Bogues Equation - Calculation and Derivation

PPT Suranjan

1. Bogue’s Equations - Calculation and Derivation

2. The Bogue calculation is used to calculate the approximate proportions of the four main minerals in Portland cement clinker. The standard Bogue calculation refers to clinker, rather than cement, but it can be adjusted for use with cement.

3. Name of Compound Formula Abbreviation Percentage Tricalcium Silicate 3 CaOSiO2 C3S 30-55 Dicalcium Silicate 2CaOSiO2 C2S 20-45 Tricalcium Aluminate 3CaOAl2O3 C3A 05-12 Tetracalcium Aluminoferrite 4CaO.Al2O3.Fe2O3 C4AF 06-15 The calculation assumes that the four main clinker minerals are pure minerals with compositions :

4. BOGUE’S EQUATIONS C 3 S = 4.07 CaO - 7.60 SiO 2 - 1.43 Fe 2 O 3 - 6.72 Al 2 O 3 C 2 S = 2.86 SiO 2 - 0.75 C 3 S C 3 A = 2.65 Al 2 O 3 - 1.69 Fe 2 O 3 C 4 AF = 3.04 Fe 2 O 3

5. SOME ASSUMPTIONS IN BOGUE’S CALCULATION

6. According to the assumption, C4AF is the only mineral to contain iron. The iron content of the clinker therefore fixes the C4AF content. The C3A content is fixed by the total alumina content of the clinker, minus the alumina in the C4AF phase. This can now be calculated, since the amount of C4AF has been calculated.

7. It is assumed that all the silica is present as C2S and the next calculation determines how much lime is needed to form C2S from the total silica content of the clinker. There will be a surplus of lime. The lime surplus is allocated to the C2S, converting some of it to C3S.

8. MOLECULAR WEIGHT OF OXIDES Oxides Abbreviation Molecular Weight CaO C 56.08 SiO2 S 60.08 Al2O3 A 101.96 Fe2O3 F 159.69

9. Bogue’s Compound Abbreviation Molecular Weight 3CaO.SiO2 C3S 228.32 2CaO.SiO2 C2S 172.24 3CaO.Al2O3 C3A 270.20 4Cao.Al2O3.Fe2O3 C4AF 485.97 MOLECULAR WEIGHT OF BOGUES COMPOUNDS

10. TOTAL SILICA CONTENT  

11. TOTAL ALUMINA CONTENT  

12. TOTAL IRON CONTENT  

13. TOTAL LIME CONTENT LIME (CaO) is available in all C3S, C2S, C3A and C4AF 3 CaO in C3S 2 CaO in C2S 3 CaO in C3A 4 CaO in C4AF

14. TOTAL LIME CONTENT  

16. Solving SiO 2  = 0.2631 x C 3 S + 0.3488 x C 2 S we get C 2 S = (SiO 2  – 0.2631 x C 3 S ) / 0. 3488 C 2 S = 2.867 x SiO 2  – 0.7544 x C 3 S

17. Solving SiO 2  = 0.2631 x C 3 S + 0.3488 x C 2 S we get C 2 S = (SiO 2  – 0.2631 x C 3 S ) / 0. 3488

18. Now solving Fe 2 O 3  = 0.3286 x C 4 AF we get C 4 AF = Fe 2 O 3 / 0.3286 C 4 AF = 1 / (0.3286 x Fe 2 O 3 ) C 4 AF = 3.043 x Fe 2 O 3

19. Now solving Fe 2 O 3  = 0.3286 x C 4 AF we get C 4 AF = Fe 2 O 3 / 0.3286 C 4 AF = 1 / (0.3286 x Fe 2 O 3 ) C 4 AF = 3.043 x Fe 2 O 3

20. Now substituting C 4 AF = 3.043 x Fe 2 O 3 in Al 2 O 3  = 0.3774 x C 3 A + 0.2098 x C 4 AF we get Al 2 O 3  = 0.3774 x C 3 A + 0.2098 x (3.043 x Fe 2 O 3 ) Al 2 O 3  = 0.3774 x C 3 A + 0.6384 x Fe 2 O 3 C 3 A = (Al 2 O 3  – 0.6384 x Fe 2 O 3 ) / 0.3774 C 3 A = 2.650 x Al 2 O 3  – 1.692 x Fe 2 O 3

21. Now substituting C 4 AF = 3.043 x Fe 2 O 3 in Al 2 O 3  = 0.3774 x C 3 A + 0.2098 x C 4 AF we get Al 2 O 3  = 0.3774 x C 3 A + 0.2098 x (3.043 x Fe 2 O 3 ) Al 2 O 3  = 0.3774 x C 3 A + 0.6384 x Fe 2 O 3 C 3 A = (Al 2 O 3  – 0.6384 x Fe 2 O 3 ) / 0.3774 C 3 A = 2.650 x Al 2 O 3  – 1.692 x Fe 2 O 3

22. Finally, substituting the values of C 2 S ,C 3 A and C 4 AF  in 𝑪𝒂𝑶=𝟎.𝟕𝟑𝟔𝟗 ×𝑪𝟑𝑺 + 𝟎.𝟔𝟓𝟏𝟐 ×𝑪𝟐𝑺×𝟎.𝟔𝟐𝟐𝟔 × 𝑪𝟑𝑨 + 𝟎.𝟒𝟔𝟏𝟔 ×𝑪𝟒𝑨𝑭 and solving, we get CaO = (0.7369 x C 3 S) + [0.6512 x {(2.867 x SiO 2 ) – (0.7544 x C 3 S)}] + {0.6226 x (2.650 x Al 2 O 3  – 1.692 x Fe 2 O 3 )} + (0.4616 x 3.043 x Fe 2 O 3  )

23. CaO = 0.7369 x C3S + 1.867 x SiO2 - 0.4913 x C3S + 1.6499 x Al2O3 - 1.0534 x Fe2O3 + 1.4046 x Fe2O3 CaO = 0.2456 x C3S + 1.867 x SiO2 + 1.6499 x Al2O3 + 0.3512 x Fe2O3

24. C3S = (CaO - 1.867 x SiO2 - 1.6499 x Al2O3 - 0.3512 x Fe2O3) / 0.2456 C3S = 4.07 x CaO – 7.60 x SiO2 – 6.72 x Al2O3 – 1.43 x Fe2O3

25. C3S = 4.07 x CaO – 7.60 x SiO2 – 6.72 x Al2O3 – 1.43 x Fe2O3

26. C 3 S = 4.07 CaO - 7.60 SiO 2 - 1.43 Fe 2 O 3 - 6.72 Al 2 O 3 C 2 S = 2.86 SiO 2 - 0.75 C 3 S C 3 A = 2.65 Al 2 O 3 - 1.69 Fe 2 O 3 C 4 AF = 3.04 Fe 2 O 3 Thus we get all the four Bogue’s Equations

27. It was just a Mathematical Derivation Soon be available on the same topic but with more clear understanding with the help of Pictures and Geometry and above all Narration D oubt ??? Please Ask on Comment Box