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Material balance reaction - Compound forming

Reference

Himmelblau DM. 1989. Basic Principles and Calculations in Chemical Engineering, 5th edition. Prentice-Hall International, Inc., Singapore.

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Stoichiometry

• Refers to quantities of reactants and products in a balanced chemical reaction.

aA + bB cC + dD i.e. a moles of A react with b moles of B to give c

moles of C and d moles of D. a,b,c,d are stoichiometric coefficient the stoichiometric factor = stoichiometric moles

reactant required per mole product

Limiting reactant/excess reactant

• In practice a reactant may be used in excess of the stoichiometric quantity for various reasons.

• In this case the other reactant is limiting (i.e., it will limit the yield of product(s))

• Limiting/excess reactant usually applied for reversible reaction

aA + bB cC + dD

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Paul Ashall, 2008

Limiting reactant/excess reactant

A reactant is in excess if it is present in a quantity greater than its stoichiometric proportion.

% excess =

(moles supplied – moles required)moles required x 100

Conversion • Fractional conversion = amount reactant

consumed/amount reactant supplied

% conversion = amount reactant consumedamount reactant supplied x 100

Note: conversion may apply to single pass reactor conversion or overall process conversion

Yield = moles productmoles limiting reactant suppliedx 100

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Examples

A B i.e. stoichiometric coefficients a = -1; b = 1 100 kmol fresh feed A; 90 % single pass

conversion in reactor; unreacted A is separated and recycled and therefore overall process conversion is 100%

reactor separation Feed

Recycle

Produk

Overall process

Reaction Systems

Process input output

input through system

boundaries

accumulation within

the system

output through system

boundaries

generation within

the system

consumption within

the system

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Reaction Systems

• Batch (closed) system mass in = mass out + accumulation • Continuous (flow/open) system

mass inper time = mass out

per time + accumulationper time

Mass Balance at Steady State: Total mass in = Total mass out

Process input output

Combustion Process of Flow System

• Reaction: Fuel + Oxygen Combustion gases CH4 + 2 O2 CO2 + 2 H2O C2H6 + O2 CO2 + H2O C8H18 + O2 CO2 + H2O

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Examples of Material Balance with Chemical Reaction

• If 300 kg of air and 24.0 kg of carbon are fed to a reactor at 600 °F and after complete combustion no material remains in the reactor, how many pounds of carbon will have been removed? How many pounds of oxygen? How many pounds total?

• How many moles of carbon and oxygen enter? How many leave the reactor?

• How many total moles enter the reactor and how many leave the reactor?

Examples of Material Balance with Chemical Reaction

• Basis: 300 kg air • Calculate the mol of carbon, oxygen and nitrogen

enter (input):

C: . . ⁄ = 2.00molC

O2: . ⁄ ×

= 2.17molO

N2: 2.17molO ×

= 8.17molN

Reactor 24.0 kg C

? 300 kg air

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Examples of Material Balance with Chemical Reaction

Reaction: C + O2 CO2

Stoichiometry: 2.00 mol C requires 2.00 mol O2 and produces 2.00 mol CO2

• O2 out = O2 in – O2 consumption O2 out = 2.17 – 2.00 = 0.17 mol

• CO2 out = CO2 in + CO2 generation CO2 out = 0 + 2.00 = 2.00 mol

Examples of Material Balance with Chemical Reaction (Pure Compounds)

Input Output mass (kg) mol mass (kg) mol

C 24.0 2.00 0 0 O2 69.5 2.17 5.5 0.17 N2 230.5 8.17 230.5 8.17 CO2 0 0 88.0 2.00 Total 324.0 12.34 324.0 10.34

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Type of balance

Equality required for input and output of steady-state process

Without chemical reaction

With chemical reaction

Total balances Total mass Total moles

Component balances Mass of a pure compound Moles of a pure compound Mass of an atomic species Moles of an atomic species

Element Balance

Mole balance C CO2 O2 C CO2 O2 C 2 + 0 + 0 = 0 + 2 + 0 O 0 + 0 + 2.17

(2) = 0 + 2(2) + 0.17

(2) or O2 0 + 0 + 2.17 = 0 + 2 + 0.17 Mass balance C 24 + 0 + 0 = 0 + 24 + O 0 + 0 + 69.5 = 0 + 64 + 5.5 or O2 0 + 0 + 69.5 = 0 + 64 + 5.5

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• Air is a mixture of gases and the total mass can be estimated by adding the weight of all major components as shown below:

http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html

Combustion and Dry Ice Production

• In the combustion of heptane, CO2 is produced. Assume hat you want to produce 500 kg of dry ice per hour and that 50% of the CO2 can be converted into dry ice. How many kilograms of heptane must be burned per hour?

