Economic dispatch dengan iterasi lamda secara manual dan mathlab
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This document provides information about three power generators (C1, C2, C3) including their minimum and maximum outputs and fuel costs. It then shows the calculations to determine the optimal dispatch of the three generators to meet a total load of 500 MW in the cheapest way, by minimizing total fuel costs. The manual calculations determine the optimal dispatch is 229.25 MW for generator C1, 73.28 MW for generator C2, and 197.11 MW for generator C3. A program is then used to verify the same optimal dispatch.
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This document provides specifications and information for components of an excavator. It includes:
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- 1. R A N D I A P A N D I 4 4 2 1 6 0 2 4
- 2. CARA MANUAL
- 3. KARAKTERISTIK PEMBANGKIT INPUT - OUTPUT Min (MW) Max (MW) Fuel Cost C1 = 225 + 8.4P1 + 0.0025P12 45 350 0.8 C2 = 729 + 6.3P2 + 0.0081P22 45 350 1.02 C3 = 400 + 7.5P3 + 0.0025P32 47.5 450 0.9
- 4. 𝑑𝑐𝑖 𝑑𝑝𝑖 = 2γ𝑖𝑃𝑖 + β𝑖 • 𝑑𝑐1 𝑑𝑝1 = 2γ1 𝑃1 + β1 = λ 𝑑𝑐2 𝑑𝑝2 = 6.426 + 0.016524 = λ • 𝑑𝑐2 𝑑𝑝2 = 2γ2 𝑃2 + β2 = λ 𝑑𝑐1 𝑑𝑝1 = 6.72 + 0.004𝑃1 = λ 𝑑𝑐3 𝑑𝑝3 = 6.75 + 0.0045 = λ • 𝑑𝑐3 𝑑𝑝3 = 2γ3 𝑃3 + β3 = λ
- 5. 𝜆 = 𝑃𝑑+ 𝑖=1 𝑛 𝛽𝑖 2𝛾𝑖𝑃𝑖 1 2𝛾𝑖𝑃𝑖 𝜆 = 500+ 6.72 0.004 + 6.426 0.016524 + 6.75 0.0045 1 0.004 + 1 0.016524 + 1 0.0045 Total Beban = 500 MW 𝜆 = 500+1680+388.89+1500 250+60.518+222.22 𝜆 = 4068.89 532.738 𝜆 =7.637 $/MWh
- 6. Pembagian Pembebanan 𝑃𝑖 = 𝜆 − 𝛽𝑖 2𝛾𝑖𝑃𝑖 • 𝑃1 = 𝜆− 𝛽1 2𝛾1 𝑃1 • 𝑃3 = 𝜆− 𝛽3 2𝛾3 𝑃3 • 𝑃2 = 𝜆− 𝛽2 2𝛾2 𝑃2 = 7.637 − 6.72 0.004 = 229.25 𝑀𝑊 = 7.637 − 6.426 0.016524 = 7.637 − 6.75 0.0045 = 197.11 𝑀𝑊= 73.28 𝑀𝑊 P1 = 229.25 𝑀𝑊 P2 = 73.28 𝑀𝑊 P3 = 197.11 𝑀𝑊
- 7. PROGRAM
- 8. Script
- 9. Script eld
- 10. HASIL
- 11. PERBANDINAGN CARA MANUAL DAN PROGRAM P1 = 229.25 𝑀𝑊 P2 = 73.28 𝑀𝑊 P3 = 197.11 𝑀𝑊 Program Manual
- 12. T H A N K Y O U !