CHAPTER 3- Material balance for SIMPLE reactive system
CHEMICAL REACTION STOICHIOMETRY
BALANCES REACTIVE PROCESSES
SEPARATION & RECYCLE
COMBUSTION REACTIONS
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
1) Stoichiometry Stoichiometry – theory of the proportions in which chemical species combine with one another. *Note: Refer Felder pp.1 pp.116. 16.
The Stoichiometric Equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction – must be balanced (no. of atoms of each atomic species must be the same on both sides of the equation). Example: 2 SO2 + O2
→
2 SO3
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
1) Stoichiometry The Stoichiometric ratio two molecular species participating in a reaction – ratio of their stoichiometric coefficients in the balanced equation.
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
2) Limiting and Excess Reactants, Fractional Conversion and Extent of Reaction The limiting reactant – reactant that would run out if a reaction proceeded to completion, other reactants are termed excess reactant. *Note: Refer Felder pp.118. A reactant reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactants.
Fractional excess of reactant is the ratio of the excess to the stoichiometric requirement. *Note: Refer Felder pp.120 for Example 4.6 – 1.
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
2) Limiting and Excess Reactants, Fractional Conversion and Extent of Reaction Fractional Fractional excess of A Percentage Percentage excess of A
=
=
(n A ) feed − (n A ) stoich (n A ) stoich 100 x
Fractio Fractional excess of A
Fractional conversion conversion of a reactant is the ratio: Fractional Fractional conversio n of A Percentage Percentage conversio n of A
moles reacted =
=
moles feed 100 x
Fractio Fractional conver sion of A
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
3) Multiple Reactions Yield and Selectivity In chemical processes, some reactants can usually combine in more than one way, and the product once formed may react to yield something less desirable. *Note: Refer Felder pp.123 and Example 4.6 – 3 pp 124.
For example, ethylene can be produced by the dehydrogenation of ethane: C2H6
→
C2H4 + H2
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
3) Multiple Reactions Yield and Selectivity Once hydrogen is produced, it can react with ethane to produce methane: C2H6 + H2
→
2 CH4
• Moreover, ethylene can react with ethane to form propylene and methane: C2H4 + C2H6
→
C3H6 + CH4
CHAPTER 3- Material balance for SIMPLE reactive system CHEMICAL REACTION STOICHIOMETRY
Yield : Yield
moles of d esired product forme d =
moles that would have been for med if the re were no side re actions and the limiting reac tan t had reacted completely
Selectivity : Selectivit y
moles of d esired product formed =
moles of undesired ndesired product for med
CHAPTER 3- Material balance for SIMPLE reactive system
MOLECULAR SPECIES BALANCE BALANCE
ATOMIC SPECIES BALANCE
EXTENT OF REACTION
CHAPTER 3- Material balance for SIMPLE reactive system Example Methane is burned with air in a continuous steady state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water. The reaction taking place are: CH4 + 3/2 O2 CH4 + 2 O2
→
→
CO + 2 H2O CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH 4 , 19.4 % O2 , and 72.8% N2. The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO2/mol CO. Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for SIMPLE reactive system HO W ? ? ?
Analyze the information….
Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol Percentage conversion of CH4: 90.0% @ 0.90 Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for SIMPLE reactive system Basis of calculation: 100 mol of feed Process flow chart:
100 mol 0.078 mol CH4/mol 0.194 mol O2/mol 0.728 mol N2/mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2 n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Write system equation and outline a solution procedure
100 mol 0.078 mol CH4/mol 0.194 mol O2/mol
Additional information – fractional conversion CH 4 = 0.90 f
=
moles CH 4 reacted moles CH 4 fed
Moles CH 4 reacted
=
=
moles CH 4 reacted 0.078 × 100
=
=
Process
n2 mol O2
unit
n3 mol N2 n4 mol CO 8n4 mol CO2 n5 mol H2O
0.90
0.078 × 100 × 0.90 7.02 mol
Moles of unreacted CH 4 (n1 ) n1
0.728 mol N2/mol
n1 mol CH4
=
=
7.80 − 7.02
0.780 moles CH 4
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2 n3
= 72.8 mol N 2
CHAPTER 3- Material balance for SIMPLE reactive system CH 4 Balance – From first reaction:- CH4 + 3/2 O2
CO + 2 H2O
→
n1 mol CH4
100 mol
Process
n2 mol O2
unit
n3 mol N2
0.078 mol CH4/mol
n4 mol CO
0.194 mol O2/mol
1 mol CH4
→
1 mol CO
x mol CH4
→
x mol CO = n4
8n4 mol CO2
0.728 mol N2/mol
x = n4
→
→
n5 mol H2O
Eq.(1)
CH 4 Balance – From second reaction:- CH4 + 2 O2
1 mol CH4
→
→
CO2 + 2 H2O
1 mol CO2
(7.02 – x) mol CH4 →
8n4 = 7.02 – x
→
Insert Eq. 1 into 2
→
(7.02 – x) mol CO2 = 8n4 Eq.(2)
8n4
=
7.02 − n4
7.02 n
4
=
9
=
→
9n4
0.78 mol CO
=
7.02
CHAPTER 3- Material balance for SIMPLE reactive system O2 Balance – From first reaction:- CH4 + 3/2 O2
CO + 2 H2O
→
100 mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2
0.078 mol CH4/mol 0.194 mol O2/mol
3/2 mol O2
→
0.78 mol CO
1 mol CO
→
0.728 mol N2/mol
0.78 x 3/2 mol O2 = 1.17 mol O2 consumed
O2 Balance – From second reaction:- CH4 + 2 O2
2 mol O2
→
→
1 mol CO2
6.24 mol CO2 n2
=
n2
=
CO2 + 2 H2O
→
6.24 x 2 mol O2 = 12.48 mol O2 consumed
O2 Fed - O2 consumed 5.75 mol O2
=
0.194x100 - 1.17 - 12.48
n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system H 2O Balance – From first reaction:- CH4 + 3/2 O2
CO + 2 H2O
→
100 mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2
0.078 mol CH4/mol 0.194 mol O2/mol
2 mol H2 O
1 mol CO
→
0.78 mol CO
0.728 mol N2/mol
→
n4 mol CO 8n4 mol CO2
0.78 x 2 mol H2 O = 1.56 mol H2 O produced
H 2O Balance – From second reaction:- CH4 + 2 O2
2 mol H2 O
CO2 + 2 H2O
→
→
6.24 mol CO2
1 mol CO2
→
6.24 x 2 mol H2 O = 12.48 mol H2 O produced
produced ∑ H O produced
n5
=
n2
= 14.04
2
mol mol H 2 O
=
1.56 + 12.48
n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol
R A NS W E
0.078 mol CH4/mol 0.194 mol O2/mol 0.728 mol N2/mol
Component
Mole
Composition (%)
O2
5 .7 5
5 .7 2
CO
0 .7 8
0 .7 8
CO2
6 .2 4
6 .2 2
CH4
0 .7 8
0 .7 8
H2O
14 . 0 4
1 3 .9 9
N2
7 2 .8
7 2 .5 1
Total
100.39
100%
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2 n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Example Methane is burned with air in a continuous steady state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water. The reaction taking place are: CH4 + 3/2 O2 CH4 + 2 O2
→
→
CO + 2 H2O CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH 4 , 19.4 % O2 , and 72.8% N2. The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO2/mol CO. Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for SIMPLE reactive system HO W ? ? ?
Analyze the information….
Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol Percentage conversion of CH4: 90.0% @ 0.90 Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for SIMPLE reactive system Basis of calculation: 100 mol of feed Process flow chart:
