material balance 2011

7/28/2019 Material Balance 2011

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Material & Energy Balances : Refresher 2011

Prof Sunil S Bhagwat, [email protected] Engineering Department


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Topics in Material Balance

Dimensions and UnitsMole concept, CompositionsReaction StoichiometryBehavior of Gases and VaporsHumidity and SaturationSimple Material Balances without ReactionsMaterial Balances with ReactionComplex Material Balances


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Topics in Energy Balance

Thermophysics - Specic Heat VariationThermochemistryFuelsStagewise OperationUnsteady State Differential BalancesChemical Processes


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Dimensions and Units

Dimensions: Mass, Length, Time, Temperature, Charge ...Units: centimeter, second, ...

Pressure Force / AreaEnergy Force x Distance

Force / Area x VolumePower Energy / Time

Force x Distance / TimeForce Mass x Acceleration

Force/Area - Gauge concept


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Conversion of Units

System of Units: CGS, MKS, SI, FPS

Pressure psia, psig, ata, atg, mm-Hg, Pa, Kgf/ cm 2

Energy lit.atm, J, cal, cm 3 -atm, erg, KcalPower Watt, Any energy / time, Horsepower unitTemperature K , 0C , 0R , 0F


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Conversion of Units 2

760 mmHg =760 mm ( 1 m 1000 mm ) 13 .6 g cm 3

1 kg 1000 g ( 100 cm 1m )3 9 .80665 m s 2

= 1 .01325 10 5 kg m / s 2

m 2 or N

m 2 or Pa = 101 .325 kPa

1.987 cal = 1 .987 cal ( 4.186 J cal ) ( 1Pa m

3

1J ) ( 1atm 1 .01325 x 10 5 Pa ) ( 1000 lit 1m 3 ) ( 1000 cm 3

1lit )= 82 .06 cm 3 .atm


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760 mmHg =760 mm ( 1 m 1000 mm ) 13 .6 g cm 3

1 kg 1000 g ( 100 cm 1m )3 9 .80665 m s 2

= 1 .01325 10 5 kg m / s 2

m 2 or N

m 2 or Pa = 101 .325 kPa

1.987 cal = 1 .987 cal ( 4.186 J cal ) ( 1Pa m

3

1J ) ( 1atm 1 .01325 x 10 5 Pa ) ( 1000 lit 1m 3 ) ( 1000 cm 3

1lit )= 82 .06 cm 3 .atm


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760 mmHg =760 mm ( 1 m 1000 mm ) 13 .6 g cm 3 1 kg 1000 g ( 100 cm 1m )3 9 .80665 m s 2= 1 .01325 10 5 kg m / s

2

m 2 or N

m 2 or Pa = 101 .325 kPa

1.987 cal

= 1 .987 cal ( 4.186 J cal ) ( 1Pa m 3

1J ) ( 1atm 1 .01325 x 10 5 Pa ) ( 1000 lit 1m 3 ) ( 1000 cm 3

1lit )= 82 .06 cm 3 .atm


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The Mol concept

gmol = mass in g molecular weight

Avogadros Number : 1 mol=6.02 x 10 23 moleculesMolecular Weight : g/mol

* C12 and O standardassigns a mass of exactly 12 to the carbon isotope 12 C .


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Compositions

wt%wt/wt g/kg-solvent

wt/vol g/lit-solutionvol%vol/volmol/wt mol/kg-solvent molalitymol%mol/vol mol/lit-solution molarity

Solids & Liquids usually % wt. usually % vol. (=%mol)


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Compositions 2

Example: 53 % wt HNO 3 - Wateri.e. 53 wt HNO 3 in 100 wt mixture

BASIS: 100 wt units mixture100 g mixture contains :HNO 3=53g and H 2O =47g(1) Find Mol% HNO 3(i .e . mol HNO 3mol Mixture x 100)mol.wt. of HNO 3=63 g / mol mol HNO 3 in Basis=53 g / 63 g / mol = 0.8413 mol

mol H 2O in Basis = 47 g /18g / mol = 2.611 molHNO 3= 0.8413 x 100/(2.611 + 0.8413) = 24.37 % mol(2) Find mol/lit HNO 3 (i.e. mol HNO 3 / vol. of solution) (density= 1.46 g/ cm 3)mol HNO 3 / lit soln. = (0.8413 mol HNO 3 / 100g solution) * (1.46

