Notes 5 Reactive System 3
Descrição do Produto
SKF 1113 - Material Balances
Balances on Reactive Process (Part C – Combustion Reaction) Azeman Mustafa, PhD Kamarul ‘Asri Ibrahim, PhD
© copyright PCS-FKKKSA
Material Balance
Page 4-1
Combustion Reaction Rapid oxidation of a fuel accompanied by the release of heat and/or light together with the formation of combustion products Fuel + oxidant
heat/light + combustion products
Definition of combustion as quoted from Webster’s dictionary “rapid oxidation generating heat, or both heat and light ; also, slow oxidation accompanied by relatively little heat and no light”
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-2
SKF 1113 - Material Balances
Combustion of Fossil Fuel
Chemical reaction between hydrogen and carbon atoms (contained in the fuel) with oxygen atoms (usually comes from the air), resulting in the heat release and the formation of combustion products(*)
(*) mainly water vapor and carbon dioxide and a certain amount combustion by-products depending on combustion process
© copyright PCS-FKKKSA
Material Balance
Page 4-3
Simplified Main Processes of Combustion
Carbon + Oxygen ( C + O2
heat + carbon dioxide
Heat + CO2 )
Hydrogen + Oxygen
Heat + Water
( H 2 + ½ O2
Heat + H2O )
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-4
SKF 1113 - Material Balances
Combustion By-Products Carbon monoxide (CO) Aldehydes (e.g. H) Unburned Fuel Radicals
mainly due to incomplete combustion
Oxides of nitrogen (NOx)
– reaction between O2 (in air) and nitrogen (present in air or fuel)
Oxides of sulphur (SOx)
– only for Sulphurcontaining fuel
© copyright PCS-FKKKSA
Material Balance
Page 4-5
Categories of Combustion Process
Complete combustion Stoichiometric Excess air or fuel lean Incomplete or partial combustion Excess fuel or fuel rich or deficient air In practice, combustion will never be complete even though at stoichiometric or excess air conditions…….due to nonuniformity of fuel and air mixture and complexity of combustion reaction
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-6
SKF 1113 - Material Balances
Stoichiometric Combustion Relative (chemically-correct) proportion of fuel and air quantities that are the theoretical minimum needed to give complete/perfect combustion (i.e., no unburned fuel and residual oxygen present in combustion products)
CH4 + 202 Æ C02 + 2H20 means that 1 mole of methane to be proportionately (and molecularly) mixed with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water vapor
or
1 cubic metre (m3) of methane requires 2 cubic metre (m3) of oxygen for complete combustion and will produce 1 cubic metre (m3) of carbon dioxide and 2 cubic metre (m3) of water vapor © copyright PCS-FKKKSA
Material Balance
Page 4-7
Excess Air or Oxygen Combustion When oxygen or air is supplied more than the stoichiometric proportion
CH4 + 302 Æ C02 + 2H20 + 02 means that 1 mole of methane to be molecularly mixed with 3 moles of oxygen to produce 1 mole of carbon dioxide, 2 moles of water vapor and 1 mole of un-reacted oxygen
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-8
SKF 1113 - Material Balances
Incomplete or Partial Combustion When fuel is supplied more than the stoichiometric proportion
Insufficient amount of oxygen or air available to burn in the fuelrich mixture caused incomplete combustion
CH4 + 02 Æ C0 + 2H20 + (other products of combustion)
incomplete
means that 1 mole of methane to be molecularly mixed with 1 mole of oxygen to produce 1 mole of carbon monoxide, 2 moles of water vapor and other products of incomplete combustion such as unburned fuel, aldehydes etc © copyright PCS-FKKKSA
Page 4-9
Material Balance
Source of Oxygen
Main of source of oxygen comes from atmospheric air Atmospheric air requirement for combustion reaction is assumed to have the following composition
Air %
By volume
By weight (mass)
O2 N2
21 79
23 77
