dr. f. iskanderaniche 201 spring 2003/2004 example1 ( how to carry elemental or atomic balance on a...
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Dr. F. Iskanderani ChE 201 Spring 2003/2004
FURNACE
Mass % CO 40 S 30 C2H5OH 10 CH4 15 O2 5
100
Mol fr. O2 0.21 N2 0.79
F Kg
A moles
P mol
EXAMPLE1 ( How to carry elemental or atomic balance on a stream) Fuel F is burned with air. Let F=25 Kg, and carry atomic balance on F in moles
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Atomic Balance In Stream F
C (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol C 28 kg CO 1 kgmol CO + (0.1 x 25) Kg C2H5OH x 1kgmol C2H5OH x 2 kgmol C 46 kg C2H5OH 1 kgmol C2H5OH + (0.15 x 25) Kg CH4 x 1kgmol CH4 x 1 kgmol C 16 kg CH4 1 kgmol CH4
H (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 6kgmol H . 46 kg C2H5OH 1kgmol C2H5OH + (0.15 x 25) Kg CH4 x 1kgmol CH4 x 4 kgmol H 16 kg CH4 1 kgmol CH4
Dr. F. Iskanderani ChE 201 Spring 2003/2004
O (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol O 28 kg CO 1 kgmol CO
+ (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 1 kgmol C . 46 kg C2H5OH 1 kgmol C2H5OH
+ (0.05 x 25) Kg O2 x 1kgmol O2 x 2 kgmol O
32 kg O2 1 kgmol O2
S (0.3 x 25) Kg S x 1kgmol S 32 kg S
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Atomic Balance In Stream F
C (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol C x 12 Kg C
28 kg CO 1 kgmol CO 1kgmol C
+ (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 2 kgmol C x 12kg 46 kg C2H5OH 1 kgmol C2H5OH kgmol C
+ (.15 x 25) Kg CH4 x 1kgmol CH4 x 1 kgmol C x 12 kg C 16 kg CH4 1 kgmol CH4 kgmol C
EXAMPLE 2( How to carry elemental or atomic balance on a stream) Fuel F is burned with air. Let F=25 Kg, and carry atomic balances on F in MASS
Dr. F. Iskanderani ChE 201 Spring 2003/2004
H (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 6kgmol H 1 kg H .
46 kg C2H5OH 1kgmol C2H5OH kgmol H
+ (0.15 x 25) Kg CH4 x 1kgmol CH4 x 4 kgmol H x 1kg H 16 kg CH4 1 kgmol CH4 kgmol H
O (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol O x 16 kg O
28 kg CO 1 kgmol CO kgmol O
+ (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 1 kgmol O x 16 kg O 46 kg C2H5OH 1 kgmol C2H5OH kgmol O
+ (0.05 x 25) Kg O2 x 1kgmol O2 x 2 kgmol O x 16 kg O 32 kg O2 1 kgmol O2 kgmol O
S (0.3 x 25) Kg S
Dr. F. Iskanderani ChE 201 Spring 2003/2004
H (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 6kgmol H 1 kg H .
46 kg C2H5OH 1kgmol C2H5OH kgmol H
+ (0.15 x 25) Kg CH4 x 1kgmol CH4 x 4 kgmol H x 1kg H 16 kg CH4 1 kgmol CH4 kgmol H
O (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol O x 16 kg O
28 kg CO 1 kgmol CO kgmol O
+ (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 1 kgmol O x 16 kg O 46 kg C2H5OH 1 kgmol C2H5OH kgmol O
+ (0.05 x 25) Kg O2 x 1kgmol O2 x 2 kgmol O x 16 kg O 32 kg O2 1 kgmol O2 kgmol O
S (0.3 x 25) Kg S
Dr. F. Iskanderani ChE 201 Spring 2003/2004
H (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 6kgmol H 1 kg H .