g soln/ cm 3soln) * ( 1000 cm 3 /1 lit)=


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Compositions 2

Example: 53 % wt HNO 3 - Wateri.e. 53 wt HNO 3 in 100 wt mixture

BASIS: 100 wt units mixture100 g mixture contains :HNO 3=53g and H 2O =47g(1) Find Mol% HNO 3(i .e . mol HNO 3mol Mixture x 100)mol.wt. of HNO 3=63 g / mol mol HNO 3 in Basis=53 g / 63 g / mol = 0.8413 mol

mol H 2O in Basis = 47 g /18g / mol = 2.611 molHNO 3= 0.8413 x 100/(2.611 + 0.8413) = 24.37 % mol(2) Find mol/lit HNO 3 (i.e. mol HNO 3 / vol. of solution) (density= 1.46 g/ cm 3)mol HNO 3 / lit soln. = (0.8413 mol HNO 3 / 100g solution) * (1.46

g soln/ cm 3soln) * ( 1000 cm

3 /1 lit)

=


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Average Molecular Weight

For Gases Vol% = mol%Example:Air: 79% mol N 2 , 21% mol O 2Basis: 100 mol Air

N 2=79 molO 2=21 mol

Avg. mol. wt. = wt. of 1 mol gas= wt. of all components in 1 mol mixture= 79 mol N 2 x 28 g/mol N 2 + 21 mol O 2 x32 g/mol O 2

= 2884 g in 100 mol mixture = 28.84 g / mol


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Problem: Composition

Waste acid Composition:21% HNO 3 , 55% H 2SO 4 , 24% H 2O

Desired Product Composition:28% HNO 3 , 62% H 2SO 4

Concentrated sulfuric acid is available at 93% concentration

Concentrated nitric acid is available at 90% concentration1000 kg desired product.


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Problem: Composition

Waste acid Composition:21% HNO 3 , 55% H 2SO 4 , 24% H 2O

Desired Product Composition:28% HNO 3 , 62% H 2SO 4

Concentrated sulfuric acid is available at 93% concentration

Concentrated nitric acid is available at 90% concentration1000 kg desired product.


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Solution

1000 kg desired mixture

Sulfuric Acid y kg

Waste Acid x kg

Nitric Acid z kg

(1)

(2)

(3)

(4)

Basis: 1000 kg of desired mixture


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Solution

[1]. Overall material balance: (units=kg)Input:(1). x kg waste acid(2). y kg Sulfuric Acid(3). z kg Nitric AcidOutput:(4) 1000 kg desired mixture

In Out = 0

x + y + z 1000 = 0 ...(1)


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Contd..

Similarly, [2]. Material balance wrt HNO 3 : (units=kg HNO 3)

0 .21 x + 0 + 0.9z 280 = 0 ...(2)

[3]. Material balance wrt H 2SO 4 : (units=kg H 2SO 4)

0 .55 x + 0 .93 y 620 = 0 ...(3)

The three equations we get are:

x + y + z 1000 = 0 ...(1)

0 .21 x + 0 + 0.9z 280 = 0 ...(2)

0 .55 x + 0 .93 y 620 = 0 ...(3)


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Contd..


0 .21 x + 0 + 0.9z 280 = 0 ...(2)


0 .55 x + 0 .93 y 620 = 0 ...(3)


x + y + z 1000 = 0 ...(1)

0 .21 x + 0 + 0.9z 280 = 0 ...(2)

0 .55 x + 0 .93 y 620 = 0 ...(3)


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Contd..


0 .21 x + 0 + 0.9z 280 = 0 ...(2)


0 .55 x + 0 .93 y 620 = 0 ...(3)


x + y + z 1000 = 0 ...(1)

0 .21 x + 0 + 0.9z 280 = 0 ...(2)

0 .55 x + 0 .93 y 620 = 0 ...(3)


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Problem : Compositions

Required: 1000 kg of 50% H 2SO 4Available: (1) 70% H 2SO 4 Rs. 2400/ton

(2) 96% H 2SO 4 Rs. 3400/tonTransportation cost: Rs. 400/ton


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Solution:

Basis: 1 ton of required acid==> 500 kg H 2SO 4and 500 kg Water.

Option(1): 70% H 2SO 4 Rs. 2400/ton

x kg X 70 kg H 2 SO 4100 kg acid = 500 kg H 2SO 4Hence, x = 714.29 kg of acid needed.

Cost of Acid = 714.29 kg X2400 Rs .

ton X1 ton

1000 kg =1714.30 Rs.

Transportation Cost = 714.29 kg X 400 Rs .ton X1 ton

1000 kg =285.72 Rs.