• Reaction: C7H16 + O2 CO2+ H2O • MW C7H16 = 100 kg/kg mol C7H16

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Engine Gas C7H16

CO2 gas (50%)

• Basis: 500 kg/hr of dry ice • CO2 solid:

500 푘푔푑푟푦푖푐푒 ℎ⁄0.5 푘푔푑푟푦푖푐푒 1푘푔퐶푂 × 44 푘푔퐶푂 푘푔푚표푙퐶푂⁄⁄

= 22.72푘푔푚표푙퐶푂 • Reaction: C7H16 + 11 O2 7 CO2+ 8 H2O CO2 solid (dry ice) 3.25 ~ 22.72 kg C7H16 = 3.25 kg mol C7H16 100 푘푔 푘푔푚표푙퐶 퐻⁄ kg C7H16 = 325 kg C7H16

Other products

CO2 solid (50%), 500 kg/hr

Corrosion of Pipes in Boilers

• Corrosion of pipes in boilers by oxygen can be alleviated through the use of sodium sulfite. Sodium sulfite removes oxygen from boiler feedwater by the following reaction:

2 Na2SO3 + O2 2 Na2SO4

• How many pounds of sodium sulfite are theoretically required (for complete reaction) to remove the oxygen from 8.33 x 106 lb of water containing 10.0 ppm of dissolved oxygen and at the same time maintain a 35% excess of sodium sulfite?

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• Basis: 8.33x106 lb of water with 10 ppm O2

• MW Na2SO3 = 126

• O2: 8.33x106 lb H2O×

= 83.3 lb O2 (2.6 lb mol O2)

• Reaction: 2 Na2SO3 + O2 2 Na2SO4 5.2 ~ 2.6

• Na2SO3 : 5.2 lb mol Na2SO3×

×1.35

= 884.5 lb Na2SO3

H2O: 8.33x106 lb 10 ppm O2

Na2SO3

H2O: 8.33x106 lb no O2

Combustion Product

• Flue or stack gas (wet basis) all the gases resulting from a

combustion process, including the water vapor

• Orsat analysis (dry basis) all the gases resulting from the

combustion process, excluding the water vapor

CO2 CO O2 N2

SO2 H2O

• Flue or stack gas (wet basis)

• Orsat analysis (dry basis)

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Combustion • Theoretical/required air (or theoretical oxygen) the amount of air (or O2) required to be brought into the process

for complete combustion • Excess air (or excess oxygen) the amount of air (or O2) in excess of that required for complete

combustion

%푒푥푐푒푠푠푎푖푟 =푒푥푐푒푠푠푎푖푟푟푒푞푢푖푟푒푑푎푖푟 × 100%

%푒푥푐푒푠푠푎푖푟 =푒푥푐푒푠푠푂 0.21⁄ 푟푒푞푢푖푟푒푑푂 0.21⁄ × 100%

%푒푥푐푒푠푠푎푖푟 =푂 푒푛푡푒푟푖푛푔푝푟표푐푒푠푠 − 푂 푟푒푞푢푖푟푒푑

푂 푟푒푞푢푖푟푒푑 × 100%

Example of Excess Air

• Compressed propane has been suggested as a source of economic power for vehicles. Suppose that in a test 20 lb of C3H8 is burned with 400 lb of air to produce 44 lb of CO2 and 12 lb of CO. What was the percent excess air?

• Reaction: C3H8 + O2 CO2 + H2O

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Example of Excess Air

• Basis: 20 lb of C3H8 • Reaction: C3H8 + 5 O2 3 CO2 + 4 H2O mass 20 MW 44 32 44 18 mol 0.45 ~ 2.27

Example of Excess Air

• The entering O2:

O2: . ⁄ ×

= 2.90molO

• The excess O2: 2.90 – 2.27 = 0.63 mol 02

• % excess air: . ..

× 100% = 27.8%

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Material Balance with Combustion

• A salesperson comes to the door selling a service designed to check “chimney rot”. He explains that if the CO2 content of the gases leaving the chimney rises above 15%, it is dangerous to your health, is against the city code, and causes your chimney to rot. On checking the flue gas from the furnace, he finds it is 30% CO2. Suppose that you are burning natural gas which is about 100% CH4 and that the air supply is adjusted to provide 130% excess air. Do you need his service?

Material Balance with Combustion

• Reaction: CH4 + 2 O2 CO2 + 2 H2O • Basis: 1.00 mole of CH4

• Stoichiometry: 1.00 mol CH4 requires 2.00 mol O2 and produces 1.00 mol CO2 and 2.00 mol H2O

Chimney Feed: CH4 100%

Air: O2 21% Air: N2 79%

Product: CO2 ?% Product: H2O ?%

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Material Balance with Combustion

• Required air: × 2푚표푙푂 = 9.52푚표푙 composed of 2.00 mol O2 and 7.52 mol N2

• Excess air: 9.52푚표푙 = 12.4푚표푙 composed of 2.60 mol O2 and 9.80 mol N2

• Total air entering process: 12.4 + 9.52 = 21.92 mol composed of 4.60 mol O2 and 17.32 mol N2

Chimney Feed: CH4 100%

Air: O2 21% Air: N2 79%

Product: CO2 ?% Product: H2O ?%

Output mol %

CH4 0 0 O2 2.60 11.4 N2 17.3 75.5 CO2 1.00 4.4 H2O 2.00 8.7 Total 22.9 10.34

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Input Output mol mass % mol mass %

CH4 1 16 4.3 0 0 O2 4.6 147.2 20.1 2.60 83.2 11.4 N2 17.3 488.4 75.6 17.3 488.4 75.5 CO2 0 1.00 44 4.4 H2O 0 2.00 36 8.7 Total 22.9 651.6 100 22.9 651.6 100

Combustion of Ethane

• Ethane is initially mixed with oxygen to obtain a gas containing 80% C2H6 and 20% O2 that is then burned in an engine with 200% excess air. Eighty percent of the ethane goes to CO2, 10% goes to CO, and 10% remains unburned. Calculate the composition of the exhaust gas on a wet basis and dry basis.

