MATERIAL BALANCES CHBI 201

CONSERVATION OF MASS Mass is neither created nor destroyed Distillation Heat Exchanger Reactor Seperator 1 2 6 7 8 13 14 12 11 10 4 5 3 9 {Input} + {Genn} - {Consumption} – {Output} = {Accumulation} CHBI 201

SYSTEMS Systems Closed System Open System OPEN or CLOSED Any arbitrary portion of or a whole process that you want to consider for analysis Reactor, the cell, mitochondria, human body, section of a pipe Closed System Material neither enters nor leaves the system Changes can take place inside the system Open System Material can enter through the boundaries CHBI 201

Rate of addition = Rate of removal STEADY-STATE Steady-State Nothing is changing with time @ steady-state accumulation = 0 500 kg H2O 100 kg/min H2O 100 kg/min H2O Rate of addition = Rate of removal Unsteady-State (transient system) {Input} ≠ {Output} CHBI 201

PROCESSES Batch Process Continuous Process Semibatch Process Feed is fed at the beginning of the process Continuous Process The input and outputs flow continuously throughout the duration of proces Semibatch Process Any process neither batch nor continuous CHBI 201

Balances on Continuous Steady-state Processes Input + Generation = Output + Consumption If the balance is on a nonreactive species, the generation and consumption will be 0. Thus, Input = Output Example Input of 1000 kg/h of benzene+toluene containing 50% B by mass is separated by distillation column into two fractions. B: the mass flow rate of top stream=450 kg/h T: the mass flow rate of bottom stream=475 kg/h Distillation 1000 kg /h Benzene + Toluene %50 Benzene by mass 475 kg Toluene/h M2 kg Benzene/h m1 kg Toluene/h 450 kg Benzene/h CHBI 201

Balances on Continuous Steady-state Processes Solution of the example Input = Output Benzene balance 1000 kg/h · 0.5 = 450 kg/h + m2 m2 = 50 kg/h Benzene Toluene balance 1000 kg/h · 0.5 = 475 kg/h + m1 m1 = 25 kg/h Toluene . . . . CHBI 201

BALANCES ON BATCH PROCESSES Initial Input + Generation = Final Output + Consumption Objective: generate as many independent equations as the number of unknowns in the problem F = B + D F.xF = D.xD + B.xB F.yF = D.yD + B.xB x: mole fraction of W y: mole fraction of A F (W+A) B D CHBI 201

EXAMPLE (Batch Process) Centrifuges are used to seperate particles in the range of 0.1 to 100 µm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has a density of 1 g/cm3. Feed (broth) 1000 L/hr 500 mg cells/L feed ( d= 1 g/cm3) Centrifuge Concantrated cells P(g/hr) 50 % by weight cells Cell-free discahrge D(g/hr) CHBI 201

EXAMPLE (Batch Process) Centrifuges are used to seperate particles in the range of 0.1 to 100 µm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has a density of 1 g/cm3. Feed (broth) 1000 L/hr 500 mg cells/L feed (d= 1 g/cm3) Centrifuge Concantrated cells P(g/hr) 50 % by weight cells Cell-free discharge D(g/hr) Cell balance Fluid balance Input: (106 – 500) g/h fluid Output 1: 1000g/h . 0.5 = 500 g/h fluid Output 2: D(g/h) = (106 – 500)g/h – 500 g/h = (106 -103)g/h fluid CHBI 201

FLOW CHARTS Boxes and other symbols are used to represent process units. Write the values and units of all known streams Assign algebraic symbols to unknown stream variables 100 mol C3H8 Combustion Chamber Condenser 50 mol C3H8 750 mol O2 3760 mol N2 150 mol CO2 1000 mol O2 3760 mol N2 200 mol H2O CHBI 201

