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CHBI 201 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)
CHBI 201 2
CHAPTERCHAPTER 4 4
MATERIAL BALANCESMATERIAL BALANCES
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CONSERVATION OF MASS
Distillation
Heat Exchanger
Reactor
Seperator
1
2
6
7
8
13
14
12
1110
4
53
9
{Input} + {Genn} - {Consumption} – {Output} = {Accumulation}
Mass is neither created nor destroyed
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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
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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 H2O
100 kg/min H2O
100 kg/min H2O
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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
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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
M2 kg Benzene/h
m1 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/hT: the mass flow rate of bottom stream=475 kg/h
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Solution of the exampleInput = 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
Balances on Continuous Steady-state Processes
..
..
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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 + DF.xF = D.xD + B.xB
F.yF = D.yD + B.xB
x: mole fraction of W
y: mole fraction of A
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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.
CentrifugeFeed (broth) 1000 L/hr
500 mg cells/L feed
( d= 1 g/cm3)
Concantrated cells P(g/hr)
50 % by weight cells
Cell-free discahrge D(g/hr)
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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.
CentrifugeFeed (broth) 1000 L/hr
500 mg cells/L feed
(d= 1 g/cm3)
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
g/hr 1000 P
P[g/hr] . P g 1
cells g 0.5
mg 1000
g 1 .
feed L 1
cells mg 500. feed L 1000
h
g
L
dm
dm
cm
cm
g
h
L 6
3
3
310
1)
1
10(
1 1000
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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
CombustionChamber
Condenser
100 mol C3H8
1000 mol O2
3760 mol N2
200 mol H2O
50 mol C3H8
750 mol O2
3760 mol N2
150 mol CO2
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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)
0.2 n1 mol O2/min
n1 mol air/min
0.21 mol O2/mol
0.79 mol N2/mol
20 cm3 H2O /min
n2 mol H2O/min
n3 mol/min
0.015 mol H2O/mol
y mol O2/mol
(0.985 – y ) mol N2/mol
AB
C
. .
.
.
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EXAMPLE
n2 = 20 cm3 H2O/min . 1 g H2O/cm3 . 1 mol/18.02 g
n2 = 1.11 mol H2O/min
H2O Balancen2 mol H2O/min = n3 mol/min . 0.015 mol H2O/mol
n3 = 74.1 mol/min
Total Mole Balance0.2 n1 + n1 + n2 = n3
n1 = 60.8 mol/min
N2 Balancen1 mol/min . 0.79 mol N2/mol = n3 mol/min . (0.985-y) mol
N2/mol
y = 0.337 mol O2/mol
EXAMPLE (Flow chart)
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FLOWCHART SCALING
A
n1
n2
n3
A
100 n1
100 n2
100 n3
Scale factor: 100
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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
DEGREE OF FREEDOM ANALYSIS (df)
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EXAMPLE 1
Example
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.
Humid air
(n0) O2
(n1) N2
(n2) H2O
(n3) H2O
225 L/h
Condenser Dry air(n4) O2
(n5) N2
(n6) H2O
ρH20 is given
In the condenser, 95% of H2O in the inlet air is condensed.
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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
EXAMPLE 2
MNaOHH2O
Product
System boundary
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EXAMPLE 2 - continue
Streams
Species
FEED WATER PRODUCT
NaOH FNaOH WNaOH PNaOH
H2O FH2O WH2O PH2O
Total F W P
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
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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 are1) 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?
EXAMPLE 3
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Unknowns: 3 xi and 4 Fi
EXAMPLE 3 - continue
F1
CH4 0.2N2 0.8
F2
C2H6 0.1N2 0.9
F4
CH4 xCH4
N2 xN2
C2H6 xC2H6
F3
N2 1
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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.
EXAMPLE 3 - continue
4
Ffor 1ix
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Balances on Multiple-unit Processes
1 3
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
Q3
x3
Q2
x2
Q1
x1
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Balances on Multiple-unit Processes
Q : mass flow ratexA : mass fraction of A
1-xA : mass fraction of B
Number of unknowns = 6Number 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/hrx3 = 0.083
You should treat any junction as a process unit!
1
3
2
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CHBI 201 26
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
rxn Sep.Feed Product
Recycle
Process Unit
Feed
Bypass stream
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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.
EXAMPLE (pg 110)
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EXAMPLE - continue
AIR CONDITIONER
n1 (mol)
0.04 W
0.96 DA
100 mol
0.983 DA
0.017 W(v)
n2 (mol)
0.977 DA
0.023 W(v)
n3 mole W(ℓ)
n5 (mol) 0.983 DA, 0.017 W
n4 (mol)
0.017 W
0.983 DA
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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 the solution
1 balance equation Degree of freedom = 1
EXAMPLE - continue
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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
EXAMPLE - continue
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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
CHEMICAL REACTION STOICHIOMETRY
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If the reactants are not in stoichiometric proportion one of them will be excess, the other will be limiting
LIMITING & EXCESS REACTANTS
iν
i0n-
in
)( reaction of Extend
fed molesreacted moles A of conversion Fractional
stoich.)
A(n
]stoich)
A(n -
feed)
A(n [
A of excess Fractional
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C3H6 + NH3 + 3/2 O2 C3H3N + 3 H2O
Feed: 10 mol % of C3H6, 12 mole % NH3 and 78 mole % airA fractional converison of limiting reactant = 30%Taking 100 mol of feed as a basis, determine which reactantis limiting, and molar amounts of all product gas constituentsfor a 30% conversion of the limiting reactant.
EXAMPLE (pg 120)
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
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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 + i 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
EXAMPLE – continue
Moles reacted Moles fed
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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
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CO (g) + H2O (g) CO2 (g) + H2 (g)
Given @ T=1105 K, K=1nCO= 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) =
EXAMPLE
OHCO
HCO
yy
yy
2
22
CHBI 201 37
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
EXAMPLE – continue