Hadi - Unsolved - Solved [PDF]

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1. The fuel cost functions for two 800 MW thermal plants in



C1 =400+6.0 P1 +0.004 P



per hour are given by



2 1



C2 =500+ P2+ P 22 where P1, and P2 are in MW. a) The incremental cost of power  is 8/MWh, when the total power demand is 550 MW. Neglecting losses, determine the optimal generation of each plant. b) The incremental cost of power  is 10/MWh, when the total power demand is 1300 MW. Neglecting losses, determine the optimal generation of each plant. c) From the results of a) and b), find the fuel cost coefficients  and  of the second plant. Solution a) The incremental cost of power  is 8/MWh, when the total power demand is 550 MW. P1 + P2 = PD  550 MW The optimal dispatch



P 1=



−❑1 8−6 = =250 MW 2❑1 2× 0.004



Therefore P2 = PD – P1 = 550 – 250 = 300 MW



P2=



−❑2 8−❑2 = =300 MW 300× 2❑2 =8−❑2 600❑2=8−❑2 2❑2 2❑2



b) The incremental cost of power  is is 1300 MW. P1 + P2 = PD  1300 MW The optimal dispatch



P 1=



10/MWh, when the total power demand



−❑1 10−6 = =500 MW 2❑1 2× 0.004



Therefore P2 = PD – P1 = 1300 – 500 = 800 MW



P 2=



−❑2 10−❑2 = =800 MW 1600 ❑2 =10−❑2 2❑2 2❑2



c) From a) and b)



600❑2=8−❑2 o r 600❑2 +❑2 =8 1600❑2=10−❑2∨1600❑2+❑2=10



Solving for 2 and 2



[



][ ] [ ]



600 1 ❑2 = 8 1600 1 ❑2 10



gives 2 = 6.8 and 2 =0.0020 2. The fuel cost functions for three thermal plants in



C1 =350+7.2 P1 +0.0040 P



per hour are given by



2 1



C2 =500+7.3 P 2+0.0025 P22 2



C3 =600+6.74 P3 +0.0030 P3 where P1, P2 and P3 are in MW. The governors are set such that the generation share the load equally. Neglecting line losses and generator limits, find the total cost in per hour, when the total load is a) PD = 450 MW b) PD = 745 MW c) PD = 1335 MW Solution a) PD = 450 MW



P1=P2=P3 =



450 =150 MW 3



The total fuel cost



Ct =C 1+C 2 +C3



Ct =350+ ( 7.2 ×150 ) + ( 0.004 ×1502 ) +500+ ( 7.3 ×150 )+ ( 0.0025× 1502 ) +600+ ( 6.74 × 150 ) + ( 0.003 ×1502 ) = 4849.75 /hour b) PD = 745 MW



P1=P2=P3 =



745 MW 3



The total fuel cost



(



Ct =350+ 7.2×



Ct =C 1+C 2 +C3 =¿



)(



2



( ))



745 745 + 0.004 × 3 3



7310.46 /h



(



+500+ 7.3×



c) PD = 1335 MW



P1=P2=P3 =



1335 =445 MW 3



The total fuel cost



Ct =C 1+C 2 +C3



)(



2



( ) )+600+(6.74 × 7453 )+(0.00



745 745 + 0.0025 × 3 3



Ct =350+ ( 7.2 × 455 ) + ( 0.004 × 4552 ) +500+ ( 7.3 × 455 ) + ( 0.0025 × 4552 ) +600+ ( 6.74 × 455 ) + ( 0.003× 4552 3. Neglecting line losses and generator limits, determine the optimal scheduling of generation for each loading condition in Problem 2 i) by analytical technique ii) by using iterative method. Start with an initial estimate of  = 7.5 per MWh. iii) Find the savings in per hour for each case compared to the costs in Problem 2 when the generators shared load equally. Solution a) i) PD = 450 MW 3



P D +∑ ¿



i=1



βi 2 γi



3



∑ 21γ i=1 i



7.2 7.3 6.74 + + 2× 0.004 2× 0.0025 2× 0.003 1 1 1 + + 2× 0.004 2× 0.0025 2 ×0.003



450+ =



= 8 /MWh



The optimal dispatch



P 1= P 2=



−β1 8−7.2 = =100 MW 2× γ 1 2 ×0.004



−β2 8−7.3 = =140 MW 2× γ 2 2 ×0.0025



P3=



−β3 8−6.74 = =210 MW 2× γ 3 2 ×0.003



ii) PD = 745 MW 3



P D +∑ ¿



i=1



3



βi 2 γi



∑ 21γ i=1 i



7.2 7.3 6.74 + + 2 ×0.004 2 ×0.0025 2 × 0.003 1 1 1 + + 2× 0.004 2 × 0.0025 2× 0.003