Analyses of solid / liquid fuels are normally reported on a mass basis, while gaseous fuels are normally analysed on a volume basis
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-10
SKF 1113 - Material Balances
Composition analysis Fuel composition analysis – conversion from a composition by mass to a molar composition or vice-versa (refer to page 51 in the textbook) Stack or flue gases composition analysis Fuel Air
Combustor / Reactor
CO2 , H2O , O2 , N2 , CO H2 , CxHy, SO2 etc
Wet basis composition :- component mole fractions of gas with the presence of water Dry basis composition :- component mole fractions of the same gas without the presence of water
© copyright PCS-FKKKSA
Material Balance
Page 4-11
Fuel composition analysis The mole composition of a sample of gases fuel is the following : H2 = 50%, CH4 = 20%, C2H4 = 2%, CO2=5%, CO = 16% and N2 = 7% Determine the average molecular weight and also the composition by weight (mass)
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-12
SKF 1113 - Material Balances
Flue gas composition analysis A flue gas contains 5 mole % H2O. Calculate a) b) c)
kmol flue gas/ kmol H2O kmol dry flue gas / kmol flue gas kmol H2O/kmol dry flue gas
Basis : 100 kmol flue gas (contains 5 kmol H2O and 95 kmol dry flue gas) a)
kmol flue gas kmol H 2O
b)
kmol dry flue gas 95 kmol dry flue gas kmol dry flue gas = = 0.95 kmol flue gas 100 flue gas kmol flue gas
c)
kmol H 2O 5 kmol H 2O = kmol dry flue gas 95 kmol dry flue gas
=
100 kmol flue gas 5 kmol H 2O
© copyright PCS-FKKKSA
= 20
kmol flue gas kmol H 2O
= 0.0526
kmol H 2O kmol dry flue gas
Material Balance
Page 4-13
Flue gas composition analysis A sample of gases fuel contains 2.29 mole %, O2 , 9.53 mole %, CO2, 1.06 mole %, 73.01 mole %N2 and the balance H2O. Determine the molar composition and molecular weight of the gas on a dry basis.
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-14
SKF 1113 - Material Balances
Theoretical and Excess Oxygen (Air) Theoretical Oxygen (air) The amount of chemically-corrected (stoichiometric) amount of oxygen (air) required for complete combustion of a given quantity of a specific fuel) Excess Oxygen (Air) The actual amount of oxygen (air) required for combustion of a specific fuel
The theoretical oxygen (air) required to burn a given quantity of fuel does not depend on how much fuel is actually burned. The fuel may not react completely and it may react to form both CO and CO2, but the theoretical air is still that which would be required to react with all of the fuel to form CO2 only.
© copyright PCS-FKKKSA
Page 4-15
Material Balance
Percent theoretical Oxygen (Air) (moles oxygen) required − (moles oxygen) theroretical
(moles
oxygen )theoretical
(moles air) required − (moles air) theroretical
(moles
air )theoretical
× 100%
× 100%
The value of the percent excess air depends on the theoretical air and the air feed rate not much much oxygen is consumed in the reactor or whether combustion is complete or partial © copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-16
SKF 1113 - Material Balances
Working Session X - Theoretical and Excess Oxygen (Air) 100 mol/h methane is fed to a reactor and burns in the reaction CH4 + 2O2 Æ CO2 + 2H2O CH4 + 1.5O2 Æ CO + 2H2O a) What is the theoretical O2 flow rate if the complete combustion occurs in the reactor? b) What is the theoretical O2 flow rate if only 70% of the methane reacts? c) What is the theoretical air flow rates? d) If 100% excess air is supplied, what is the flow rate of air entering the reactor? e) If the actual flow rates of air is such that 300 mol/h O2 enters the reactor , what is the percent excess air?
© copyright PCS-FKKKSA
Material Balance
Page 4-17
Material Balances Involving Combustion Reaction
The procedures for writing and solving material balances for a combustion reactor is the same as that for any other reactive system. What information should be in a process flow chart?
