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Energy Systems II

by: Fredy Okuneva

18

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5

Energy Systems II ECE 420

Fredy Okuneva
UI
GPA 3.81

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

Popular in ELECTRICAL AND COMPUTER ENGINEERING

This 5 page Class Notes was uploaded by Fredy Okuneva on Thursday October 22, 2015. The Class Notes belongs to ECE 420 at University of Idaho taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/227735/ece-420-university-of-idaho in ELECTRICAL AND COMPUTER ENGINEERING at University of Idaho.

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Date Created: 10/22/15
Example of per unit system problem The one line diagram of a three phase power system is shownin Figure 3 29 Select a common base of 100 MVA and 22 W on the generator side Draw an impedance diagram with all impedances including the load impedance marked in per unit The man ufacturers39 data for each device is given as follows 1 T1 gt T3 T1 T2 T3 T4 Lo ad Line SbaseG 90MVA SbaseTl 50MVA SbaseTZ 40MVA Sbaserm 40MVA SbaseT4 40MVA SbaseM 665MVA SL 57MVA thel 484ohm T 4 2 3 2 E ILine1 E T 5 6 4 E ILineZI VbaseG 22kV VbaseTlL 22kv VbaseTlH 220kV VbaseTZL llkv VbaseTZH 220kv VbaseTBL 22kv VbaseT3H IIOkV VbaseT4L llkv VbaseT4H 1 10kv VbaseM 1045kV VL 1045kV XII1162 6543ohm XG 018 XTl 010 XT2 0060 XT3 0064 XT4 0080 XM 0185 pr 06 MVA 106VA lagging j1 1 Find base impedances of each item We need find only one for each transformer 2 VbaseG ZbaseG39 Sb G ase 2 7 VbaseTlL ZbaseTlL39T Sb T1 ase 2 VbaseTZH ZbaseTZH39 Sb T2 ase 2 VbaseTBH ZbaseTBH39 Sb T3 ase 2 VbaseT4L ZbaseT4L39 Sb T4 ase 2 VbaseM ZbaseM39 Sb M ase zbaseG 5378 Q ZbaseTlL 968g 3 zbaseTZH 121 X 10 Q ZbaseT3H 3025 Q ZbaseT4L 3025 Q zbaseM 1642 Q First we write out a consistent set of bases The power base is given as 100MVA throughout The voltage base is given as 22kV at the generator Therefore the voltage base at bus 1 is 22 W as given The transformerturns ratio gives the voltage bases as follows SBase 100MVA VBI 22kV v baseTlH V132 VB139 VbaseTlL V1331 V132 2 2 VB1 2 4 849 B1 B1 SBase 2 VB2 VB2 220kV 2B2 2B2 484g SBase 2 V133 vB3 220 kV 2B3 2B3 484 o SBase VbaseT2L 2 V V V V 11kV B4 B4 B3 B4 VbaseTZH 2B4 2B4 121 Q SBase V V 2 baseTBH B5 VBS VBl v VBS 110kV 2B5 2B5 121 Q baseTSL SBase 2 VB6 VB6 VB5 VBG 110kV 2B6 2B6 121 Q SBase Second we set up the impedanoes ZbaseG XG XGZ XG 02 B1 Z baseTlL XTI XTlZ XTI 02 B1 ZbaseT2H XT2 XTzZ XT2 015 B2 ZbaseTBH XT3 XT3Z XT3 016 B5 Z baseT4L XT4 XT4Z XT4 02 B4 ZbaseM XM XMZ XM 0251 B4 Line impedanceswill be expressed on ZBZ and ZBS They are found as follows leen thel 4849 thez 6543 Q X 1 1 LIne 1 thel thel 01 ZB2 X1 2 Llne 2 xhne2 me thez 0541 ZB5 The load is expressed in per unit as follows sL 57 MVA vL 1045 kV pr 06 6 acospr 6 5313 deg vL2 ZLe zL115153319 gt ZL ZL Z zL 095 12671 B4 Application problem The motor runs at full load with a 080 power factor leading Determine the voltage at Bus 1 Restate the given vbaseM 1045 kV VB4 11 kV pr 080 Calculate the motor terminal voltage in per unit VbaseM V134 Vmotor 095 V motor lts terminal voltage is also the Bus 4 voltage V4 Vmotor Find the motor current from its voltage and power output sbaseM 665 MVA 0M iacospr 9M 73637 deg SbaseM JI39eM e S motor SBase Smotor Imotor v Imotor 056 0421 motor Current drawn by the load is V4 1L Z IL 036 7 0481 L Total current drawn at bus 4 is 14 1mOtor IL 14 092 7 006i The equivalent impedance ofthe parallel branches is J39XTl t J39thel t J39XT239J39XT3 t J39XIinez t J39XT4 z z 031 arallel arallel p J39XTI t J39Xh39nel t J39XTZ t J39XTB t J39Xh nez t J39XT4 p Generator terminal voltage is v1 v4 zpmuelI4 v1 0968 0276i lvll 1007 argV1 159 deg

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