CaseStudy CH140

 

SCHOOL OF CHEMICAL, BIOLOGICAL AND MATERIALS ENGINEERING AND SCIENCES  4th Semester SY 2020-2021

Members:

Arzobal, Mark Edison De Leon, Khenette Dimatulac, Aaron Joseph Estrada, Chris Wilfred Lagatierra, Rezel Jourence Mallere, Reginald Sison, Maria Alekxandra

Date: August, 2021

CH140: CHEMICAL PROCESS INDUSTRIES Case Study

Given: The following Data were obtained from a test on a rotary kiln Average flue-gas analysis: G, kmol 1410 ᵒF

25.39 % CO2 0.89 % O2 0.08 % CO 73.64 % N2

Clinker Analysis: 100 kg

Raw-Mix Analysis: F, kg

Coal Analysis: C, kg [HV=1280 BTU/lb]

63.42 % CaO 3.28 % MgO 11.42% Fe2O3, Al2O3 20.80 % SiO2 0.63 % SO2

33.34 % CO2 41.97 % CaO 13.00 % SiO2

64.59 % C 5.07 % H 1.76 % S

7.00 % Fe2O3, Al2O3 2.50 % MgO 0. 0.665 % Gra raph phit itee 0.56 % FeS2 0.71 % H2O

1.71 % N 5.97 % O 0.8 0.86 % Mois Moistu ture re 20.04 % Ash

Air: A, kmol

21 % O 2 79 % N 2

Diagram: Air, A, kmol Coal, C, kg Hv = 1280 BTU/lb

21 % O2 79 % N2

64.59 % C 5.07 % H 1.76 % S 1.71 % N 5.97 % O 0.86 % Moisture 20.04 % Ash

│ Case Study│ August 2021

CH140  of 4

 

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SCHOOL OF CHEMICAL, BIOLOGICAL AND MATERIALS ENGINEERING AND SCIENCES  4th Semester SY 2020-2021

KILN

Raw-Mix, F, kg

33.34 % CO2 13.00 % SiO2 2.50 % MgO 0.56 % FeS2

41.97 % CaO 0.71 % H2O 0.65 % Graphite 7.00 % Fe2O3, Al2O3

1410 ᵒ Flue Gas, G, kmol

25.39 % CO2 0.89 % O2 0.08 % CO 73.64 % N2

Clinker, K, kg

63.42 % CaO O3.28 % MgO 11.42% Fe2O3, Al2O3 20.80 % SiO2 0.63 % SO2 Assumptions: 1. In estimatin estimatingg the compositi composition on of the clinker clinker,, the following following items items will be consid considered ered negligible: a. Free Cao  b. Combined MgO c. The amo moun untt of of FFee2O3 in clinker, other than that derive from pyrites in the raw mix and coal. d. Comp Compou ounnd of of SO SO3 2. All All iron iron pre ressent ent in the the clilink nker er is consi onsiddered red to be pr preesent sent as tetr tetrac acal alcciu ium m aluminoferrite, (4CaO · Al2O3Fe2O3), the remaining alumina as tricalcium aluminate, Ca3Al2O6, and the remaining calcium combined with silica as dicalcium silicate, 2CaO·SiO2, and tricalcium silicate, 3CaO·SiO2 3. The followi following ng assumpt assumption ionss are emp employ loyed ed in estima estimatin tingg the compos compositi ition on of the raw mix: a. All the MgO is in the form form of of MgCO MgCO3  b. The rest of the carbonate carbonate is CaCO3 c. Any CaO CaO not not present present as as carbona carbonate te is pprese resent nt a s CaO.SiO CaO.SiO2 d. The amou amounnt ooff Fe Fe2O3 in the fraction reported as Fe 2O3 and Al2O3 is negligible e. All Al2O3 is present as Al2O3.SiO2 4. In the the Ash assump assumptio tions ns aare re as follow follow,, 5. Fe2O3 is present as Fe2O3.SiO2 6. CaO CaO and and rema remain inin ingg SiO SiO2 are partially combines as CaO.SiO2 and 2CaO.SiO2 Solutions:  Required: Calculate the 100 per kg of clinker, the kmol of flue gas, the kg of RawMix and the kg of Coal Assume: S burned reported with CO2 O2 Balance: ( 0.21 ) A + 0.0597 │ Case Study│ August 2021

CH140  of 4

( )  1 32

C + 0.3334

( )  1 44

 F =( 0.2539 ) G + ( 0.0089 ) G + ( 0.0008 ) ( 2 ) G

 