745+ =



The optimal dispatch



P1= P 2=



−β1 8.6−7.2 = =175 MW 2× γ 1 2 ×0.004



−β2 8.6−7.3 = =260 MW 2× γ 2 2 ×0.0025



P 3= iii) PD = 1335 MW



−β3 8.6−6.74 = =310 MW 2× γ 3 2 ×0.003



= 8.6 /MWh



3



P D +∑ i=1



¿



3



βi 2 γi



∑ 21γ i=1 i



7.2 7.3 6.74 + + 2 ×0.004 2 ×0.0025 2 ×0.003 1 1 1 + + 2× 0.004 2 × 0.0025 2× 0.003



1335+ =



= 9.8 /MWh



The optimal dispatch



P 1= P 2= P 3=



−β1 9.8−7.2 = =325 MW 2× γ 1 2 ×0.004



−β2 9.8−7.3 = =500 MW 2× γ 2 2 ×0.0025



−β3 9.8−6.74 = =510 MW 2× γ 3 2 ×0.003 (1)



❑ =7.5 /MWh



b) i) The initial value of The optimal dispatch



P1(1)= P2(1)= P3(1)=



−β 1 7.5−7.2 = =37.5000 MW 2 ×γ 1 2 ×0.004



−β 2 7.5−7.3 = =40.0000 MW 2 ×γ 2 2 × 0.0025



−β 3 7.5−6.74 = =126.6666 MW 2 × γ 3 2× 0.003



PD = 450 MW, the error P 3



∆ P =PD −∑ Pi=450− ( 37.5+ 40+126.6666 )=245.8333 (1 )



i=1



(1)



❑ =



P(1) 245.8333 245.8333 = = =0.5 3 1 1 1 491.6666 1 ∑ 2❑ 2 ×0.004 + 2× 0.0025 + 2 ×0.003 i=1 i



The new value of  is (2) = (1) + (1) = 7.5 + 0.5 = 8 /MWh Continuing the process for second iteration (2 )



(2)



P1 =



❑ −β 1 8−7.2 = =100 MW 2 ×γ 1 2 ×0.004 (2)



P2(2)=



❑ −β 2 8−7.3 = =140 MW 2 ×γ 2 2 ×0.0025



(2)



P



❑ −β 3 8−6.74 = = =210 MW 2 ×γ 3 2 × 0.003



(2) 3



3



∆ P =PD −∑ Pi =450−( 100+ 140+210 )=0 (2 )



i=1



Since P(2) = 0, the equality constraint is met in two iterations. ii)



INCOMPLETE http://www.academia.edu/8232987/Power_System_Analysis_Hadi_Saadat_Sol ution_manual



4. The fuel cost functions for two thermal plants in



C1 =320+6.2 P1 +0.004 P



per hour are given by



2 1 2



C2 =200+6.0 P2+ 0.003 P2 where P1, and P2 are in MW. The total load, PD is 180 MW. Plant output are subject to the following limits in MW 50 < P1 < 250 50 < P2 < 350 The per unit system real power loss with generation expressed in per unit on a 100 MVA base is given by PL(pu) = 0.0125 P12(PU) + 0.00625 P22(PU) The total load is 412.35 MW. Determine the optimal dispatch of generation. Start with an initial estimate of  = 7 per MWh. Solution



(



PL( MW)= 0.0125



P1 2 P2 +0.00625 100 100



( )



2



( ) ) x 100 MW



PL( MW)=0.000125 P21 +0.0000625 P22 MW ❑1 = 7 per MWh. −❑1 7−6.2 P11= = =82.0513 MW 2 ( ❑1+ B11 ) 2 ( 0.004+7 × 0.000125 ) 1



P 2=



−❑2 2 ( ❑2+ B22 )



=



7−6.0 =145.4545 MW 2 ( 0.003+7 × 0.0000625 )



The real power loss



P1L =0.000125 ( 82.0513 )2 +0.0000625 ( 145.4545 )2=2.1639 Since PD =412.35 MW, the error P1 = 412.35 + 2.1639 – 82.0513 – 145.4545 = 187.0080 2



