3.电力系统的有功-频率静态特性frequency static characteristic of power systemsf=f2-f<0AP=-KGAfAP,=KAf△PDo +△P, =Ppo + KDAf△Ppo + △P, = △PGPpo = △PG -△Pp = -(KG +Kp)Af = -Kf12SCU-SEEI-Tqliu2025/8/7
12 SCU-SEEI-Tqliu 2025/8/7 3.电力系统的有功-频率静态特性 frequency static characteristic of power systems f = f 2 − f 1 0 P K f G = − G P K f D = D P P P K f D0 + D = D0 + D PD0 + PD = PG P P P K K f K f D0 = G − D = −( G + D ) = −
系统的功率-频率静特性系数(单位调节功率)APDoPONAPDOPGNK= KG+K, =K+KD*TfNfNAfAf两端除以人Pp/得:APDOPDNPGNAPDOK个G*PONAf.AfIfNPDPD△Ppo*K, =k,KG+KD+ =Af.tPGN备用系数PDN0P13SCU-SEEI-Tqliu2025/8/7
13 SCU-SEEI-Tqliu 2025/8/7 两端除以 得: f P K K K D G D = + = − 0 f P f P K f P K D N DN D N GN G + = − 0 = − + = − f P f f P P K P P K D N D D N D D N G N G 0 0 − = + = f P K k K K D r G D 0 P o f 1 f 2 f A PD ' PD DN N P / f DN GN r P P k = 备用系数 系统的功率-频率静特性系数(单位调节功率)