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TECHNICAL SEMINAR-2004

Dipanwita Dash [1]

UNIT COMMITMENT

Under the guidance of

Mr. Debasisha Jena

Presented by

Dipanwita Dash

Roll # EE200157176

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TECHNICAL SEMINAR-2004

Dipanwita Dash [2]

INTRODUCTION

Committing a generating unit

Unequal distribution of industrial load

Problem of unit commitment in electrical power systems

The problem and methods for its solution – described in following sections

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Dipanwita Dash [3]

UNIT COMMITMENT PROBLEM

It is not economical to run all the units available all the time Optimum allocation (commitment) of generators (units) at each generating station at various load levels To determine the units of a plant that should operate for a particular load– problem of UCThere should be least operating costThis problem is important for thermal plants

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Dipanwita Dash [4]

CONSTRAINTSSpinning reserve: It makes up the loss of the most heavily loaded unit in a given period of time.

 Thermal Unit Constraint:

Minimum Up Time

Minimum down time

Crew constraint

start-up cost

Must-run: Some units are given this status

Fuel constraint

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Dipanwita Dash [5]

Lets postulate the following situation:A loading pattern must be established for M periods There are N units to commitAny one unit or a combination of units can supply the loads.

The total number of combinations to try each hour is C (N, 1) + C (N, 2) + …+ C (N, N-1) + C (N, N) = 2N–1

C (N, j) is the combination of N items taken j at a time.Maximum number of possible combinations is (2N-1) M

SOLUTION METHODS

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TECHNICAL SEMINAR-2004

Dipanwita Dash [6]

The techniques for the solution of the unit commitment problem are as follows:

Priority-list scheme: the most efficient unit is loaded first

Dynamic Programming (DP):

Forward DP approach

Backward DP approach Mixed Integer Linear Programming (MILP)

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TECHNICAL SEMINAR-2004

Dipanwita Dash [7]

Backward DP Approach:The solution starts at the last interval and proceeds back the initial point

Fcost(K, I) = Min [Pcost (K, I) + Scost(I, K: J,K+1) + Fcost(K+1,J)]

where

Fcost (K, I) = minimum total fuel cost

Pcost (K, I) = minimum generation cost

Scost (I, K: J, K+1) = incremental start-up cost. {J} = set of feasible states in interval K+1.

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TECHNICAL SEMINAR-2004

Dipanwita Dash [8]

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TECHNICAL SEMINAR-2004

Dipanwita Dash [9]

Forward DP Approach The initial conditions are easily specified Previous history of the unit can be computed at each stage Fcost (K, I) = Min [Pcost (K, I) + Scost (K-1, L: K, I) +

Fcost (K-1, L)] where Fcost (K, I) =least total cost to arrive at state (K, I) Pcost (K, I) = production cost for state (K, I). Scost (K-1, L: K, I) = transition cost for state (K-1, L)

to state (K, I) where state (K, I) is the Ith combination in hour K.

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TECHNICAL SEMINAR-2004

Dipanwita Dash [10]

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TECHNICAL SEMINAR-2004

Dipanwita Dash [11]

EXAMPLE OF DPThe problem is to find out the minimum cost from A to N At the terminal of each stage there is a set of choices of nodes {Xi} to be chosen

The symbol Va (Xi, Xi+1) represents the cost of traversing stage a (=1…V)

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Dipanwita Dash [12]

fI(X1) : Minimum cost for the 1st stage is obvious :

fI(B) : VI(A, B) = 5.

fI(C) : VI(A, C) = 2.

fI(D) : VI(A, D) = 3.

fII(E)= min [fI(X1) + VII(X1, E)]

{X1}

= min [5+11, 2+8, 3+ ] =10

X1 =B =C =D

fII(F) = min [, 6, 9] = 6, X1 = C

fII(G) = min [, 11, 9] = 9,X1 = D

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TECHNICAL SEMINAR-2004

Dipanwita Dash [13]

(X2) E F GfII (X2) 10 6 9Path X0X1 AC AC ADTracing back, the path of minimum cost is found as

follows: Stage {Xi} fi

1 B, C, D 5, 2, 3 2 E, F, G 10, 6, 9 3 H, I, J, K 13, 12, 11, 13 4 L, M 15, 18 5 N 19

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Dipanwita Dash [14]

CONCLUSION

By optimal scheduling of generating units, we can save time, power and cost

Important for industrial application

Dynamic programming method gives a reliable solution

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TECHNICAL SEMINAR-2004

Dipanwita Dash [15]