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CHBI 2011 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)

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Presentation on theme: "CHBI 2011 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)"— Presentation transcript:

1 CHBI 2011 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 million 2 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 = million 3 = 10 18  quadrillion = quad+(m)illion = million 4 = 10 24  centillion = cent+(m)illion = million 100 = 10 600 The American naming seems to work on the principle 10 3+(number×3)

2 CHBI 2012 CHAPTER 4 MATERIAL BALANCES

3 CHBI 2013 CONSERVATION OF MASS Distillation Heat Exchanger Reactor Seperator 1 2 6 7 8 13 14 12 11 10 4 53 9 {Input} + {Gen n } - {Consumption} – {Output} = {Accumulation} Mass is neither created nor destroyed

4 CHBI 2014 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 CHBI 2015 STEADY-STATE  Steady-State Nothing is changing with time @ steady-state accumulation = 0 Rate of addition = Rate of removal  Unsteady-State (transient system) {Input} ≠ {Output} 500 kg H 2 O 100 kg/min H 2 O

6 CHBI 2016 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 CHBI 2017 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 Distillation 1000 kg /h Benzene + Toluene %50 Benzene by mass 475 kg Toluene/h M 2 kg Benzene/h m 1 kg Toluene/h 450 kg Benzene/h 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 CHBI 2018  Solution of the exampleInput = Output  Benzene balance 1000 kg/h · 0.5 = 450 kg/h + m 2 m 2 = 50 kg/h Benzene  Toluene balance 1000 kg/h · 0.5 = 475 kg/h + m 1 m 1 = 25 kg/h Toluene Balances on Continuous Steady-state Processes....

9 CHBI 2019 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 (W+A) B D F = B + D F.x F = D.x D + B.x B F.y F = D.y D + B.x B x: mole fraction of W y: mole fraction of A

10 CHBI 20110 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/cm 3. Centrifuge Feed (broth) 1000 L/hr 500 mg cells/L feed ( d= 1 g/cm 3 ) Concantrated cells P(g/hr) 50 % by weight cells Cell-free discahrge D(g/hr)

11 CHBI 20111 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/cm 3. Centrifuge Feed (broth) 1000 L/hr 500 mg cells/L feed (d= 1 g/cm 3 ) Concantrated cells P(g/hr) 50 % by weight cells Cell-free discharge D(g/hr)  Cell balance  Fluid balance Input: (10 6 – 500) g/h fluid Output 1: 1000g/h. 0.5 = 500 g/h fluid Output 2: D(g/h) = (10 6 – 500)g/h – 500 g/h = (10 6 -10 3 )g/h fluid

12 CHBI 20112 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 Combustion Chamber Condenser 100 mol C 3 H 8 1000 mol O 2 3760 mol N 2 200 mol H 2 O 50 mol C 3 H 8 750 mol O 2 3760 mol N 2 150 mol CO 2

13 CHBI 20113 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 cm 3 /min, B: Air, C: Pure oxygen (with a molar flow rate one-fifth of the molar flow rate of stream B) 0.2 n 1 mol O 2 /min n 1 mol air/min 0.21 mol O 2 /mol 0.79 mol N 2 /mol 20 cm 3 H 2 O /min n 2 mol H 2 O/min n 3 mol/min 0.015 mol H 2 O/mol y mol O 2 /mol (0.985 – y ) mol N 2 /mol A B C....

14 CHBI 20114 EXAMPLE n 2 = 20 cm 3 H 2 O/min. 1 g H 2 O/cm 3. 1 mol/18.02 g n 2 = 1.11 mol H 2 O/min  H 2 O Balance n 2 mol H 2 O/min = n 3 mol/min. 0.015 mol H 2 O/mol n 3 = 74.1 mol/min  Total Mole Balance 0.2 n 1 + n 1 + n 2 = n 3 n 1 = 60.8 mol/min  N 2 Balance n 1 mol/min. 0.79 mol N 2 /mol = n 3 mol/min. (0.985-y) mol N 2 /mol y = 0.337 mol O 2 /mol EXAMPLE (Flow chart)

15 CHBI 20115 FLOWCHART SCALING A n1n1 n2n2 n3n3 A 100 n 1 100 n 2 100 n 3 Scale factor: 100

16 CHBI 20116  n df = n unknowns – n indep.eqn’s  If n df = 0 Problem can be solved (determined)  If n df > 0 Unknowns > knowns (underspecified)  If n df < 0 overspecified (no solution)  Material balances,  Energy balances,  Process specificaitons,  Physical props&laws,  Physical constraints DEGREE OF FREEDOM ANALYSIS (df)

