Actual Problem

1. Composite Washing of Coals fromMultiple Sourcesdr kalyan sen, Director,Central Fuel Research Institute,Dhanbad, 2001

2. Actual Problem Best Quality More Quantity Customer Management Washery Engineer

3. Maximize or Minimize Objective Function what is optimization ? To maximize or minimize the objective function Max --- Yield, Quantity, Profit, etc. Min --- Undesired product, Loss, Production cost, etc

4. Optimizationparameters Variable that governs the objective function. Like, temperature, pressure, concentration, mass, density, etc. Obj. function = function (Opt. Parameters)

5. TypesofProblem • Linear • Non-linear • Unconstrained • Constrained • Multi level multi variable

6. Problemsin theCoal preparation plants

7. Problems • Variation of raw coal sources • Deterioration of feed coals • Evaluation of Wash. Charac. & Size distribution • Frequent variation in feed • Control of Individual Eqpt.

8. Resulting • Variation in product • Quality • Quantity

9. Howitissolvedpresently ?

10. ExistingProcedure Graphical method can handle : • Upto three coals or sizes at a time • Requires skilled personnel • Time consuming

11. Limitations • Frequent evaluation & control • Practical limitation imposed from the equipment (e.g. Jig cannot be run at low gravity or HMC / HMB difficult to run at higher gravities)

12. Whattodo ?

13. Feed1 Ratio R1 Sp. Gr. d1 Overall Y, A Sp. Gr. d2 Ratio R2 Feed2 Problem

16. Question ? At what condition Overall yield (Y) is maximum at the desired quality (Ash, A)

17. Solution

18. From material balance, Y (m1 +m2) = Y1 m1 + Y2 m2 Y1 m1 + Y2 m2Y = --------------------- (1) (m1 +m2) From ash balance, Y A (m1 +m2) = Y1 A1 m1 + Y2 A2 m2 (2) Y1 A1 m1 + Y2 A2 m2A = ------------------------------ (3) Y1 m1 + Y2 m2

19. when the ratio of two coals are given and quality of overall product is fixed, we may assume that, m1 and m2 = constant and A = constant we are to find the relation between Y1 and Y2 so that the overall yield Y is maximum. L = (Y1m1 + Y2m2) A - Y1 A1 m1 - Y2 A2 m2 = 0 …… (4)

20. using differential equations as per Lagrange’s theorem for maximization, dY / dY1 + a dL / dY1 = 0 dY / dY2 + a dL / dY2 = 0 where, a = Lagrangian multiplier

21. Differentiating equations (1) & (4) andsubstituting the derivatives, we get, {1/(m1+m2)} + a [A - (d(Y1 A1)/dY1] = 0 {1/(m1+m2)} + a [A - (d(Y2 A2)/dY2] = 0 But we know that, d(Y1 A1)/dY1 = l1 d(Y2 A2)/dY2 = l2

22. Thus, {1/(m1+m2)} + a [A - l1] = 0 {1/(m1+m2)} + a [A - l2] = 0 or, l1 = l2 and at this condition a maximum overall yield can be achieved. But, it is to be noted that individual overall ash of the two coals may not be equal.

23. Conclusion Necessary condition : Equalization of elementary ash at the cut points

25. Advantage • Any number of feed coals & sizes • Any number & type of restrictions • Operable by washery supervisors • Shop floor quick decision • Accuracy & reliability of operating conditions • Very useful for modern on-line controls

26. MethodologyforOptimization • Develop equation on the basic concept (F&S test) • Mathematical formulation of optimization problem • Selection of necessary condition for optimization

27. Concept CFRI’s Publication Composite washing of coals from multiple sources : Optimization by numerical technique Int. J. Mineral Processing, 41 (1994) 147-160.

28. Basis • Petrographic study • Coke property Limit for Characteristic Ash limiting value26-28%

29. 80 70 60 50 40 Char. Ash% 30 20 10 0 1.63 1.52 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 Relative density Composite washing of Feeds Char. Ash vs. Relative density

30. Comparisonwithcoalsfrom westernorigin

31. Indian COAL WESTERN COAL 26-28% Western coal vs. Indian coal 30 80 ! ! 70 25 cum.Ash ! 60 ! 20 50 17.0% Char.Ash% Cum. Ash% CLEANS ASH 15 40 cum.Ash ! , 30 10 , ! CHAR..ASH LEVEL ? 20 9.0% , 5 , 10 0 0 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 Relative density