Conversion and reactor sizing Objective: sizing a reactor once the relationship between reaction rate ra and concentration is known

Conversion: definition Given the general reaction: A (limiting reactant) is the base for the calculation For irreversible reactions, the maximum value for X is for complete conversion For reversible reactions, the maximum value for X is at equilibrium Batch Flow                              

Conversion (X) Quantification of how a reaction has progressed Batch Reactors Continuous (or Flow) Reactors

Mole Balance for Batch Reactor Two different cases: 1. Constant Volume 2. Constant Pressure If reactor is perfectly mixed

Design equations in terms of conversion: batch reactor Mole balance for a batch reactor is Definition of conversion General equation Batch time

Flux reactors Conversion grows with raising residence time  with volume Molar flow rate is used instead of mole number: FA = FAo (1-X) Inlet Molar flow rate is the product of inlet concentration CAo times the volumetric inlet flow rate vo FAo = CAo vo CAo is: Liquids: molarity (mol/l) Gas: calculated from EOS (Ideal gas law): CAo= PAo /RTo = yAo Po/RTo

Example of calculation of CA0 A gas mixture 50% A and 50% inert at 10 atm enters the reactor with constant flow rate of 6 dm3/s at 300°F (422.2 K) Estimate the inlet concentration CA0 and volumetric flow rate FA0. Use 0.082 dm3 atm/mol K as gas constant.

Solution A gas mixture 50% A and 50% inert at 10 atm enters the reactor with constant flow rate of 6 dm3/s at 300°F (422.2 K) Estimate the inlet concentration CA0 and volumetric flow rate FA0. Use 0.082 dm3 atm/mol K as gas constant. Solution For an ideal gas

Design equation: CSTR (back mix) Mass balance in terms of conversion Introduction of ra Volume of CSTR for obtaining a given conversion

Design equation: PFR Similar equation for Packed Bed Reactor Mole balance In terms of conversion Differentail form of the design equation Integral Equation Similar equation for Packed Bed Reactor

Design equations in terms of conversion

Design equations: application Given -rA as a function of conversion -rA=f(X),… … it is possible to size any (ideal) reactor Using the Levenspiel plots … …. In which Fao / (-ra) or 1 / (-ra) is plotted as a function of X In the Levenspiel plots Fao / (-ra) vs. X, the volume of a CSTR and that of a PFR are represented by the following areas:

Levenspiel Plots PFR FAO -rA X FAO -rA X CSTR FAO -rA X=X

Example The isoptherma reaction A  B + C takes place in a CSTR. Calculate the volume of the CSTR and that of the PFR for consuming 80% of A with the following conditions: Inlet Volumetric flow rate is constant at 6 litri/sec Pressure 10 atm Initail concentration yA0 = 0.5 Temperature 422.2 K Ideal gas law is valid Data reaction rate and conversion X .1 .2 .3 .4 .5 .6 .7 .8 .85 -rA 0.0053 0.0052 0.005 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001

Solution: CSTR FAO -rA X Report data of r vs. X in the Levenspiel plot X=X Report data of r vs. X in the Levenspiel plot Calcualtion of FA0 Value of rA at X= 0.8 is 0.00125, therefore 1/rA = 800 Substituting in the design equation: Area of the rectangle is 800 * 0.8 = 640 : this should be multiplied by FA0 and gives the same result as above

Solution: PFR FAO -rA X Report data of r vs. X in the Levenspiel plot Calculation of FA0 Applying Simpson method for integration: Gives the value of V= 225 dm3 Direct measure of the area = 260 dm3 ; this should be multiplied by FA0 and gives the same result as above

Value of conversion along the reactor X V 250 1 Along the reactor: Concentration of reactant drops in the reactor Reaction rate drops too Conversion raises Calculations are made at different conversion (likewise for X=0.8) , giving the table: V (dm3) 33.4 71.6 126 225 X 0.2 0.4 0.6 0.8 -rA (mol/dm3 s) 0.0053 0.005 0.004 0.0025 0.00125

Comparison between CSTR e PFR For same conversion, Is the CSTR volume always higher than PFR volume ? It seems so, but is it ALWAYS true?? FAO -rA X X=X FAO -rA X

For same conversion, Is the CSTR volume always higher than PFR volume ? FAO -rA X VPFR VCSTR < VPFR VCSTR

For same conversion, Is the CSTR volume always higher than PFR volume ? FAO -rA X VPFR VCSTR VCSTR = VPFR

