Ch 8: Steady – State Non-isothermal Reactor Design
Energy Balances, Rationale and Overview
Calculate the volume necessary to achieve a conversion, X, in a PFR for a firstorder, exothermic reaction carried out adiabatically.
For an adiabatic, exothermic reaction the temperature profile might look something
like this:
The combined mole balance, rate law, and stoichiometry yield:
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,
energy balance can be written in the form
set X, then calculate T, -VA, and
, increment X, then plot
, the
vs. X
1
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
2
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
3
Energy Balance
1st Law of Thermodynamics:
dE = dQ – dW
Energy of the
system
(for a closed system)
Heat flow to
the system
Work done by the
system on the
surroundings
dE • •
= Q− W + Fin E in - Fout E out
123
dt
{for an open system}
rate of energy
added to the
system by mass
flow
Evaluation of the Work Term:
•
W = Wf + Ws
flow
shaft
n
n
~
= −∑ Fi ⋅ P ⋅ V i
i =1
dt
•
in
i =1
+ Ws
out
~
Hi = Ui + P Vi
Combining with
dE sys
~
+ ∑ Fi ⋅ P ⋅ V i
•
n
n
i =1
i =1
= Q+ W s − ∑ Fi ⋅ H i in + ∑ Fi ⋅ H i out
4
(1)
1.
2.
3.
4.
5.
6.
Replace Ei by Ei=Hi-PVi
Express Hi in terms of enthalpies of formation and heat capacities
Express Fi in terms of either conversion or rates of reaction
Define ∆HRX
Define ∆CP
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:
5
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:
(2)
In general,
Fi = FA0 (Θi +ν i X)
ν A = -1 ν B = -
b
a
where Θi =
νC =
c
a
νD =
Fi0
FA0
(3)
d
a
Then
n
∑H
i =1
n
i0
∆H RX
∑F
i0
⋅ Fi0 - ∑ H i ⋅ Fi = FA0 [(H A0 - H A ) + (H B0 - H B ) Θ B ] + ...........
i =1
⎤
n
n ⎡
= FA0 ∑ [(H i0 - H i ) Θ i ] - ∑ ⎢(ν i ⋅ H i ) FA0 ⋅ X ⎥
14 2 43
⎥
i =1
i =1 ⎢
⎦
⎣ ∆H RX
d
c
b
= HD + HC - HB - HA
a
a
a
H i0 − ∑ Fi H i = FA0 ∑ Θi (H i0 - H i ) - ∆H RX (T) FA0 X
(4)
(5)
(6)
⎧ dE
⎫
for st - st ⎨
= 0⎬; combining (1) and (6) :
⎩ dt
⎭
•
•
n
Q - W + FA0 ∑ Θi (H i0 - H i ) - ∆H RX (T) FA0 X = 0
(7)
i =1
If a phase change takes place, it will be included in ∆H RX .
6
Assuming no phase change:
H i = H i (TR ) + ∆H Qi
o
Enthalpy of
formation at TR
(ref. temp.)
(8)
Change in enthalpy when T
is changed from TR to T.
If there is a phase change:
TM
T
TR
TR
H i = H i (TR ) + ∫ C ps,i ⋅ dT + ∆H mi (Tm ) + ∫ C pl,i dT
o
T2
∆H Qi = ∫ C pi ⋅ dT
(if no phase change)
(9)
(10)
T1
T
H i = H i (TR ) + ∫ C pi ⋅ dT
o
(11)
TR
Cp = α i + βi ⋅ T + γ i ⋅ T 2
T
H i − H i0 =
∫C
pi
⋅ dT = C pi ⋅ [T - Ti0 ]
(12)
Ti0
Substitute into (7) :
•
•
Q - W + FA0
n
∑Θ
i =1
i
⋅ C pi ⋅ [T - Ti0 ] - ∆ H RX (T) FA0 X = 0
(13)
Combine eq’n (11) & (5)
∆H RX = ∆H RX + ∆C p (T - TR )
o
(14)
d
c
b
o
o
o
o
H D (TR ) + H C (TR ) - H B (TR ) - H A (TR )
a
a
a
(15)
where
∆ H RX =
o
∆C p =
d
c
b
C pD + C pC - C pB - C pA
a
a
a
(16)
7
Combine (13) and (14)
•
•
[
n
]
Q - W + FA0 ∑ Θi ⋅ C pi ⋅ [T - Ti0 ] - ∆H RX + ∆C p (T - TR ) ⋅ FA0 X = 0
i =1
o
(17)
Adiabatic Operation:
•
⎫
Ws = 0⎪
⎬
•
Q = 0 ⎪⎭
X=
[
∑Θ C
i
o
pi
(T - Tio )
- ∆HRX + ∆Cp (T - TR )
]
for an exothermic, adiabatic rxn
X
for an exo. rxn why does X
increase?
