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Showing posts with label Clinker. Show all posts
Showing posts with label Clinker. Show all posts

Friday, 1 March 2019

RAW MEAL TO CLINKER RATIO -THEORY & CALCULATION

Calculation , Clinker
RAW MEAL TO CLINKER RATIO -THEORY & CALCULATION

RAW MEAL TO CLINKER RATIO

Raw meal to clinker ratio is the amount of raw meal required to produce 1 kg of clinker.

If you are using kiln feed containing dust material then you have to consider this dust  into the account to calculate "raw mix to clinker ratio".


let's suppose, you start with 100 tons of fresh raw meal. In the kiln feed bin you are adding 8% kiln dust. i.e. 8 tons of dust. So for every 100 T of fresh raw meal going through the kiln, you will really measure 108 T of kiln feed. Therefor, the kiln feed to clinker ratio will be the raw meal to clinker ratio multiplied by 
=108/100 = 1.08

If your fresh raw meal to clinker ratio is 1.55. then you kiln feed to clinker ratio will be
= 1.55 *1.08 = 1.674




However, using the LOI data, you can calculate your actual fresh raw meal to clinker ratio:

 Given:
D = % Dust in kiln feed =8
K = %  LOI of kiln Feed = 35.54
L = % LOI of the Dust = 38
if R = % LOI of fresh raw meal

 Then
R = (1+D/100) * { K - (D * L/100 )}
    = ( 1+ 8/100) * { 35.4 - (8 *38/100)}
    = (1.08 * 35.4 - 0.08 * 38)
    = 35.2

 Now you know that the LOI of your fresh raw meal is 35.2%, then we can calculate the actual raw meal to clinker ratio for fresh raw meal using below formula:
= 1/ (1-R/100)
= 1/(1-35.2/100)
= 1/(1-0.352)
=1.54

 Therefor, your meal to clinker ratio for kiln feed leaving the kiln bin is really:
= 1.54 * 1.08
= 1.66

 With 4% bypass:
Assuming the LOI of the bypass dust is close to zero then the kiln feed to clinker ratio would be:
= 1.66 * 1/ (1-4/100)
= 1.73

 if the LOI of the bypass is > 10% then you have to consider the LOI while doing calculation:

 Kiln feed to clinker ratio with 4% bypass:
= 1.66 * 1 / (1-4/100* (1- 10/100))
= 1.72


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Thursday, 28 February 2019

CALCULATION OF STOICHIOMETRIC AIR OF COAL AND COMBUSTION

Calculation , Clinker
CALCULATION OF STOICHIOMETRIC AIR OF COAL AND COMBUSTION

Calculation of Stoichiometric Air of Coal and Combustion Calculation

The specifications of furnace oil from lab analysis is given below:
Constituents % By weight
Carbon 85.9
Hydrogen 12
Oxygen 0.7
Nitrogen 0.5
Sulphur 0.5
H2O 0.35
Ash 0.05
GCV of fuel: 10880 kCal/kg


Calculation for Requirement of Theoretical Amount of Air

Considering a sample of 100 kg of furnace oil. The chemical reactions are:
Element Molecular Weight
Kg / Kg mole
C 12
O2 32
H2 2
S 32
N2 28
CO2 44
SO2 64
H2O 18

C   +    O2      →  CO2
H2   + 1/2O2  →    H2O
S    +   O2      →     SO2

 Constituents of fuel

 C      +     O2   →         CO2
12     +     32  →          44

 12 kg of carbon requires 32 kg of oxygen to form 44 kg of carbon dioxide therefore 1 kg of carbon requires 32/12 kg i.e 2.67 kg of oxygen

 (85.9) C  +  (85.9 × 2.67) O2  →    315.25 CO2

2H2      +    O2   →     2H2O
4          +    32   →     36

 4 kg of hydrogen requires 32 kg of oxygen to form 36 kg of water, therefore 1 kg of hydrogen requires 32/4 kg i.e 8 kg of oxygen

 (12) H2  +  (12 × 8) O2    →       (12 x 9 ) H2O
S    +    O2   →    SO2
32   +    32  →    64

 32 kg of sulphur requires 32 kg of oxygen to form 64 kg of sulphur dioxide, therefore 1 kg of sulphur requires 32/32 kg i.e 1 kg of oxygen

 (0.5) S + (0.5 × 1) O2   →     1.0 SO2
Total Oxygen Required (229.07 + 96 + 0.5) = 325.57 kg
Oxygen already present in [100 kg fuel (given)] = 0.7 kg
Additional Oxygen Required = 325.57 - 0.7
= 324.87 kg
Therefore quantity of dry air required. = (324.87) / 0.23
(air contains 23% oxygen by wt.) = 1412.45 kg of air
Theoretical Air required = (1412.45)/100
= 14.12 kg of air / kg of fuel

Calculation of theoretical CO2 content in flue gases


Nitrogen in flue gas = 1412.45 - 324.87
= 1087.58 kg

Theoretical CO2% in dry flue gas by volume is calculated as below :


Mole of CO2 in flue gas = (314.97) / 44 = 7.16
Mole of N2 in flue gas = (1087.58) / 28 = 38.84
Mole of SO2 in flue gas = 1/64 = 0.016

