Mole Balances PDF

SMJC 3303 CHEMICAL KINETICS AND REACTOR DESIGN MOLES BALANCES DR. NURFATEHAH WAHYUNY BINTI CHE JUSOH.. innovative entrepreneurial global  MOLE BALANCES At the end of this topic, it is expected that students have the ability to: Describe an...

SMJC 3303 CHEMICAL KINETICS AND REACTOR DESIGN MOLES BALANCES DR. NURFATEHAH WAHYUNY BINTI CHE JUSOH.. innovative entrepreneurial global  MOLE BALANCES At the end of this topic, it is expected that students have the ability to: Describe and define the rate of reaction Derive the general mole balance equation Apply the general mole balance equation to the four most common types of reactors.  Recalled: Type of Reactor Batch vs. Semi-Batch vs. Continuous Catalytic vs, Non-catalytic Homogeneous vs. Heterogeneous Common types: - Batch reactor - Continuous Stirred Tank Reactor (CSTR) - Plug Flow Reactor (PFR) - Packed Bed Reactor (PBR)  LET’S BEGIN, Chemical Identity A chemical species is said to have reacted when it has lost its chemical identity. The identity of a chemical species is determined by the kind, number, and configuration of that species’ atoms. There are three ways for a species to loose its identity: 1. Decomposition CH3CH3 → H2 + H2C=CH2 2. Combination N2 + O2 → 2 NO 3. Isomerization C2H5CH=CH2 → CH2=C(CH3)2  RATE OF REACTION, -rA The reaction rate tell us how fast a number of moles of one chemical species are being consumed to form another chemical species. RATE = SPEED The RATE of reaction is the SPEED at which a reaction happens. If a reaction has a low rate that means the molecules combine at a slower speed than a reaction with a high rate.  RATE OF REACTION, -rA The rate of a reaction can be expressed as the rate of disappearance of a reactant or as the rate of appearance of a product. Consider isomerization of species A: A→ B rA = the rate of formation of species A per unit volume -rA = the rate of a disappearance of species A per unit volume rB = the rate of formation of species B per unit volume rA tells us how fast a number of moles of one chemical species A being consumed to form chemical species B.  Example: A→ B If B is being created at a rate of 0.4 moles per decimetre cubed per second, therefore The rate of formation of B is, rB = 0.4 mole/dm3 s Then A is disappearing at the same rate, -rA = 0.4 mole/dm3 s The rate of formation of A is, rA = -0.4 mole/dm3 s  Remember….rB rB is the rate of formation of species B per unit volume [e.g. mol/dm3s] rB is a function of concentration, temperature, pressure, and the type of catalyst (if any) rB is independent of the type of reaction system (batch, plug flow, etc.) rB is an algebraic equation, not a differential equation  REACTION RATE Reaction rate, r is a function of concentration, for example: − rA = kC A (First order reaction) − rA = kC A2 (Second order reaction) k = Specific reaction rate (time-1 ) For a catalytic reaction (heterogeneous), the rate of reaction –rA’, is number of moles of A reacting per unit time per mass of catalyst (mole/s.g catalyst)  SELF TEST Consider the reaction in which the rate of disappearance of A is 5 moles of A per dm3 per second at the start of the reaction. At the start of the reaction (a) What is -rA? (b) What is the rate of formation of B? (c) What is the rate of formation of C? (d) What is the rate of disappearance of C? (e) What is the rate of formation of A, rA? (f) What is -rB?  GENERAL MOLE BALANCE EQUATION To perform a mole balance - Specify boundary of the system - Specify chemical species, used A System Volume, V FA0 GA FA  GENERAL MOLE BALANCE EQUATION System Volume, V FA0 Gj FA Mole balance on species A : in A − out A + G A = AcumA Number of moles dN A of species A in Substitute all the terms : FA0 − FA + G A = the system at dt time t  GENERAL MOLE BALANCE EQUATION System Volume, V FA0 Gj FA dN A FA0 − FA + G A = dt This is a master equation for any system, any reactor.  