100 mol 0.078 mol CH4/mol 0.194 mol O2/mol 0.728 mol N2/mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2 n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Write system equation and outline a solution procedure
100 mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2
0.078 mol CH4/mol 0.194 mol O2/mol
Additional information – fractional conversion CH 4 = 0.90 f
=
moles CH 4 reacted moles CH 4 fed
Moles CH 4 reacted
=
=
moles CH 4 reacted 0.078 × 100
=
=
0.90
0.078 × 100 × 0.90 7.02 mol =
Moles of unreacted CH 4 (n1 ) n1
0.728 mol N2/mol
=
7.80 − 7.02
0.780 moles CH 4
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2 n3 = 72.8 mol N 2
n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Analyze the atomic balance for: for:
0.780 mol CH 4
100 mol
Process
n2 mol O2
unit
72.80 mol N 2
0.078 mol CH 4/mol
Atom C: 1 unk. (n4)
n4 mol CO
0.194 mol O 2/mol
8n4 mol CO2
0.728 mol N 2/mol
Atom H: 1 unk. (n5)
n5 mol H2O
Atom O: 3 unk. (n2, n4, n5)
Solve the atomic balance on C… 7.8 mol CH 4
1 mol C ×
1 mol CH 4
7.8 = 0.780 + n4
+
8n4
=
→
0.780 mol mol CH 4 × 9n4
=
1 mol C 1 mol CH 4
7.8 - 0.780
+ n
4
mol CO ×
7.02 mol →
n
4
=
9
=
1 mol C 1 mol CO
+
8n4 mol CO2 ×
0.78 mol mol
Solve the atomic balance on H… mol CH 4 7.8 mol 31.2
4 mol mol H ×
1 mol mol CH 4
3.12 + 2
→
=
2
0.780 mol mol CH 4 × 31.2 3.12
4 mol mol H 1 mol CH 4
+ n
5
mol mol H 2O ×
28.08 mol mol →
2 mol mol H 1 mol mol H 2O
14.04 mol
1 mol C 1 mol CO2
CHAPTER 3- Material balance for SIMPLE reactive system 0.780 mol CH 4
100 mol
Process
n2 mol O2
unit
72.80 mol N 2
0.078 mol CH 4/mol
0.78 mol CO
0.194 mol O 2/mol
6.24 mol CO 2
0.728 mol N 2/mol
14.04 mol H 2O
Solve the atomic balance on O…
19.4 mol O 2 ×
2 mol O 1 mol O2
=
+
38.8 = 2n2
+
n2 mol O 2 ×
2 mol O 1 mol O2
6.24 mol CO2 ×
0.78 + 12.48 + 14.04
→
+
0.78 mol CO ×
2 mol O 1 mol CO2 2n2
=
11.5
+
1 mol O 1 mol CO
14.04 mol H 2O ×
→
n2
1 mol O 1 mol H 2O
11.5 mol =
2
=
5.75 mol
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol
R A NS W E
0.078 mol CH 4/mol 0.194 mol O 2/mol 0.728 mol N 2/mol
Component
Mole
Composition (%)
O2
5 .7 5
5 .7 2
CO
0 .7 8
0 .7 8
CO2
6 .2 4
6 .2 2
CH4
0 .7 8
0 .7 8
H2O
14 . 0 4
1 3 .9 9
N2
7 2 .8
7 2 .5 1
Total
100.39
100%
0.780 mol CH 4 Process
5.75 mol O 2
unit
72.80 mol N 2 0.78 mol CO 6.24 mol CO 2 14.04 mol H 2O
CHAPTER 3- Material balance for SIMPLE reactive system Example Methane is burned with air in a continuous steady state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water. The reaction taking place are: CH4 + 3/2 O2 CH4 + 2 O2
→
→
CO + 2 H2O CO2 + 2 H2O
The feed to the reactor contains 7.80 mole% CH 4 , 19.4 % O2 , and 72.8% N2. The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO2/mol CO. Calculate the molar composition of the product stream.
*Note: Refer felder pp.131
CHAPTER 3- Material balance for SIMPLE reactive system HO W ? ? ?
Analyze the information….
Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol Percentage conversion of CH4: 90.0% @ 0.90 Product gas: 8 mol CO2/mol CO
CHAPTER 3- Material balance for SIMPLE reactive system Basis of calculation: 100 mol of feed Process flow chart:
100 mol 0.078 mol CH4/mol 0.194 mol O2/mol 0.728 mol N2/mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2 n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Write system equation and outline a solution procedure
100 mol
n1 mol CH4 Process
n2 mol O2
unit
n3 mol N2
0.078 mol CH4/mol 0.194 mol O2/mol