Inlet stream components Fuel (single or mixture) Oxygen or Air (21 mole % oxygen, 79 mole% nitrogen) Excess oxygen or air (percent excess oxygen or air that is required or already given) Outlet stream components CO2, H2O and CO (if combustion is partially complete) Un-reacted fuel (if fuel not fully consumed) Un-reacted O2 ( depends on combustion reaction or excess air) N2 ( if oxygen comes from air)
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-18
SKF 1113 - Material Balances
Material Balances Involving Combustion Reaction
Percentage of excess oxygen and excess air would have the same value.
Calculate the theoretical amount of oxygen or air required for a given amount fuel to produce complete combustion (write stoichiometric balance of fuel-oxygen combustion to form CO2 and H2O) Multiply the theoretical amount of oxygen or air with the excess air fraction (i.e 1 + fractional excess oxygen or air)
Which mass balance method is preferred?
If only one reaction is involved, all three balance methods (molecular species, atomic species or extent of reaction) are equally convenient. If only more than one reaction is involved simultaneously, atomic species balances is usually more convenient.
© copyright PCS-FKKKSA
Material Balance
Page 4-19
Combustion reaction of a mixture of hydrocarbon fuels A mixture of hydrocarbon gases containing , on a volume basis, 95 mole % methane, 2 mole % propane and 3 mole % nitrogen is completely burned with 30 mole % excess air. Calculate the molar composition of combustion products on a dry and wet basis 95 mole % CH4 2 mole % C3H8 3 mole % N2 no mol Air 21 mole % O2 79 mole % N2 30% excess
n1 mol CO2 n2 mol H2O n3 mol O2 n4 mol N2
CH4 + 2O2 Æ CO2 + 2H2O C3H8 + 5O2 Æ 3CO2 + 4H2O
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-20
SKF 1113 - Material Balances
Combustion reaction of a mixture of hydrocarbon fuels
Atomic species balances General balance at steady state Input = output Degree of freedom analysis No. unknown labeled variables (n1 , n2 ,n3 , n4 , no ) - 3 No. independent atomic species balances (C, H, O) - 1 No. molecular species balances on independent non-reactive species (N2) - 1 No. other equations relating unknown variables (% excess air) = 0
Degree of freedom
© copyright PCS-FKKKSA
Material Balance
Page 4-21
Combustion reaction of a mixture of hydrocarbon fuels Basis : 100 mol of fuel mixture Strategy : Solve the unknowns from the given information. no mol Air (30% Excess) mol O 2 (n o )theo = (95 mol CH 4 ) 12mol CH
5 mol O 2 + (2 mol C 3 H 8 ) 1 mol C H 4 3 8 1 mol Air = 1238 mol Air n o = (1.3 )( 200 mol O 2 ) 0.21 mol O 2 2
= 200 mol O 2
Nitrogen balance (Input = output) 0.79 mol N 2 n 4 = ( 1238 mol Air ) + 3 mol N 2 = 981 mol N 2 1 mol Air © copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-22
SKF 1113 - Material Balances
Combustion reaction of a mixture of hydrocarbon fuels Atomic Carbon balance (Output = Input) 1 mol C 1 mol C = (95 mol CH 4 ) 1 mol CO 2 1 mol CH 4
(n 1 mol CO 2 ) ∴
3 mol C + (2 mol C 3 H 8 ) 1 mol C H 3 8
n 1 = 101 mol CO 2
Atomic Hydrogen balance (Output = Input)
2 mol H
(n 1 mol H 2O )
4 mol H = (95 mol CH 4 ) 1 mol CH 4
( 1 mol H 2 O )
8 mol H + (2 mol C 3 H 8 ) 1 mol C H 3 8
∴ n 1 = 198 mol H 2 O
© copyright PCS-FKKKSA
Material Balance
Page 4-23
Combustion reaction of a mixture of hydrocarbon fuels Atomic Oxygen balance (Output = Input) 2 mol O + (101 mol CO 2 ) (n 3 mol O 2 ) 1 mol O 2
2 mol O + (198 mol H 2 O ) 1 mol CO 2
1 mol O 1 mol H 2 O
0.21 mol O 2 2 mol O = 1238 mol Air 1 mol Air 1 mol O 2 ∴ n 3 = 60 mol O 2
Total moles (wet)
= nw
Total moles (dry) = nd (= nw – n2)
The molar composition of combustion products ( dry basis) , yi = ni/ nd The molar composition of combustion products ( wet basis) , yi = ni/ nw © copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-24
SKF 1113 - Material Balances
Working Session XI – Partial combustion Combustion of methane can be presented by the following reactions; CH4 + 2O2 Æ CO2 + 2H2O CH4 + 1.5O2 Æ CO + 2H2O 100 mol/h of methane is fed to a reactor and completely burns with 50% excess air. The analysis of the flue gases indicates that the mole ratio of CO2 to CO is 3. Calculate the molar composition of the flue gases on a wet basis.
