(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

1

2

6

7

8

13

14

12

1110

4

53

9

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?