Total Cost = 1714.3 + 285.72 = 2000 Rs.


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Solution:





ton X1 ton

1000 kg =1714.30 Rs.


1000 kg =285.72 Rs.

Total Cost = 1714.3 + 285.72 = 2000 Rs.


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Solution:





ton X1 ton

1000 kg =1714.30 Rs.


1000 kg =285.72 Rs.

Total Cost = 1714.3 + 285.72 = 2000 Rs.


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Option(2)

: 96% H 2SO 4 Rs. 3400/ton


Cost of Acid = 520.83 kg X 3400 Rs .ton X1 ton

1000 kg =1770.83 Rs.


1000 kg

=208.33 Rs.Total Cost = 1770.83 + 208.33 = 1979 Rs.Option(1): 2000 Rs. Option(2): 1979 Rs.Ans .


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Option(2)

: 96% H 2SO 4 Rs. 3400/ton



1000 kg =1770.83 Rs.


1000 kg



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Option(2)

: 96% H 2SO 4 Rs. 3400/ton



1000 kg =1770.83 Rs.


1000 kg



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Behavior of Gases and Vapors

Ideal Gas Law:pV = nRT

Its relationship to Kinetic Theory of Gases :Volume Temperature Volume = 0 at Temperature =-273.155 CKelvin Scale: K and Not K0 C=273.155 0.005K =459.69 R

T of 1

C = T of 1 K = T of 1 .8

F = T of 1.8

R Universal Gas Constant, R

R =82.059 cm3 atm/mol K=8.314 J/mol K


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Pressure and Concentration of Gases

Concentration of gases:

mol vol

= P abs R .T

= n V

Daltons Law :

p A

= P

Amagats Law :

v A = V


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Density of Gas Mixtures

mix = Average mol . wt .P

RT

Density of Air (consisting of 79% N 2 and 21% O 2) at 1 atm and298.16 K :P = 1 atm = 1.01325 X 10 5 Pa = 1.01325 X 10 5 kg .m 1 .s 2

mix = 28 .84 g mol X 1.01325 X 105

Pa 8 .314 J / (mol K ) X 298 .16 K X 1 kg 1000 g

= 1 .2866kg m 3


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Vapors

Vapor:- A gas that exists below its critical temperature.Vapor Pressure:- Vaporization and condensation at constant temperature andpressure are equilibrium processes, and the equilibriumpressure is called the vapor pressure.

Boiling:- Total gas pressure Equilibrium vapor pressure of liquidSublimation:Melting :to become liquid.


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Analyzing Equilibrium for a Single Component (*)

P = 0 P=Vapor Pr.

BrownianMotion

Liquid Liquid

Time t=0 At Equilibrium

=Rate of Capture

Rate of Escape

Rate of Capture=0

Figure: Brownian Motion of a single component in Two phases


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Correlations

Semi-Empirical form of a given Pressure dependence ontemperature is used for correlating data:

lnP = A

B

T or A

B

T + C Antoine Correlation:data are listed in handbooks.Vapor liquid equilibrium (ideal)

Raoults Law:

P = p s A.x A + p s B .x B + ...

p A = p s A

.x A


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Humidity and Saturation

Partial pressure: Composition of a gas phase

Vapor pressure: Property of a liquid

Mixture of vapour and non-condensible (permanent) gas i.e.above critical conditionsSurface of Liquid A exposed to Gas

p A increases upto p s Ap A < p s A : partial saturation

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Denitions

Relative Saturation :

RS =p Ap s A

Percentage Relative Saturation:= 100 x p Ap s APercentage Saturation:= 100 x Mol vapor / MOl vap free gas

(Mol vapor / MOl vap free gas )at saturation = 100 x p A/ (p

p A)p S A / (p p

s A)


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Problem: Humidication (*)

It is desired to maintain the air entering a building at a constanttemperature of 25 0C and a percentage humidity of 40. This isdone by passing the air through a series of water sprays in

which it is cooled and saturated with water. The air leaving thespray chamber is then heated to 25 0C .(a) Assume that the water and air leave the spray chamber atthe same temperature. What is the temperature of the waterleaving?

(b) Estimate the water content of the air in the building in kg ofwater per kg of dry air.0C and percentage humidity of 65, how much water wil beevaporated or condensed in the spray chamber per kmol ofmoisture-free air?

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Solution (*)

(1) Outside Airt1=32 C

.

