Electric Power Generation, Transmission, and Distribution
Example 21.5 Determine the three-phase shunt admittance matrix for the concentric neutral line of Example 21.2. Solution Rb¼R¼ 0 ...
Substitute intoEq. (21.54): Yts¼j 77 : 586 ln Rb Ra ¼j^77 :^586 ln 0 : 0452 0 : 0153 ¼j 71 : 8169 mS=mile 21.1.3 Line Segmen ...
½¼B ½þU 1 2 ½Zabc½Yabc 1 ½Zabc (21:63) ½¼U 100 010 001 2 4 3 (^5) (21:64) In many cases the shunt admittance is so sma ...
SolveEq. (21.71)for the load voltages: ½VLGabcm¼½VLGabcn 1 : 5 ½Zabc½Iabcn¼ 6761 : 10 ff 2 : 32 6877 : 7 ff 122 : 43 6836 ...
Solution The sequence impedance matrix is zseq ¼ 0 : 7735 þj 1 : 9373 0 0 00 : 3061 þj 0 : 6270 0 000 : 3061 þj 0 : 6270 2 4 ...
Since the type B regulator is more common, the remainder of this section will address the type B step- voltage regulator. The ta ...
VS¼ 1 N 2 N 1 Vl IL¼ 1 N 2 N 1 IS (21:81) VS¼aRVL IL¼aRIS (21:82) aR¼ 1 N 2 N 1 (21:83) Equations (21.82) and (21.83 ...
The defining voltage and current equations for a regulator in the lower position are as follows: Voltage equations Current equat ...
It is important to understand that the value ofRline_ohmsþjXline_ohmsis not the impedance of the line between the regulator and ...
where ½¼aRVABC ½aRVabc^1 (21:98) 21.1.4.6 Current Equations IA IB IC 2 4 3 (^5) ¼ 1 aRa 00 0 1 aRb 0 00 1 aRc 2 6 6 (^66) (^6 ...
21.1.4.7 Closed Delta Connected Regulators Three single-phase regulators can be connected in a closed delta as shown in Fig. 21. ...
or ½¼Iabc ½AIDABC½IABC (21:108) also ½¼IABC ½AIDabc½Iabc (21:109) where ½¼IADabc ½IADABC^1 (21:110) The closed delta co ...
½¼IABC ½aRIabc½Iabc (21:113) ½¼Iabc ½aRIABC½IABC (21:114) The matrices for the three open connections are defined as follo ...
Phases CA and BA ½¼aRVabc aRB 00 aRB 0 aRC 00 aRC 2 4 3 (^5) (21:123) ½¼aRVABC 1 aRB 00 1 aRB 0 1 aRC 00 1 aRC 2 6 (^66) ...
21.1.5 Transformer Bank Connections Unique models of three-phase transformer banks applicable to radial distribution feeders hav ...
When the ‘‘ladder technique’’ or ‘‘sweep’’ iterative method is used, the ‘‘forward’’ sweep is assumed to be from the source work ...
wherets¼ 1 ffiffiffi 3 p ff 30 .Matrix to convert line-to-line voltages to equivalent line-to-neutral voltages: ½W ¼½As½Ts½A ...
IS¼ kVAbase ffiffiffi 3 p kVLLsource (21:150) ZS¼ kVLL^2 source 1000 kVAB (21:151) .The load side base values are computed by kV ...
½¼At 1 nt 10 1 110 0 11 2 6 4 3 7 5 ½¼Bt Zta 00 0 Ztb 0 00 Ztc 2 (^64) 3 (^75) 21.1.5.4.2 Ungrounded Wye–Delta Power-flo ...
21.1.5.4.3 Grounded Wye–Delta Power-flow equations: Backward sweep: ½¼VLGABC ½at½þVLNabc ½bt½Iabc ½¼IABC ½ct½þVLNabc ½d ...
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