1 (θ ) = y 2 (θ )
which yields θ max
(17a)
and dy1
≤
d θ
dy 2 d θ
θ max
for avoiding the hot spot
(17b)
θ max
This means that at the maximum temperature, θ max , the slope of the heat generation line, y1 (θ ) , must be less than the slope of the heat removal line, y 2 (θ ) . The condition at which the slopes are equal corresponds to using the equality sign in eq (17b) and is illustrated by line 2 in Figure 2. This defines the maximum allowable maximum temperature θ max 2 . Beyond this point excessive hot spots may arise. We now apply eq (17a) and (17b) to eq (15) to get eθ / (1+θ / γ )
=
κ δ
θ
(18a)
and 1
⎛ 1 + θ ⎞ ⎜ ⎟ ⎝ δ ⎠
2
e
θ / (1+θ / γ )
+θ / γ )
θ /(1 By eliminating e
≤
κ δ
(18b)
κ
θ , as per eq (18a), and by using the δ equality sign in eq (18b), we can develop an equation for the maximum permissible temperature. Do that for an exercise by solving the resulting quadratic equation for θ max . Substitution of the in equation (18b) in terms of
result for θ max in the inequality of eq (18b) determines the parameter space that guarantees the absence of hot spots. The result is not pretty looking. To get a clear and easy formula to remember an approximate analytical solution is developed via θ /(1+θ / γ ) θ a conservative approach in which one replaces e by a rising exponential e . This is ' ' α T equivalent to replacing k=k oe-E/RT , i.e the Arrhenius temperature dependence with k = k o e dk dk ' = and demanding that at T = To, k = k ' and at T = T o i.e at that point of T = T o both dT dT
6
ChE 471
Lecture 9
Fall 2005
θ
θ
+θ
γ
/(1 / ) functions are equal and their derivatives are equal. Indeed e and e have the same value and the same value of the first derivative at θ = 0 . In Figure 3 we sketch the approximation of θ +θ γ θ y1 = e /(1 / ) with e
eθ eθ (1 +θ γ ) e
y1
θ
1
θ
Figure 3: Approximation of the Arrhenius dependence of the rate constant with an exponential function.
Since we are interested in limiting the θ max the two curves are pretty close for small θ . Clearly, θ if we replaced y1 (θ ) with e in Figure 2 we would lose the intersection with y 2 (θ ) for Case 3 but this is not of concern since we are not interested in θ > θ max 2 , and for Case 2 the intersection θ
of y 2 (θ ) with e is close to that of y 2 (θ ) with y1 (θ ) in Figure 2. The advantage of this approximation is that the criteria of equations (17a) a nd (17b) applied to κ θ y1 (θ ) = e and y 2 (θ ) = θ yield δ eθ
=
κ
θ
≤
κ
δ
θ
(19a)
(19b)
and e
δ
From which it follows: θ max
= 1
(20)
7
ChE 471
Lecture 9
Fall 2005
Substitution of this into the inequality described by eq (19b) yields the parameter space that guarantees absence of hot spots e≤
κ δ
(21)
Replacing the above dimensionless quantifies in terms of actual physical variables we get from eq (20) the maximum permissible temperature that can be tolerated in absence of hot spots T max
⎛ 1 ⎞ ≤ ⎜⎜ + 1⎟⎟T o ⎝ γ ⎠
(22)
From eq (21) we get the conditions that the system parameters must satisfy to guarantee absence of hot spots. 4 U R T o2
(
k oe
− E / RT o
C Ao
n
)(−Δ
)
H r A E d t
≥ e
(23)
where R is the ideal gas constant in proper units. This equation is usually written in terms of tube diameter that will guarantee absence of hot spots by providing sufficient heat transfer area per unit volume 2
d t ≤
(
4U R T o e−1
)
E −Δ H r A k oe
− E / RT o
C Ao
n
(24)
Clearly, other permutations of eq (23) are possible. Given kinetic and reaction parameters, feed temperature and tube diameter one can calculate the maximum permissible feed concentration that guarantees safe operation, etc. Approximate analysis, using the approximate exponential dependence of the rate constant, but accounting for the effect of reactant depletion, is possible for an n-th order reaction and yields the following result which is a little less conservative than equ ation (23). κ δ
≥
e
1 + 2.703n 2 / 3δ −2 / 3
(25)
This result is due to Chandler Barkelew, Chem. Engr. Progress Symp. Ser. No. 25, Vol. 55, p. 37 (1959) and is one of the most useful expressions for nonisothermal tubular or batch reactors. The graph showing the parameter space that is safe, based on Barkelew's and similar criteria, is appended as Figure 3.
