episode 61 : material balance for reacting system

SAJJAD KHUDHUR ABBAS

Ceo , Founder & Head of SHacademyChemical Engineering , Al-Muthanna University, IraqOil & Gas Safety and Health Professional – OSHACADEMYTrainer of Trainers (TOT) - Canadian Center of Human Development

Episode 61 : MATERIALBALANCE FOR REACTING

SYSTEM

COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS

System

CONTROLVOLUME

Fi1

wi1k

Fi2

ENVIRONMENT

Fo1

wo1k

Fo2

wo2k

Fo3

wo3kwi2k

j1 j1

wo1k Fo1 wo2k Fo2 wo3k Fo3 wi1k Fi1 wi2k Fi2 M k rk

Mkrk

Reactor

Compressor

Phase

separatior

Distillation

M krk

M L

Fojwokj Fijwikj

COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS

System

CONTROLVOLUME

Ni1

xi1k

Ni2

ENVIRONMENT

No1

xo1k

No2

xo2k

No3

xo3kxi2k

rk

Reactor

Compressor

Phase

separator

Distillation

rk

M L

j1 j1

xo1k No1 xo2k No2 xo3k No3 xi1k Ni1 xi2k Ni2 rk

Noj xokj N ij xikj

RATE OF CHEMICAL REACTION• Rate of chemical reaction of component k : rk

• Net rate of generation of component kcomponent k per unit time

in units of moles of

• Obtained from stoichiometric balance of chemical reactions

• Stoichiometry = relative proportions of chemical componentsparticipating in a chemical reaction

• Stoichiometric equation of chemical reaction:

– Showing the relative number of molecules/moles of components participating in the chemical reaction

• Reactants– components that react with each other in a chemical reaction

• Products – components that are produced by a chemical reaction

• Chemical reactor- equipment in which chemical reactions occur

CHEMICAL REACTOR

Ammonia ReactorAcid Nitric Reactor

EXAMPLE• Example 3.1 Stoichiometric

equations

• SO3 synthesis

reaction 2SO2 +

O2 2SO3

• 2 moles of SO2 reacting with 1 mole of O2 to produce 2 molesof SO3

• Ammonia synthesis

reaction N2 +

3H2 2NH3

• 1 mole of N2 reacting with 3 moles of H2 to produce 2 moles of NHO3

Reactants Product

2SO2 + (1)O2 2SO3

STOICHIOMETRIC BALANCE• Stoichiometric equation

• S = total number of components

• k = stoichiometric coefficient

•

•

Ck = molecular formula of component k

Sign Convention: k + for products & - for reactant

k Ck 0k1

S

2SO2 + (1)O2 2SO3

2SO3 - 2SO2 - (1)O2 = 0

N2 + 3H2 2NH3

2NH3 - N2 - 3H2 = 0

STOICHIOMETRIC BALANCE• Material balance

Total mass of reactants = mass of products in Stoich. Equation Conservation of mass

S

• Mk = MW of components k

2SO2 + (1)O2 2SO3

2MST – 2MSD - (1)MO = 0

2(64) – 2(48) - (1)(32) = 0

N2 + 3H2 2NH3

2MA – MN - 3MH = 0

2(17) - (28) - 3(2) = 0

0k M k

k1

STOICHIOMETRIC BALANCE• Elemental balance= total element in reactants is equal to the

total element in the product in the stoichiometric equationS

•k1

• mkl = number of element atom in a molecule of komponent k.

2SO2 + (1)O2 2SO3

Balance on S: 2mSTS – 2mSDS - (1)mOS = 0

2(1) – 2(1) - (1)(0) = 0

Balance for o2: 2mSTO – 2mSDO - (1)mOO = 0

2(3) – 2(2) - (1)(2) = 0

0 k mkl

STOICHIOMETRIC BALANCEN2 + 3H2 2NH3

Balance of N: 2mAN – mNN – 3mHN = 0

2(1) – 1(2) - 3(0) = 0

Balance of H: 2mAH – mNH - 3mHH = 0

2(3) – 1(0) - 3(2) = 0

• Element balance can be used to determine the stoichiometric coefficients provided that both the reactants and the products are known

• If L elements are involved in the stoichiometric equations, then there are L independent element balance equation a

• If S components and L elements are involved the stoichiometric equations , degree of freedom = S – L

EXAMPLEExample 3.2 Balance the stoichiometric equations of a reaction between As2S5 and HNO3.

