LSF (Lime Saturation Factor) is the theoretical point in the C-S-A-F system where there is just enough CaO present to react completely to form C3S from C2S at 1450oC under equilibrium conditions. At this point the C2S content of the clinker would theoretically be = 0.
Therefore, one can only expect C3S, C3A and C4AF to exist on the C-S-A-F quaternary phase diagram. While this is certainly true for C3S and C4AF, it turns out that a significant amount of SiO2 is incorporated into the C3A at this point and its actual composition is closer to C3.31 AS0.39 (i.e. 3.31 moles of CaO, one mole of Al2O3 and 0.39 moles of SiO2)
Therefore, assuming that the minerals existing at the point of lime saturation are C3S, C4AF and C3.31 AS0.39 . we can say that the sum of the oxides, SiO2, Fe2O3 and Al2O3 in these minerals, expressed as a molar fraction of the CaO will be equal to unity.
Now, given that the minerals existing at the point of lime saturation are C3S, C4AF and C3.31 AS0.39 we can calculate the molar weight ratios in the respective minerals.
For C3S
For C4AF
For C3.31 AS0.39
Let the relative multipliers for SiO2, Al2O3 and Fe2O3 required for lime saturation be “a”, “b” and “c” respectively. Now, using the molar weight ratio, we can write three equations (one for each mineral) which are all true at the lime saturation point
The three equations take the form of
For C3S, (a x 0.357) + (b x 0) + (c x 0) = 1 —————–Eq(1)
For C4AF, (a x 0) + (b x 0.455) + (c x 0.714) = 1 ————- Eq(2)
For C3.31 AS0.39 , (a x 0.126) + (b x 0.549) + (c x 0) = 1 ——————-Eq(3)
Solving the equations
a = 2.8, b = 1.18 and c = 0.65
The equation becomes
Rearranging the same, we get Lime saturation factor