GENERAL MOLE BALANCE EQUATION The Concept of Generation GA GA : mol of A being generated per time Generation + if produced or - if being reacted If all the system variables are spatially uniform throughout the system volume, Then, G A = rAV What if our rate of reaction or volume changes through the system? Not spatially uniform, We have to calculate each Generation individually and then add them up for the Total generation V G A =  rA dV rA1 rA 2 V dN A G A1 = rA1V1 FA 0 − FA +  rA dV = G A 2 = rA 2 V2 dt Actual Master Equation for Molar Balances in Reactors  TYPES OF REACTORS Batch reactors Perfectly mixed batch reactor (Batch) Continuous-flow reactors Continuous stirred tank reactor (CSTR) Plug flow reactor (PFR) Packed bed reactor (PBR)  BATCH REACTOR Description All reactants are supplied to the reactor at the outset. The reactor is sealed and the reaction is performed. No addition of reactants or removal of products during the reaction. Vessel is kept perfectly mixed. This means that there will be uniform concentrations. Composition changes with time. The temperature will also be uniform throughout the reactor - however, it may change with time.  BATCH REACTOR Advantages High conversion Testing new process Preparation of expensive products Small scale production Disadvantages High labour cost Variation in the product per batch Difficulty in large scale production TYPICAL COMERCIAL BATCH REACTOR  BATCH REACTOR Mole Balances Batch dN A FA0 − FA +  rA dV = dt FA0 = FA = 0 Well-Mixed  r dV = r V A A dNA = rAV dt 21  BATCH REACTOR EXAMPLE Lets suppose we have A→ B Calculate the time needed to achive certain amount of NA dNA = rAV dt Integrating dt = dNA when t = 0 N A = N A0 rAV t = t NA = NA  𝑁𝐴 𝑑𝑁𝐴 Time necessary to reduce the number of 𝑡= න 𝑟𝐴 𝑉 moles of A from NA0 to NA. 22  𝑁𝐴0  BATCH REACTOR EXAMPLE 𝑁𝐴 Lets used -rA 𝑑𝑁𝐴 𝑡= න 𝑟𝐴 𝑉 𝑁𝐴0 𝑁𝐴 −𝑑𝑁𝐴 𝑡= න −𝑟𝐴 𝑉 𝑁𝐴0 𝑁𝐴 𝑑𝑁𝐴 𝑡=− න −𝑟𝐴 𝑉 𝑁𝐴0 𝑁𝐴0 𝑁𝐴 𝑑𝑁𝐴 Similar 𝑑𝑁𝐴 𝑡= න 𝑡= න −𝑟𝐴 𝑉 𝑟𝐴 𝑉 23 𝑁𝐴 𝑁𝐴0  BATCH REACTOR Mole Balances NA NA0 dN A t=  N A0 − rAV NA NA1 0 t1 t Continuous-Flow Reactor  CONTINUOUS-STIRRED TANK REACTOR (CSTR) Description There is at least one inlet for reactant and one outlet of product No accumulation in the tank (will not spill) Steady state with assumed perfect mix Liquid phase reaction Temperature and concentrations are the same in all the vessel Relatively easy to maintain good temperature control The conversion of reactant per volume of reactor is the smallest of the flow reactors - very large reactors are necessary to obtain high conversions CONTINUOUS-STIRRED TANK REACTOR (CSTR) Cutaway view of a Pfaudler CSTR/ Batch Reactor  CSTR Mole Balances dNA FA 0 − FA +  rA dV = dt Steady State, dNA =0  No accumulation dt   CSTR Mole Balances Well Mixed  r dV = r V A A FA 0 − FA + rAV = 0 FA 0 − FA V=  −rA CSTR volume necessary to reduce the molar flow rate from FA0 to FA.   PLUG-FLOW REACTOR (PFR) Description Also known as tubular reactors Cylindrical pipe reactors Operate in steady state Often gas-phase reactions Reactants are consumed as they pass through pipe Simplest form of reactor Concentration varies across the pipe length. Usually produces the highest conversion per reactor volume Industrial PFRs  PFR Mole Balances FA FA0 dN A FA0 − FA +  rA dV = dt dN A Steady State, =0 No accumulation dt FA0 − FA +  rA dV = 0  PFR Mole Balances FA FA0 V FA0 − FA = −  rA dV 0 There are two problems: FA0 and FA rA varies with length  PFR Mole Balances FA FA0 Need to analyze this ”disk” V  In  Out  Generation at V  − at V + V  + in V =0 FA  FA       FA V − FA V + V + rA V =0 V V + V  PFR Mole Balances Rearrange and take limit as ΔV→0 FA V + V − FA V lim = rA V → 0 V Taking the limit as ΔV approach zero, the dFA differential form of steady state mole = rA balance on PFR is, dV  PFR Mole Balances 𝑑𝐹𝐴 Differientiate with respect to V 𝑑𝑉 = 𝑟𝐴 Integrate with limit at V=0, the FA=FA0 and at V=V1, then FA=FA1 𝐹𝐴1 𝐹𝐴0 𝑑𝐹𝐴 𝑑𝐹𝐴 The integral form is: 𝑉1 = න 𝑉1 = න 𝑟𝐴 −𝑟𝐴 𝐹𝐴0 𝐹𝐴1 This is the volume necessary to reduce the entering molar flow rate (mol/s) from FA0 to the exit molar flow rate of FA.  