EXAMPLE (Flow charts) Humidification and Oxygenation Process in the Body: An exp. on the growth rate of certain organisms requires an environemnt of humid air enriched in oxygen. Three input streams are fed into an evaporator to produce an output stream with the desired composition. A: liquid water, fed at a rate of 20 cm3/min, B: Air, C: Pure oxygen (with a molar flow rate one-fifth of the molar flow rate of stream B) . . n3 mol/min 0.015 mol H2O/mol y mol O2/mol (0.985 – y ) mol N2/mol 0.2 n1 mol O2/min C . B A n1 mol air/min 0.21 mol O2/mol 0.79 mol N2/mol . 20 cm3 H2O /min n2 mol H2O/min CHBI 201

EXAMPLE (Flow chart) EXAMPLE n2 = 20 cm3 H2O/min . 1 g H2O/cm3 . 1 mol/18.02 g n2 = 1.11 mol H2O/min H2O Balance n2 mol H2O/min = n3 mol/min . 0.015 mol H2O/mol n3 = 74.1 mol/min Total Mole Balance 0.2 n1 + n1 + n2 = n3 n1 = 60.8 mol/min N2 Balance n1 mol/min . 0.79 mol N2/mol = n3 mol/min . (0.985-y) mol N2/mol y = 0.337 mol O2/mol CHBI 201

FLOWCHART SCALING A A Scale factor: 100 n1 n3 n2 100 n1 100 n3 100 n2 CHBI 201

DEGREE OF FREEDOM ANALYSIS (df) ndf = nunknowns – nindep.eqn’s If ndf = 0 Problem can be solved (determined) If ndf > 0 Unknowns > knowns (underspecified) If ndf < 0 overspecified (no solution) Material balances, Energy balances, Process specificaitons, Physical props&laws, Physical constraints CHBI 201

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EXAMPLE 1 Example ρH20 is given In the condenser, Condenser (n4) O2 95% of H2O in the inlet air is condensed. Humid air (n0) O2 (n1) N2 (n2) H2O (n3) H2O 225 L/h Condenser Dry air (n4) O2 (n5) N2 (n6) H2O 7 unknowns (n0 -> n6) 7 equations needed 3 independent material balance n3 = ρ.V n0/n1 = 21/79 n3 = 0.95 n2 One more equation is needed Volume is not conserved! Use consistent units (mole, kg) Do not make mole balances in reactive processes. CHBI 201

EXAMPLE 2 EXAMPLE A continuous mixer mixes NaOH with H2O to produce an aqueous solution of NaOH. Determine the composition and flow rate of the product, if the flow rate of NaOH is 1000 kg/hr and the ratio of the flow rate of H2O to the product solution is 0.9. Nsp = number of species Ns = number of streams Nu = total number of variables System boundary H2O NaOH M Product CHBI 201

EXAMPLE 2 - continue Nu = 3(2+1) = 9 Streams Species FEED WATER PRODUCT NaOH FNaOH WNaOH PNaOH H2O FH2O WH2O PH2O Total F W P Nu = 3(2+1) = 9 Last row in the table Specifications: ratio of two streams the % conversion in a reaction the value of each concentration, flow rate, T, P, ρ, V, etc. a variable is not present in a stream, hence ,it is 0 CHBI 201

EXAMPLE 3 EXAMPLE A cylinder containing CH4, C2H6, and N2 has to be prepared containing a CH4 to C2H6 mole ratio of 1.5 to 1. Avaliable to prepare the mixture are 1) a cylinder containing a mixture of 80% N2 and 20% CH4 2) a cylinder containing a mixture of 90% N2 and 10% C2H6 3) a cylinder containing a mixture of pure N2 What is the number of degrees of freedom? CHBI 201

EXAMPLE 3 - continue Unknowns: 3 xi and 4 Fi F4 F1 CH4 xCH4 CH4 0.2 N2 xN2 C2H6 xC2H6 F3 N2 1 Unknowns: 3 xi and 4 Fi CHBI 201