%H=65

(2) Saturated

t2Water Spray Chamber

(4) Water

t2

t3=25 C

%H=40

a. t2=?b. kg H2O / kg dry air= ?c. kg Water condensed or evaporated = ?

Heater

To Building

(3) Desired Air

Basis: 1 kmol dry air inlet and P = 1.01325 bar At 32 C p H 2 O = 0.04712 bar , and

At 25 C

p H 2 O = 0.03136 bar


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Solution (*)

(a). Temperature of stream 2

kmol H 2O in stream 2 is: 0.0133 kmolStream 2 is saturated p H 2 O = p H 2 O = 0 .0133 bar p H 2 O = 0 .0133 bar = 284 .4 K Ans a (b). kg H 2O /kg DA in exit

0.0133 kmol H 2O /kmol DA =

18 kg kmolH 2 O 0.0133 kmolH 2 O

28 .84 kg kmol DA 1kmol DA= 0.0083 kg H 2O /kg DA Ans b

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(c). Water Condensed or evaporated (*)

(1) Outside Airt1=32 C

.

%H=65

(2) Saturated

t2

(4) Water

t2

t3=25 C

%H=40Heater

To Building

(3) Desired Air

Water Spray Chamber0.034 kmol

Water0.0133 kmol

Water 0.0133 kmolwater

c. kg Water condensed or evaporated = ?

Material balance around spray chamber:Water going in = 0.0340 kgmol Water going out = 0.0133 kgmol

Ans c


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Problem Solving Strategy

* Read Available Information carefully and understand what is

required* Determine data needed and locate the same* Draw a simplied picture Pick a basis (starting informationrequired answers)* Decide on formulae needed and plan the strategy.* Check units consistency while calculating

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Problem: Hydrocarbon

Fuel gas exhaust composition:CO 2 6.36%, CO 1.12%, O 2 8.04%, N 2 84.48%1. Guess which is the hydrocarbon ( CH n ) in the fuel?

Find the ratio of air used to that required theoretically?

(ii) Air: c gmol

(iii) Dry Flue Gas

2(iv) H O : b gmol

(i) Fuel: a gatom C and x gatom H

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Solution

Basis: Dry fuel gas 100 mol Hydrocarbon: CH n [1].Elemental balance wrt Carbon in (gatom)

Input:(i). a gatom C

Output:(iii).

100 gmol DFG 6.36 gmol of CO 2100 gmol DFG

1gatom C

gmol CO 2+ 1.12 gmol of CO 2

100 gmol DFG

1gatom C

gmol CO

a (6.36 + 1 .12 ) = 0 ..[1]

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Solution

[2] Elemental balance wrt Hydrogen (gatom H) x + 0 = 0 + b 2[3].Elemental balance wrt Oxygen (gatom O)

0 + c . 0 .21 2 [12 .72 + 1 .12 + 16 .08 + b ] = 0 ...[3]

[4].Elemental balance wrt Nitrogen (gatom)

c 0.79 2 84 .48 2 = 0 ...[4]

Solving eqn [1] (gives a ), [4] (gives c ),[3] (gives b ),[2] (gives n )

a = 7.51 gatom C , c = 106 .94 mol air , b = 14 .99 mol H 2O

x = 2b = 29 .98 atom H , x a = 4 H drocarbon = CH 4

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(2). Percentage Excess Air:

7.51 mol CH 4 requires (2 X 7.51) mol oxygenwhich is supplied by (2 X 7.51 / 0.21 ) mol air.

=71.52 mol air.

Supplied air = 106.94 mol

excess = supplied - required = (106.94 - 71.52)=35.42 mol air%excess = excess required X100 % =35 .4271 .52 X 100 % = 49.52 %

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Material Balance for a Reactor

i rate of

reactant

ow into element of volume

=

ii rate of

reactant

ow out of element of volume

+

iii rate of reactant

loss due to

chemical reaction within the element of volume

iv

+

rate of accumulation

of reactant in element of volume

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Material Balance for a Reactor - 2

* The volume element may be a differential element of reactoror the reactor as a whole.* i & ii zero: batch reactor* iv zero: steady state ow reactor

: plug ow / mixed ow

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Material Balances with Reaction

Problem: Recycle and PurgingN 2 / H 2 mixture in feed = 1:3; Conversion of ammonia = 25 %Argon in Feed = 0.2 kmol per 100 kmol of N 2 H 2 mixtureToleration of limit of argon entering the reactor = 5 parts per100 parts of N