-1As2S5 - 2HNO3 3H3AsO4 + 4H2SO4 + 5H2O + 6NO2

The stoichiometric equation is rewritten as:

1As2S5 + 2HNO3 + 3H3AsO4 + 4H2SO4 + 5H2O + 6NO2 = 0

There are 6 species & 5 elements. Degree of freedom= 6 – 5 = 1

Balance of each element

O 32 + 43 + 44 + 5 + 26 = 0

As 21 +

3

= 0

S 51 +

4

= 0

H 2 +

33

+24 + 25 = 0

N 2 + 6 = 0

STOICHIOMETRY OF BIOCHEMICAL REACTIONS

• Biotechnological products are produced in fermentationproses involving cell growth and bioproduiction

• Bioreactor/fermentor

• Biochemical transformation processes involved thousandsof biochemical reactions in the cell.

• Its stoichiometry is represented by a simple pseudochemical reaction equation

• Stoichiometric balance of pseudochemical reaction:

– Elemental balance = ordinary chemical reactions

– Electron balance = different from ordinary chemical reactions = involves energy transition

– Yield coefficient of biomass

– Yield coefficient of product

BIOREACTOR/FERMENTOR

Bioreactor/fermentor


• Biochemical reaction involves

– Substrate = glucose (CHmOn), oxygen & ammonia

– Products: cell mass (CHON), biochemical product (CHxOyNz),water and & carbon dioxide

-1CHmOn - 2O2 -3NH3 4CHON +5CHxOyNz + 6H2O + 7CO2

• Value of coefficients m and n depends on substrate

•

•

•

•

Example: glucose m = 2 and n = 1.

Value of coefficients , and depends on microbe

Example: yeast, = 1.66, = 0.13 and = 0.40.

Divide the stoichiometric equations with 1

-CHmOn - ’1O2 -’2NH3 ’3CHON +’4CHxOyNz + ’5H2O + ’6CO2

• ’j = j-1/1 and j = 1, 2, 3, …6.

• Number of elements = 4; Number of components/stoichiometric

coefficients = 6Degree of freedom of biochemical Stoichiometry = 6 – 4 = 2


• Biochemical transformation involves electron transfer determined by an electron balance

• Additional independent balance equations!

• Degree of reduction of component k, k is used in the electron balance

• Degree of reduction k = number of equivalents of availableelectrons per atom C

• Available electrons = electrons transferred to oxygen afterorganic compound is oxidized to carbon dioxide, water andammonia in biochemical reactions

• Degree of reduction of organic compounds = sum of all the product of element valency and element atomic number divided by the number of C atoms in the compound

•Vl = element valency in component l, mkC = number of carbon atoms in component

L

vlmkl mkC

k1

k


• Degree of reduction of several common organic materials: :

Methane CH4 = [1(4) + 4(1)]/1 = 8

= [6(4) + 12(1) + 6(-2)]/6 = 24/6 = 4Glucose C6H12O6

Ethanol C2H5OH = [2(4) + 6(1) + 1(-2)]/2 = 12/2 = 6

Glucose CHmOn

Cell mass CHON

s = [1(4) + m(1) + n(-2)]/1 = 4 + m - 2n

b = [1(4) + (1) + (-2) + (-3)]/1 = 4 + -

- 2 - 3

p = [1(4) + x(1) + y(-2) + z(-3)]/1 = 4 + x- 2y – 3z

Product CHxOyNz

• The degree of reduction of water, ammonia & carbon dioxide = 0

• The degree of reduction of oxygen = -4

• S

•

Electron balance equation:

Additional independent equations! 'k k 0k1


• Respiratory quotient RQ molar basis

• RQ 7 2 '6 '1

• Yield of cell biomass mass basis

•

• Mb = formula MW of biomass & Ms = formula MW of substrate

• Yield of product mass basis

•

•

•

Mp = formula MW of product

Value of RQ is obtained from experiment

M S 4M B 1M S '3 M BYX / S

M S 5M P 1M S '4 M PYP / S


• Element balance equations (4 equations) plus

– Electron balance equation

– Respiratory quotient equation

– Yield of biomass

– Yield of product

• 8 equations & 6 unknown variables

•

•

Degrees of freedom = 6-8 = -2

Two equations are not independent and can be used to check the balance stoichiometry