Levenspiel plot for bacteria growth

Reactors in series

PFR in series x3 FAO -rA x2 x1 Vsingle = V1+V2+V3 FAO X=0 FA1 X=X1 FA2 Let us compare two scenarios single reactor achieves X3 3 reactors in series achieve X3 How are the total volume of 3 reactors in series related to single reactor ?? FAO -rA x2 x1 Vsingle = V1+V2+V3

CSTR in series FAO -rA V1 + V2 + V3 < Vsingle Vsingle V3 X3 X=X1 FA2 X=X2 FA3; X=X3 FAO -rA V1 + V2 + V3 < Vsingle Vsingle V3 X3 Can we model PFR as a series of “n” equal volume CSTRs?? X2 X1 V2 V1

Reactors in series: numerical example Same data of the previous example. Determine: The total volume of 2 CSTR in series for a total conversion of 80% with the first reactor has a conversion of X=0.4. The total volume of 2 PFR in series for a total conversion of 80% with the first reactor has a conversion of X=0.4. Solution CSTR FA0= 0.867 mol/s V1=86.7; V2= 277.4  V= V1+V2 = 364 l Only one CSTR V= 555 l (see previous example) Solution PFR V1=71.6; V2= 153  V=V1+V2= 225 l (=to previous example)

Order of the sequence for intermediate X=.5 Case A PFR equation Integration from 0 to 0.5 gives V1= 97 l CSTR V2= 0.867 (0.8 – 0.5) (800) = 208 l Vtot= V1+V2= 305 l Case B CSTR equation V1= 0.867 (0.5 – 0) (303) = 131.4 l PFR V2 = 130.9 l Vtot= V1+V2= 262 l FAO FA1 X1 X2 FAO FA1 X1 X2

Modelling of Hippopotamus Digestive System

Blood perfusion Stomach VG = 2.4 l Gastrointestinal tG = 2.67 min Liver Alcohol VL = 2.4 l tL = 2.4 min Central VC = 15.3 l tC = 0.9 min Muscle & Fat VM = 22.0 l tM = 27 min It is the process in which the blood is moved to the muscles and fat (using capillary vases). Iteractions are given by arrows. VG, VL, VC, e VM are the volumes of tissues-water representing gastrointestinal, liver and muscles. VS is the volume of the stomach.

Some more definitions Space time (or Mean residence time) () is the Time required to process 1 reactor volume of fluid at inlet conditions Space velocity: Other conditions for flow rate v0 (in terms of velocity) LHSV - Liquid Hourly Space Velocity GHSV - Gas Hourly Space Velocity Actual Residence Time: The time actually spent by fluid inside the reactor. Plug flow

More definitions Relative rates of reaction aA + bB  cC + dD How is (-rA) related to (-rB), (rC) and (rD) ?

Exapmple The exothermic reaction A  B + C was carried out adiabatically and the following data recorded: The entering molar flow rate of A was 300 mol/min What PFR volume is necessary to achieve 40% conversion What conversion can be achieved if the PFR in part(a) is followed by a 2.4 L CSTR?

Solution h1 w Design Equation Method-1: Area of Trapezoid Area under curve = w x [h1+h2]/2 h2 w = 0.4 x [0.1+0.02]/2 = 0.4 x 0.06 = 0.024 L·min/mol V = 0.024 x 300 L = 7.2 L

Solution h Design Equation Method-2: Simpson’s Rule Area under curve = [h/3]x [f(XO) + 4f(X1) +f(X2)] where, h = [X2-XO]/2 & X1=XO+h Area = [0.4/3] x [0.1 + 4x0.06 + 0.02] = [0.2/3] x [0.36] = 0.024 L·min/mol V = 0.024 x 300 L = 7.2 L

Solution VCSTR = 2.4 L Part(b) Conversion in CSTR in series with PFR FAO X=0 FA1 X=0.4 FA2 = ? X=X2=? VCSTR = 2.4 L Design Equation Step-1: Evaluate if X2<0.6 Area under flat portion of curve = [0.6-0.4] x 0.02 = 0.004 V = 300 x 0.004 = 1.2 L Therefore, we know that X2>0.6

Solution Part(b) Conversion in CSTR in series with PFR Step-2: for X>0.6 find relationship between (1/rA) and X 1/(-rA) = f(X) for X>0.6; f(X) = 0.02 +0.3(X - 0.6) = 0.3 X -0.16 Area under curve = f(X2).[X2-X1] V = Area under curve x FAO =2.4 L = [0.3 X2 - 0.16] [X2 -0.4] 300 L Solving for X2, we get X2 =0.6425