∑( i=1



2



∑( i=1



Pi 1 ❑1+ B 11 ❑1 ❑2+ B 22 ❑2 = + 2 2 ❑ 2 (❑1 +B 11 ) 2 (❑2 +B 22)



)



Pi 1 0.004+0.000125 ×6.2 0.003+0.0000625 ×6 = + =243.2701 2 2 ❑ 2 ( 0.004+7 × 0.000125 ) 2 ( 0.003+ 7 ×0.0000625 )



)



❑1 =



1



P 187.0080 = =0.7687 1 2 243.2701 Pi ∑ ❑ i=1



( )



Therefore, the new value of , 2 = 1 + 1 = 7 + 0.7687 = 7.7687 7 per MWh. Continuing the process, 2



2 1



P=



2 2



P=



❑ −❑1 2



2 ( ❑1+❑ B11 ) ❑2−❑2 2



2 ( ❑2+❑ B22 )



=



7.7687−6.2 =157.7824 MW 2 ( 0.004+7.7687 × 0.000125 )



=



7.7687−6.0 =253.7194 MW 2 ( 0.003+7.7687 × 0.0000625 )



The real power loss 2



2



2



PL =0.000125 ( 157.7824 ) +0.0000625 ( 253.7194 ) =7.1353 P2 = 412.35 + 7.1353 – 157.7824 – 253.7194 = 7.9835 2



∑( i=1



2



∑( i=1



Pi 2 ❑1+ B 11 ❑1 ❑2+ B 22❑2 = + 2 2 ❑ 2 ( ❑1 +❑2 B11 ) 2 (❑2 +❑2 B 22 )



)



Pi 2 0.004+0.000125 ×6.2 0.003+0.0000625 ×6 = + =235.5143 2 2 ❑ 2 ( 0.004+7.7687 × 0.000125 ) 2 ( 0.003+ 7.7687× 0.0000625 )



)



❑2=



2



P 7.9835 = =0.0339 2 2 235.5143 Pi ∑ ❑ i=1



( )



Therefore, the new value of , 3 = 2 + 2 = 7.7687 + 0.0339 = 7.8026 per MWh 3 1



P=



❑3−❑1 3



2 ( ❑1+❑ B11 )



=



7.8026−6.2 =161.0548 MW 2 ( 0.004+ 7.8026× 0.000125 )



=



7.8026−6.0 =258.4252 MW 2 ( 0.003+ 7.8026× 0.0000625 )



3



3 2



P=



❑ −❑2 3



2 ( ❑2 +❑ B22 )



The real power loss 3



2



2



PL =0.000125 ( 161.0548 ) +0.0000625 ( 258.4252 ) =7.4163 P3 = 412.35 + 7.4163 – 161.0548 – 258.4252 = 0.2863 2



∑( i=1



2



∑( i=1



Pi 3 ❑1 +B 11 ❑1 ❑2 +B 22❑2 = + 2 2 ❑ 2 ( ❑1 +❑3 B 11 ) 2 (❑2 +❑3 B 22 )



)



3



Pi 0.004+0.000125 ×6.2 0.003+0.0000625 ×6 = + =235.1810 2 2 ❑ 2 ( 0.004+7.8026 × 0.000125 ) 2 ( 0.003+ 7.8026× 0.0000625 )



)



❑3 =



P3 0.2863 = =0.0012 3 2 235.1810 Pi ∑ ❑ i=1



( )



Therefore, the new value of , 4 = 3 + 3 = 7.7687 + 0.0012 = 7.8038 per MWh Since 3 is small, the equality constraint is met in four iterations, and optimal dispatch for  = 7.8038 per MWh are 4 1



P=



4 2



P=



❑4 −❑1 4



2 (❑1 +❑ B11 ) ❑4−❑2 4



2 (❑2 +❑ B22 )



=



7.8038−6.2 =161.1705 MW 2 ( 0.004+ 7.8038× 0.000125 )



=



7.8038−6.0 =258.5917 MW 2 ( 0.003+ 7.8038× 0.0000625 )



The real power loss



1



2



2



PL =0.000125 ( 161.1705 ) +0.0000625 ( 258.5917 ) =7.4263 and the total fuel cost



Ct =C 1+C 2=320+ 6.2 P1 +0.004 P21 +200+6.0 P2 +0.003 P22 2



Ct =320+6.2 ×161.1705+0.004 × ( 161.1705 ) +200+6.0 ×258.5917+ 0.003× ( 258.5917 )



Ct =3375.3



¿



h



2