EXAMPLE 3 - continue Equations: Material balance (CH4, C2H6, N2) One specified ratio xCH4/xC2H6 = 1.5 One summation of mole fractions 5 independent equations Ndf = 7 – 5 = 2 If you pick a basis as F4=1, one other value has to be specified in order to solve the problem. CHBI 201

Balances on Multiple-unit Processes 30 kg/hr 0.6 kg A/kg 0.4 kg B/kg 40 kg/hr 0.9 kg A/kg 0.1 kg B/kg 1 Q1 x1 Q2 x2 3 Q3 x3 100 kg/hr 0.5 kg A/kg 0.5 kg B/kg 2 30 kg/hr 0.3 kg A/kg 0.7 kg B/kg 4 CHBI 201

Balances on Multiple-unit Processes Q : mass flow rate xA : mass fraction of A 1-xA : mass fraction of B Number of unknowns = 6 Number of equations = 2+2+2 = 6 Therefore, solution exists 100 = 40 + Q1 Q1 = 60 kg/hr 100.(0.5) = 40.(0.9) + 60.(x1) x1 = 0.233 30 + Q1 = Q2 Q2 = 90 kg/hr x2 = 0.256 30 + Q3 = Q2 Q3 = 60 kg/hr x3 = 0.083 You should treat any junction as a process unit! 1 2 3 CHBI 201

RECYCLE & BYPASS STREAM It is rare that a chemical reaction A  B proceeds to completion in a reactor. Its efficiency is never 100. Some A in the product ! To find a way to send the “A” back to feed you need a seperation and recycle equipment, this would decrease the cost of purchasing more A. If a fraction of the feed to a process unit is diverted around the unit and combined with the output stream, this process is called bypass. rxn Sep. Feed Product Recycle Process Unit Feed Bypass stream CHBI 201

Fresh air is combined with a recycle stream of dehumidified air. EXAMPLE (pg 110) Feed: Fresh air with 4 mole% H2O(v) is “cooled” and “dehumidified” to a water content of 1.7 mole% H2O. Fresh air is combined with a recycle stream of dehumidified air. The blended stream entering unit contains 2.3 mole% H2O. In the air conditioner some of the water is removed as liquid. Take 100 mole of dehumidified air delivered to the room, calculate moles of feed, water condensed, dehumidified air recycled. CHBI 201

EXAMPLE - continue n5 (mol) 0.983 DA, 0.017 W n1 (mol) 0.04 W 0.96 DA AIR CONDITIONER n4 (mol) 0.017 W 0.983 DA 100 mol 0.983 DA 0.017 W(v) n3 mole W(ℓ) n2 (mol) 0.977 DA 0.023 W(v) CHBI 201

EXAMPLE - continue Overall system: 2 variables (n1, n3) 2 balance equations (two species) Degree of freedom = 0  (n1, n3) are determined!!! Mixing point: 2 variables (n2, n5) Cooler: 2 variables (n2, n4) Splitting point: 2 variables (n4, n5) Donot use SP in the solution 1 balance equation Degree of freedom = 1 CHBI 201

EXAMPLE - continue Overall DA balance: 0.96 n1 = 0.983 (100)  n1 = 102.4 mol fresh feed Overall mole balance: n1 = n3 + 100  n3 = 2.4 mol H2O condensed Mole balance on Mixing point: n1 + n5 = n2 Water blance on Mixing point: 0.04n1 + 0.017n5 = 0.023n2 n2 = 392.5 mol n5 = 290 mol recycled CHBI 201

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CHEMICAL REACTION STOICHIOMETRY If there is a chemical reaction in a process  More complications The stoichiometric ratios of the chemical reactions  Constraints The stoichiometric equation 2SO2 + O2  2SO3 2 molecules of SO2 reacts with 1 molecule of O2 and yields 2 molecules of SO3 2, 1 and 2 are stoichiometric coefficients of a reaction CHBI 201

LIMITING & EXCESS REACTANTS If the reactants are not in stoichiometric proportion  one of them will be excess, the other will be limiting CHBI 201