2 H

2mixture by volume

NH3 liquid

CondenserReactor

(1)(3)

(4)

(5)

Purge stream

y kgmoles N2+H2

(6)

(2)Fresh feed100 kgmole N2+H2

Argon0.2 kgmole

Recycle x kg moles N2+H2

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Solution

Basis: 100 kmol N 2 H 2 mixture in fresh feedLet, x = kmol of N 2 and H 2 recycled to reactor

y = kmol of N 2 and H 2 purged

N 2 + 3H 2 2NH 3

[1]. Overall balance of N 2 and H 2 :Input:(1). 100 kmol

Output:(6). y kmol(4). 0.25(100+x) kmol N 2 and H 2 reacted

100 y + 0 .25 (100 + x ) = 0 ...(1)

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Solution

[2]. Overall balance of Argon:Input:(1). 0.2 kmolOutput:

(6). y kmol N 2+ H 2 X 0.05 (100 + x )kmol argon 0.75 (100 + x )kmol N 2 + H 2 =0.0667y(* Toleration limit of argon entering the reactor=0.05 kmol perkmol of N 2and H 2)

0.2 0.667 y = 0 ...(2)

y = 3 .00 kmol

x = 288 kmol

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Summary of solution,

Fresh N 2 and H 2=100 kmol

Recycle N 2 and H 2=288 kmolPurge N 2 and H 2=3 kmolAmmonia formed = 48.5 kmolArgon = 0.2 kmolRecycle Ratio = 288100 = 2.88Purge Ratio = 3288 = 0.0104

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Problem: Absorption Column (*)

9% SO 3 is fed in the Absorption Column. All of which isabsorbed in incoming liquid stream of 95% Sulfuric acid.Water is added and mixed with 98 % H 2SO 4 recycle stream tomake 95 % H 2SO 4 stream.SO 3 / day and B, C, D, E in TPD as shown in ow diagram ?

9% SO3

98% H2SO4

H2O

95% H2SO4

B

C D

E

A kmol SO3/da

(1)

(2)(3)

(4)

(6)

Mixer

(7)

NO SO3

Absorption

Column

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l ( )


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Solution (*)

Basis: 1000 TPD 98 % H 2SO 4productA is in kmol SO 3 per dayB, C, D, E are in TPD of total stream

[1]. Material balance across total system involving stream(1), (2), (5) and (7) in TPD:

Input:(1). A kmol SO 3 /day X

80 kg SO 3kmol SO 3 X

ton 1000 kg

(7). E TPD

Output:(5). 1000 TPD(2). 0

0.08 A + E 1000 = 0 ...(1)

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S l i (*)


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Solution (*)

[2]. Material balance across column involving streams (1), (2),(3), (4) in TPD:

0 .08 A + B C = 0 ...(2)

[3]. Material balance across mixer involving stream (6), (7) and(3) in TPD:

D + E B = 0 ...(3)

[4]. Elemental material balance wrt S across the system

(units=katom S)Input:(1). AOutput:(5). 1000X10 3kg 98 % H 2SO 4 /day X 1 kmol S 98 kg H 2 SO 4

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S l i (*)


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Solution (*)

[5]. Elemental mat bal wrt H across system (units=katom H)Input: E X 1000 kg H 2O X 2 katom H 18 kg H 2 O Output:980000 kg H 2SO 4X 2 katom H 98 kg H 2 SO 4 + 20000 kg H 2O X

2 katom H 18 kg H 2 O

E = 200 TPD water ..(5)

[6]. Elemental mat. bal. wrt S across column (units=katom S)

10 C = 9 .7B + 10000 ..(6)

Liquid Stream Input to Absorption column B = 6533 TPD Liquid Stream Output from the column C = 7333 .3 TPD Recycle Stream D = 6333 TPD

E = 200 TPD

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P bl Bl t F M t i l B l


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Problem: Blast Furnace Material Balance

kg of ux per 100 kg of pig iron produced. The compositions ofthese materials are:

Ore(212.7)kg % Charcoal (110)kgFe 2O 3 54.93 C 86.89FeO 8.48 O 3.15CaO 9.58 H 0.45