PACKED BED REACTOR (PBR) Description Heterogeneous reactions Fluid-solid phase Ideal for catalyst bed reaction Based on reactor’s catalyst mass (not reactor volume) Steady state operation, no accumulation If gases; there is pressure drop Concentration of product change with length Industrial PBRs  PACKED BED REACTOR (PBR) Operation Typically a clean catalyst is placed The catalyst bed is fixed so it does not moves as fluid passes by The inlet is open, fluid starts entering the reactor The fluid interact with the catalyst bed There is reaction, and the product go to outlet The catalyst is sometime saturated; it must be changed The catalyst may be poisoned so it must be changed as well Industrial PBRs  PBR We are talking about ”Mass of catalyst” Now we analyze –r’A –rA = moles of A/ Vol.time –r’A = moles of A/ mass catalyst.time –rA x Volume of reactor = moles of A per unit time –r’A x Mass of catalyst = moles of A per unit time  PBR Mole Balances W PBR FA0 FA W W + W  dN A FA0 − FA + rA W = dt Steady State dN A =0 dt  PBR Mole Balances W FA FA W W + W   dN A dN A FA W − FA W + W + rA W = Steady State =0 dt dt FA W + W − FA W lim = rA W → 0 W  PBR Mole Balances Rearrange: dFA = rA dW The integral form to find the catalyst weight is: 𝐹𝐴 𝐹𝐴0  𝑑𝐹𝐴 𝑑𝐹𝐴 𝑊= න ′ 𝑊= න 𝑟𝐴 −𝑟𝐴′ 𝐹𝐴0 𝐹𝐴 PBR catalyst weight necessary to reduce the entering molar flow rate FA0 to molar flow rate FA.  REACTOR MOLE BALANCE SUMMARY The GMBE applied to the four major reactor types (and the general reaction A→B) Reactor Differential Algebraic Integral NA 𝑁𝐴0 dN A 𝑑𝑁𝐴 NA Batch dN A = rAV t=  rAV 𝑡= න 𝑁𝐴 −𝑟𝐴 𝑉 dt N A0 FA 0 − FA t V= CSTR −rA FA 𝐹𝐴0 FA dFA 𝑑𝐹𝐴 PFR dFA = rA V =  drA 𝑉= න −𝑟𝐴 dV FA 0 𝐹𝐴1 V  FA 𝐹𝐴0 FA PBR dFA dFA 𝑑𝐹𝐴  dW = rA W =  FA 0 rA 𝑊= න −𝑟𝐴′ 𝐹𝐴 W  Flow vs. Concentration We have done our balances using flow rates Flow rate = gmol of A/ time In reactor Engineering, specificly in lab scale, we use a lot of Concentration terms Concentration = gmol of A/ volume of solution FA with CA The relationship between them is volume/time (Volumetric flowrate) N A = C AV FA = vC A Batch Continous flow rate  From Moles to Concentration Batch dNA 1. Design equation of a Batch reactor = rAV dt 2. Substituting this formula N A = C AV 3. Get this equation  d (C AV ) = rAV dt 4. If volume is constant, take it out VdC A = rAV dt 5. You get the first order differential dC A = rA equation dt  From Flow to Concentration CSTR FA 0 − FA 1. Design equation of a CSTR V= −rA 2. Substituting this formula FA = vC A 3. Get this equation v0C A0 − vC A V=  − rA 4. If inlet and outlet volumetric flow rates are v(C A0 − C A ) the same V= − rA V C A0 − C A 5. You will end up with Conc. terms = v − rA  From Flow to Concentration PFR dFA 1. Design equation of a PFR = rA dV 2. Substituting this formula FA = vC A 3. Get this equation d (vC A )  = rA dV 4. If inlet and outlet volumetric flow rates are the same vdC A = rA dV 5. You will end up with Conc. terms dC A rA = dV v  From Flow to Concentration PFR 6. If you continue to develop the equation dC A 1 = dV rA v 7. And integrate to find out the limits CA dC A V 1 C rA = 0 v dV A0 8. You end up with ths equation for volume C A0 dC A V  CA = − rA v  From Flow to Concentration PBR dFA 1. Design equation of a PBR = rA dW 2. Substituting this formula FA = vC A vdC A 3. If volumetric flow is constant, put it out of the = rA derivative  dW dC A 1 4. If you continue to develop it = dW rA v C A0 dC A W 5. Final equation  CA rA = v