Composite Washing of Coals from Multiple Sources dr kalyan sen, Director, Central Fuel Research Institute, Dhanbad, 2001 Composite Washing of Coals from Multiple Sources dr kalyan sen, Director, Central Fuel Research Institute, Dhanbad, 2001

Actual Problem Washery Engineer Management Customer More Quantity Best Quality

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

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

Types of Problem  Linear  Non-linear  Unconstrained  Constrained  Multi level multi variable

Problems in the Coal preparation plants

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

Variation in product  Quality  Quantity Resulting

How it is solved presently ?

Existing Procedure Graphical method can handle :  Upto three coals or sizes at a time  Requires skilled personnel  Time consuming

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)

What to do ?

Problem Sp. Gr.  2 Ratio R 2 Feed 1 Feed 2 Ratio R 1 Sp. Gr.  1 Overall Y, A

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

Solution

From material balance, Y (m 1 +m 2 ) = Y 1 m 1 + Y 2 m 2 Y 1 m 1 + Y 2 m 2 Y = (1) (m 1 +m 2 ) From ash balance, Y A (m 1 +m 2 ) = Y 1 A 1 m 1 + Y 2 A 2 m 2 (2) Y 1 A 1 m 1 + Y 2 A 2 m 2 A = (3) Y 1 m 1 + Y 2 m 2

when the ratio of two coals are given and quality of overall product is fixed, we may assume that, m 1 and m 2 = constant and A = constant we are to find the relation between Y 1 and Y 2 so that the overall yield Y is maximum. L = (Y 1 m 1 + Y 2 m 2 ) A - Y 1 A 1 m 1 - Y 2 A 2 m 2 = 0 …… (4)

using differential equations as per Lagrange’s theorem for maximization, dY / dY 1 +  dL / dY 1 = 0 dY / dY 2 +  dL / dY 2 = 0 where,  = Lagrangian multiplier

Differentiating equations (1) & (4) and substituting the derivatives, we get, {1/(m 1 +m 2 )} +  [A - (d(Y 1 A 1 )/dY 1 ] = 0 {1/(m 1 +m 2 )} +  [A - (d(Y 2 A 2 )/dY 2 ] = 0 But we know that, d(Y 1 A 1 )/dY 1 = 1 d(Y 2 A 2 )/dY 2 = 2

Thus, {1/(m 1 +m 2 )} +  [A - 1 ] = 0 {1/(m 1 +m 2 )} +  [A - 2 ] = 0 or, 1 = 2 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.

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

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

Methodology for Optimization  Develop equation on the basic concept (F&S test)  Mathematical formulation of optimization problem  Selection of necessary condition for optimization

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

Limit for Characteristic Ash Basis Petrographic study Coke property limiting value 26-28%

Composite washing of Feeds Char. Ash vs. Relative density

Comparison with coals from western origin

CHAR..ASH LEVEL ?,,,, ! ! ! ! ! ! Relative density Cum. Ash% Char.Ash% Western coal vs. Indian coal Indian COAL WESTERN COAL 26-28% CLEANS ASH 17.0% 9.0% cum.Ash