Recalculate the resulting λ and P1, P2, and P3 in Example 3.3 (Module 3.4) with the following starting values: P1 = 450
Posted: Tue May 24, 2022 9:04 am
Recalculate the resulting λ and P1, P2, and P3 in Example 3.3 (Module 3.4) with the following starting values:
P1 = 450 MW
P2 = 270 MW
P3 = 130 MW
Use a tolerance of 1 MW.
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
P1 = 450 MW
P2 = 270 MW
P3 = 130 MW
Use a tolerance of 1 MW.
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resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284
resulting in 1 = 9.5215 $/MWh and P₁ = 433.94 MW P₂ = 300.11 MW P₂ = 131.74 MW : The table below summarizes the iterative process used to solve this problem. Losses Interation P(MW) P₂ (MW) P, (MW) (MW) 2($/MWh) Start 400.00 300.00 150.00 15.60 9.5252 1 440.68 299.12 125.77 15.78 9.5275 2 433.94 300.11 131.74 15.84 9.5285 3 435.87 299.94 130.42 15.83 9.5283 4 435.13 299.99 130.71 15.83 9.5284