Additional information – fractional conversion CH 4 = 0.90 f
=
moles CH 4 reacted moles CH 4 fed
Moles CH 4 reacted
=
=
moles CH 4 reacted 0.078 × 100
=
=
0.90
0.078 × 100 × 0.90 7.02 mol =
Moles of unreacted CH 4 (n1 ) n1
0.728 mol N2/mol
=
7.80 − 7.02
0.780 moles CH 4
Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2 n3 = 72.8 mol N 2
n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol
From equation : ni
=
ni0
+
∑ vijξ j
0.78 mol CH 4 Process
n2 mol O2
unit
72.8 mol N2
0.078 mol CH 4/mol 0.194 mol O 2/mol 0.728 mol N 2/mol
From first reaction:- CH4 + 3/2 O2
ξ1
CO + 2 H2O
→
From second reaction:- CH4 + 2 O2 nCH
4
=
n1
→
=
7.80 + [(− 1× ξ 1 ) + (− 1× ξ 2 )] = 7.80 - ξ 1 - ξ 2
From calculatio n,
n1
7.80 - ξ 1 - ξ 2 = 0.780 nO
=
n2
n
=
n
2
ξ2
CO2 + 2 H2O
=
0.780 mo moll
→
ξ 1 + ξ 2
=
7.02 mol → Eq. (1)
⎡⎛ 3 ⎤ 3 ⎞ 19.4 + ⎢⎜ − × ξ 1 ⎟ + (− 2 × ξ 2 )⎥ = 19.4 - ξ 1 - 2ξ 2 → Eq. (2) 2 ⎠ ⎣⎝ 2 ⎦ = 0 + (+ 1× ξ ) = ξ → n = ξ → Eq. (3)
=
n4 mol CO 8n4 mol CO2 n5 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system nCH 4
=
n1 = 7.80 + (− 1 × ξ1 ) + (− 1 × ξ2 )
=
7.80 - ξ1 - ξ2
From calculation, n1 = 0.780 mol 7.80 - ξ1 - ξ2 = 0.780 → ξ1 + ξ2 = 7.02 mol → Eq. (1) nO2
⎡⎛ 3 ⎤ 3 ⎞ n2 = 19.4 + ⎢⎜ − × ξ1 ⎟ + (− 2 × ξ2 )⎥ = 19.4 - ξ1 - 2ξ2 → Eq. (2) 2 ⎠ ⎣⎝ 2 ⎦ = n = 0 + (+ 1 × ξ ) = ξ → n4 = ξ1 → Eq. (3) 4 1 1
=
nCO nCO2
=
8n4
nH 2O
=
n5
=
=
0 + (+ 1 × ξ2 ) = ξ2
→ 8n4 = ξ2
→ Eq. (4 )
0 + [(+ 2 × ξ1 ) + (+ 2 × ξ2 )] = 2ξ1 + 2ξ2 → n5
=
2ξ1 + 2ξ2
→ Eq. (5 )
Substitute Eq (3) and (4) into Equation (1) ξ1
ξ2
+
nCO
=
nCO2 nO2
n4
=
ξ2
7.02 mol → n4 =
ξ1
=
=
8n4
+
8n4
=
7.02 mol 9n4
=
7.02 mol → n4
0.78 mol =
8 × 0.78 = 6.24 mol
3 3 n2 = 19.4 - ξ1 - 2ξ2 = 19.4 - (0.78 ) - 2(6.24 ) = 5.75 mol 2 2 = n = 2ξ + 2ξ = 2(0.78 ) + 2(6.24 ) = 14.04 mol 1
=
nH O
=
=
0.78 mol
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol
R A NS W E
0.078 mol CH 4/mol 0.194 mol O 2/mol 0.728 mol N 2/mol
Component
Mole
Composition (%)
O2
5 .7 5
5 .7 2
CO
0 .7 8
0 .7 8
CO2
6 .2 4
6 .2 2
CH4
0 .7 8
0 .7 8
H2O
14 . 0 4
1 3 .9 9
N2
7 2 .8
7 2 .5 1
Total
100.39
100%
0.780 mol CH 4 Process
5.75 mol O 2
unit
72.80 mol N 2 0.78 mol CO 6.24 mol CO 2 14.04 mol H 2O
CHAPTER 3- Material balance for SIMPLE reactive system SEPARATION & RECYCLE
1) Overall Conversion and Single Pass Conversion Conversion Overall Conversion : Overall Conversion
=
reactant input to the process process − reactant output from the process process reactant input to the process process
Single – Pass Conversion : Single - Pass Pass Conversion
=
reactant input to the reactor − reactant output from the process process reactant input to the reactor
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.135)
Dehydrogenation of propane Propane is dehydrogenated to form propylene in a catalytic reactor: C3 H8
→
C3 H6 + H2
The process is to be designed for a 95% overall conversion of propane. The reaction products are separated into two streams: 1st : contains H2 , C3H6 and 0.555% of the the propane that leaves the reactor, reactor, is taken off as product 2nd : contains balance of the unreacted propane and 5% of the propylene in the first stream. stream. Calculate the composition of the product, the ratio (moles (moles recycled)/(mole fresh feed) and the single pass conversion.
CHAPTER 3- Material balance for SIMPLE reactive system HO W ? ? ?
Basis of calculation..
100 mol mol C 3 H 8 fed to the reactor
Analyze the information..
95% overall conversion of propane
Sketch and insert the known and unknown values….