© copyright PCS-FKKKSA
Page 4-25
Material Balance
Solution to Working Session XI 100 mol/h CH4
n1 mol/h CO2 n2 mol/h CO n3 mol/h H2O n4 mol/h O2 n5 mol/h N2
no mol/h Air 21 mole % O2 79 mole % N2 50 % excess
Given (flue gases): n1 /n2 = 3 mol CO2 / mol CO
CH4 + 2O2 Æ CO2 + 2H2O CH4 + 1.5O2 Æ CO + 2H2O
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-26
SKF 1113 - Material Balances
Solution to Working Session XI
Atomic species balances General balance at steady state Input = output Degree of freedom analysis No. unknown labeled variables (n1 , n2 ,n3 , n4 , no ) - 3 No. independent atomic species balances (C, H, O) - 1 No. molecular species balances on independent non-reactive species (N2) - 1 No. other equations relating unknown variables (% excess air) = 0
Degree of freedom
© copyright PCS-FKKKSA
Material Balance
Page 4-27
Solution to Working Session XI Basis : 100 mol/h of methane Strategy : Solve the unknowns from the given information. Theoretical Air
Percent excess air (50%)
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-28
SKF 1113 - Material Balances
Solution to Working Session XI Nitrogen balance (Input = output)
Atomic Hydrogen balance (Output = Input)
© copyright PCS-FKKKSA
Material Balance
Page 4-29
Solution to Working Session XI Atomic Carbon balance (Input = Output)
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-30
SKF 1113 - Material Balances
Solution to Working Session XI Atomic Oxygen balance (Output = Input)
© copyright PCS-FKKKSA
Page 4-31
Material Balance
Solution to Working Session XI mol dry gas h mol Total moles of flue gases (wet) , n w = n 1 + n 2 + n 3 + n 4 + n 5 = 1540 wet gas h n The molar compositio n of flue gases on a wet basis, i.e. y i = i nw Total moles of flue gases (dry) , n d = n 1 + n 2 + n 4 + n 5 = 1340
y CO 2 = y O2 =
n1 nw
n4 nw
y CO = y N2 =
n2 nw
y H 2O =
n3 nw
n5 nw
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-32
SKF 1113 - Material Balances
Analysis of combustion products Methane is burned with atmospheric air in a combustor. The analysis of the combustion products on a dry basis is as follows; CO2 O2 CO N2
10.00 % 2.37 % 0.53 % 87.10 %
Calculate a) the molar composition of combustion products on a wet basis. b) the percentage of excess air required. c) fractional conversion of methane to carbon monoxide. © copyright PCS-FKKKSA
Material Balance
Page 4-33
Analysis of combustion products Process flowchart Basis : 100 mol of dry flue gases
10.0 mol CO2 0.53 mol CO 2.37 mol O2 87.1 mol N2 nw mol H2O
nf mol CH4 no mol Air 0.21 mol O2 / mol 0.79 mol N2 / mol X % excess
CH4 + 2O2 Æ CO2 + 2H2O CH4 + 1.5O2 Æ CO + 2H2O © copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-34
SKF 1113 - Material Balances
Analysis of combustion products Basis : 100 mol/h of methane Strategy : Solve the unknowns from the given information. Nitrogen balance (Input = output)
