LSF

Raw Mix Composition

& Quality Control1.1 Hydraulic Modulus

HM =CaO

= 1.7 --- 2.3SiO2 + AL2O3 + Fe2O3

It was found that with increasing HM, more heat is required for clinker

burning : the strengths , especially the initial strengths step up and also

the heat of hydration rises and the resistance to chemical attack decreases.

1.2 Silica Ratio

SR =SiO2

= 1.9 --- 3.2 Al2O3 + Fe2O3

It was found that with increasing SR impairs the burnability of the clinker,

by reducing the liquid phase content and tendancy towards formation of

coating in the kiln, also causes a slow setting and hardening of the cement

with decreasing SR the content of liquid phase increases: this improves the

burnability of clinker and formation of coating in the kiln.

1.3 Alumina Ratio

AR =Al2O3

= 1.5 --- 2.5Fe2O3

the Alumina Ratio determines the composition of liquid phase in the clinker

a high AR together with a low silica ratio results among other things,

in a fast setting of cement: this requires the addition of higher gypsum rate

to control the setting time.

1.4 Lime Saturation Factorto obtain complete lime saturation factor in clinker, the total silica must be

combined as C3S, all iron onide must combine with the equivalent amount

of alumina to form C4AF, and the remaining alumina must combine to C3A.

the lime saturation factor is based on the assumption that from the sintering temperature the clinker cools down at such slow rate that during

crystallization the liquid phase can achieve equilibrium with solid phases.

However, this is not the case with clinkers containing C3A. At the sintering

temperature af about 1450 C, the silicate minerals C3S and C2S and

possibly not transposed free lime are in a solid state, whereas C3A and

C4AF are in a state of fusion.

how ever, the liquid phase is lower in lime that it would result from

participation in C3A to complete the C3A the lime deficiency can be

restored by extracting the lacking lime during the crystallization process

from the solid phases, which simultaneously are the completed during fast

technical clinker cooling: partically, the liquid aluminate cannot bind more

lime than it already has absorbed at sentering temperature,experimental

investigation showed that the most lime saturated liquid aluminate binds

practically two molecules CaO per Al2O3. Therefore, under technical

conditions, this is attainable lime limit, that called "standarad lime"

CaO = 2.8 SiO2 + 1.1 Al2O3 + 0.7 Fe2O3

stand

the lime saturation factor is the ratio of the real lime content to the

standard lime:

L.S.F =100 CaO

2.8 SiO2 + 1.18 Al2O3 + 0.65 Fe2O3

The Lime Saturation Factor is also part of the British Standard Specification

and serves for the definition of admissible lime content.

L.S.F = CaO - 0.7 SO3

= 0.66 --- 1.22.8 SiO2 + 1.2 Al2O3 + 0.65 Fe2O3

in this formula LSF refers to the finished cement. The factor 0.7 SO3 is the

CaO - content which equals analytically estimated SO3 - content , should

be substracted from the total CaO - content.

it is assumed that the total SO3 comes from the added Gypsum and not

from the clinker.

a high lime standard normally causes high cement strength 90 --- 95.

2. Allegation Alternate Method.

the simplest course of calculation for solving blending problems. This

method allows the determination of proportion of two raw materials

in this case, only required lime content is fixed as a setpoint, so that the

91 45

76

clay 31 15

comp clinker

Raw

mix

Raw

mix

Raw

mix

Raw

mix

Raw

mix

NO.1 NO.2 NO.3 NO.4

proportion of both components can be determined.

Example : 1

what mixing proportion is required for limestone with 91% CaCO3 to get a

raw mix with a CaCO3 content 76% . The procedure is as follows,

limestone

76 - 31 = 45

parts lime stone

91 - 76 = 15

parts clay

to get a raw mix with CaCO3 contents of 76% , 45 parts of lime stone

should be mixed with 15 parts of clay. Thus the proportion of the

components in the raw mix

limestone : clay

45 : 15

or 3 : 1

3. Calculation based on the Hydraulic Moodulus.this method is appliccable to two raw material components, with the

hydraulic module selected for clinker. To simplify the following calculations

symbols are used for the designations of the clinker components,

Table 1

CaO C Cm C1 C2

SiO2 S Sm S1 S2 S3

Am A1 A2 A3

C3 C4

S4

A4

Fe2O3 F Fm F1 F2 F3 F4

Al2O3 A

applying these symbols, the Hydraulic Moduli for clinker and raw mix will be.

so:

Cm =C

( for clinker )S + A + F

HM =Cm

( for raw mix )Sm + Am + Fm

since both moduli are of the same value, one can equate.