46 kg C2H5OH 1kgmol C2H5OH kgmol H
+ (0.15 x 25) Kg CH4 x 1kgmol CH4 x 4 kgmol H x 1kg H 16 kg CH4 1 kgmol CH4 kgmol H
O (0.4 x 25) Kg CO x 1kgmol CO x 1 kgmol O x 16 kg O
28 kg CO 1 kgmol CO kgmol O
+ (0.1 x 25) Kg C2H3OH x 1kgmol C2H5OH x 1 kgmol O x 16 kg O 46 kg C2H5OH 1 kgmol C2H5OH kgmol O
+ (0.05 x 25) Kg O2 x 1kgmol O2 x 2 kgmol O x 16 kg O 32 kg O2 1 kgmol O2 kgmol O
S (0.3 x 25) Kg S
Dr. F. Iskanderani ChE 201 Spring 2003/2004
FURNACE
Mole % CO 40 S 30 C2H5OH 10 CH4 15 O2 5
100
Mol fr. O2 0.21 N2 0.79
F
A moles
P mol
40% excess air
mol
CO2 x1
SO2 x2
H2O x3
O2 x4
N2 x5
P
Let us change the problem as follows:
Example3 (How to carry elemental or atomic balance on a stream) Fuel F is burned with 40% excwss air. Let F=25 Kgmoles , and
carry atomic balance in moles
Basis: F=100 kgmol
Dr. F. Iskanderani ChE 201 Spring 2003/2004
CO + 1/2 O2 CO2
S + O2 SO2
CH4 + 2 O2 CO2 + 2H2O
C2H5OH + 3 O2 2CO2 + 3H2O
moles O2 required Excess O2 O2 EnteringCO
S
CH4
C2H5OH
(O2)
40
30
15
10
20
30
30
30
8
12
12
12
28
42
42
42 (-5)149
40% excess air
Dr. F. Iskanderani ChE 201 Spring 2003/2004
C (0.4 x 100) Kgmol CO x 1 kgmol C 1 kgmol CO + (0.1 x 100) Kgmol C2H5OH x 2 kgmol C . 1 kgmol C2H5OH + (0.15 x 100) Kgmol CH4 x 1 kgmol C 1 kgmol CH4
H (0.1 x 100) Kgmol C2H5OH x 6kgmol H . 1kgmol C2H5OH + (0.15 x 100) Kgmol CH4 x 4 kgmol H 1 kgmol CH4
= x1 * 1
= x3 * 2
In exit stream
THUS
x1 = 40+ 20 + 15 = 75 moles = moles CO2 OUT
x3 = (60 + 60)/2 =60 moles = moles H2O OUT
In stream FAtomic bal
Dr. F. Iskanderani ChE 201 Spring 2003/2004
O (0.4 x 100) Kgmol CO x 1 kgmol O 1 kgmol CO + (0.1 x 100) Kgmol C2H5OH x 1 kgmol O . 1 kgmol C2H5OH + (0.05 x 100) Kgmol O2 x 2 kgmol O
1 kgmol O2
+149 * 2
In exit streamIn Feed stream In Air stream
= x1 * 2 + x2 * 2 + x3 * 1 + x4 *2
Dr. F. Iskanderani ChE 201 Spring 2003/2004
S Balance
0.3 * 100 kgmol S = x2 mol SO2 * 1 mol S
Thus x2 = 30 moles = moles SO2 OUT
Now we can find x4
1 Mol SO2
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Example : 20 Kg of Propane (C3H8) is burned with 400 kg of air toproduce 44 kg of CO2 and 6 Kg of CO. What was the % excess air?
C3H8 + 5 O2 3 CO2 + 4 H2O ( the complete combustion rxn )
C3H8 + 3.5 O2 3 CO + 4 H2O ( the partial/incomplete combustionrxn)
FURNACE
Mol fr.O2 0.21N2 0.79
20 Kg C3H8
400 kg Air
CO2
COH2O
Kg4412?