1mol Air = 110.25 mol Air n o = ( 87.1 mol N 2 ) 0.79 mol N 2 Atomic Carbon balance (Input = Output) 1 mol C 1 mol C = (10 mol CO 2 ) + (0.53 CO 1 mol CH 4 1 mol CO 2
(n f mol CH 4 )
mol C ) 1 1mol CO
∴ n f = 10.53 mol CH 4
© copyright PCS-FKKKSA
Material Balance
Page 4-35
Analysis of combustion products Atomic Hydrogen balance (Output = Input) 2 mol H 4 mol H = (10.53 mol CH 4 ) 1 mol H 2O 1 mol CH 4
(n w mol H 2O )
∴ n w = 21.06 mol H 2O
Theoretical Air 2 mol O 2 = 21.06 mol O 2 = (10.53 mol CH 4 ) 1 mol CH 4 (n air )theo = ( 21.06 mol O 2 ) 1 mol Air = 100.3 mol Air 0.21 mol O 2
(n )
o 2 theo
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-36
SKF 1113 - Material Balances
Analysis of combustion products The molar composition of combustion products on a wet basis Total moles of flue gases (wet) n T = 10 + 0.53 + 2.37 + 87.1 + 21.06 = 121.06 mol wet flue gas The molar compositio n of flue gases on a wet basis, i.e. y i = y CO 2 = 0.0826
y CO = 0.0044
y O 2 = 0.0196
y N 2 = 0.720
ni nT
y H 2O = 0.174
Percentage of excess air :Fractional conversion of methane to carbon monoxide :-
© copyright PCS-FKKKSA
Page 4-37
Material Balance
Working Session XII – Unknown Hydrocarbon Composition An unknown hydrocarbon gaseous fuel is burned with excess atmospheric air in a combustor. The analysis of the combustion products on a dry basis is as follows; 5.63% CO2 9.69 % O2 CO 1.88 % 82.38% N2 Unburned fuel (measured as CH4 )
0.42%
Calculate a) the ratio of hydrogen to carbon in the fuel gas. b) the percentage of excess air required.
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-38
SKF 1113 - Material Balances
Solution to Working Session XII Process flowchart Basis : 100 mol of dry flue gases 5.63 mol CO2 1.88 mol CO 9.69 mol O2 82.38 mol N2 0.42 mol CH4 nw mol H2O
nC mol C nH mol H no mol Air 0.21 mol O2 / mol 0.79 mol N2 / mol X % excess
Stoichiometric balance of reactive species
© copyright PCS-FKKKSA
C + O2 Æ CO2 C + O2 Æ 2CO 4H + O2 Æ 2H2O
Material Balance
Page 4-39
Solution to Working Session XII Basis : 100 mol/h of dry flue gas Strategy : Solve the unknowns from the given information. Nitrogen balance (Input = output)
Atomic Carbon balance (Input = Output)
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-40
SKF 1113 - Material Balances
Solution to Working Session XII Atomic Oxygen balance (Input = Output)
© copyright PCS-FKKKSA
Material Balance
Page 4-41
Solution to Working Session XII Atomic Hydrogen balance (Input = Output)
Theoretical Air
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-42
SKF 1113 - Material Balances
Solution to Working Session XII The ratio of hydrogen to carbon in the fuel gas.
The fuel may be described by the formula (CH3.1)
n
Percentage of excess air (X%) :-
© copyright PCS-FKKKSA
Instructor: Dr Azeman Mustafa
Material Balance
Page 4-43
Lihat lebih banyak...
Comentários
