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Composite Washing of Coals fromComposite Washing of Coals fromMultiple SourcesMultiple Sources
dr kalyan sen, Director,dr kalyan sen, Director,Central Fuel Research Institute,Central Fuel Research Institute,
Dhanbad, 2001Dhanbad, 2001
what is optimizationwhat is optimization ? ?
To maximize or minimize the objective functionTo maximize or minimize the objective function
Objective Function
Maximize
or
Minimize
Max --- Yield, Quantity, Profit, etc.Max --- Yield, Quantity, Profit, etc.
Min --- Undesired product, Loss, Production cost, etcMin --- 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)
Problems
Variation of raw coal sources
Deterioration of feed coals
Evaluation of Wash. Charac. & Size distribution
Frequent variation in feed
Control of Individual Eqpt.
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)
Two Coals -
feed proportion - m1 : m2
Yield of coals - Y1 and Y2
Overall ash content - A1 and A2
Elementary ash content - and
Cumulative Yield & Ash of two coals
1.35 1.45 1.55 1.65 1.75 1.85 1.95 2.05 2.15
Relative density
0
20
40
60
80
100Cumulative % (Yield or Ash)
0
20
40
60
80
100
Coal 1 Yield Coal 2 Yield Coal 1 Ash Coal 2 Ash
From material balance,
Y (m1 +m2) = Y1 m1 + Y2 m2
Y1 m1 + Y2 m2 Y = --------------------- (1) (m1 +m2)
From ash balance,
Y A (m1 +m2) = Y1 A1 m1 + Y2 A2 m2 (2)
Y1 A1 m1 + Y2 A2 m2 A = ------------------------------ (3) Y1 m1 + Y2 m2
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)
using differential equations as per Lagrange’s theorem for maximization,
dY / dY1 + dL / dY1 = 0
dY / dY2 + dL / dY2 = 0
where, = Lagrangian multiplier
Differentiating equations (1) & (4) and
substituting the derivatives, we get,
{1/(m1+m2)} + [A - (d(Y1 A1)/dY1] = 0
{1/(m1+m2)} + [A - (d(Y2 A2)/dY2] = 0
But we know that,
d(Y1 A1)/dY1 = 1
d(Y2 A2)/dY2 = 2
Thus,
{1/(m1+m2)} + [A - 1] = 0
{1/(m1+m2)} + [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.
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
ConceptCFRI’s Publication
Composite washing of coals from multiple sources :
Optimization by numerical technique
Int. J. Mineral Processing, 41 (1994) 147-160.
Limit for Characteristic AshLimit for Characteristic Ash
Basis Petrographic study
Coke property
limiting value 26-28%
Composite washing of FeedsChar. Ash vs. Relative density
1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2
Relative density
0
10
20
30
40
50
60
70
80
Ch
ar.
Ash
%
1.52 1.
63