Theoretical CO2 % by Volume = (Moles of CO2) x 100
Total moles (dry)
= 7.16
7.16 + 38.84 + 0.016
= 15.5%

Calculation of constituents of flue gas with excess air:



% Excess air   =  [(15.5/10)-1] x 100
                        = 55%

Theoretical air required for 100 kg  of fuel burnt  =  1412.45 kg 
Total quantity of air supply required with 55% excess 
                                                                         air  = 1412.45 X 1.55 
                                                                              = 2189.30 kg
Excess Air quantity = 2189.30 - 1412.45 
                                = 776.85 kg  

                             O2 = 776.85 x 0.23 = 178.68  kg
                            N2 = 776.85 - 178.68 = 598.17 kg

The final constitution of flue gas with 55% excess air for every 100 kg fuel.

CO2 = 314.97 kg
H2O = 108.00 kg
SO2 = 1 kg
O2 = 178.68 kg
N2 = 1087.58 + 598.17 = 1685.75 kg

Calculation of Theoretical CO2% in Dry Flue Gas By Volume

Moles of CO2 in flue gas = 314.97/44 = 7.16
Moles of SO2 in flue gas = 1/64 = 0.016
Moles of O2 in flue gas = 178.68 / 32 = 5.58
Moles of N2 in flue gas = 1685.75 / 28 = 60.20

     =  (7.16 * 100) / (7.16 + 0.016 + 5.58 + 60.20)

    =  7.16 *100/ 72.956
  = 10%

 Theoretical O2 % by volume = 5.58 *100 *100/72.956
                                              = 7.5%


Optimizing Excess Air and Combustion:

For complete combustion of every one kg of fuel oil 14.1 kg of air is needed. In practice, mixing is never perfect, a certain amount of excess air is needed to complete combustion and ensure that release of the entire heat contained in fuel oil. If too much air than what is required for completing combustion were allowed to enter, additional heat would be lost in heating the surplus air to the chimney temperature. This would result in increased stack losses. Less air would lead to the incomplete combustion and smoke. Hence, there is an optimum excess air level for each type of fuel.
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Sunday, 10 February 2019

HEAT OF REACTION OF CLINKER OF THEORETICAL HEAT REQUIREMENT TO PRODUCE 1 KG OF CLINKER

Clinker , Online Calculator
HEAT OF REACTION OF CLINKER OF THEORETICAL HEAT REQUIREMENT TO PRODUCE 1 KG OF CLINKER

 HEAT OF REACTION TO PRODUCE 1KG OF CLINKER

You can calculate heat required to produce 1kg of clinker by using following inputs:
1. CaCO3 content in kiln feed, %
2. Kiln feed:Clinker ratio, -
3. MgCO3 content in kiln feed, %
4. LOI content in kiln feed, %
5. Al2O3 content in clinker, %
6. Water of crystalisation in kiln feed, %
7. SiO2 content in clinker, %
8. Fe2O3 content in clinker, %
9. K2O content in clinker, %
10. Na2O content in clinker, %
11. Carbon content in kiln feed, %
12. Sulphur as pyrites in kiln feed, %



Calculate Heat of Reaction of Clinker
Input Data:
CaCO3 content in kiln feed, % CaCO3
Kiln feed:Clinker ratio, - RMCK
MgCO3 content in kiln feed, % MgCO3
LOI content in kiln feed, % LOI
Al2O3 content in clinker, % Al2O3
Water of crystalisation in kiln feed, % CH2O
SiO2 content in clinker, % SiO2
Fe2O3 content in clinker, % Fe2O3
K2O content in clinker, % K2O
Na2O content in clinker, % Na2O
Carbon content in kiln feed, % C
Sulphur as pyrites in kiln feed, % S
Formula:
Heat of Reaction of Clinker in kcal/kg-clinker
( 4.282 * CaCO3 * RMCK ) + ( 3.085 * MgCO3 * RMCK )
+ ( 222 * Al2O3 / ( 100 - LOI ) ) + ( 11.6 * CH2O * RMCK )
- ( 511.6 * SiO2 / ( 100 - LOI ) ) - ( 59 * Fe2O3 / ( 100 - LOI ) )
- ( 1000 * ( K2O + Na2O ) / ( 100 - LOI ) )
- ( 79 * C * RMCK ) - ( 32 * S * RMCK )
HEAT OF REACTIONO TO PRODUCE 1 KG CLINKER
INPUT
SN DESCRIPTION UNIT VALUE
1 CaCO3 content in kiln feed, % %
2 Kiln feed:Clinker ratio, - %
3 MgCO3 content in kiln feed, % %
4 LOI content in kiln feed, % %
5 Al2O3 content in clinker, % %
6 Water of crystalisation in kiln feed, % %
7 SiO2 content in clinker, % %
8 Fe2O3 content in clinker, % %
9 K2O content in clinker, % %
10 Na2O content in clinker, % %
11 Carbon content in kiln feed, % %
12 Sulphur as pyrites in kiln feed, % %
OUTPUT
RESULT HEAT OF REACTION OF CLINKER Kcal/kg-clinker
Target



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