CHAPTER 3- Material balance for SIMPLE reactive system
100 mol C3H8 n1 mol C3H8 n2 mol C3H6
Reactor
n3 mol C3H8 n4 mol C3H6 n5 mol H2
0.99445 n3 mol C3H8 0.05 n4 mol C3H6
Separator
5.5x10-3 n3 mol C3H8 0.95 n4 mol C3H6 n5 mol H2
CHAPTER 3- Material balance for SIMPLE reactive system Overall conversion of propane = 0.95 0.95
100 − 5.55 ×10
−3
n3
=
100
n3
=
900.90 mol C 3 H 8
Mol balance at fresh feed – recycle on propane: 100 + 0.99445n3
(
=
n1
100 + 0.99445 900.90
n1
=
mol 995.9 mol
)
=
n1
CHAPTER 3- Material balance for SIMPLE reactive system Mol balance at fresh feed – recycle on propylene: 0.05n4
=
n
2
Atomic Balance at Reactor: C: n4 H : n5 , n4 Atomic Balance on C:
()
n1 3
+
()
0.05n4 3
()
() (3) 900.9(3)
=
0.05n4
995.9 3
+
n4
mol mol C 3 H 6
= 100
0.05n4
n2
=
=
()
n3 3 =
+
n4 3
+
()
n4 3
n2
0.05 ×100
=
5 mol C 3 H 6
CHAPTER 3- Material balance for SIMPLE reactive system Atomic Balance on H:
()
( ) n (8) n (6) n (2) 995.9(8) 5(6 ) 900.9(8) 100(6) n1 8
+
n2 6
=
+
n5
=
+
3
+
4
=
5
+
+
( )
n5 2
95 mol mol H 2
Product gas outlet: C 3 H 8
=
C 2 H 6
=
H 2
=
n5
5.55 ×10 0.95n4 =
−3
=
n3
=
(
5.55 × 10
0.95 100
95 mol
)
=
−3
(900.9)
95 mol
=
5 mol C 3 H 8
CHAPTER 3- Material balance for SIMPLE reactive system COMBUSTION REACTIONS
1) Combustion Chemistry Combustion – the rapid reaction of a fuel with oxygen. *Note: Refer Felder pp.142.
When a fuel is burned, carbon in the fuel reacts to form either CO2 or CO, hydrogen forms H2O, and sulfur form SO2. Partial or incomplete combustion – a combustion reaction in which CO is formed from a hydrocarbon.
CHAPTER 3- Material balance for SIMPLE reactive system COMBUSTION REACTIONS
Example: C + O2
CO2
→
C3 H8 + 5O2
→
C3 H8 + 3.5O2
Complete combustion of carbon
3CO2 + 4H2 O
→
Complete combustion of propane
3CO + 4H2 O Incomplete combustion of propane
Composition on a wet basis – mole fraction of a gas that contains water. Composition on a dry basis (Orsat Analysis)– mole fraction of the same gas that without the water. Stack gas or flue gas – product gas that leaves a combustion furnace.
CHAPTER 3- Material balance for SIMPLE reactive system COMBUSTION REACTIONS
2) Theoretical and Excess Air Theoretical Oxygen – the moles (batch) or molar flow rate (continuous) of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is oxidized to CO 2 and all hydrogen is oxidized to H 2O. Theoretical Air – the quantity of air that contains theoretical oxygen. Percent Excess Air:
(moles of air ) fed (moles of air )theoretical % Excess Air 100% (moles of air )theoretical −
=
*Note: Refer Felder pp.145.
×
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.143) A stack gas contains 60.0 mole% N 2 , 15.0% CO2 , 10.0% O2 , and the balance H2O. Calculate the molar composition of the gas on a dry basis. HO W ? ? ?
Basis of calculation.. Comp.
O2
Mol (wet basis)
100 100 mol wet gas gas
Mol (Dry % Mol basis) (Dry basis)
10
10
11.8
CO2
15
15
17.7
N2
60
60
70.5
H2O
15
%O 2
10 =
85
×
100
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.144) An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: 65.0 mole% N2 , 14.0% CO2 , 11% CO and 10.0% O2. A humidity measurement shows that the mole fraction of H 2O in the stack gas is 0.0700. Calculate the the stack gas component on a wet basis.
HO W ? ? ?
Basis of calculation..
100 100 mol dry dry basis
CHAPTER 3- Material balance for SIMPLE reactive system
Comp.
Mol (Dry Mol (Wet % Mol basis) basis) (Dry basis)
O2
10
10
CO2
14
14
N2
65
65
CO
11
11
H2O
0
x
0.0700
100 + x
1.00
Total
y H 2O
100
=
nH 2O ntotal
7 + 0.07x = x 0.93x = 7
→
→
→
0.070 =
x 100 + x
x - 0.070x = 7 x = 7.527 mol
CHAPTER 3- Material balance for SIMPLE reactive system
Comp.
O2
Mol (Dry Mol (Wet basis) basis)
Mol fraction (Wet basis)
10
10
0.093
CO2
14
14
0.130
N2
65
65
0.605
CO
11
11
0.102
H2O
0
7.527
0.070
107.527
1.00
Total
100
yO2
=
10 107.527
=
0.093
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.145) 100 mol/h of butane (C4H10) and 5000 mol/h of air are fed into a combustion reactor. reactor. Calculate the percentage of excess air. C 4 H 10
13 +
2
O2
→
4 CO2
+
5 H 2O
HO W ? ? ?
