Non Isothermal CSTR
Chemical Reaction Engineering I
Aug 2011 - Dec 2011
Dept. Chem. Engg., IIT-Madras
Overview
• Notation
• CSTR design equation (mass balance)
• Heat Balance
• Accounts for heat removal/addition
• Methodology given
Notation
•
Usual notation applies.
– FA = molar flow rate of A, FT = total molar flow rate
– V = volume of reactor, Q = volumetric flow rate
– T = temperature
– x = conversion.
– k = rate constant, CA = concentration of A
– Ref = Reference temperature where the heat of reaction is given
 DH = heat of reaction (exothermic reactions will have negative values)
– Subscript “in” indicates inlet. E.g. FA-in is molar flow rate of A at the inlet.
We consider only liquid phase reactions because they are the
most common ones in CSTR
CSTR design equation- Solve Mass Balance and
Energy Balance simultaneously
• Mass Balance
F A  in  F A  o u t   rA  o u t V C S T R
• Rate constant is
k  k0e
 E 


 RT 
• Energy balance equation
• Heat released by reaction at operating conditions = Heat
used to raise the feed to operating conditions
• The rest are just details.
• Heat released by reaction at operating conditions
• If Cp doesn’t depend on temperature, it is given by
 D H T
• What is DCp?
  DH
T
R ef
 D C p  T  T R ef

Heat of Reaction
If the reaction is given by
a A  b B  cC  d D
And DHRef is given per mol ‘A’ reacted, then
DC p 
cC p  C  dC p  D  aC p  A  bC p  B
a
e.g. Sulfuric acid + sodium hydroxide react and the heat of reaction
is given “per one mole sulfuric acid reacted”
H 2 SO 4  2 NaOH  Na 2 SO 4  2 H 2 O
(Although NaOH is a solid, we will consider the case where it is
dissolved in water. Water is an inert material. We will also
neglect the heat of dissolution, although it is actually high)
D C p  C p  N a 2 SO 4  2 C p  H 2 O  C p  H 2 SO 4  2 C p  N aO H
Heat of Reaction
• In a more general case, the specific heat capacity is a
function of temperature. The heat of reaction at any
temperature T is given by
 DH
T
  DH
T

R ef

T
TR ef
D C p dT
• Example. Reaction A B+C
D H 273   10000 J/mol A reacted
C p  A  3 0  0 .0 1T  0 .0 0 0 0 3T
2
J/m o l/K
C p  B  21  0.006 T  0.00001T
2
J/m ol/K
C p  B  10  0.002 T  0.00001T
2
J/m ol/K
D H 500  ? J/m ol A reacted
Heat of Reaction
D H 500   D H
500
 2 7 3  2 7 3 D C p dT
D C p  1  0.002 T  0.00001 T
2
500
D H 500   D H
 273
2
3

T
T 
  T  0.002
 0.00001

2
3  273

D H 500   9947 J/mol A reacted
• In general, the Heat of reaction at any temperature
would be a cubic equation in Temperature
DH
   T  T   T
2
T
3
Energy Balance Equation
• Energy released by reaction / unit time
• = Number of moles reacted per unit time * Heat released per
mole
F A  in x
 D H T
• Energy needed (/ unit time )to bring feed to a temperature T
• If Cp is a constant, then
F
i  in
C p i
T
 Tin 
i
• If Cp varies with temperature, then
T
F
i
i  in
C
p i
dT
Tin
• Note: To calculate DHT, we use Cp of reactants and products.
Data on inert present in the feed is not necessary
• To calculate ‘energy needed to bring feed to a temperature of
T’, we need data on all materials (including inert) in feed.
Energy Balance
•
Under steady state conditions
T
F A  in x
  D H T

F
i
•
C
i  in
p i
dT
Tin
This can be re-written as
x
A cubic equation in T
Another cubic equation in T
•
If Cp is a constant, then it can be written as a
linear equation
x   + T
•
Note: When we substitute actual values we
find that “x vs T” is roughly linear even when
Cp is a function of T
Non-Isothermal Case
•
If heat is removed (e.g. cooling coils), then energy balance
becomes
•Heat released by reaction at operating conditions = {Heat
used to raise the feed to operating conditions + Heat
removed by cooling coils}
•
Heat removed by cooling coils = U A (T-Tc) where
Tc is the temperature of cooling fluids in the
coils.
•
Note: For endothermic reactions, with heat
added by coils, the equation will be similar.
•Heat ABSORBED by reaction at operating conditions = {Heat
released by LOWERING the feed to operating conditions +
Heat SUPPLIED by heating coils}
•
Still, “x vs T” is a linear equation (or a roughly linear
equation) for energy balance
Pictorial View
• ‘X vs T’: Energy Balance equation gives roughly a straight line
• Mass balance equations (for positive order reactions) gives a
‘non liner’ (i.e. curvy) line, which starts at X = 0 and ends at X =
1.
• Note:
– For energy balance equation, there is no ‘inherent’ check to ensure
that X will be between 0 and 1
– If the reaction order is 0, there is no inherent check to ensure
that X will be between 0 and 1.
Choices
1. If the conversion desired (X) is known, then find V and T for adiabatic
conditions
2. If Volume is known, find X and T
3. If Volume is known, and heat is removed to maintain the temperature
below certain value (e.g. exothermic reaction), find the conversion
4. If Volume is known, and critical temperature (i.e. maximum allowed
temperature) is known, find the conversion and the heat load (U A
Del-T).
1. Solution:
–
If X is known, then using heat balance equation, find T
•
–
Constant Cp: the linear equation can be solved easily. Variable Cp. The cubic
ratios has to be solved, using a numerical method (e.g. fsolve in Matlab).
Using T and X, find V
Solutions
2. To find X and T, given V (adiabatic) :
Solution:
–
–
Substitute for V in the mass balance equation
Solve for T and X simultaneously using numerical methods
–
Alternative: Graphically solve for T and X simultaneously
•
•
•
i.e. Plot X vs T using energy balance equation alone for various assumed values of
T
Plot X vs T using mass balance equation alone, for various assumed values of T
Find where these two lines intersect. That would give the X and T
3. If Volume is known, and heat is removed to maintain the
temperature below certain value (e.g. exothermic reaction),
find the conversion
Solution:
Use the same method as above, but the heat balance equation
has one more term to accommodate heat removal
Solution
4. If Volume is known, and critical temperature (i.e. maximum allowed
temperature) is known, find the conversion and the heat load (U A
Del-T).
Solution:
T = maximum allowed temperature
Using mass balance equation (since V is known and T is known),
find X
Using heat balance equation (now X and T are known), find the
heat load: i.e. UA Del-T