• Balance of elements

C

H -m + 3’2 + ’3 + x’4 + 2’5 = 0

N

O

-1+’3 + ’4 + ’6 = 0

’2 + ’3 + z’4 = 0

-n + 2’1 + ’3 + y’4 + ’5 + 2’6 = 0


• Electron balance

•

•

•

•

•

-s - 4’1 + b’3 + p’4 = 0

s = degrees of reduction of substrate

b = degrees of reduction of biomass

p = degrees of reduction of product

H and O element balances involve water and there is so much water

• Both balances are difficult to use

• Only the C, N and electron balances are used

C -1+’3 + ’4 + ’6 = 0

N

-s

’2 + ’3 + z’4 = 0

- 4’1 + b’3 + p’4 = 0

EXAMPLEExample 3.3 Aerobic growth of S. cerevisiae (yeast) on ethanol

-CH3O0.5 - ’1O2 -’2NH3 ’3CH1.704O0.408N0.149 + ’5H2O + ’6CO2

• Determine the values of ’1, ’2, ’3, and ’6 if RQ = 0.66, Yield of

biomass on substrate & Yield of biomass on oxygen

• Degree of reduction of substrate & biomass

Ethanol

Biomass

CH3O0.5

CH1.704O0.408N0.149

S = [1(4) + 3(1) + (0.5)(-2)]/1 = 6

B = [1(4) + 1.704(1) + 0.408(-2) +

0.149(-3)]/1 = 4.441

• Element balance of C & N, electron balance and RQ.

• C

• N

• Electron

• RQ

-1+’3 + ’6 = 0

’2 + 0.149’3 = 0

-6 - 4’1 + 4.41’3 = 0

'6 0.66 '1

EXAMPLE

• 4 unknowns & 4 equations & degree of freedom= 4 – 4 = 0

• Substitute last equation with first equation’3 - 0.66’1 = 1

and

• Then

4.41’3 - 4’1 = 6

and

• Yield of biomass on oxygen

Formula biomass MW M B

• Formula ethanol MW M S

• Yield of biomass on substrate

1 4 6 0.66 '3 1 4 4.41 0.66

0.0367 '1 1 0.0367 0.66 1.4595

'2 0.1490.0367 0.0055 '6 0.661.4595 0.96327

12 1.7041 0.14914 0.40816 22.318

12 31 0.516 23

M 0.036722.318 23 0.0356 g g -1X / S 3 BY ' M S

Y '3 M B '1 M O 0.036722.318 1.459532 0.0175 g g1X / O2

6

1

11 0.663

4.41 4

MATERIAL BALANCE WITH SINGLEREACTION

• COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMS Molar Mass

• Rate of chemical reaction of component k rk

• Ammonia synthesis reaction :

•

•

•

•

N2 + 3H2 2NH3

If rate of reaction of nitrogen = -rN

Rate of reaction of hydrogen = -rH

(negative: nitrogen is used)

= (H /N)(-rN )= (-3/-1)(-rN)

Rate of reaction of ammonia = rA = (A /N)(-rN )= (2/-1)(-rN)

Then

Rate of reaction r is fixed for a given reaction stoichiometric equation

• Rate of reaction of component

• COMPONENT MATERIAL BALANCE FOR REACTING SYSTEMSMolar Mass

rk M krk

M L M L

j1 j1 j1Noj xokj N ij xikj Fojwokj Fijwikj

j1

rrA

rH rN

rk

A H N k

rk kr

M kk rFojwokj Fijwikj

M L

j1 j1

k rNoj xokj N ij xikj

M L

j1 j1

MATERIAL BALANCE WITH SINGLE REACTION• Example 3.4 Lets say for the SO3 synthesis reaction, 15 mole h1 O2 (A)

40 mole h-1 SO2 (B) and 0 mole h-1 SO3 (C) is fed into a reactor. If the flow rate out of O2 is 8 mole h-1 calculate the flow rates of other components

• Basis 55 mole j-1 feed O2 + 2SO2 2 SO3

• 3 components & 3 independent material balance equations

• Degree of freedom= 3 – 3 = 0

• Choose component mole balance that has most information to get r :O2

• SO2 mole balance

• SO3 mole balance

• Substituting r inSO2 & SO2 balances:

NiA = 15 mole h-1

NiB = 40 mole h-1

NiC = 0 mole h-1

NoA =8 mole h-1

NoB

NoC

Reactor SO3

NoA Ar

NoB Br

NoC Cr

NiA

N iB

N iC

N 40 2r 40 27 26 mole h1oB

NoC 2r 27 14 mole h1

1r 7 mole h15 8 1r

40 NoB 2r

0 NoC 2r

EXAMPLE

• Example 3.5 Growth of S. cerevisiae on glucose is described by the following equation

•

• In a batch bioreactor of volume 105 L, the yeast concentration required

C6H12O6 + 3O2 + 0.48NH3 0.48C6H10NO3 + 4.32H2O + 3.12CO2

is 50 g dry mass L-1.