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Engine

• O2 21% • N2 79%

• CO2 Proct: • CO • C2H6 • O2 • N2 • H2O

Fuel Gas Exhaust Gas

Air, 200% excess

• C2H6 80 lb mole • O2 20 lb mole

• Basis: 100 lb mole of fuel • Reaction:

C2H6 + O2 2 CO2 + 3 H2O 80 %

C2H6 + O2 2 CO + 3 H2O 10 %

• Total O2 entering: 3.00 times required O2 (100 % required + 200 % excess) O2 for complete combustion:

= 80 lbmol C2H6 .

= 280 lb mol O2

C2H6 + O2 2 CO2 + 3 H2O

lb mol 80 ~ 280 Required O2: 280 – 20 = 260 lb mol O2

O2 entering with air: 3 (260 lb mol O2) = 780 lb mol O2 N2 entering with air: 780 lb mol O2

= 2934 lb mol N2

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224

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• Stoichiometry:

C2H6 + O2 2 CO2 + 3 H2O 80 %

80 128 192

C2H6 + O2 2 CO + 3 H2O 10 %

80 16 24 • Unburned: 80 lb mole C2H6 (0.1) = 8 lb mole C2H6 • O2 available in system: 780 + 20 = 800 lb mol O2

• O2 “used up” by reaction: 224 + 20 = 244 lb mol O2

• O2 out: 556 lb mol O2

• H2O out: 192 + 24 = 216 lb mol H2O

Exhaust Gas in Wet Basis

lb mol Percent in exhaust gas Component Fuel Air Exhaust gas

C2H6 80 - 8 0.21 O2 20 780 556 14.41 N2 - 2934 2934 76.05 CO2 - - 128 3.32 CO - - 16 0.41 H2O - - 216 5.60 Total 100 3714 3858 100.00

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Exhaust Gas in Dry Basis

lb mol Percent in exhaust gas Component Fuel Air Exhaust gas

C2H6 80 - 8 0.22 O2 20 780 556 15.27 N2 - 2934 2934 80.56 CO2 - - 128 3.51 CO - - 16 0.44 Total 100 3714 3642 100.00

Multiple Units in Combustion

• In the face of higher fuel costs and the uncertainty of the supply of a particular fuel, many companies operate two furnaces, one with natural gas and the other with fuel oil. Each furnace had its own supply of oxygen; the gas furnace used air and the fuel oil furnace used a gas stream that analyzed O2 20%; N2 76%, and CO2 4%, but the stack gases went up a common stack.

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Multiple Units in Combustion

• During blizzard, all transportation was cut off and the reserve of fuel oil was only 1000 bbl. How many hours could the company operate before shutting down if no additional fuel oil? How many lb mol/hr of natural gas were being consumed? The minimum heating load when translated into the stack gas output was 6205 lb mol/hr of dry stack gas. Analysis the fuels and stack gas at this time.

Gas Furnace

Oil Furnace

Natural Gas (G) CH4: 96% C2H2: 2% CO2: 2%

Fuel Oil (F) C: 50% H2: 47% S: 3%

Air/Gas stream (A*) O2: 20% N2: 76% CO2: 4%

Air (A) O2: 21% N2: 79%

Water Vapor (W) H2O: 100%

Stack Gas (P): 6205 lb mol/hr O2: 4.13% N2: 84.93% SO2: 0.10% CO2: 10.84%

(Orsat Analysis)

Problem: F & G in lb mol/hr? F in bbl/hr?

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• Basis: P = 6205 lb mol/hr • MW of fuel = 7.91 lb/lb mol • Density of fuel = 7.578 lb/gal • Solving by using elemental mole balances

In Out H2 G((2)(0.96) + 0.02) + F(0.47) = W(1) N2 A(0.79) + A*(0.76) = 6205(0.8493) O2 G(0.02) + A(0.21) + A*(0.24) = 6205(0.0413+0.001+0.1084)

+ W(0.5) S F(0.03) = 6205(0.001) C G(0.96 + (2)(0.02) + 0.02) +

F(0.5) + A*(0.04) = 6205(0.1084)

• F = 207 lb mol/hr • G = 498 lb mol/hr • The fuel oil consumption:

= × .

. ×= 5.14 bbl hr⁄

If the fuel oil reserve was only 1000 bbl, it could last:

=

.= 195hr