This is from E balance, not
mole balance.
∆H RX > ∆C p (T - TR )
o
T
so X vs T is linear!!!
Adiabatic Tubular Reactor
Rearrange (18):
T=
[
]
X ⋅ - H RX + ∑ Θ i ⋅ C pi To + X ⋅ ∆C p ⋅ TR
o
∑Θ ⋅C
i
pi
+ X ⋅ ∆C p
(19)
Combine with differential mole balance:
FA0
dX
= - rA (X, T)
dV
(20)
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 Æ -rA as a function of X.
Look at Table 8-2A.
8
Steady – State Tubular Reactor with Heat Exchange:
•
•
∆ Q = U ⋅ ∆A (Ta - T) = Ua ⋅ (Ta - T) ⋅ ∆V
FA0
∑F ⋅H
i
To
V
•
∆ Q+ ∑ Fi ⋅ H i
V
− ∑ Fi ⋅ H i
∑F ⋅H
T
i
assume Ws = 0
i
FAe
i
Te
V+∆V
V + ∆V
=0
(1)
•
The heat flow to the reactor ∆ Q
Overall heat transfer coefficient (U)
Heat exchange area (∆A)
Difference between ambient temperature (Ta) and reactor temperature (T)
•
∆ Q = U ⋅ ∆A (Ta - T) = Ua ⋅ ∆V ⋅ (Ta - T)
A 4
=
V D
(2)
Diameter of reactor
If ∆V Æ 0
d ∑ (Fi ⋅ H i )
Ua (Ta - T) =
(3)
dV
mole balance :
dFi
= ri = ν i (-rA )
dV
(4)
diff. Eq' n (3)
Ua (Ta - T) =
∑ dFi
dV
H i = C pi ⋅ dT
⋅ H i - ∑ Fi
dH i
dV
(5)
dT
dH i
= C pi ⋅
dV
dV
(6)
Combine (5) & (4) & (6)
dT
Ua (Ta - T) = ∑ν i ⋅ H i ⋅ (-rA ) - ∑ Fi ⋅ C pi
1 4 2 43
dV
(7)
∆H RX
9
Æ Rearrange:
6Heat
4 generated
7 48 6 Heat
4 7removed
48
dT rA ⋅ ∆H RX − Ua (T - Ta)
=
dV
∑ Fi ⋅ Cpi
(8)
Fi = FA0 ⋅ (Θ + ν i X)
Substitute into (8)
dT
Ua (Ta - T) + rA ⋅ ∆H RX
=
dV FA0 ⋅ (∑ Θ i + ν i X + ∆C p ⋅ X )
Ua
for a PBR {dW = ρb dV}
ρ
dT
= b
dW
{for a PFR}
⋅ (Ta - T) + (rA ') ⋅ ∆H RX
∑F ⋅C
i
These eq’ns will be coupled with mole balance eq’ns
pi
dX - rA
=
dV FA0
Balance on the Coolant Heat Transfer Fluid:
Heat transfer fluid
Tao
R2
R1
FA0, To
V
V + ∆V
Reactants
The fluid will keep the rxn temperature constant for endo/exo – thermic rxns
You might have
A. Co – current Flow
B. Counter – current Flow
10
A. Co – Current Flow
The Energy Balance
•
•
E in
Ta: coolant temperature
mc, Hc
•
+ Q conduction = 0
- E out
FA, T
V + ∆V
V
•
•
+ Ua (T - Ta) ∆V = 0
mc Hc - mc Hc
FA, T
V
V + ∆V
V
mc, Hc
V + ∆V
Divide by ∆V and take limit as ∆V Æ 0
•
mc ⋅
dH c
+ Ua (T - Ta) = 0
dV
dH c
dT
= Cp a
dV
dV
⇒
dTa Ua (T - Ta)
= •
dV
mc ⋅ Cp
Tao
exothermic
Tao
endothermic
V
B. Counter – Current Flow
Ta2
Ta
FAo, To
T
V
dTa Ua (Ta - T)
= •
dV
m c ⋅ C pc
V + ∆V
V=0
V = Vf
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)
If Ta (V = Vo ) - Ta2 < ε → Ta (V = Vo ) = Ta2
Else assume another Ta2
11
Equilibrium Conversion
As T
X
{For endothermic rxns}
As T
X
{For exothermic rxns}
Exothermic Rxns:
Xe =
Kc
1 + Kc
{1st order rxn}
From Le Chaltlier’s Law
(Kc
as T
Kc =
if ∆H<0) exothermic
Kp
(RT )γ
