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

CHBI 201 3

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

CHBI 201 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

CHBI 201 5

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

CHBI 201 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

CHBI 201 7

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

CHBI 201 8

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

..

..

CHBI 201 9

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

CHBI 201 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.

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)

CHBI 201 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.

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

CHBI 201 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

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

CHBI 201 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)

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

. .

.

.

CHBI 201 14

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)

CHBI 201 15

FLOWCHART SCALING

A

n1

n2

n3

A

100 n1

100 n2

100 n3

Scale factor: 100

CHBI 201 16

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)

CHBI 201 17

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.

CHBI 201 18

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

CHBI 201 19

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

CHBI 201 20

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

CHBI 201 21

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

CHBI 201 22

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

CHBI 201 23

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

CHBI 201 24

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

CHBI 201 25

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

CHBI 201 27

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)

CHBI 201 28

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

CHBI 201 29

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

CHBI 201 30

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

CHBI 201 31

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

CHBI 201 32

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

CHBI 201 33

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

CHBI 201 34

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

CHBI 201 35

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

CHBI 201 36

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