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Unformatted text preview: EEL 6266 Power System Operation and Control Chapter 9 Control of Generation © 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 2 Tieline Model c The power flow across a tieline can be modeled using a linear load flow approach r steadystate or nominal flow quantity: r deviation from the nominal tieline flow c where ∆ δ must be in radians for ∆ P tie to be in per unit r using the relationship for speed and ∆ δ : then c where T = 377 / X tie for a 60Hz system ( 29 2 1 flow tie 1 δ δ = tie X P ( 29 ( 29 [ ] ( 29 ( 29 ( 29 2 1 flow tie 2 1 2 1 2 2 1 1 flow tie flow tie 1 1 1 δ δ δ δ δ δ δ δ δ δ ∆ ∆ = ∆ ∆ ∆ + = ∆ + ∆ + = ∆ + tie tie tie X P X X P P ϖ ϖ δ ∆ ⋅ = ∆ s ( 29 ( 29 2 1 2 1 flow tie ϖ ϖ ϖ ϖ ϖ ∆ ∆ = ∆ ∆ = ∆ s T X s P tie © 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 3 Tieline Model c Simplified control for two interconnected areas 1 1 1 G T s + ∆ ϖ 1 ∆ P mech 1 + _ + _ Σ 1 1 R P ref1 1 1 1 CH T s + 1 1 1 D s M + Σ load reference set points 2 1 1 G T s + ∆ ϖ 2 ∆ P mech 2 ∆ P L 2 + _ + _ Σ 2 1 R P ref2 2 1 1 CH T s + 2 2 1 D s M + Σ governors prime movers ∆ P L 1 Σ _ + + _ s T ∆ P tie © 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 4 Tieline Model c Consider two areas each with a generator r the two areas are connected with a single transmission line r the line flow appears as a load in one area and an equal but negative load in the other area r the flow is dictated by the relative phase angle across the line, which is determined by the relative speeds deviations r let there be a load change ∆ P L 1 in area 1 r to analyze the steadystate frequency deviation, the tieflow deviation and generator outputs must be examined © 2002, 2004 Florida State University EEL 6266 Power System Operation and Control 5 Tieline Model r after the transients have decayed, the frequency will be constant and equal to the same value in both areas r applying substitutions yields ( 29 ( 29 2 2 mech 2 tie 2 mech 1 1 mech 1 1 tie 1 mech 2 1 2 1 d d d d R P D P P R P D P P P t t L ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ∆ = ∆ ⋅ ∆ = ∆ + ∆ ∆ = ∆ ⋅ ∆ = ∆ ∆ ∆ = ∆ = ∆ → ∆ = ∆ = ∆ + ∆ = ∆ + + ∆...
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This note was uploaded on 07/15/2011 for the course EEL 6266 taught by Professor Thomasbaldwin during the Fall '04 term at FSU.
 Fall '04
 THOMASBALDWIN

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