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Remember : the basis of the calculation of excess air iscomplete combustion
Change to kg moles : C3H8 is 0.454 , air is 13.36 CO2 is: 1 and CO is 0.215
O2 theoretical = 0.454 kgmol C3H8 5 kgmol O2= 2.27 kg mol O2
1 kgmol C3H8
O2 entering = 13.36 kgmol air 0.21 kgmol O2 = 2.90 kgmol O2
Kgmol air
% excess air = 100 x O2 entering O2 theoreticalO2 theoretical
= 100 x 2.90 kgmol O2 2.27 kgmol O2 = 28%2.27 kgmol O2
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Example : 20 Kg of Propane (C3H8) is burned incompletely with200% excess air to produce CO2 and 14 Kg of CO. What is theanalysis of the stack gas produced?FIGURE : SAME AS ABOVE WITH DIFFERENTAMOUNTS OF FLOWS.Unknown variables: nCO2
P , nH2OP nN2
P, nO2P
Atomic Balances : C, H, OComponent Balances: N2
(not reacting), and total mass balance
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Solution 1 ( using the stoichiometric equations)Step1: Calculate O2 entering and N2 entering
O2 theoretical = 0.454 kgmol C3H8 5 kgmol O2= 2.27 kg mol O2
1 kgmol C3H8
O2 entering = 2.27 kgmol O2 + (200/100) 2.27 kgmol O2
= 2.27 kgmol O2 (1 + 200/100) = 6.81 kgmol O2
N2 entering =O2 entering x 0.79/0.21 = 25.62 kgmol = N2 Out
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Step2 : calculate from stoichiometry CO2 , H2O and O2 out
CO out = 14 kg/28 = 0.5 kgmol
O2 used up to produce CO = 0.5 kgmolCO 3.5 kgmol O2 = 0.583 kgmol 3 Kgmol CO
C3H8 reacted to produce CO = 0.5 kgmolCO 1 kgmol C3H8 = 0.167 kgmol3 Kgmol CO
C3H8 reacted to produce CO2 = 0.454 – 0.167 = 0.287 kgmol
C3H8 + 5 O2 3 CO2 + 4 H2O ( the complete combustion rxn )
C3H8 + 3.5 O2 3 CO + 4 H2O ( the partial/incomplete combustion rxn)
Dr. F. Iskanderani ChE 201 Spring 2003/2004
O2 reacted to produce CO2 = 0.287 kgmol C3H8 5 kgmol O2 = 1.435 kgmol1Kgmol C3H8
O2 Out = O2 in – O2 that have reacted in both reactions = 6.81-(1.435 + 0.583) = 4.792 kgmol
CO2 Out = 0.287 x 3 = 0.861 kgmol 1H2O Out = 0.287 x 4 + 0.167 x 4 = 1.816 kgmol
1 1
C3H8 + 5 O2 3 CO2 + 4 H2O ( the complete combustion rxn )
C3H8 + 3.5 O2 3 CO + 4 H2O ( the partial/incomplete combustion rxn)
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Step3 : Calculate the analysis of stack gas
Component Kgmoles % kgmolCO2 0.861 2.6
CO 0.5 1.5
H2O 1.816 5.4
N2 25.62 76.3
O2 4.792 14.3
Total 33.634 100.0
Dr. F. Iskanderani ChE 201 Spring 2003/2004
Solution 2 Using atomic balances
Step1: Same as step 1 above
Step 2 : C, H, O Atomic balances in moles ( note: we do not usethe chemical equations)
Given : CO out = 14 kg/28 = 0.5 kgmol
IN = OUTC balance 3 x 0.454 + 0 = 1 x nCO2 + 1 x nCO
H balance 8 x 0.454 + 0 = 2 x nH2O
O balance 0 +2 x 6.81 = 2 x nCO2 + 1 x nCO + 1 x nH2O
+ 2 x nO2
Solve : nH2O = 1.816 kgmol ; nCO2 = 1.362 – 0.5 = 0.862kgmol
And nO2 =13.62 – 2 (0.861) – 0.5 – (1.816) = 4.791 kgmol