National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [1] UNIT COMMITMENT Under the guidance of Mr. Debasisha Jena Presented by Dipanwita Dash Roll # EE

National Institute of Science & Technology 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

National Institute of Science & Technology TECHNICAL SEMINAR-2004 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 UC There should be least operating cost This problem is important for thermal plants

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [4] CONSTRAINTS Spinning 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

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [5] Lets postulate the following situation: A loading pattern must be established for M periods There are N units to commit Any 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) = 2 N –1 C (N, j) is the combination of N items taken j at a time. Maximum number of possible combinations is (2 N -1) M SOLUTION METHODS

National Institute of Science & Technology 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)

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [7] Backward DP Approach: The solution starts at the last interval and proceeds back the initial point F cost (K, I) = Min [P cost (K, I) + S cost (I, K: J,K+1) + F cost (K+1,J)] where F cost (K, I) = minimum total fuel cost P cost (K, I) = minimum generation cost S cost (I, K: J, K+1) = incremental start-up cost. {J} = set of feasible states in interval K+1.

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [8]

National Institute of Science & Technology 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 F cost (K, I) = Min [P cost (K, I) + S cost (K-1, L: K, I) + F cost (K-1, L)] where F cost (K, I) =least total cost to arrive at state (K, I) P cost (K, I) = production cost for state (K, I). S cost (K-1, L: K, I) = transition cost for state (K-1, L) to state (K, I) where state (K, I) is the I th combination in hour K.

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [10]

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [11] EXAMPLE OF DP The 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 {X i } to be chosen The symbol V a (X i, X i +1) represents the cost of traversing stage a (=1…V)

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [12] f I (X 1 ) : Minimum cost for the 1st stage is obvious : f I (B) : V I (A, B) = 5. f I (C) : V I (A, C) = 2. f I (D) : V I (A, D) = 3. f II (E)= min [f I (X 1 ) + V II (X 1, E)] {X 1 } = min [5+11, 2+8, 3+  ] =10 X 1 =B =C =D f II (F) = min [ , 6, 9] = 6, X 1 = C f II (G) = min [ , 11, 9] = 9,X 1 = D

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [13] (X 2 ) E F G f II (X 2 ) Path X 0 X 1 ACACAD Tracing back, the path of minimum cost is found as follows: Stage{X i } f i 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

National Institute of Science & Technology TECHNICAL SEMINAR-2004 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

National Institute of Science & Technology TECHNICAL SEMINAR-2004 Dipanwita Dash [15]