HM =C

=Cm

S + A + F Sm + Am + Fm

according to this method of calculation, it is assumed that X parts of the

first material apportioned to one part of the second raw material.

under this assumption, the quantities of the particular raw material can be

calculated by using the following formulas :

Cm =XC1 + C2

Sm =XS1 + S2

X + 1 X + 1

Am =XA1 + A2

Fm =XF1 + F2

X + 1 X + 1

inserting the values Cm , Sm , Am , Fm into the formula for Hydraulic

Moduli, we get

HM =

XC1 + C2

X + 1

XS1 + S2+

XA1 + A2+

XF1 + F2

X + 1 X + 1 X + 1

since the oxide components are known from the chemical analysis of raw

material,and the Hydraulic Module is selected according to the quality

requirements,the only remaining unknown is X .after transformation of the

above formula, to calculate the value for X ,the following formula appears:

X =HM ( S2+A2+F2 ) - C2

=C2 - HM (S2+A2+F2)

C1 - HM (S1+A1+F1) – C1 - HM (S1+A1+F1)

i.e

1 2 3 4 5 6

limestone marl 76.87% 23.13% 100%

SiO2 8.70 33.01 6.69 7.64 14.33 21.94

Al2O3 2.35 7.31 1.81 1.69 3.50 5.36

Fe2O3 1.32 4.83 1.01 1.12 2.13 3.26

CaO 47.80 30.22 36.75 6.99 43.74 66.92

MgO 1.50 0.66 1.15 0.15 1.30 2.00

SO3 0.30 0.20 0.23 0.05 0.28 0.44

LOI 37.96 23.77 29.18 5.49 34.67 ..

balance 0.01 .. 0.05 .. 0.05 0.08

total 100.00 100.00 100.00 23.13 100.00 100.00

Example 2

two raw materials are given with the following composition (see column 1

and 2 of table 2 ).

calculate the composition of the raw mix, assuming a hydraulic module of

HM = 2.2

solution:

X =2.2 ( 33.01 + 4.31 + 4.83 ) - 30.22

Ξ 3.32447.80 - 2.2 (8.72 + 2.35 +1.32 )

to get a clinker with a HM = 2.2 we have to mix 3.324 parts lime stone with

one part of marl, or expressing as % , thus the raw mix consists of : 76.87%

limestone, and 23.13 % marl.

Table 2

comp

limestone marlraw

mixclinker

in the table 2 the calculated raw mix components appear in column 3 and 4, and

the composition of raw mix is given in column 5 (column 3 + 4 = column 5 )

column 6 contains the calculated clinker composition as raw mix of raw of

column 5, free of loss on iginition. From column 6 we obtain the value for the

hydraulic module HM = 2.2

4. Calculation based on Lime Saturation Factor.

1 2 4

Lst clay clinker

X + 1 X + 1

LSF =100 CaO

2.8 SiO2 + 1.18 Al2O3 + 0.65 Fe2O3

Example 3

given 2 raw materials (see table 3, column 1 and 2 ).

XF1 + F2

X + 1 X + 1

To obtain a raw mix with a given LSF of 0.92, let us assume that X part of lime

stone will apportioned to 1 part of clay.

LSF for

Cm =XC1 + C2

Sm =XS1 + S2

2.8 (XS1 + S2)+

1.18 (XA1 + A2)

Am =XA1 + A2

Fm =

+0.65 (XF1 + F2)

X + 1 X + 1 X + 1

After transformation of the above formula, to calculate the value of X, the

LSF =

100XC1 + C2

X + 1

following formula appears :

X =100C2 - LSF (2.8S2 + 1.18A2 + 0.65F2)

LSF (2.8S1 + 1.18A1 + 0.65F1) - 100C1

accordingly we get

X =100*1.4 - 0.92(2.8*62.95 + 1.18*18.98 + 0.65*7.37)

Ξ 3.850.92(2.8*1.42 + 1.18*0.48 + 0.65 *0.38) -100*52.6

thus 3.85 parts of limestone are apportioned to one part of clay

raw mix consists of :

limestone clay

79.40% 20.60%

Table 3

components

3

raw mix

1.42 62.95 21.96

0.48 18.98 6.68

0.38 7.37 2.84

52.6 1.4 65.51

1.11 0.98 1.69

0.85 0.85 1.32

43.16 7.47 ..

