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1
Ch 8: Steady – State Non-isothermal Reactor DesignEnergy Balances, Rationale and Overview
Calculate the volume necessary to achieve a conversion, X, in a PFR for a first-order, exothermic reaction carried out adiabatically.
The combined mole balance, rate law, and stoichiometry yield:
For an adiabatic, exothermic reaction the temperature profile might look something like this:
To solve this equation we need to relate X and T. We will use the Energy Balance to relate X and T. For example, for an adiabatic reaction, , the energy balance can be written in the form
set X, then calculate T, -VA, and , increment X, then plot vs. X
2
1. Adiabatic (Q = 0); CSTR, PFR, Batch and PBR:
2. CSTR with heat exchanger, UA(Ta-T) and large coolant flow rate.
3 . PFR/PBR with heat exchange
3A. In terms of conversion, X
3B. In terms of molar flow rates, Fi
3
4. For Multiple Reactions
5. Coolant Balance
These are eq’ns we will use to solve rxn engineering problems with heat changes.
3. Energy Balance
Typical units for each term are J/s; i.e. Watts
4
Energy Balance
1st Law of Thermodynamics:
dE = dQ – dW (for a closed system)
Energy of the system
Heat flow to the system
Work done by the system on the surroundings
system}open an {for E F - E F WQdtdE
outout
flowmassby system
the toaddedenergy of rate
inin 321+−=••
Evaluation of the Work Term:
s1
i
~
i1
i
~
i
sf
WV P FV P F
WWW
+⋅⋅+⋅⋅−=
+=
∑∑==
•
n
i out
n
i in
flow shaft
∑∑==
••
⋅+⋅−+=n
iout
n
iins
1ii
1ii
sys H FH FWQdt
dE
Combining with i
~
ii VP U H +=
5
(1)
1. Replace Ei by Ei=Hi-PVi
2. Express Hi in terms of enthalpies of formation and heat capacities 3. Express Fi in terms of either conversion or rates of reaction4. Define ∆HRX
5. Define ∆CP
6. Manipulate so that the overall energy balance is either in terms of the Equations above 1.A, 1.B, 2, 3A, 3B, or 4 depending on the application
Step 1:Substitute
and,
into equation (1) to obtain the General Energy Balance Equation.
General Energy Balance:
For steady state operation:
6
We need to put the above equation into a form that we can easily use to relate X and T in order to size reactors. To achieve this goal, we write the molar flow rates in terms of conversion and the enthalpies as a function of temperature. We now will "dissect" both Fi and Hi.
Flow Rates, Fi
For the generalized reaction:
In general,
ad
ac
ab- 1-
FF where X) ( F F
DCBA
A0
i0iiiA0i
====
=Θ+Θ=
νννν
ν
(2)
(3)
[ ]
[ ]
∑∑∑
∑∑
∑∑
∆Θ=−
+=∆
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡⋅⋅Θ=
+Θ+=⋅⋅
= ∆=
==
X F (T)H - )H - (H F H F H F
H - H ab - H
ac H
ad H
X F )H ( - )H - (H F
........... )H - (H )H - (H F F H - F H
Then
A0RXii0iA0iii0i0
ABCDRX
n
1iA0
H
ii
n
1iiii0A0
n
1iBBB0AA0A0ii
n
1ii0i0
RX
434 21ν (4)
(5)
(6)
.Hin included be it will place, takeschange phase a If
0 X F (T)H -)H - (H F W - Q
:(6) and (1) combining ;0dtdEst -st for
RX
n
1iA0RXii0iA0
∆
=∆Θ+
⎭⎬⎫
⎩⎨⎧ =
∑=
•• (7)
7
Assuming no phase change:
QiRo
ii H )(TH H ∆+=
Enthalpy of formation at TR(ref. temp.)
Change in enthalpy when T is changed from TR to T.
If there is a phase change:
dT C )(TH H
change) phase no (if dT C H
dT C )(TH dT C )(TH H
T
TpiR
oii
T
TpiQi
T
Tipl,mmi
T
Tips,R
oii
R
2
1
R
M
R
∫
∫
∫∫
⋅+=
⋅=∆
+∆+⋅+=
(8)
(9)
(10)
(11)
2iiip T T C ⋅+⋅+= γβα
0 X F (T)H -]T -[T C F W - Q
:(7) into Substitute
]T -[T CdT C H H
n
1iA0RXi0piiA0
i0pi
T
Tpii0i
i0
=∆⋅⋅Θ+
⋅=⋅=−
∑
∫
=
••(13)
(12)
Combine eq’n (11) & (5)
pApBpCpDp
Ro
ARo
BRo
CRo
Do
RX
Rpo
RXRX
C - C ab - C
ac C
ad C
)(TH - )(TH ab - )(TH
ac )(TH
ad H
where
)T - (T C H H
+=∆
+=∆
∆+∆=∆ (14)
(15)
(16)
8
[ ] 0 X F )T - (T C H -]T -[T C F W - Qn
1iA0Rp
oRXi0piiA0 =⋅∆+∆⋅⋅Θ+ ∑
=
••(17)
Combine (13) and (14)
[ ] )T - (T C H-)T - (T C
X 0Q
0W
Rpo
RX
iopiis
∆+∆
Θ=
⎪⎭
⎪⎬⎫
=
= ∑•
•
Adiabatic Operation:
T
Xfor an exothermic, adiabatic rxn for an exo. rxn why does X
increase?