• Calculate the yield of biomass/substrate YX/S Yield of biomass /oxygenYX/G and respiratory quotient RQ. Calculate the required concentration and total amount of glucose and (NH3)2SO4 in the nutrient media.

• How much oxygen is required and carbon produced by the bioreaction ?

• If the growth rate at exponent phase is r = 0.7 gdm L-1 h-1, determine the rate of oxygen utilization.

• MW glucose = 180, MW oxygen = 32, MW ammonia = 17, MW (NH4)2SO4 = 116, MW biomass = 144, MW carbon dioxide = 44 and MW water = 18.

EXAMPLE

CoO = 4 mg L-1

CoC = 2 mg L-1

Batch Bioreactor

Glucose feedCiG

CiO = 0

CiC = 0

CiB = 0

Gas exhaust

Feed (NH4)2SO4CiA

Biomass productCoB= 50 gdm L-1

CoG = 5.0 g L-1

CoA = 1.0 g L-1

O2 Feed

EXAMPLE• 6 components and six independent mass balance equations

• Water balance is not used because the presence of a lot of water

• Degrees of freedom = 6 – 5 = 1

• Yield of biomass/glucose YX/S

• Yield of biomass/oxygen YX/O2.

• Respiratory quotient

•

• Choose Basis =500 kg dry biomass = 50 gdm L-1 in a 105 L bioreactor

• Choose component balance with the most information to get r

• Biomass balance

0.481441180

1 0.384 g g

MYX / S

G G

BM B

M0.48144

332

1

X / O2 0.72 g gY O O

B B

M

RQ C

3.121.04 mole mole1

O 3

NoB BrVN iB

EXAMPLE• Biomass balance

• Hence the rate of reaction

• Glucose balance

• Total amount of glucose required = 13,520.8 kg

144 0.48 0.772 mole L150

r

CiBV CoBV rV

M MB

B B

5010 0.48r105

5

1440

N NoG GrViG

C 105 5105 50105

180 1440.48180iG

CiGV

CoGV rV

M MG

G G

50180 5

1440.48 5 130.208 135.208 g L

1CiG

requires 1/2 mole of (NH4)2SO4.

•

• Total (NH4)2SO4 required = 2,113.9 kg

• O2 balance

• Total O2 utilization Nio - Noo = 217.026 kmole oxygen

50116

21441

1 20.139 21.139 g LCiA 1

N N A rViA oA2

EXAMPLE

• One mole NH3

• (NH4)2SO4

C 105 0.485010521440.48116116

iA

CiAV

CoAV A rV

M A M A 2

OrVM O

C VNiO NoO oO

0.00410 35010 1440.48

217.026x103

55

32NiO NoO

11

05

EXAMPLE

• CO2 balance

• NoCO2 = 225.689 kmole carbon dioxide = 9930.316 kg carbon dioxide

• The dissolved gas concentrations are very small and will be neglected infermenter balances

CrVM C

CoCV NoC NiC

0.00210 3.125010 1440.48

355

225.689x1044

NiC NoC

CONVERSION & LIMITING REACTANT

• Common measure of course of reaction is the fractional conversion / conversion of the limiting reactant

• Conversion links the outlet flow rate with the inlet flow rate of the same component = additional independent equation!

• Reaction rate r

•

•

The limiting reactant finishes first if the reaction is left to react by itself

If the reaction is left to react, the rate of reaction r increases to reach

the value of rlimiting when Nok = 0

• Reactant with the lowest value for Nik/(-k) finishes first

• Limiting reactant=reactant that has the lowest Nik/(-k)

• Other reactants= excess reactant

Excess fraction of component k

ik

N ik Nok

kN

X N ik X k N ik N ok

k

r N ik X k

k

Nik

Limitingr

k ip p

k N ip p

N ik

E k

N

N ik X k k r

EXAMPLE

• Example 3.6 The reaction between ammonia (A) and oxygen (O) on Pt catalyst produces nitric acid and water (W). The stoichiometric equation is given by

4NH3 + 5O2 4NO + 6H2O•

• Under certain conditions, conversion of NH3 into NO (N) can achieve90% at ammonia flow rate NH3 40 mole h-1 and O2 60 mole h-1 . Calculate the other flow rate

• Basis is 100 mole h-1 feed.