100 100 100

Note :

S

A + F

79.4 Lst + 20.6 clay

SiO2 14.09

Al2O3 4.29

Fe2O3 1.82

CaO 42.05

MgO 1.08

SO3 0.85

LOI 35.81

total 100

clinker % =100

× Raw mix100 - LOI

5. Calculation based on Lime Saturation Factor and

Silica Ratio

Example 4

to calculate a raw mix composed of limestone, slag and pyrite. The required

LSF is 0.92 and S.M is 2.5 . The analysis of raw materials are shown in table 4

let's assume that x parts of limestone will be apportioned to y parts and 1 part

of pyrite.

L.S.F = 100 C

SM =2.8 S + 1.2 A + 0.65 F

the values of C,S,A and F are inserted in the above expressions :

L.S.F =100( xC1+yC2 + C3)

2.8(xS1+yS2+S3)+1.18(xA1+yA2+A3) + 0.65 (xF1+yF2+F3)

SM =xS1+yS2+S3

(xA1+yA2+A3) + (xF1+yF2+F3)

Expliciting x and y in the above equations we get :

x [ LSF (2.8 S1 +1.18 A1 +0.65 F1) - 100C1 ] + y [ LSF (2.8 S2 +1.18 A2 +0.65 F2) - 100C2 ]

Ξ 100C3 - LSF (2.8 S3 +1.18 A3 +0.65 F3)

x [ SM (A1 + F1 ) - S1 ] + y [ SM ( A2 + F2 ) - S2 ] Ξ S3 - SM (A3 + F3 )

c1 - a1x c2 - a2x

b1 b2

c1 - b1y c2 - b2y

a1 a2

Now applying the values of SiO2, Al2O3,Fe2O3 and CaO from table 4 we get :

to simpify let's assume

a1 = LSF (2.8 S1 +1.18 A1 +0.65 F1) - C1 a2 = SM (A1 + F1 ) - S1

b1 = LSF (2.8 S2 +1.18 A2 +0.65 F2) - C2 b2 = SM ( A2 + F2 ) - S2

c1 = C3 - LSF (2.8 S3 +1.18 A3 +0.65 F3) c2 = S3 - SM (A3 + F3 )