Find the theoretical air needed 1 mol mol C 4 H 10 100 mol mol C 4 H 10
13 →
→
2
mol mol O2
100 ×
13 2
mol O2
=
650 mol O2
CHAPTER 3- Material balance for SIMPLE reactive system O2 theoretical, nO2 0.21 (ntheoretical air ) ntheoretical air
=
=
650 mol
nO2
=
650 mol =
0.21
=
Percentage of excess air
650 mol
3095 mol
=
nexcess air − ntheoretical air × 100% ntheoretical air
5000 − 3095 × 100% Percentage of excess air 3095 Percentage of excess air 61.6% =
=
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.147) Ethane is burned with 50% excess air. The percentage conversion of the ethane is 90%; of the ethane burned, 25% react to form CO and the balance react to form CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to to dry stack gas. 7
C 2 H 6
+
C 2 H 6
+
2 5 2
O2
→
2 CO2
O2
→
2 CO + 3 H 2O
+
3 H 2O
CHAPTER 3- Material balance for SIMPLE reactive system HO W ? ? ?
Analyze the information….
Feed: 50% excess air Percentage conversion of C2H6: 90.0% @ 0.90 – 25% form CO, balance form CO2
CHAPTER 3- Material balance for SIMPLE reactive system Basis of calculation: 100 mol of C 2H6 fed Process flow chart: 100 mol C2H6 50% excess air
n1 mol C2H6 Process
n2 mol O2
unit
n3 mol N2
n0 mol air
n4 mol CO
0.79 mol N2/mol
n5 mol CO2
0.21 mol O2/mol
n6 mol H2O
CHAPTER 3- Material balance for SIMPLE reactive system Write system equation and outline a solution procedure
100 mol C 2H6 50% excess air
Additional information – fractional conversion C2H 6 = 0.90
n1 mol C2H6 Process
n2 mol O2
unit
n3 mol N2
n0 mol air
n4 mol CO
0.79 mol N2/mol
n5 mol CO2
0.21 mol O2/mol
n6 mol H2O
f 0.90 =
Moles C 2 H 6 fed 100 mol =
Moles C 2 H 6 reacted 100 × 0.90 =
=
90 mol
Moles of unreacted C 2 H 6 (n1 ) 100 × 0.1 10 mol =
n1
=
=
10 mol C 2 H 6
Find the amount of air fed into the process unit: Calculate the theoretical O 2 needed Calculate the theoretical air needed From % excess air, calculate the amount of excess air fed to the process unit
CHAPTER 3- Material balance for SIMPLE reactive system Calculate the theoretical O 2 needed
100 mol C 2H6 50% excess air
For complete combustion : 1 mol C 2 H 6
→
100 mol C 2 H 6
→
7 mol mol O2 2 7 100 × mol O2 2
=
Calculate the theoretical air needed 0.21(ntheoretical air ) ntheoretical air
=
350 mol O2
350 mol O2 =
0.21
=
1667 mol
n1 mol C2H6 Process
n2 mol O2
unit
n3 mol N2
n0 mol air
n4 mol CO
0.79 mol N2/mol
n5 mol CO2
0.21 mol O2/mol
n6 mol H2O
350 mol mol O2 Theoretical O2 =
CHAPTER 3- Material balance for SIMPLE reactive system Calculate the amount of excess air fed to the 100 mol C H process unit 50% excess air 2
%excess air n0
−
n0 =
n0
=
ntheoretical air
ntheoretical air
ntheoretical air
ntheoretical air
−
=
0.50
→
×
100%
n0
−
1667
1667
=
2500mol
N 2 in = N 2 out = 0.79 x 2500 = 1975 mol N 2 n3 = 1975 mol N 2
n1 mol C2H6
6
Process
n2 mol O2
unit
n3 mol N2
n0 mol air
n4 mol CO
0.79 mol N2/mol
n5 mol CO2
0.21 mol O2/mol
n6 mol H2O
0.50
CHAPTER 3- Material balance for SIMPLE reactive system
100 mol C2H6
10 mol C2H6 Process
n2 mol O2
unit
1975 mol N2
50% excess air
Given 75% ethane reacted form CO 2
Amount C 2 H 6 react to form CO2
=
2500 mol air
n4 mol CO
0.79 mol N 2/mol
n5 mol CO2
0.21 mol O 2/mol
n6 mol H2O
0.75 × 90 mol 67.5 mol =
From reaction 1 : 1 mol C 2 H 6 → 2 mol CO2 67.5 mol C 2 H 6 n5
=
→
67.5 × 2 135 mol CO2 formed formed =
135 mol CO2
Given 25% ethane reacted form CO Amount Amount C 2 H 6