This is from E balance, not mole balance.
)T - (T CH Rpo
RX ∆>∆
so X vs T is linear!!!
Adiabatic Tubular Reactor
Rearrange (18):
[ ]∑∑
∆⋅+⋅Θ
⋅∆⋅+⋅Θ+⋅=
ppii
Rpopiio
RX
C X CT C X T CH-X
T (19)
(20) T) (X,r- dVdX F AA0 =
Combine with differential mole balance:
To obtain T, X and conc’n profiles along the reactor!
Use (19) to construct a table of T vs X.
Obtain k(T) as a function of X -rA as a function of X.
Look at Table 8-2A.
9
TFA0
To
FAe
Te
∑ ⋅ ii HF ∑ ⋅ ii HF
V V+∆V
0 Ws assume V T) - (Ta Ua T) - (TaA U Q =∆⋅⋅=∆⋅=∆••
Steady – State Tubular Reactor with Heat Exchange:
0H FH FQVViiVii =⋅−⋅+∆ ∑∑ ∆+
•
•
∆Q
(1)
The heat flow to the reactor
Overall heat transfer coefficient (U)
Heat exchange area (∆A)
Difference between ambient temperature (Ta) and reactor temperature (T)
T)-(Ta V Ua T) - (TaA UQ ⋅∆⋅=∆⋅=∆•
(2)
DVA 4= Diameter of reactor
If ∆V 0
dVdHF - H
dVdF
T)-(Ta Ua
(3)n Eq' diff.
)(-r r dVdF :balance mole
dV)H (F d
T)-(Ta Ua
iii
i
Aiii
ii
∑∑
∑
⋅=
==
⋅=
ν
(3)
(4)
(5)
dVdT C
dVdH
dT C H
pii
pii
⋅=
⋅=
(6)
Combine (5) & (4) & (6)
dVdTCF - )(-r H T)-(Ta Ua piiA
H
ii
RX
⋅⋅⋅= ∑∑∆
434 21ν (7)
10
Rearrange:
∑ ⋅−∆⋅
=pii
removedHeat generatedHeat
RXA
C FTa) -(T UaHr
dVdT
4 84 76484 76
(8)
X) ( F F iA0i ν+Θ⋅=
( ) { }PFR afor XC X F
Hr T) -(Ta Ua dVdT
piA0
RXA
∑ ⋅∆++Θ⋅∆⋅+
=νi
Substitute into (8)
for a PBR {dW = ρb dV}( )
∑ ⋅
∆⋅+⋅=
pii
RXAb
C F
H'r T) -(Ta Ua
dWdT ρ
These eq’ns will be coupled with mole balance eq’nsA0
A
Fr-
dVdX
=
Tao
FA0, To
V V + ∆V
R1R2
Reactants
Heat transfer fluid
The fluid will keep the rxn temperature constant for endo/exo – thermic rxns
You might have
A. Co – current Flow
B. Counter – current Flow
Balance on the Coolant Heat Transfer Fluid:
11
FA, T
V V + ∆V
FA, T
Ta: coolant temperature
mc, Hc mc, Hc
0 V Ta)-(T UaH m -H m
0 Q E -E
VVcc
Vcc
conductionVV
outV
in
=∆+
=+
∆+
••
•
∆+
••
Divide by ∆V and take limit as ∆V 0
A. Co – Current Flow
The Energy Balance
Cm
Ta) - (T Ua dVdT
dVdTC
dVdH
0 Ta) - (T Ua dVdHm
p
aap
c
c
⋅=⇒=
=+⋅
•
•
c
cexothermic
endothermic
Tao
Tao
V
TFAo, To
V V + ∆V
Ta2 Ta
V = 0 V = Vf
Cm
T) - (Ta Ua dVdT
pc
a
⋅= •
c
At the entrance X = 0; V = 0; Ta = Ta2
At the exit V = Vf; Ta = Tao
The sol’n to the counter-current flow problem to find {T X} is a trial & error procedure.
