TBT 1653-1996 牵引供电系统电能损失的计算条件和方法

TBTB/T1653199619961001199704011.25kV2.2.1.2.2.mTTtNmViii∑==1()(1)N()t(min)T1440(min)V2.3.∑∑===VigiiViiitNtNa11(2):tgi(min)3.3.1.()3.2.3.3.4.4.1.jA∆4.1.1.1a)BT()421013121.176.8−×−+=∆mamLrIAj(kWh)(3)(m-1)0∆(kWh)(4)amLrIAj2818.4=L(km)r(km)I(A)∑∑===ViiiViiitNANI114.2(A)(5):A(kVAh)b)AT()()[]{}612103121778.201.9365−=×′+−−+=∆∑llVigiijrnnpnrLtNIA(kWh)(6)nATrl(km)rlAT(km)PATnTtNPVigii∑==1(7)4.1.2.a)BT()()+++−++=∆48532435)15.0(3)2(1.176.8lglglg2rrmrrmrraLmIAj(kWh)(8)rlg(km)BTrlg0b)AT()()42222212102096.151314107.521096−=×+−−+′+=∆∑LLTTLLVigiijTPInnnDTPInTTtNLIA(Kwh)(9)2DAT(km)4.2.rA∆4.2.1.CA∆CNCPA∆=∆8760(kWh)(10)(kW)CNP∆4.2.2.tA∆a)YN1d11tNNcabcabtPIIIIA∆⋅++=∆2222238760(kWh)(11)(kw)tNP∆I2N(A)IabIbcIca(A)b)tNtPNItIA∆=∆2228760()hkW⋅(12)It(A)c)tNtPmKA∆+=∆)2446.0T2(0.554K8760)(hkW⋅(13)TNTTIIK=(14)MNMmIIK=(15)ITN,IMNTM(A)IT,IMTM(A)d)tNtPNIxIxIA∆+=∆222)2212(8760()hkW⋅(16)355NNSI=(A)(17)IX1IX2(A)SN(kVA)4.2.3.tCTAAA∆+∆=∆(18))(hkW⋅4.3.AatattatcatAAA∆+∆=∆(19))(hkW⋅CNatcPA∆=∆8760(20))(hkW⋅tNNatatattPIIA∆=∆228760(21))(hkW⋅Iat(A)INat(A)Aatc(kWh)Aatt(kWh)4.4.A4.4.1.BTTjAAA∆+∆=∆(22))(hkW⋅4.4.2.AT∑=∆+∆+∆=∆kiatiTjAAAA1)(hkW⋅(23)kAT4