react to form CO 0.25 × 90 mol 22.5 mol =
=
From reaction 2 : 1 mol C 2 H 6 → 2 mol CO 22.5 mol C 2 H 6 n4
=
45 mol CO
→
22.5 × 2 45 mol CO formed formed =
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol C2H6
Analyze the atomic balance for: for:
50% excess air
Process
n2 mol O2
unit
1975 mol N2
2500 mol air
45 mol CO
0.79 mol N 2/mol
135 mol CO 2
0.21 mol O 2/mol
n6 mol H2O
Atom C: 0 unk. Atom H: 1 unk. (n6)
10 mol C2H6
Atom O: 2 unk. (n2, n6)
Solve the atomic balance on H… 100 mol mol C 2 H 6 × 600 = 60 + 2n6
6 mol mol H 1 mol C 2 H 6 →
2n6
=
=
10 mol mol C 2 H 6 ×
600 - 60
→
6 mol mol H 1 mol mol C 2 H 6 n6
+
n6 mol mol H 2O ×
540 mol mol =
2
=
270 mol mol
2 mol mol H 1 mol H 2O
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol C2H6
10 mol C2H6 Process
n2 mol O2
unit
1975 mol N2
50% excess air
Solve the atomic balance on O…
0.21× 2500 mol O2 ×
2 mol O 1 mol O2
=
n2 mol O2 ×
135 mol CO2 × 1050 = 2n2
+
45 + 270 + 270
→
2n2
=
2500 mol air
45 mol CO
0.79 mol N 2/mol
135 mol CO 2
0.21 mol O 2/mol
n6 mol H2O
2 mol O +
1 mol O2
45 mol CO ×
2 mol O 1 molCO 2
465
→
+
n2
1 mol O 1 mol CO
270mol H 2O ×
465 mol =
2
=
+
1 mol O 1 mol H 2O
232.5 mol
CHAPTER 3- Material balance for SIMPLE reactive system 100 mol C2H6
R A NS W E
50% excess air
0.21 mol O 2/mol
270 mol H 2O
2 3 2 .5
0 .0 9 7
CO
45
0 .0 1 9
Moles ratio of water to dry stack gas:
=
10
0 .0 0 4
N2
19 7 5
0 .8 2 4
Total
2397.5
100%
1975 mol N2 135 mol CO 2
O2
C2H6
unit
0.79 mol N 2/mol
Composition (y (y)
0 .0 5 6
232.5 mol O2 45 mol CO
Mole
13 5
Process 2500 mol air
Component
CO2
10 mol C2H6
=
270 mol H 2O 2397.5 mol dry stack gas mol H 2O 0.113 mol dry stack gas
CHAPTER 3- Material balance for SIMPLE reactive system Example (Felder pp.149) A hydrocarbon gas is burned with air. The dry basis product gas composition is 1.5 mol%CO, 6.0 mol% CO2 , 8.2 % O2 , and 84.3% N2. There is no atomic oxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on what the fuel might be. Then calculate the percent excess air fed to the reactor. HO W ? ? ?
Analyze the information….
Dry basis: 0.015 mol CO/mol, 0.06 mol CO2/mol, 0.082 mol O2/mol, 0.843 mol N2/mol
CHAPTER 3- Material balance for SIMPLE reactive system Basis of calculation: 100 mol of dry basis product Process flow chart: n1 mol C n2 mol H n3 mol air
1.5 mol CO Process
8.2 mol O2
unit
84.3 mol N 2
0.79 mol N2/mol
6.0 mol CO 2
0.21 mol O 2/mol n4 mol H2O
Calculate the air fed to the unit N 2 in = N 2 out = 84.3 mol N 2 0.79(n3 ) n3
=
84.3 mol
84.3 mol =
0.79
=
106.70 mol
CHAPTER 3- Material balance for SIMPLE reactive system n1 mol C
Analyze the atomic balance balance for:
n2 mol H 106.70 mol air
Atom C: 1 unk. (n1)
0.79 mol N2/mol
Solve the atomic balance on C…
n1
=
1.5 mol CO ×
1.5 + 6
→
n1
=
1 mol C 1 mol CO 7.5 mol
8.2 mol O2
unit
84.3 mol N2 6.0 mol CO2 n4 mol H2O
Atom O: 1 unk. (n4)
=
Process
0.21 mol O2/mol
Atom H: 2 unk. (n2, n4)
n1 mol C
1.5 mol CO
+
6 mol CO2 ×
1 mol C 1 mol CO2
CHAPTER 3- Material balance for SIMPLE reactive system n1 mol C n2 mol H 106.70 mol air
1.5 mol CO Process
8.2 mol O2
unit
84.3 mol N2 6.0 mol CO2
0.79 mol N2/mol
Solve the atomic balance on O…
0.21 mol O2/mol n4 mol H2O
0.21× 106.70 mol O2 ×
2 mol O 1 mol O2
8.2 mol O2 ×
=
2 mol O 1 mol O2
1.5 mol CO × 44.814 = 16.4 + 12 + 1.5 + n4
→
n4
=
n2 mol H n2
=
=
14.9 mol H 2O ×
29.8 mol
2 mol H 1 mol H 2O
6 mol CO2 ×
1 mol O 1 mol CO
14.9 mol