Mn 3O 4 4.97 N 0.51Al 2O 3 3 H 2O 7

MgO 1.83 Ash 2SiO 2 4.92 100%H 2O 4.48CO 2 7.81

100%

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Contd


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Contd

Flux % Pig Iron % Clean gas %(13.9kg) (100kg) moisture free

SiO 2 78.38 C 3.12Al 2O 3 13.99 Si 1.52 CO 2 12.62CaO 0.53 Mn 2.22 CO 25.56

Fe 2O 3 3.9 Fe 93.14 CH 4 0.69

H 2O 3.2 100% H 2 1.34100% N 2 59.79

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Solution


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Solution

Charcoal 110.0 kg

Flux 13.9 kg

Ore 212.7 kg

Air 455.2 kg(15.70 moles)

Slag 63.8 kg

Water vapor11.9 kg

(0.66 mole)

Dry gas 617.2 kg

(20.77 moles)

Material balance of blast furnace

Pig iron 100.0 kg

(1)(2)(3)

(4)

(5)

(6)

(7)

T=25 C

T=300 C

173 C

1360 C


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Solution


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Solution

2. Weight of dry blast-furnace gas formedCarbon balance involving streams (1), (3),(7) and (5): (katomC):Carbon in 1kmol dry blast-furnace gas=0.3887 katomDry furnace gas = 20.774 kmol3. Weight of Air suppliedNitrogen balance (kmol N 2)Dry air=455.22 kg4. Weight of water vapor in the blast-furnace gasHydrogen balance (katom H)H 2O in blast-furnace gas= 11.92 kg

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Problem: Adiabatic Flame Temperature


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Problem: Adiabatic Flame Temperature

1 mol of Methane is reacted with stoichiometric amount of airand complete combustion takes place.

CH4 1 mol

298.16 K

Air 9.5238mol

(Stoichiometric)

T= ?

Adiabatic FlameTemperature

(1)

(2)

(3)

N2=7.5238 mol

CO2H2O

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Solution


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Solution

Basis: 1 mol CH 4

Stoichiometric amount of Air=2 mol O 2+7.5238 mol N 2Reference Temperature 298 K

For complete conversion:

CH 4+ 2O 2 CO 2 + 2H 2O


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Solution..


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Solution..

Now, Cp i = A + BT + CT 2 + DT 3

T

298 .16 (Cp CO 2 + 2 Cp H 2 O + 7 .5238 Cp N 2 )dT =

T

298 .16 ( m i Cp i )dT

Simpliying eq (A) for T using Newton Raphason Techniquewith initial guess of T=2607 K

T = 2350 K

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Problem: Fuel Gas composition (*)


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p ( )

A mixture of CH 4 + N 2 is burned using 10 mol Air to 1 molmixture. Complete Combustion takes place and naltemperature is 1500 K.

Air 10mol

CH4+N2 Mixture

1 mol

CO2

298.16 K1500 K

H2O

O2N2

Calculate Com osition of Fuel Gas

298.16 K

(1)

(2)

(3)

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Solution (*)


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( )

Basis: 10 mol Air and 1 mol of CH 4 + N 2 MixtureReference Temperature 298 K

Assume x mol of CH 4 in inlet

For complete conversion:

CH 4+

x

2O 2

2x

CO 2

x

+ 2H 2O

2x Stream (2): (units:mol)CH 4x molN 2(1-x) mol

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Solution.. (*)


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( )

From material balance, stream (3): (units:mol)N 2 7.9+(1-x) molO 2 2.1-2x molCO 2 x molH 2O 2x mol

Enthalpy of a stream = m i ( H 0 + T

Tref

Cp i dT + i )

i = 0, if compound is in standard state H 0 = 0 for elements

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Solution.. (*)


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( )

Enthaply of stream (1):E 1 = 10 mol air (0.21 mol O 2mol air (0 + 0 + 0)+ 0.79

mol N 2mol air (0 + 0 + 0))

Stream (2):E 2 = x mol CH 4 ( H 0fCH 4 + 0 + 0) + ( 1 x )mol N 2 (0 + 0 + 0)Stream(3):E 3 = x mol CO 2 ( H 0fCO 2 +

1500298 .16 Cp CO 2 dT + 0)

+ 2x mol H 2O ( H 0fH 2 O +

1500

298 .16 Cp H 2 O dT + Vap H 2 O )

+( 2 .1 2x )mol O 2 (0 + 1500298 .16 Cp O 2 dT + 0)+( 8.5 x )mol N 2 (0 + 1500

298 .16 Cp N 2 dT + 0)

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Solution.. (*)


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Now, taking material balance of all the streamsE 1 + E 2 = E 3Rearranging,x ( H 0fCH 4 H 0fCO 2 2 H 0fH 2 O 2 Vap H 2 O )

= 1500

298 .16 [x Cp CO 2 + 2x Cp H 2 O + ( 2 .1 2x )Cp O 2+( 8.5 x )Cp N 2 ]dT ..(A)

First Estimate: Take Cp i = 7cal / mol for all compounds, we

get,

x 0 .54

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Solution.. (*)


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Now,

Cp i = A + BT + CT 2 + DT 3Cp CO 2 = ACO 2 + B CO 2 T + C CO 2 T

2 + D CO 2 T 3

and Similarly for others..