8
ChE 471
Lecture 9
e A
x
o A
C v a
U
Figure 3:
Runaway diagram
9
Fall 2005
SAFE OPERATION OF TUBULAR (PFR) ADIABATIC REACTORS (CHE 471)
M.P. Dudukovic Chemical Reaction Engineering Laboratory (CREL), Washington University, St. Louis, MO
ChE 471
Lecture 10
Fall 2005
SAFE OPERATION OF TUBULAR (PFR) ADIABATIC REACTORS In an exothermic reaction the temperature will continue to rise as one moves a long a plug flow reactor until all of the limiting reactant is exhausted. Schematically the adiabatic temperature rise as a function of space time, measured from the reactor entrance, takes the form shown in Figure 1.
T ad T
T inf T 0
τ *
τ
FIGURE 1: Temperature as a function of space time in an adiabatic PFR with
exothermic reaction.
The final adiabatic temperature T ad :
⎡
(− Δ H ) C
⎢⎣
ρ C p T o
T ad = T o ⎢1 +
r A
Ao
⎤ ⎥ = T o [1 + β ] ⎥⎦
(1)
is often excessive for highly exothermic systems ( β large) and needs to be avoided. In such situations the old rule of thumb suggests that we should operate in such a way that the inflection ⎛ dT ⎞ point, i.e. the point of maximum temperature rise ⎜ ⎟ , is never reached. This implies that τ d ⎝ ⎠ max we operate with τ < τ * where τ * is the value of space time at which the inflection point
⎛ d 2T ⎞ ⎜⎜ 2 = 0 ⎟⎟ occurs. ⎝ d τ ⎠
1
ChE 471
Lecture 10
Fall 2005
Let us now develop a convenient and simple formula for use by practicing engineers which would guarantee safe operation. Consider an n-th order irreversible reaction:
− R A = k o e − E RT C Ao n (1 − x A )n
(2)
At adiabatic conditions temperature and conversion are related by the adiabatic line equation T = T o (1 + β x A )
(3)
If we define, as in the case of wall cooled reactors, the dimensionless temperature by θ = γ
T − T o T o
(4)
then conversion in Eq. (3) can be expressed in terms of dimensionless temperature as: x A =
θ δ
(5)
where δ = β γ
(6)
γ = E RT o
(7)
with
The rate of reaction evaluated at adiabatic conditions in terms of dimensionless temperature is obtained by using the adiabatic line eq (5) to replace conversion. The result is:
(− R A ) ad = (− R A )o
e
γθ (θ + γ )
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
n
(8)
where the rate evaluated at the feed condition is:
(− R A )o = k o e E RT
o
n
C Ao
(9)
The energy balance for adiabatic PFR operation is: ρ C p
dT d τ
= (− Δ H r ) (− R A )ad
(10)
A
Written in terms of dimensionless temperature it becomes:
2
ChE 471
Lecture 10
d θ d τ
δ
=
γθ (θ + γ )
e
τ R
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
Fall 2005
n
(11)
where the characteristic reaction time is: C Ao
τ R =
(− R A )o
(12)
The initial condition is τ = 0
θ = 0
(13)
We can rewrite Eq. (11) in the most compact way by defining the Damkohler number as the ratio of characteristic process time and reaction time, i.e. Da =
τ τ R
=
(− R A )o τ C Ao
(14)
Equation (11) and initial condition (13) become: d θ d ( Daδ )
At Daδ = 0
=e
γθ (θ + γ )
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
n
(15)
θ = 0
(16)
Now we want to have the reactor “short” enough (limit the conversion achievable) so that we can 2 d θ keep the inflection point = 0 out of the reactor i.e. we do not let it occur in the reactor. 2 d τ The critical temperature at inflection point, θ inf , is obtained by setting 2
d θ d ( Daδ )
2
= 0
(17)
which results in an equation for θ = θ inf yielding: θ inf =
[ γ 2n γ
θ inf = δ ,
2
]
+ 4n (γ + δ ) − 2n − γ ; n ≠ 0
n = 0
(18a) (18b)
3
ChE 471
Lecture 10
Fall 2005
The result for the zeroth order reaction, eq (18b), can be obtained by applying the L’Hospital rule to eq (18a). A better approach is to notice that for n = 0 equation (15) indicates that d θ d (Daδ ) > 0 always, so that there is no inflection point as the rate of temperature rise keeps
rising until all reactant is depleted. Then at x A = 1, from eq (5) it follows that θ max = δ and therefore T max = T ad . We can integrate the differential equation (15) by separating the variables to obtain the critical value of Daδ * and, hence, of τ * * Daδ
θ inf
e−
∫ d ( Daδ ) = ∫ ⎛
Daδ = *
0
0
γθ (θ + γ )
θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
n
d θ
(19)
Clearly for given values of parameters γ , β , δ = βγ , and reaction order n, we can evaluate θ inf from eq (18a) and then calculate the value of Daδ * by numerically evaluating the integral in eq (19). Using then the definitions of Da and δ we get the criterion for safe operation Daδ < Daδ * which can be expressed as: k o e
− E RT o
C Ao
n
(− Δ H ) E τ
*
r A
R ρ C p T o
2
≤ Daδ *
(20)
To get a convenient, easy to remember value of Daδ * , the following approximations are often made. First, the Arrhenius dependence of the rate constant on temperature is replaced by an exponential dependence, in effect e
γθ (θ + γ )
≈ eθ
(21)
Substituting this approximation in eq (15) yields via eq (17) to a new approximate value of the temperature at inflection point
(θ inf )app = δ − n
(22)
Substituting eq (21) and eq (22) into eq (19) yields
( Daδ ) *
app
δ − n
=
∫ 0
e
−θ
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
δ
d θ = δ e n
n
−δ
e
∫u n
4
u n
du
(23)
ChE 471
Lecture 10
Fall 2005
The final approximation (which is conservative in nature as it assumes the worst possible case of zeroth order reaction) ignores the slowdown of the temperature rise due to the consumption of the reactant, which is the same as taking n = 0 in eq (23). This yields δ
( Daδ )
= e
*
app , 0
−δ
∫
( )
e u du = e −δ e u
δ 0
0
= e −δ [e δ − 1] = 1 − e −δ = 1
(24)
(for large enough δ )
(
For highly exothermic reactions δ ≥ 10 and clearly Daδ *
)
app , o
= 1.