then the system of two equations quated above will take the following form

a1x +b1y = C1 y = .=

a2x +b2y = C2 x = .=

x = c1b2 - b1c2

y = a1c2 - a2c1

a1b2 - a2b1 a1b2 - a2b1

a1 = 0.92 ( 2.8 x 6.75 +1.18 x 0.71 +0.65 x 1.47 ) - 49.80 = - 30.762164

b1 = 0.92 ( 2.8 x 39.45 +1.18 x 9.67 +0.65 x 0.67 ) - 42.09 = 70.431612

c1 = 0.87 - 0.92 ( 2.8 x 11.21 +1.18 x 1.57 +0.65 x 83.72 ) = - 79.775912

a2 = 2.5 (0.71 + 1.47 ) - 6.75 = - 1.3

b2 = 2.5 (9.67 + 0.67 ) - 39.45 = - 13.6

c2 = 11.21 - 2.5 (1.57 + 83.72 ) = - 13.7

x = c1b2 - b1c2

.=15313.1945

.= 30.03019792a1b2 - a2b1 509.926526

y = a1c2 - a2c1

.=6110.709875

.= 11.98351045a1b2 - a2b1 509.926526

limestone slag pyrite

69.8154% 27.8597% 2.3249%

Table 4constituent

1 2 3 4 5 6 7 8

0.69815 0.27859 0.02324

L.stone slag pyrite

constituent L.stone slag pyrite raw mix clinker

SiO2 6.75 39.45 11.21 4.71 10.99 0.26 15.96 22.07

Al2O3 0.71 9.67 1.57 0.50 2.69 0.04 3.23 4.47

Fe2O3 1.47 0.67 83.72 1.03 0.19 1.95 3.17 4.38

CaO 49.80 42.09 0.87 34.77 11.73 0.02 46.52 64.33

MgO 1.48 7.36 0.64 1.03 2.05 0.01 3.09 4.27

SO3 0.10 0.70 1.36 0.07 0.20 0.03 0.30 0.42

LOI 39.65 0.00 0.63 27.68 0.00 0.01 27.69 0.00

rest 0.04 0.06 0.00 0.00 0.00 0.00 0.04 0.06

LSF 0.92 0.92

SM 2.50 2.50

total 100.00 100.00 100.00 69.79 27.85 2.32 100.00 100.00

the calculated figures of the above table prove the correctness of the method of calculation

6. Calculation based on Lime Saturation Factor ,

Silica Ratio and Alumina Modulus

to calculate a raw mix composed of four raw materials, the mechanism is the same applied

to the previous problem.

x = parts of limestone , y = parts of clay , z = parts of slag or silica sand or bauxite,

apportioned to 1 part of iron ore = w.

performing the calculations we get

a1 = LSF (2.8 S1 +1.18 A1 +0.65 F1) - C1

b1 = LSF (2.8 S2 +1.18 A2 +0.65 F2) - C2

c1 = LSF (2.8 S3 +1.18 A3 +0.65 F3) - C3

d1 = C4 - LSF (2.8 S4 +1.18 A4 +0.65 F4)

a2 = SM ( A1 + F1 ) - S1

b2 = SM ( A2 + F2 ) - S2

c2 = SM ( A3 + F3 ) - S3

d2 = S4 - SM (A4 + F4 )

a3 = AM . F1 - A1

b3 = AM . F2 - A2

c3 = AM . F3 - A3

d3 = A4 - AM . F4

x =d1 ( b2c3 - b3c2 ) - d2 ( b1c3 - b3c1 ) + d3 (b1c2 - b2c1)

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

y =a1 ( d2c3 - d3c2 ) - a2 ( d1c3 - d3c1 ) + a3 (d1c2 - d2c1)

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

z =a1 ( b2d3 - b3d2 ) - a2 ( b1d3 - b3d1 ) + a3 (b1d2 - b2d1)

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

w = 1

7. Calculation based on KH , Silica Ratio

and Alumina Modulus

KH =100 ( CaO - 1.65 Al2O3 - 0.35 Fe2O3 )

2.8 SiO2

to calculate a raw mix composed of four raw materials, the mechanism is the same applied

to the previous problem.

x = parts of limestone , y = parts of clay , z = parts of slag or silica sand or bauxite,

apportioned to 1 part of iron ore = w.

performing the calculations we get

a1 = KH (2.8 S1 ) - ( C1 - 1.65 A1 - 0.35 F1 )

b1 = KH (2.8 S2 ) - ( C2 - 1.65 A2 - 0.35 F2 )

c1 = KH (2.8 S3 ) - ( C3 - 1.65 A3 - 0.35 F3 )

d1 = ( C4 - 1.65 A4 - 0.35 F4 ) - KH ( 2.8 S4 )

a2 = SM ( A1 + F1 ) - S1

b2 = SM ( A2 + F2 ) - S2

c2 = SM ( A3 + F3 ) - S3

d2 = S4 - SM (A4 + F4 )

a3 = AM . F1 - A1

b3 = AM . F2 - A2

c3 = AM . F3 - A3

d3 = A4 - AM . F4

x =d1 ( b2c3 - b3c2 ) - d2 ( b1c3 - b3c1 ) + d3 (b1c2 - b2c1)

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

y =a1 ( d2c3 - d3c2 ) - a2 ( d1c3 - d3c1 ) + a3 (d1c2 - d2c1)

w = 1

y =a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )

z =a1 ( b2d3 - b3d2 ) - a2 ( b1d3 - b3d1 ) + a3 (b1d2 - b2d1)

a1 (b2c3 - b3c2 ) - a2 (b1c3 - b3c1 ) + a3 ( b1c2 - b2c1 )