1500

298 .16 (x Cp CO 2 + 2x Cp H 2 O +( 2.1 2x )Cp O 2 +( 8 .5 x )Cp N 2 )dT =

1500

298 .16( m

i Cp

i )dT

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Solution.. (*)


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Using the rst estimate of x=0.544 in this integration we get,

1500

298 .16( m

i Cp

i )dT = 437199 J

Putting this value in equation (A) and solving for x we get,

x = 0.577

So, Composition of fuel gas:CH 4=0.577 mol N 2= 0.423 mol

ICT - SSB

Problem: Sulfur dioxide (*)


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Molten sulfur at 115 0C is fed in the adiabatic reactor andreacted with air coming in at 25 0C , complete conversion takesplace and SO 2 is produced.Output composition is:

8.5 % SO 2 , 12.5 % O 2 , 79 % N 2Find the amount of air fed into the reactor?

O 2 12.5 molN 79 mol

Air

Solid

Sulfur

T=25 C

8.5 mol

SO2

8.5 mol(1)

(2)2

(3)

T=?T=115 C

Molten

Sulfur

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Solution (*)


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Basis: 100 mol of output, Ref. T= 25 0C

S + O 2 SO 2

Sulfur fed in the reactor = 8.5 molOxygen fed in reactor = 8.5 mol + 12.5 mol = 21 molAir fed in reactor = 21 mol + 79 mol = 100 mol

Enthalpy of a stream = m i ( H 0 +

T

Tref Cp i dT + i )

i = 0, if compound is in standard state H 0 = 0 for elements


82/98

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Solution.. (*)


83/98

Now, taking material balance of all the streamsE 1 + E 2 = E 3Rearranging,8.5(

388 K

298 .16 Cp S dT + Fusion S H 0fSO 2

vap S )

= T 298 .16 (8.5Cp SO 2 + 12 .5Cp O 2 + 79 Cp N 2 )dT ..(A)First Estimate: Take Cp i = 7cal / mol for all compounds, weget,2.56 10 6J = 100 7 4.18 [T 298 .16 ]

T = 1173 K

Now, Cp i = A + BT + CT 2 + DT 3

T

298 .16 (8 .5Cp SO 2 + 12 .5 Cp O 2 + 79 Cp N 2 )dT =T

298 16( m i Cp i )dT

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Problem: Catalytic Converter


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Feed stream at 400 0C contains: (% by volume)8.5 % SO 2 , 12.5 % O 2 , 79 % N 2Products are coming out of the reactor at 550 0C

SO2 8.5%O2 12.5%N2 79%

% by volumeSO2

SO3

400 C 550 C

Catalytic

Converter

O2N2

Find the Conversion ?


85/98

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Solution..


86/98

E 1 =8.5 ( H 0fSO 2 +

673

298 Cp SO 2 dT ) + 0.125 ( H 0fO 2 +

673

298 Cp O 2 dT

+ 0 .79 ( H 0fN 2 + 673298 Cp N 2 dT E 1 = 8 .5 ( H 0SO 2 +

673298 (8 .5Cp SO 2 + 12 .5Cp O 2 + 79 Cp N 2 )dT

Let, z=mole of SO 2converted to SO 3E 2 = z H 0fSO 3 + ( 8 .5 z ) H

0fSO 2

+ 823298 zCp SO 3 + ( 8 .5 z )Cp SO 2 + ( 12 .5 z / 2)Cp O 2 + 79 Cp N 2 dT

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Solution ..


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Now, E1-E2=0 z H 0fSO 3 + z H

0fSO 2

+ 673

823 (8.5Cp SO 2 + 12 .5Cp O 2 + 79 Cp N 2 )dT -(A)+ z

823298 ( Cp SO 3 + Cp SO 2 + 0.5Cp O 2 )dT -(B)

= 0Evaluating Integration (A)So,

673823 (8 .5Cp SO 2 + 12 .5Cp O 2 + 79 Cp N 2 )dT = 1.1848 X 10 5 Cal -(A)

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Solution..


