Question

Economic dispatch. The fuel cost function in dollars / hour of two
thermal plants is
C1=320+6.2 P1+0.004 P1^2
C2=200+6.0 P2+0.003 P2^2
where P1 and P2 are in MW. Plant outputs are subject to the following constraints:
50 ≤ P1 ≤ 250
50 ≤ P2 ≤ 350
On a 100MVA base, the per unit real power loss is
PL = 0.0125 P1^2+ 0.00625 P2^2
The total load is 412.35MW. Determine the optimal dispatch of generation.

Answer 1

☺ - 320+6.1P, +0.004842 C2 = 200 + 6-08, +0.003Pi PL :0.01257, +0.00625 P? 50 sp, ¢ 250 - 50 € P2 & 350 Power Demand= 412.35 Using method of legrange multiplier d=rl 7 dCi T1 - d P T dpi T api for Pe O PL: 2x0.0125 P = 0.025P, ap, da_ 6.1+ 0.008 ?, _ di is 1 : 1 (6.1 +0.008P,) 1 -0.025 Pi for P2 ale = 0.012512 ap2 dla = 6.0+ 0.00612 di2 1 : (6.0+0.006P2) - © 1-0.012512 Equating es ☺ and eq@ ' 6:1+0.008P, - 6.0+0.00612 1-0.025P, = -0.0125P2 r6ilto. ODBP.) 11-0.0125P2) = ( 60+0.006P2) 71-0.025 6:1-0.07625 P2 +0.008P, -0.0001P, PZ 60-0.15P, +0.006P2 - 0.00015 PP

6.1.-6.0 -0.076251,-0.006Pqf0.008P+0:15 Pi -0.00012, P+0.00015P, P2 : 0 - © -0.08225 P2 +0:1588 to 00005 P, PR-0 Also Pitp2 = 412.35 _ . fie 41235 - Pz Substituting P, in eq ② 0.1-0.08225 P + Dice141235-) +0.00005 (412-35-P2) (P)=0 -0.0822592 +65-1573 -0.158P2 +0.0206175P2.-0.0000SPP=0 -_-0.00005P2-0-21963.25P27 65.2573=0 son 0.0000SP2+0.2196325P5-65-2573-0 Solving the above quadratic equation P.?+ 4392-658-1305026= 0 D= 4392.65²- 4 (-1305026) (1) D = 24515478.0225 r= -b two and r2 = -b-ro 2a za o ri: -4392.65 + 4957-310 - 279.330 2 r2 = -4392.65- 4957310 – 4671.980 2 Neglecting the negative value - P2 = 279.330 MW . This value is within the limits of Generator P2 . P, = 412.35-279.330 in P, - 133-0196 MW Hence optimal dispatch P = 133.0196 MW 2 = 279.330 mW