Power System Equations Kansas State University（堪萨斯州立大学电力系统方程）.pptVIP

A Power System Example Starrett Mini-Lecture #3 Power System Equations Start with Newton again .... T = I a We want to describe the motion of the rotating masses of the generators in the system The swing equation 2H d2 d = Pacc wo dt2 P = T w a = d2d/dt2, acceleration is the second derivative of angular displacement w.r.t. time w = dd/dt, speed is the first derivative Accelerating Power, Pacc Pacc = Pmech - Pelec Steady State = No acceleration Pacc = 0 = Pmech = Pelec Classical Generator Model Generator connected to Infinite bus through 2 lossless transmission lines E’ and xd’ are constants d is governed by the swing equation Simplifying the system . . . Combine xd’ XL1 XL2 jXT = jxd’ + jXL1 || jXL2 The simplified system . . . Recall the power-angle curve Pelec = E’ |VR| sin( d ) XT Use power-angle curve Determine steady state (SEP) Fault study Pre-fault = system as given Fault = Short circuit at infinite bus Pelec = [E’(0)/ jXT]sin(d) = 0 Post-Fault = Open one transmission line XT2 = xd’ + XL2 XT Power angle curves Graphical illustration of the fault study Equal Area Criterion 2H d2 d = Pacc wo dt2 rearrange multiply both sides by 2dd/dt 2 dd d2d = wo Pacc dd dt dt2 H dt =  d {dd}2 = wo Pacc dd dt {dt } H dt Integrating, {dd}2 = wo Pacc dd{dt} H dt For the system to be stable, d must go through a maximum = dd/dt must go through zero. Thus . . . dm wo Pacc dd = 0 = { dd }2 H { dt } do The equal area criterion . . . For the total area to be zero, the positive part must equal the negative part. (A1 = A2) Pacc dd = A1 = “Positive” Area Pacc dd = A2 = “Negative” Area For the system to be stable for a given clearing angle d, there must be sufficient area under the curve for A2 to “cover” A1. In-class Exercise . . . Draw a P-d curve For a clearing angle of 80 degrees is the