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CHAPTER 4

Billion: The British say that a billion is a million million (1,000,000,000,000). American say that a billion is a thousand million (1,000,000,000) and insist that a million million is actually a trillion. The Canadian Press agrees with the Americans http://www.scit.wlv.ac.uk/~jphb/american.html

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CHAPTER 4

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  1. Billion: The British say that a billion is a million million (1,000,000,000,000). American say that a billion is a thousand million (1,000,000,000) and insist that a million million is actually a trillion. The Canadian Press agrees with the Americans • http://www.scit.wlv.ac.uk/~jphb/american.html • Please note that "tonne" is not a British spelling of "ton" but a quite separate metric unit equal to 1000 kg as distinct from the British ton of 2240 lbs (= 1016.96 kg). • Billion: thousand million The old British usage in which a billion was a million2 is now largely obsolete and most British speakers would assume the American meaning. Careful users avoid the words altogether and use exponent notation. The usage continued • trillion = tri+(m)illion = million3 = 1018 • quadrillion = quad+(m)illion = million4 = 1024 • centillion = cent+(m)illion = million100 = 10600 • The American naming seems to work on the principle 103+(number×3)

  2. CHAPTER 4 MATERIAL BALANCES

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

  4. SYSTEMS • Systems • 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

  5. 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}

  6. PROCESSES • Batch 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

  7. m1 kg Toluene/h 450 kg Benzene/h Distillation 1000 kg /h Benzene + Toluene %50 Benzene by mass 475 kg Toluene/h M2 kg Benzene/h 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

  8. 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 . . . .

  9. D F (W+A) B 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

  10. 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. Concantrated cells P(g/hr) 50 % by weight cells Centrifuge Feed (broth) 1000 L/hr 500 mg cells/L feed ( d= 1 g/cm3) Cell-free discahrge D(g/hr)

  11. 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. Concantrated cells P(g/hr) 50 % by weight cells Centrifuge Feed (broth) 1000 L/hr 500 mg cells/L feed (d= 1 g/cm3) 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

  12. 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

  13. 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

  14. 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

  15. n1 A n3 n2 100 n1 A 100 n3 100 n2 FLOWCHART SCALING Scale factor: 100

  16. 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

  17. Condenser Humid air (n0) O2 (n1) N2 (n2) H2O Dry air (n4) O2 (n5) N2 (n6) H2O (n3) H2O 225 L/h EXAMPLE 1 • Example ρH20 is given In the condenser, 95% of H2O in the inlet air is condensed. 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.

  18. 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

  19. EXAMPLE 2 - continue 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

  20. 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?

  21. EXAMPLE 3 - continue F4 CH4 xCH4 N2 xN2 C2H6 xC2H6 F1 CH4 0.2 N2 0.8 F3 N2 1 F2 C2H6 0.1 N2 0.9 Unknowns: 3 xi and 4 Fi

  22. 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.

  23. 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 3 Q2 x2 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

  24. 1 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 + Q1Q1 = 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! 2 3

  25. Feed Product Feed Process Unit rxn Sep. Recycle Bypass stream 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.

  26. 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.

  27. 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)

  28. 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) 2 balance equations (two species) Degree of freedom = 0 • Cooler: 2 variables (n2, n4) 2 balance equations (two species) Degree of freedom = 0 • Splitting point: 2 variables (n4, n5) Donot use SP in thesolution 1 balance equation Degree of freedom = 1

  29. 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

  30. 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

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

  32. nC3H6 nNH3 nO2 nN2 nC3H3N nH2O 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 REACTOR 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.

  33. 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 + niξ => 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

  34. 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

  35. 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) =

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

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