Max: 100 MW Cost: $20/MW Load: 50 MW G x1=0.2 p.u Cap: 50 MW. x2=0.2 Cap: 50 MW. x3=0.2 p.u. Cap: 50 MW. p.u. x4=0.2 p.u. Cap: 100 MW. 4 Max: 200 MW G Cost: $40/MW Load: 100 MW Load: 150 MW G Max: 250 MW Cost: $50/MW Example • Let's formulate the optimization problem for the following system. Max: 70 MW Cost: $30/MW Max: 80 MW Cost: $50/MW G x=0.1 p.u. Cap = 50 MW x=0.1 p.u. Cap = 50 MW x=0.1 p.u. Cap = 50 MW The VA base value is 100MVA 1 b = - = -10 p.u. x Objective: Subject to: Objective: Minimize (30P₁ +50P₂) Subject to: 0 ≤ P₁ ≤ 70 0 ≤ P₂ ≤ 80 −50 ≤ F₁3 ≤ 50 -50 ≤ F12 ≤ 50 -50 ≤ F23 ≤ 50 F1310 x 100(0₁ - 03) F12 10 x 100(0₁ - 0₂) F23 = 10 × 100(0₂ – 03) F12 = 0 P₁-F13 P₂ + F12 F23 = 0 F13 + F23 = 100 0₁ = 0 Load: 100 MW Minimize (30P₁ + 50P₂) 0 ≤ P₁ ≤ 70 0 ≤ P₂ ≤ 80 -50 ≤ F13 ≤ 50 -50 ≤ F12 ≤ 50 -50 ≤ F23 ≤ 50 F13 10 X 100 (0₁ - 03) F12 10 × 100(0₁ – 0₂) F23 = 10 x 100(02 - 03) F12 = 0 P₁ F13 P₂ + F12 Identify the variables and their ranges ● F23 = 0 F13 + F23 = 100 0₁ = 0 • Variables in this formulation: P1 P₂ F13 F12 • F23 0₁ (use a range of -1 to 1) 02 (use a range of -1 to 1) • 03 (use a range of -1 to 1)

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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Question

Please write down the formulation of the DCOPF problem for the 4-bus system provided in the first figure. I have provided an example of the formulation in Figure 2. Hope that helps

Max: 100 MW
Cost: $20/MW
Load: 50 MW
G
x1=0.2 p.u
Cap: 50 MW.
x2=0.2
Cap: 50 MW.
x3=0.2 p.u.
Cap: 50 MW.
p.u.
x4=0.2 p.u.
Cap: 100 MW.
4
Max: 200 MW
G Cost: $40/MW
Load: 100 MW
Load: 150 MW
G Max: 250 MW
Cost: $50/MW
Transcribed Image Text:Max: 100 MW Cost: $20/MW Load: 50 MW G x1=0.2 p.u Cap: 50 MW. x2=0.2 Cap: 50 MW. x3=0.2 p.u. Cap: 50 MW. p.u. x4=0.2 p.u. Cap: 100 MW. 4 Max: 200 MW G Cost: $40/MW Load: 100 MW Load: 150 MW G Max: 250 MW Cost: $50/MW
Example
• Let's formulate the optimization problem for the following system.
Max: 70 MW
Cost: $30/MW
Max: 80 MW
Cost: $50/MW
G
x=0.1 p.u.
Cap = 50 MW
x=0.1 p.u.
Cap = 50 MW
x=0.1 p.u.
Cap = 50 MW
The VA base value is 100MVA
1
b = - = -10 p.u.
x
Objective:
Subject to:
Objective: Minimize (30P₁ +50P₂)
Subject to: 0 ≤ P₁ ≤ 70
0 ≤ P₂ ≤ 80
−50 ≤ F₁3 ≤ 50
-50 ≤ F12 ≤ 50
-50 ≤ F23 ≤ 50
F1310 x 100(0₁ - 03)
F12 10 x 100(0₁ - 0₂)
F23 = 10 × 100(0₂ – 03)
F12 = 0
P₁-F13
P₂ + F12
F23 = 0
F13 + F23 = 100
0₁ = 0
Load: 100 MW
Minimize (30P₁ + 50P₂)
0 ≤ P₁ ≤ 70
0 ≤ P₂ ≤ 80
-50 ≤ F13 ≤ 50
-50 ≤ F12 ≤ 50
-50 ≤ F23 ≤ 50
F13 10 X 100 (0₁ - 03)
F12 10 × 100(0₁ – 0₂)
F23 = 10 x 100(02 - 03)
F12 = 0
P₁ F13
P₂ + F12
Identify the variables and their ranges
●
F23 = 0
F13 + F23 = 100
0₁ = 0
• Variables in this formulation:
P1
P₂
F13
F12
• F23
0₁ (use a range of -1 to 1)
02 (use a range of -1 to 1)
• 03 (use a range of -1 to 1)
Transcribed Image Text:Example • Let's formulate the optimization problem for the following system. Max: 70 MW Cost: $30/MW Max: 80 MW Cost: $50/MW G x=0.1 p.u. Cap = 50 MW x=0.1 p.u. Cap = 50 MW x=0.1 p.u. Cap = 50 MW The VA base value is 100MVA 1 b = - = -10 p.u. x Objective: Subject to: Objective: Minimize (30P₁ +50P₂) Subject to: 0 ≤ P₁ ≤ 70 0 ≤ P₂ ≤ 80 −50 ≤ F₁3 ≤ 50 -50 ≤ F12 ≤ 50 -50 ≤ F23 ≤ 50 F1310 x 100(0₁ - 03) F12 10 x 100(0₁ - 0₂) F23 = 10 × 100(0₂ – 03) F12 = 0 P₁-F13 P₂ + F12 F23 = 0 F13 + F23 = 100 0₁ = 0 Load: 100 MW Minimize (30P₁ + 50P₂) 0 ≤ P₁ ≤ 70 0 ≤ P₂ ≤ 80 -50 ≤ F13 ≤ 50 -50 ≤ F12 ≤ 50 -50 ≤ F23 ≤ 50 F13 10 X 100 (0₁ - 03) F12 10 × 100(0₁ – 0₂) F23 = 10 x 100(02 - 03) F12 = 0 P₁ F13 P₂ + F12 Identify the variables and their ranges ● F23 = 0 F13 + F23 = 100 0₁ = 0 • Variables in this formulation: P1 P₂ F13 F12 • F23 0₁ (use a range of -1 to 1) 02 (use a range of -1 to 1) • 03 (use a range of -1 to 1)
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