Assume a coolant temp at the entrance (Ta2)
Solve ODEs to calculate X, T and Ta as a function of V:
Find Ta(V = Vo)
a2
a2oaa2oa
Tanother assume Else
T)V(VT T-)V(VT If ==→<= ε
B. Counter – Current Flow
12
Equilibrium Conversion
As T X {For endothermic rxns}
As T X {For exothermic rxns}
Exothermic Rxns:
rxn}order {1st Kc 1
Kc Xe+
=
From Le Chaltlier’s Law
(Kc as T if ∆H<0) exothermic γ)(RTK
K pc =
To find the max X in an exothermic rxn carried adiabatically:
To1To
Xe
Xe1
Adabatic temperature
To1 >To
Energy balance
⎭⎬⎫
⎩⎨⎧
∆−
⋅⋅Θ=∑
X T as T tochanged is T if
)T(H)T-(T C
X
e
o1o
RXN
opiie
11R
)(THexp )(TK )(TK
0C if
T R)T-T(C )(TH
T R(T)H
dTdlnK
21
RRXNo
1p2p
~
2R
~
RRXNo
2RXNp
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−
∆=
=∆
⋅∆+∆
=⋅
∆=
TT
p
p
13
Xe
Xeb
To increase the conversion in an exothermic rxn, use multiple reactors with interstate cooling:
For Endothermic Rxn (you need heating)
Xe
3
2
1
350 500 600
X
T, K
As T Xe but the (-rA) decreases!
(so the conversion is achieved at the end of the reactor)
So there must be an optimum temperature to achieve max X.
Curve A: Rxn rate slow, rxn dictated by rate of rxn and reactor volume
As T , r , X
Curve B: Rxn rate very rapid. Virtual equilibrium reached in X dictated by equilibrium conversion
Optimum Feed Temperature
Adiabatic Reactor of fixed size
Reversible & Exothermic Rxn T ; X as 0H if
XCH-T T
rx
pA
rxo
<∆
⋅∆
=
14
Adiabatic Rxn Algorithm
Suppose
1. Choose X
Calculate T
Calculate k
Calculate T/To
Calculate CA
Calculate CB
Calculate KC
Calculate -rA
2. Increment X and then repeat calculations.
3. When finished, plot vs. X or use some numerical technique to find V.
Levenspiel Plot for anexothermic, adiabatic reaction.
15
Consider:
For an exit conversion of 70%For an exit conversion of 40%PFR Shaded area is the volume.
For an exit conversion of 70%For an exit conversion of 40%
CSTR Shaded area is the reactor volume.
We see for 40% conversion very little volume is required.
CSTR+PFR
16
For an intermediate conversion of 40% and exit conversion of 70%
Looks like the best arrangement is a CSTR with a 40% conversion followed by a PFR up to 70% conversion.
Evaluating the Heat Exchanger Term
Energy transferred between the reactor and the coolant:
Assuming the temperature inside the CSTR, T, is spatially uniform:
17
At high coolant flow rates the exponential term will be small, so we can expand the exponential term as a Taylor Series, where the terms of second order or greater are neglected, then:
Since the coolant flow rate is high, Ta1 Ta2 Ta:
Multiple Steady States (MSS)
where Heat generated term
Heat removed term
18
R(T):
Varying Entering Temperature
Vary non-adiabatic parameter κ:
if you increase FA0 (molar flow rate) or decrease heat – exch area then κ will decrease
Increase To
Slope = Cpo (1+κ)
T
R(T)
κ decrease
T
R(T)κ=∞
κ= 0
Ta To
κκ
κ
+⋅+
=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
=
1T T T
C FUa
a0c
p0A0
Low E
High EG(T)
T
increasing TG(T)
T
G(T)
For a first order rxn:
kkX⋅+
⋅=
ττ
1
19
Extinction temp
Ignition temp
Lower st-st
Upper st-st
Ignition – Extinction Curve:
The point of intersection of G(T) and R(T) give Tst-st.
By plotting Tst-st vs To, we obtain ignition – extinction curve.
As To Tst-st
Runaway Rxns in a CSTR: R(T),
G(T)
Tc T* Tangency pointκκ+⋅+
=1
T T T a0c
E
*2
c*
rcTR T-TT ⋅
==∆
Reactor T
if this diff. is exceeded, transition to the upper st – st will occur.
At this high temp, it is undesirable or even dangerous.
20
PFR
rxn) single a(for C F
(T))H)(-(-rT)-(T UadVdT
m
1ipii
RxnAa
∑=
∆+=
When q multiple rxns occur with m species:
[ ]species :j
rxn : i
C F
(T)H-)(-rT)-(T Ua
dVdT
m
1jpjj
1ijRxn,ija
∑
∑
=
=
∆+=
q
i
Ex:
[ ] [ ]
C FC FC F(T)H-)(-r(T)H-)(-rT)-(T Ua
dVdT
pCCpBBpAA
Rxn,2B2BRxn,1A1Aa
++∆+∆+
=
C B
B A
:1Rxn
2
1
k
k
⎯→⎯
⎯→⎯
CSTR
[ ] [ ] rxn) single a(for 0 Vr(T)H -]T -[T C F W - Qn
1iARXN0piiA0s =⋅∆⋅⋅Θ+ ∑
=
••
0 (T)HrV-]T -[T C F T)-UA(T
0 (T)HrV-]T -[T C F W - Q
n
1i 1ijRXN,ij0piiA0a
n
1i 1ijRXN,ij0piiA0s
=∆⋅⋅⋅Θ+
=∆⋅⋅⋅Θ+
∑ ∑
∑ ∑
= =
= =
••
q
i
q
i
When q multiple rxns take place0