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SCHOOL OF CHEMICAL, BIOLOGICAL AND MATERIALS ENGINEERING AND SCIENCES  4th Semester SY 2020-2021

 N2 Balance: ( 0.79 ) A + ( 0.0171 )

( )  1

28

C =( 0.7364 ) G

Coal + Sulfur Balance: ( 0.6459 )

( )  1

12

C + ( 0.0176 )

( )  1

32

C + ( 0.3334 )

( )  1

44

 F + ( 0.0056 ) (2 )

(

  1 119.98

)

 F + ( 0.6459 )

( )  1

12

 F =( 0.0063 ) (

 Non-Volatile Oxide Balance: Balance: ( 0.2004 ) C + ( 0.4197 ) F +( 0.1300 ) F + ( 0.07 ) F +( 0.0250 ) F =( 0.2080 ) ( 100 ) + ( 0.1142) ( 100 ) + ( 0.0328 ) ( 100 )

Values obtained using material balances:

Coal, C Air, A Raw-Mix, F Flue Gas, G

57.378 kg 15. 412 kg 135. 589 kg 16.587 kg G

kmol of flue gas / 100 kg clinker,  K  G 16.587 kmol flue gas =  K  100 kgclinker

 

 F 

kg of raw-mix / 100 kg clinker,  K   F 

  K =

kgraw w mix mix 135.589 kgra clinkerr 100 kg clinke C 

kg of coal / 100 kg clinker,  K 

 

coal al C  57.378 kg co  K = 100 kgclinker

Clinker Analysis:

 Basis: 100 kg  Components CaO  MgO  Fe2 O3  Al2 O3 S iO2 S O2

Mass

Molecular Mass

kmole

63.42 3.28 0.5053 11.42 20.80 0.63

56.0774 40.3044 159.69 101.96 60.08 64.066

1.130937 0.081381 0.003164 0.112005 0.346205 0.009834

Reaction: │ Case Study│ August 2021

CH140  of 4

 

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SCHOOL OF CHEMICAL, BIOLOGICAL AND MATERIALS ENGINEERING AND SCIENCES  4th Semester SY 2020-2021

2 Fe S 2+

11 2

O 2 → Fe 2 O 3 + 4 S O 2

 Fe2 O3 mass =

[

(

0.0056 135.589

)kgFeS

2

(

  1 kmol Fe S2 119.98 kg Fe S2

)(

 )(

  )]

kmol ol Fe2 O3 1 km

159.69 kg Fe 2 O 3

kmo ol Fe S 2 2 km

kmol ol Fe 2 O3 1 km

 Fe2 O3 mass =¿ 0.5053 kg

4CaO * Al2O3 * Fe2O3 mol = 0.003164 mol 3CaO * Al2O3 mol = 0.112005-0.003164 0.112005-0.003164 = 0.108841 mol Remaining CaO mol = 1.130937 – 3[(0.108841) – 4(0.003164) = 0.791758 mol

+ 2 y =0.791758  x + y =0.346205  x =0.099348  y = 0.246857 3 x

Components

Mole

Molar Mass

Mass

% Mass

3CaO * SiO2 2CaO * SiO2 3CaO * Al2O3 4CaO * Al2O3 *

0.099348 0.246857 0.108839 0.003166

228.3122 172.2348 270.1922 485.9596

22.6824 42.5174 29.40745 1.538548 3.28 0.63

22.67 42.49 29.39 1.54 3.28 0.63

Fe O  MgO S O2 2

3

In Clinker:

Solve for 4CaO * Al2O3 * Fe2O3 kmol: 0.5056 kg Fe 2 O 3

(

  1 kmo kmoll Fe2 O3 159.69 kg Fe2 O 3

)

=0.003166 kmol 4 CaO∗ Al 2 O 3∗ Fe2 O 3

Solve for kmol of 3CaO * Al 2O3 mol:

[

0.108841 kg Al 2 O 3

(

  1 kmol

Al2 O3

101.96 kg

Al 2 O3

)

− 0.003164 km kmol ol Fe O

Solve for remaining CaO:

[

1.130 1.13 0 937 kmolCaO

−0.002097 kmol 3 CaO∗ Al 2 O 3

│ Case Study│ August 2021

CH140  of 4

2

(

 

( )]= 1

3

1

3 kmolCaO

∗ Al 2 O 3

1 kmol 3 CaO

∗ Al 2 O 3

0.002097 kmol 3 CaO

)−

0.003164

kmol 4 CaO∗ Al 2 O

 

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