• 4 components & 4 independent material balance equations .

• Degree of freedom= 4 - 4 = 0

NiA = 40 mole j-1

NiO = 60 mole h-1

NiN = 0 mole h-1

NiW = 0 mole h1

NoA

NoO

NoN

NiW

Acid nitricReactor

Conversion 90%

EXAMPLE

• Stoichiometric coefficient

• Use conversion of ammonia

to get Rate of reaction r

• Component mole balance

• NH3

• O2

• NO

• H2O

400.9

41

9 mole h

r NiA X A

A

NoA ArN iA1

4 mol hNoA

NoO OrN iON 15 mol h 1

oO

N rNiN NoNN 36 mol h1

oN

NoW W rN iW1

54 mol hNoW

40 NoA 49

60 NoO 59

0 NoNO 49

0 NoW 69

• NH3 A = -4 O2 O = -5

• NO N = 4 H2O W = 6

EXAMPLE

Example 3.7 If the reaction in Example 3.6 achieves 80% conversion withequimolar ammonia and oxygen feed that is fed at 100 mole h-1. Calculatethe flow rate out of all components

• Stoichiometric equation is given by

•

•

4NH3 + 5O2 4NO + 6H2O

Choose basis 100 mole h-1 feed

• Determination of the limiting reactant

• Limiting reactant= Oxygen because it has the smallest Nik/(-k)

• Conversion information is for conversionm of oxygen!

NiA = 50 mole j-1

NiO = 50 mole j-1

NiN = 0 mole j-1

NiW = 0 mole j-1

NoA

NoO

NoN

NiW

Acid nitricreactor

Conversion 80%

12.5 4

50

A

NiA

510

50

O

NiO

EXAMPLE

• Use conversion of oxygen to get the rate of reaction:

• Component balance

• NH3

• O2

• NO

• H2O NiW

500.8 5

1 8 mole h

r

O

NiO XO

NoA ArN iA

NoO OrN iO

NoN N r

NoW W r

NiN

N 50 48 18 mole h 1oA

N 50 58 10 mole h1

oO

N 0 48 32 mole h1oN

0 68 48 mole h1NoW

EXAMPLEExample 3.8 Acrylonitrile (C) is produced by the following reaction:

C3H6 + NH3 + (3/2)O2 C3H3N + 3H2O

The feed contains 10% mole propylene (P), 12% mole ammonia (A) and 78% mole air. Conversion of limiting reactant is 30%. By choosing 100 mole h-1 feed as the basis, determine the limiting reactant, fractional excess of other reactants and flow rate out of all components.

6 unknown & 6 independent material balance equations Degree of freedom= 6 – 6 = 0Determine the limiting reactant

Propylene is the limiting reactant

Ni = 100 mole h-1

xiA = 0.12

xiP = 0.10

NiO = (0.21)(0.78)

NiN = (0.79)(0.78)

NoA

NoP

NoO

NiN

NoC

NoW

AcrylonitrileReactor

Conversion 30%

12

112

W P

NiW 10

1 10

NiP16.38

1.5 10.92

O

NiO

EXAMPLE

• Fractional excess of other reactants

• NH3

• O2

• From conversion, calculate rate of reaction

• NH3

• O2

• C3H3N

• H2O

• N2

P 12 (1)10 1(1)10 1

0.2NiA ANiP

EA A iP P N

P 16.38 (1.5)10 1(1.5)10 1

0.092O2NiP P

NiO O NiP

EO

0.310

11

3 mole h

r X P NiP

P

NoA ArN iA

NoO OrN iO

CrN iC NoC

1 N iN 61.62 mole hNoN

NoW W rN iW

N 12 13 9 mole h1

oA

N 16.38 1.53 11.88 mole h1

oO

0 13 3 mole h1NoC

N 0 33 9 mole h1

oW

EXAMPLEExample 3.9 Natural gas containing methane only is burnt in anincinerator CH4 + 2O2 CO2 + 2H2O

Calculate all the outgoing molar flow rate of components, total molar flow rate and fractional excess ofair

4 unknown & 4 independent material balance equationsDegree of freedom= 4 – 4 = 0

• Convert the flow rate units into mole units.