17 CHBI 20117 EXAMPLE 1  Example 7 unknowns (n 0 -> n 6 )  7 equations needed 3 independent material balance n 3 = ρ.V n 0 /n 1 = 21/79 n 3 = 0.95 n 2  One more equation is needed  Volume is not conserved!  Use consistent units (mole, kg)  Do not make mole balances in reactive processes. Humid air (n 0 ) O 2 (n 1 ) N 2 (n 2 ) H 2 O (n 3 ) H 2 O 225 L/h Condenser Dry air (n 4 ) O 2 (n 5 ) N 2 (n 6 ) H 2 O ρ H20 is given In the condenser, 95% of H2O in the inlet air is condensed.

18 CHBI 20118 EXAMPLE  A continuous mixer mixes NaOH with H 2 O 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 H 2 O to the product solution is 0.9. Nsp = number of species Ns = number of streams Nu = total number of variables EXAMPLE 2 M NaOH H2OH2O Product System boundary

19 CHBI 20119 EXAMPLE 2 - continue Streams Species FEEDWATERPRODUCT NaOHF NaOH W NaOH P NaOH H2OH2OF H2O W H2O P H2O TotalFWP Nu = 3(2+1) = 9 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 Last row in the table

20 CHBI 20120 EXAMPLE  A cylinder containing CH 4, C 2 H 6, and N 2 has to be prepared containing a CH 4 to C 2 H 6 mole ratio of 1.5 to 1. Avaliable to prepare the mixture are 1) a cylinder containing a mixture of 80% N 2 and 20% CH 4 2) a cylinder containing a mixture of 90% N 2 and 10% C 2 H 6 3) a cylinder containing a mixture of pure N 2 What is the number of degrees of freedom? EXAMPLE 3

21 CHBI 20121 Unknowns: 3 x i and 4 F i EXAMPLE 3 - continue F 1 CH 4 0.2 N 2 0.8 F 2 C 2 H 6 0.1 N 2 0.9 F 4 CH 4 x CH4 N 2 x N2 C 2 H 6 x C2H6 F 3 N 2 1

22 CHBI 20122 Equations: Material balance (CH 4, C 2 H 6, N 2 ) One specified ratio x CH 4 /x C 2 H 6 = 1.5 One summation of mole fractions 5 independent equations N df = 7 – 5 = 2 If you pick a basis as F 4 =1, one other value has to be specified in order to solve the problem. EXAMPLE 3 - continue

23 CHBI 20123 Balances on Multiple-unit Processes 13 2 4 100 kg/hr 0.5 kg A/kg 0.5 kg B/kg 30 kg/hr 0.3 kg A/kg 0.7 kg B/kg 40 kg/hr 0.9 kg A/kg 0.1 kg B/kg 30 kg/hr 0.6 kg A/kg 0.4 kg B/kg Q3x3Q3x3 Q2x2Q2x2 Q1x1Q1x1

24 CHBI 20124 Balances on Multiple-unit Processes Q : mass flow rate x A : mass fraction of A 1-x A : mass fraction of B Number of unknowns = 6 Number of equations = 2+2+2 = 6 Therefore, solution exists 100 = 40 + Q 1 Q 1 = 60 kg/hr 100.(0.5) = 40.(0.9) + 60.(x 1 )x 1 = 0.233 30 + Q 1 = Q 2 Q 2 = 90 kg/hr x 2 = 0.256 30 + Q 3 = Q 2 Q 3 = 60 kg/hr x 3 = 0.083 You should treat any junction as a process unit! 1 3 2

25 CHBI 20125

26 CHBI 20126  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. RECYCLE & BYPASS STREAM rxnSep. FeedProduct Recycle Process Unit Feed Bypass stream

27 CHBI 20127  Feed: Fresh air with 4 mole% H 2 O(v) is “cooled” and “dehumidified” to a water content of 1.7 mole% H 2 O. Fresh air is combined with a recycle stream of dehumidified air. The blended stream entering unit contains 2.3 mole% H 2 O. 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. EXAMPLE (pg 110)

28 CHBI 20128 EXAMPLE - continue AIR CONDITIONER n 1 (mol) 0.04 W 0.96 DA 100 mol 0.983 DA 0.017 W(v) n 2 (mol) 0.977 DA 0.023 W(v) n 3 mole W(ℓ) n 5 (mol) 0.983 DA, 0.017 W n 4 (mol) 0.017 W 0.983 DA