To find the max X in an exothermic rxn carried adiabatically:
Energy balance
To1 >To
Xe
Xe1
To
Xe =
∑Θ
i
⋅ C pi ⋅ (T - To )
− ∆H RXN (T )
⎧if To is changed to To1 ⎫
⎨
⎬
Xe
⎩as T
⎭
To1
Adabatic temperature
~
⎫
⎧ dlnK
o
H
(T)
H
∆
∆
RXN (TR ) + ∆ C p (T - TR )
p
RXN
⎪
⎪
=
=
⎪
⎪ dT
R ⋅ T2
R ⋅ T2
⎪⎪
⎪⎪
~
if
C
0
∆
=
p
⎬
⎨
⎪
⎪
o
⎪
⎪K (T ) = K (T ) exp⎧ ∆H RXN (TR ) ⎛⎜ 1 − 1 ⎞⎟⎫
⎨
⎬
p
1
⎟
⎜
⎪
⎪ p 2
R
⎝ T1 T2 ⎠⎭
⎩
⎪⎭
⎪⎩
12
To increase the conversion in an exothermic rxn, use multiple reactors with
interstate cooling:
Xe
Xeb
For Endothermic Rxn (you need heating)
Xe
Optimum Feed Temperature
T = To -
Adiabatic Reactor of fixed size
Reversible & Exothermic Rxn
∆H rx
⋅X
C pA
if ∆H rx < 0 as X
X
2
3
350
As T
Xe
;T
1
500
600
T, K
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
13
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 an
exothermic, adiabatic reaction.
14
Consider:
PFR Shaded area is the volume.
For an exit conversion of 40%
CSTR
For an exit conversion of 70%
Shaded area is the reactor volume.
For an exit conversion of 40%
For an exit conversion of 70%
We see for 40% conversion very little volume is required.
CSTR+PFR
15
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:
16
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
17
R(T):
Varying Entering Temperature
R(T)
Slope = Cpo (1+κ)
Increase To
T
Vary non-adiabatic parameter κ:
if you increase FA0 (molar flow rate) or decrease heat – exch area then κ will
decrease
R(T)
⎧⎪
Ua ⎫⎪
⎬
⎨κ =
FA0 C p0 ⎪⎭
⎪⎩
T + κ ⋅ Ta
Tc = 0
1+ κ
κ=∞
κ= 0
κ decrease
Ta
T
To
G(T)
G(T)
High E
G(T)
increasing T
Low E
T
T
For a first order rxn:
X=
τ ⋅k
1+τ ⋅ k
18
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.
Upper st-st
As To
Tst-st
Lower st-st
Ignition temp
Extinction temp
Runaway Rxns in a CSTR:
R(T),
G(T)
Tc =
T0 + κ ⋅ Ta
1+ κ
∆Trc = T * - Tc =
Tc
T*
Tangency point
R ⋅ T *2
E
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.
19
PFR
dT Ua (Ta - T) + (-rA )(-∆H Rxn (T))
=
m
dV
∑ Fi Cpi
(for a single rxn)
i =1
When q multiple rxns occur with m species:
[
q
dT
=
dV
Ua (Ta - T) + ∑ (-rij ) - ∆H Rxn,ij (T)
]
i : rxn
j : species
i =1
m
∑F C
j=1
j
pj
Rxn 1 :
Ex:
k1
A ⎯⎯→
B
k2
B ⎯⎯→
C
dT Ua (Ta - T) + (-r1A )[- ∆H Rxn,1A (T) ] + (-r2B )[- ∆H Rxn,2B (T) ]
=
dV
FA C pA + FB C pB + FC C pC
CSTR
•
•
Q - W s + FA0
n
∑Θ
i =1
i
⋅ C pi ⋅ [T - T0 ] - [∆ H RXN (T) ]⋅ [rA V ] = 0
(for a single rxn)
When q multiple rxns take place
•
•
0
Q - W s + FA0
q
n
∑Θ
i =1
UA(T a - T) + FA0
i
⋅ C pi ⋅ [T - T0 ] - V ⋅ ∑ rij ∆ H RXN, ij (T) = 0
i =1
n
q
i =1
i =1
∑ Θ i ⋅ C pi ⋅ [T - T0 ] - V ⋅ ∑ rij ∆H RXN, ij (T) = 0
20