•

•

• FU NU

NUO =

NUN =

FG NG

Air MW= 0.21(32) + 0.79(28) = 28.84

0.21(10.4) = 2.184 kmole h-1

0.79(10.4) = 8.216 kmole h-1

FG = 16 kg h-1

wGM = 1.0

No

NoC

NoN

NoO

NoW

Naturalgas burner

FU = 300 kg h-1

xUO = 0.21

xUN = 0.79

41 kmole h -1 CH

4

16 kg 1 kmole CH4

h 16 kg CH

10.40kmole h airh 28.84 kg air

300 kg 1 kmole air -1

EXAMPLE• Basis is 1.0 kmole h-1 natural gas

• Stoichiometric coefficients

• CH4 CH4 = -1 O2 O2 = -2 CO2

•

CO2 = 1 H2O H2O = 2

Assume complete combustion = all methane reacted.Rate of methane reaction

• Component mole balances

• N2

• O2

• CO2

• H2O

•

•

•

Total flow rate out No = 11.4 kmole h-1

Air fractional excess:

or 9.2%

N 1.0G

M 11

1.0 kmole hr

N N 8.216 kmole h1

oNUN

NoO Or

NoC Cr

NoW W r

NUO

NGC

NGW

M 2.184 (2)1.0 1(2)1.0 1

0.092NUO O NG

EO O G M N

N 2.184 21 0.184 kmole h1

oO

0 11 1.0 kmole h1NoC

N 0 21 2.0 kmole h1

oW

EXAMPLEExample 3.10 Yeast in example 3.2 reacts with glucose, oxygen and ammonium sulfate according to the same stoichiometry in a chemostat bioreactor with volume V = 105 L in the figure below.

The rate of ventilation is 50,000 L min-1. Dilution rate D = Qi/V =0.1 j-1.

Feed

Qi = 10000 L h-1

CiACiG

CiB = 0

ProductQo = Qi = 10000 L h-1

CoG = 5 g L-1

CoB = 50 gdm L-1

CoA= 1 g L-1

Air feedQiU = 50,000 L min-1

NiU = 133928.6 mole h-1

N = 28125.0 mole h-1iO

N = 105803.6 mole h-1iN

N =0iC

Exhaust airNoU

QoU

NoO

NoN

NoC

EXAMPLE• Chemostat = continuous bioreactor

• At steady sate, substrate, other nutrient and oxygen are fed and products are withdrawn at the same volumetric flow rate

• Volumetric flow rate Qi = VD

•

•

•

= (105)(0.1) = 10000 L h-1

5 equations for 5 unknowns & Degree of freedom= 5 – 5 = 0

Basis 10000 L h-1 volumetric flow rate

Chose component balance with most information to get r: Biomass

• Then

144

r = (50)/[(10)(144)(0.48)] = 0.072338 mole L-1 h-1

NoB BrVNiB

0 104 50 0.48r105

BrVM

QCiB Q

CoB

M B B

ii

EXAMPLE• Glucose

• Then CiG = 5 + (188)(10)(0.072338) = 140.99 g L-1

• One mole NH3 requires 1/2 mole of (NH4)2SO4. Then (NH4)2SO4

balance

• Then CiA = 1 + (116)(0.48)(10)(0.072338)/2 = 21.14 g L-1

• Molar flow rate of air feed

• Then

NoG GrVN iG

N N A rV

iA oA2

GrVM

QCiG Q

CoG

M G G

ii

104Cig 1045 0.072338105188 188

104CiA 10 1 0.480.0723382

10

116 116

54

A rV

M A 2Q

CiA Q CoA

M A

i i

N 133928.571 mole h -1

min 1 j 22.4 L

60min molQ 50000L

iUiU

EXAMPLE• Hence

•

• Oxygen

NiO = (0.21)( 133928.571) = 28125.0 mole h-1

NiN = (0.79)( 133928.571) = 105803.6 mole h-1

or

oO

NoO =28125.0–21701.4 = 6423.6 mole oxygen h-1

• Nitrogen: NoN = NiN = 105803.6 mole h-1

• Carbon dioxide

• Rate of CO2 production NoCO2 = 22569 mole carbon dioxide h-1

• Rate of gas out

NoU=NoO+NoN+NoC=6423.6+105803.6+22569.0=134795 mole h-1

NoO OrVNiO

28125.0 N 30.072338105

NoC CrVNiC

0 N 3.120.072338105oC

22.4 L 50323.47 L min -1

h 60 min mole

1hQ 134795mole

oU

Thanks for WatchingPlease follow me / SAJJAD KHUDHUR ABBAS