PRESENTATION ON

UNIT COMMITMENTSubmittd by : Debasish ChoudhuryPRESENTATION ONOUTLINEWhy Unit Commitment ?What is Unit COMMITMENT ?Difference between Economic Load Dispatch and Unit CommitmentUnit Commitment procedure Solving Unit Commitment Problem Why unit Commitment?What is Unit CommitmentUnit commitment plans for best set of units to be available to supply the predicted or forecast load of the system over a future time period.

Unit Commitment is therefore, one way to suggest, just sufficient number of generating units with sufficient amount of generating capacity to meet a given load economically with sufficient reserve capacity to meet any abnormal, operating condition.

Here we consider the problem of scheduling fossil fired thermal units in which the aggregate costs( such as start up cost, operating fuel costs & shut down costs ) are to be minimized over a daily load cycle.

Difference between Economic Load Dispatch And Unit CommitmentEconomic Load Dispatch Its a short term determination. Dispatch at lowest possible cost

Unit CommitmentUnit commitment aims to make power system reliable

5CONSTRAINTSSpinning reserve: It makes up the loss of the most heavily loaded unit in a given period of time.

Thermal Unit Constraint:Minimum Up TimeMinimum down timeCrew constraintstart-up cost

Must-run: Some units are given this status

Fuel constraint

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

Priority-list scheme: the most efficient unit isloaded first

Dynamic Programming (DP):Forward DP approachBackward DP approach

Mixed Integer Linear ProgrammingLets postulate the following situation:A loading pattern must be established for M periodsThere are N units to commitAny one unit or a combination of units can supply theloads.The total number of combinations to try each hour is C (N, 1) + C (N, 2) + + C (N, N-1) + C (N, N) = 2N1C (N, j) is the combination of N items taken j at a time.Maximum number of possible combinations is (2N-1) MSolution methods If we have K generating units(no two identical), then there are (2k 1) number of combinations.

For example if K=4, then there are 15 theoretically possible combinations.

Assumptions:The transmission line loss in the system is disregarded.PgT=PD PgT = Total plant output PD = Total system load

Prerequisites: Load forecasting

Xi(k) = combination Xi of interval k

Pi(k) = minimum production cost of combination Xi(k)

Tij(k) = cost of transition from combination Xi(k) to combination Xj(k+1) between intervals k and k+1

Fij(k) = Pi(k) + Tij(k) Fij(k) = The cost associated with any stage kUnit Commitment Procedure:

Dynamic ProgrammingWe use Dynamic programming to solve Unit Commitment problem.

Here we take an iterative relation embodying the principle that starting with given combination Xi* at stage k, the minimum unit commitment cost is found by minimizing the sum of the current single-stage cost fij(k) plus the minimum cumulative cost Fij(k+1) over the later stages of study.

This is one example of the principle of optimality, which states: if possible path from A to C passes through intermediate point B , then the best possible path from B to C must be corresponding part of the best path from A to C.

Computationally, we evaluate one decision at a time beginning with the final stage N and carry the minimum cumulative cost function backward in time to stage k to find the minimum cumulative cost Fi*(k) for the feasible combination.

The problem is to find out the minimum cost from Ato N At the terminal of eachstage there is a set ofchoices of nodes {Xi} tobe chosen

The symbol Va (Xi,Xi+1) represents the costof traversing stage a(=1V)Cont.The minimum cumulative cost decisions are recovered as we sweep from stage 1 to stage N searching through the tables already calculated for each stage. This computational procedure is known as Dynamic Programming.2

It involves two sweeps through each stage k.

In the first sweep, which is computationally intensive, we work backward computing and recording for each candidate combination xi of stage k the minimum Fi(k) and its associated xj(k+1).

The second sweep in the forward direction does not involve any processing since with xi*(k) identified we merely enter the table of results already recorded to retrieve the value Fi*(k) and its associated combination xj(k+1), which becomes xi*(k+1) as we move to next forward stage.

fI(X1) : Minimum cost for the 1ststage 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+ ] =10X1 =B =C =DfII(F) = min [, 6, 9] = 6, X1 = CfII(G) = min [, 11, 9] = 9,X1 = D

(X2) E F GfII (X2) 10 6 9Path X0X1 AC AC AD

Tracing back, the path of minimum cost is found asfollows:Stage {Xi} fi1 B, C, D 5, 2, 32 E, F, G 10, 6, 93 H, I, J, K 13, 12, 11, 134 L, M 15, 185 N 19Backward 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)]whereFcost (K, I) = minimum total fuel costPcost (K, I) = minimum generation costScost (I, K: J, K+1) = incremental start-up cost.{J} = set of feasible states in interval K+1.

Forward DP ApproachThe 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)]

whereFcost (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.

CONCLUSIONBy optimal scheduling of generating units, we can save time, power and costImportant for industrial applicationDynamic programming method gives a reliable solutionTHANK YOU