Substituting this into eq (20) gives the conservative criterion for safe operation. It is constructive to note that this same equation (20a) with time t replacing τ k o e
− E RT o
C Ao
n
(− Δ H ) E t
*
r A
R ρ C p T o
2
< 1
(20a)
is used to determine the so called “time of no return” or time to explosion in batch systems. This time to inflection point t * may be very long for low T o but becomes quite short if the system of high activation energy is exposed to higher T o . Hence, chemicals that may be safe to store at 25°C may be explosion prone if exposed to 40 – 50°C!
SAFE OPERATION OF ADIABATIC CSTR The mass balance for an irreversible n-th order reaction is: C Ao x A = (− R A ) τ
(25)
The adiabatic equation relates conversion and temperature x A =
T − T o
(− Δ H ) C r A
Ao
ρ C p
(26)
Upon substitution of dimensionless temperature we get from eq (25) 1 Da n δ
n
θ = e
γθ (θ + γ )
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
(27)
let G (θ ) = e
γθ (θ + γ )
⎛ θ ⎞ ⎜1 − ⎟ ⎝ δ ⎠
n
(28a)
5
ChE 471
L (θ ) =
Lecture 10 1 Da nδ
Fall 2005
θ
(28b)
We know from before that G is a sigmoidal curve in θ and represents heat generated by reaction. L is the heat removal rate (i.e., heat removed by sensible heat of the fluid that flows through the CSTR). We know that up to three intersections are possible between G and L lines. To avoid the intersection leading to excessively high temperatures we must assure that intersections at low temperatures are available. The last permissible operating condition is the one when line L is also tangent to curve G as schematically shown in Figure 2. 1
2
G (θ )
3
L(θ )
θ 1
θ 2 = θ *
θ 3
θ
FIGURE 2: Schematic of the G, L vs. θ
Clearly as Daδ increases the slope of the L line decreases so that
( Daδ )1 < ( Daδ )2 < ( Daδ )3
(29)
While operating adiabatic temperatures at θ 1 and θ 2 are acceptable, θ 3 represents too large a temperature jump. Hence, we must assure that the L line always intersects the G line at its lower temperature branch. The critical point is reached when L is also tangent to the G line. For safe adiabatic operation we therefore require
6
ChE 471
Lecture 10
L = G dL d θ
≥
Fall 2005 (30)
dG d θ
(31)
Applying the above to eqs (28a) and (28b), and using the equality sign in eq (31), we get the equation for the maximum permissible temperature θ max perm = θ * . The critical value of the space time τ * can be obtained from the critical value of Daδ * , which in turn results from substituting the expression for θ * into eq (30). To get a simple, easy to remember expression, usually we again replace e γθ (θ + γ ) with eθ . This yields θ * =
1 ⎡ δ − n + 1 − 2 ⎢⎣
(δ − n + 1)2 − 4 δ ⎤⎥ ⎦
(32)
The negative sign in front of the square root needs to be taken as we are interested in the lower of the two temperatures at which the L line could be tangent to the curve G. Then equation (30) yields:
( Daδ )
* app
⎛ θ * ⎞ = θ ⎜⎜1 − ⎟⎟ ⎝ δ ⎠
−n
*
e
−θ *
(33)
A conservative estimate, with n = 0, yields θ * = 1 and −
Daδ * ≤ e 1
(34)
For safe operation then k 0 e
− E RT o
C Ao
n
(− Δ H ) E τ
R ρ C p T o
r A
2
≤ e −1
(35)
7