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Evaluating Integration (B)z

823298 ( Cp SO 3 + Cp SO 2 + 0.5Cp O 2 )dT = z 521 .34 Cal -(B)

Putting the values of (A) and (B) in main equation:

z ( H 0fSO 3 + H 0fSO 2 ) + 1 .1848 10 5 z 521 .34 = 0Putting, H 0fSO 3 = 94450 Cal / gmol

and H 0fSO 2 = 70960 Cal / gmol

z = 5.15 gmol SO 3 reacted

Conversion = 100 5 .15 / 8.5 = 60 .58

ICT - SSB

Problem: Blast Furnace (2)Energy Balance (*)


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Air 0 kcal(3)Slag 23272 kcal

(2) Water vapor 7752 kcal

(1) Dry gas 22560 kcal

(4) Pig iron 30000 kcal

T=300 C

173 C

(13) Heat Losses = 94922 kcal

T=25 C

Flux 0 kcal

Charcoal 0 + 773850 kcal

Ore 0 kcal

1360 C

(2)+(5)

(3)

(1)

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Solution: (*)


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H In = H Out + H Rea + Heat losses Input kcal

1 Enthalpy of ore 02 Enthaply of charcoal 03 Enthalpy of ux 04 Enthalpy of hot blast 305325 Heating value of charcoal 7738506 Heat evolved in forming slag 6420

Total 810802

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Solution.. (*)


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Output kcal1 Enthalpy of dry blast furnace gas 225602 Enthalpy of water vapor in the gas 77523 Enthalpy of slag 232724 Enthalpy of pig iron 300005 Heat absorbed in decomposing iron oxides 1571386 Heat absorbed in decoposing Mn 3O 4 62777 Heat absorbed in decomposing SiO 2 111338 Heat absorbed in calcining carbonates 15859

9 Heating value of blast furnace gas 40864710 Heating value of C in pig iron 2445411 Heat absorbed forming Fe 3C 130012 Heat absorbed by cooling water 748813 Heat losses (by difference) 94922

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Dynamics


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Energy balancerate of heat

ow into element of

volume

=

rate of heat ow out of

element of volume

+

rate of disappearance

of heat by

reaction within element of volume

+

rate of accumulation of heat within

element of volume

Complex Material Balances

ICT - SSB

ENERGY BALANCES


93/98

* Thermophysics - specic heat variation* Thermochemistry

Heat of Reaction at Different TemperatureHeat of Solution, Mixing* Adiabatic Temperature Calculations* Fuels* Solid-Coal, Liquid Petroleum Based, Gases-Natural, Coke

Oven, Water Gas etc.* Adiabatic Flame Temperatures

ICT - SSB

Stagewise Operation


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Distillation, Extraction, LeachingUnsteady State Differential BalancesChemical Processes

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Problem: Unsteady state (*)


95/98

Toxic material having 30 ppm concentration enters 10 m 3 surgetank at the ow rate of 1 m 3 / hr Suddenly the input concentration increases to 100 ppm.The upper limit of output stream is 70 ppm.

Find time for C to reach 70 ppm ?

F0 = 1 m / hr3F0 = 1 m / hr

3

C=100 ppm

Surge Tank

V=10 m3

t=0, C = 30 ppm

C max=70 ppmStep Input

t=0, C = 30 ppm

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Solution.. (*)


96/98

Basis: 1 m 3 / hr of input ow rateLet, C be the concentration of toxic materialTaking, material balance of toxic material (assuming Outlet owrate =Inlet ow rate):

d (C V )dt

= C 0 F 0 C F 0

Integrating above equation, we get,

C = 70

C = 30dC C 0 C = 1

t = t

t = 0

= V F 0 = 10 hr At time t = 0, C = 30 ppm and time t = t , C = 70 ppm,C 0 = 100 ppm; Solving for t. t = 8 .472hr

ICT - SSB


97/98

ICT, Mumbai THANK YOU

ICT - SSB

Course Suggestions


98/98

Duration : 60 hrs spread over one semester as two sessions oftwo hours per week.Prerequisites: Mathematics, Physics and Chemistry upto 12th.Useful: Some exposure to chemical process equipments in use.

Textbooks:1 Basic calculations in chemical engineering. Himmelblau2 Chemical Process Principles Vol I. Houghen, Watson,

Ragatz

3 Material and Energy Balances. Reklaitis4 Stoichiometry. Bhat, Vora5 Elementary principles of chemical processes. Felder,

Rosseau

material balance 2011

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