Solve the atomic balance on H…
+
+
2 mol O 1 mol CO2
n4 mol H 2O ×
+
1 mol O 1 mol H 2O
CHAPTER 3- Material balance for SIMPLE reactive system Calculate the H/C in the fuel nH
29.8 =
nC
7.5 7.5
=
3.97 mol H/mol C ≈ 4 mol H/mol C
→
methane (CH 4 )
Percentage of excess air C
+
O2
2H +
nO
2
1 2
→
O2
theoretical
CO2 →
=
Reaction 1
H 2O
nO
2
Reaction 2
theoretical 1
nO
2
theoretical
=
7.5 mol C ×
nO
theoretical
=
14.95 mol
2
+
nO
2
theoretical 2
1 mol O2 1mol C
+
29.8 mol H ×
0.5 mol O2 2mol H
CHAPTER 3- Material balance for SIMPLE reactive system Calculate the theoretical air needed 0.21(ntheoretical air ) ntheoretical air
=
=
14.95 mol O 2
14.95 mol O2 0.21
=
71.19 mol air
Calculate the amount of excess air fed to the process unit %excess air
=
%excess air
=
n0
−
ntheoretical air
ntheoretical air 106.7 − 71.19 71.19
×
×
100%
100%
=
49.9%
CHAPTER 3- Material balance for SIMPLE reactive system Exercise A researcher burned n-Pentane (C5H12) with excess air in a continuous combustion chamber. chamber. C 5 H 12
+ 8O
2
→
5CO2
+ 6 H
2
O
The analysis on the product gas and report shows that product gas contains 0.304 mole% pentane, 5.9 mole% oxygen, 10.2 mole% carbon dioxide and the balance nitrogen on a dry basis. Based on 100 mol of dry product gas, i)) i
Dra Draw w and and labe labell a flow flowch char artt for for thi thiss proc proces ess s (2 marks)
ii)) Calc ii Calcul ulat atee th thee mo moll of n-Pentane and air fed in the combustion chamber, the percentage of excess air and the fractional conversion of n-Pentane in this process. (8 marks)
CHAPTER 3- Material balance for SIMPLE reactive system Exercise Ethy Ethyle lene ne (C2H4) has been commercially used for production of ethanol (C2H5OH) by hydration process: C 2 H 4
+ H
2
O → C 2 H 5OH
However, some of the product is converted to diethyl ether ((C 2H5)2O) in the side reaction:
(
)
2 C 2 H 5OH
→
(C H ) O 2
5
2
+ H
2
O
The feed to the reactor contains ethylene (C 2H4), steam (H2O) and inert gas (G). A sample of the reactor effluent gas is analyzed and found to contain 43.3 43 .3 mole% ethylene, 2.5 mole% ethanol, 0.14% ether, ether, 9.3% inert gas and the balance water. Based on 100 mol of effluent gas, i)) i ii)
Dra Draw w an and d label label a flowc flowcha hart rt fo forr th this is proc proces esss
(3 ma mark rks) s)
Calculate Calcul ate the mol molar ar com compos positi ition on of the reacto reactorr feed, feed, the percen percentag tagee conversio conversion n of ethylene, ethylene, the fraction fractional al yield of ethanol ethanol and the the selectivity of ethanol production relative to diethyl ether production. (9 marks)
CHAPTER 4- ENERGY balance for non reactive system Reference State A common practice is to arbitrarily designate a reference state for a substance at which U or H is declared to equal zero, and then tabulate U and/or H for the substance relative to the reference state. *Note: Refer Felder pp. 339 and 359
In Chapter 7 (Felder), U and H are state properties of a species; their values depend only on the state of the species – primarily on its temperature and state of aggregation (solid, liquid or gas) and, to a lesser extent, on its pressure (and for mixtures of some species, on its mole fraction in the mixture).
When a species passes from one state to another, both ΔU and ΔH for the process are independent of the path taken from the first state to