EXAMPLE (pg 120) C3H6 + NH3 + 3/2 O2  C3H3N + 3 H2O Feed: 10 mol % of C3H6, 12 mole % NH3 and 78 mole % air A fractional converison of limiting reactant = 30% Taking 100 mol of feed as a basis, determine which reactant is limiting, and molar amounts of all product gas constituents for a 30% conversion of the limiting reactant. REACTOR 100 mol 0.1 mol C3H6/mol 0.12 mol NH3/mol 0.78 mol air/mol 0.21 mol O2/mol air 0.79 mol N2/mol air nC3H6 nNH3 nO2 nN2 nC3H3N nH2O CHBI 201

EXAMPLE – continue Feed: nC3H6= 10 mole nNH3=12 mole nO2= 78.(0.21) =16.4 mole nNH3/nC3H6= 12/10 = 1.2  NH3 is excess (stoich. 1) nO2/nC3H6= 16.4/10 = 1.64  O2 is excess (stoich. 1.5) (nNH3)stoich.= 10 mole (nO2)stoich.= 15 mole (% excess)NH3 = (12-10) /10 x 100 = 20% excess NH3 (% excess)O2 = (16.4-15) /15 x 100 = 9.3% excess O2 (nC3H6)out=0.7 x (nC3H6)0= 7 mole C3H6 (since the fractional conversion of C3H6 is 30%) Extent of reaction = ζ = 3 mole (since ni = ni0 + ni ξ => 7= 10 - 1. ξ) nNH3 = 12- ζ =9 mole nO2=16.4 – 1.5.(ζ)= 11.9 nC3H3N= ζ = 3 mole nH2O=3.(ζ) = 9 mole nN2= (nN2)0=61.6 mole Moles reacted Moles fed CHBI 201

CHEMICAL EQUILIBRIUM If you are given a set of reactive species and reaction conditions; What will be the final (equilibrium) composition of the reaction mixture? How long will the system take to reach a specified state short of equilibrium? Chemical equilibrium thermodynamics & Chemical Kinetics A reaction can be Reversible Irreversible CHBI 201

EXAMPLE CO (g) + H2O (g) CO2 (g) + H2 (g) Given @ T=1105 K, K=1 nCO= 1 mol, nH2O= 2mol, initially no CO2 and H2 Calculate the equilibrium composition and the fractional converison of the limiting reactant. Equilibrium constant; K(T) = CHBI 201

EXAMPLE – continue nCO = 1-ζe , nH2O = 2-ζe , nCO2 = ζe , nH2 = ζe yCO = (1-ζe)/3 yH2O = (2-ζe)/3 yCO2 = ζe /3 yH2 = ζe /3 K(T) = (ζe)2 / (1-ζe) (2-ζe) = 1 ζe = 0.667 mole yCO = 0.111 yH2O = 0.444 yCO2 = 0.222 yH2 = 0.222 Limiting reactant is CO. nCO = 1-0.667 = 0.333 Fractional conversion = (1-0.333) / 1 mol feed = 0.667 CHBI 201

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Step 1. Draw a block diagram of the process. Procedure Step 1. Draw a block diagram of the process. Show each significant step as a block, linked by lines and arrows to show the stream connections and flow direction. Step 2. List all the available data. Show on the block diagram the known flows (or quantities) and stream compositions. Step 3. List all the information required from the balance. Step 4. Decide the system boundaries. Step 5. Write out all the chemical reactions involved for the main products and byproducts. Step 6. Note any other constraints, such as: specified stream compositions, azeotropes, phase equilibria, tie substances. The use of phase equilibrium relationships and other constraints in determining stream compositions and flows is discussed in more detail in. Step 7. Note any stream compositions and flows that can be approximated. Step 8. Check the number of conservation (and other) equations that can be written, and compare with the number of unknowns. Decide which variables are to be design variables; This step would be used only for complex problems. Step 9. Decide the basis of the calculation; The order in which the steps are taken may be varied to suit the problem. CHBI 201