Dr. F. IskanderaniChE 201 Spring 2003/2004 EXAMPLE1 ( How to carry elemental or atomic balance on a stream) Fuel F is burned with air. Let F=25 Kg, and

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


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


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


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


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









S (0.3 x 25) Kg S









S (0.3 x 25) Kg S


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


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


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


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


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


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?


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


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


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


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 + 3.5 O2 3 CO + 4 H2O ( the partial/incomplete combustion rxn)


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 + 3.5 O2 3 CO + 4 H2O ( the partial/incomplete combustion rxn)


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


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


Step 3: same as step 3 above

Questions:

1. When can we use any of the 2 solutions aboveto solve?

2. When can we only use solution 2 to solve?

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Dr. F. IskanderaniChE 201 Spring 2003/2004 EXAMPLE1 ( How to carry elemental or atomic balance on a stream) Fuel F is burned with air. Let F=25 Kg, and