29 CHBI 20129  Overall system: 2 variables (n 1, n 3 ) 2 balance equations (two species) Degree of freedom = 0  (n 1, n 3 ) are determined!!!  Mixing point: 2 variables (n 2, n 5 ) 2 balance equations (two species) Degree of freedom = 0  Cooler: 2 variables (n 2, n 4 ) 2 balance equations (two species) Degree of freedom = 0  Splitting point: 2 variables (n 4, n 5 ) Donot use SP in the solution 1 balance equation Degree of freedom = 1 EXAMPLE - continue

30 CHBI 20130 Overall DA balance: 0.96 n 1 = 0.983 (100)  n 1 = 102.4 mol fresh feed Overall mole balance: n 1 = n 3 + 100  n 3 = 2.4 mol H 2 O condensed Mole balance on Mixing point: n 1 + n 5 = n 2 Water blance on Mixing point: 0.04n 1 + 0.017n 5 = 0.023n 2 n 2 = 392.5 mol n 5 = 290 mol recycled EXAMPLE - continue

31 CHBI 20131  If there is a chemical reaction in a process  More complications  The stoichiometric ratios of the chemical reactions  Constraints  The stoichiometric equation 2SO 2 + O 2  2SO 3 2 molecules of SO 2 reacts with 1 molecule of O 2 and yields 2 molecules of SO 3  2, 1 and 2 are stoichiometric coefficients of a reaction CHEMICAL REACTION STOICHIOMETRY

32 CHBI 20132  If the reactants are not in stoichiometric proportion  one of them will be excess, the other will be limiting LIMITING & EXCESS REACTANTS

33 CHBI 20133  C 3 H 6 + NH 3 + 3/2 O 2  C 3 H 3 N + 3 H 2 O Feed: 10 mol % of C 3 H 6, 12 mole % NH 3 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. EXAMPLE (pg 120) REACTOR 100 mol 0.1 mol C 3 H 6 /mol 0.12 mol NH 3 /mol 0.78 mol air/mol 0.21 mol O 2 /mol air 0.79 mol N 2 /mol air n C3H6 n NH3 n O2 n N2 n C3H3N n H2O

34 CHBI 20134 Feed: n C3H6 = 10 mole n NH3 =12 mole n O2 = 78.(0.21) =16.4 mole n NH3 /n C3H6 = 12/10 = 1.2  NH 3 is excess (stoich. 1) n O2 /n C3H6 = 16.4/10 = 1.64  O 2 is excess (stoich. 1.5) (n NH3 ) stoich. = 10 mole (n O2 ) stoich. = 15 mole (% excess) NH3 = (12-10) /10 x 100 = 20% excess NH 3 (% excess) O2 = (16.4-15) /15 x 100 = 9.3% excess O 2 (n C3H6 ) out =0.7 x (n C3H6 ) 0 = 7 mole C 3 H 6 (since the fractional conversion of C3H6 is 30%) Extent of reaction = ζ = 3 mole (since n i = n i0 + i ξ => 7= 10 - 1. ξ) n NH3 = 12- ζ =9 molen O2 =16.4 – 1.5.(ζ)= 11.9 n C3H3N = ζ = 3 molen H2O =3.(ζ) = 9 mole n N2 = (n N2 ) 0 =61.6 mole EXAMPLE – continue Moles reacted Moles fed

35 CHBI 20135  If you are given a set of reactive species and reaction conditions; a)What will be the final (equilibrium) composition of the reaction mixture? b)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 CHEMICAL EQUILIBRIUM

36 CHBI 20136 CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) Given @ T=1105 K, K=1 n CO = 1 mol, n H2O = 2mol, initially no CO 2 and H 2 Calculate the equilibrium composition and the fractional converison of the limiting reactant. Equilibrium constant; K(T) = EXAMPLE

37 CHBI 20137 n CO = 1-ζ e,n H2O = 2-ζ e, n CO2 = ζ e, n H2 = ζ e y CO = (1-ζ e )/3y H2O = (2-ζ e )/3 y CO2 = ζ e /3 y H2 = ζ e /3 K(T) = (ζ e ) 2 / (1-ζ e ) (2-ζ e ) = 1 ζ e = 0.667 mole y CO = 0.111y H2O = 0.444 y CO2 = 0.222 y H2 = 0.222 Limiting reactant is CO. n CO = 1-0.667 = 0.333 Fractional conversion = (1-0.333) / 1 mol feed = 0.667 EXAMPLE – continue


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