Post on 12-Apr-2022
(Single / Multiple unit operation, reactive and non-reactive)
CONSERVATION OF MASS§ Mass is neither created nor destroyed
§ {Input} + {Genn} - {Consumption} – {Output} = {Accumulation}
Distillation
Heat Exchanger
Reactor
Seperator
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2
6
7
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1110
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qSystems üOPEN or CLOSEDüAny arbitrary portion of or a whole process that you want to
consider for analysisüReactor, the cell, mitochondria, human body, section of a pipe
qClosed SystemüMaterial neither enters nor leaves the systemüChanges can take place inside the system
qOpen SystemüMaterial can enter through the boundaries
BALANCES ON CONTINUOUS STEADY-STATE PROCESSESqInput + Generation = Output + Consumption
üIf the balance is on a nonreactive species, the generation and consumption will be 0.üThus, Input = Output
qExample
§ 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
qSolution of the example Input = OutputqBenzene balance
1000 kg/h · 0.5 = 450 kg/h + m2m2 = 50 kg/h Benzene
qToluene balance1000 kg/h · 0.5 = 475 kg/h + m1m1 = 25 kg/h Toluene
Distillation1000 kg /h
Benzene + Toluene
%50 Benzene by mass 475 kg Toluene/h
M2 kg Benzene/h
m1 kg Toluene/h
450 kg Benzene/h
BALANCES ON BATCH PROCESSESqInitial Input + Generation = Final Output + ConsumptionüObjective: 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.xBF.yF = D.yD + B.xBx: mole fraction of Wy: mole fraction of A
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 arerecovered from a broth ( a mix with cells) using tubular centrifuge.Determine the amount of the cell-free discharge per hour if 1000 L/hr isfed to the centrifuge, the feed contains 500 mg cells/L, and the productstream contains 50 wt% cells. Assume that the feed has a density of 1g/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)
qCell balance
qFluid balanceInput: (106 – 500) g/h fluid
Output 1: 1000g/h . 0.5 = 500 g/h fluidOutput 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 1cells g 0.5
mg 1000 g 1 .
feed L 1cells mg 500. feed L 1000
=
=
hg
Ldm
dmcm
cmg
hL 6
33
3101)
110(1 1000 =
FLOW CHARTSqBoxes and other symbols are used to represent process units.
üWrite the values and units of all known streamsüAssign 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
EXAMPLE (FLOW CHARTS)§ Humidification and Oxygenation Process in the Body: An exp. on the
growth rate of certain organisms requires an environemnt of humid airenriched in oxygen. Three input streams are fed into an evaporator toproduce 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 rateone-fifth of the molar flow rate of stream B)
§Draw the flow chart for above example.
n3 mol/min
0.015 mol H2O/mol
y mol O2/mol
(0.985 – y ) mol N2/molAB
C
EXAMPLE (FLOW CHARTS) CONTINUE…§ What flow charts that you will get?
§ Below is the flow charts
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/molAB
C
EXAMPLE (FLOW CHARTS) CONTINUE…Take basis: 1 g H2O / cm3
n2 = 20 cm3 H2O/min . 1 g H2O/cm3 . 1 mol/18 g
n2 = 1.11 mol H2O/min
qH2O Balancen2 mol H2O/min = n3 mol/min . 0.015 mol H2O/moln3 = 74 mol/min
qTotal Mole Balance0.2 n1 + n1 + n2 = n3
n1 = 60.74 mol/min
EXAMPLE (FLOW CHARTS) CONTINUE…qN2 Balance
n1 mol/min . 0.79 mol N2/mol = n3 mol/min . (0.985-y) mol N2/mol47.98 mol N2/min = 72.89 mol N2/min – 74y mol N2/mol47.98 mol N2/min + 74y mol N2/mol = 72.89 mol N2/min74y mol N2/mol = 24.91 mol N2/miny = 0.337 mol O2/mol
EXAMPLE (FLOW CHARTS) CONTINUE…§ Final flow charts after solve the problem should be:
12.15 molO2/min
60.74 molair/min 0.21 mol O2/mol
0.79 mol N2/mol
20 cm3 H2O /min
1.11 mol H2O/min
74 mol/min
0.015 mol H2O/mol
0.337 mol O2/mol
0.648 mol N2/molAB
C
What do we learn?