Per Unit Analysis. Single-Phase systems

Transcription

1 Per Unit Analyi The per unit method of power ytem analyi eliminate the need for converion of voltae, current and impedance acro every tranformer in the circuit. n addition, the need to tranform from 3- to - equivalent and vice vera i avoided with the ue of per unit quantitie. inle-phae ytem n the per unit ytem, any electrical quantity may be expreed in per unit a the ratio of the actual quantity and the choen value for that quantity. actual value per unit value value Four quantitie mut be conidered, that are: apparent power, voltae, current and impedance. n inle- ytem, the relationhip amon the quantitie are: With only two equation relatin the four quantitie, it i neceary to pecify two value. The apparent power and the voltae are uually choen equal to the rated value and the other two value are comted accordinly. ( ) ( k ) MA The pecified power i applicable to all part of the power ytem. However, the voltae varie acro a tranformer and o doe the current and impedance. Note that the actual voltae, current and impedance chane acro the tranformer. However, becaue the alo chane acro the tranformer, the per unit value are the ame on both ide of the tranformer. 3

2 Example A enerator upplie a load throuh a feeder whoe impedance i f + jω. The load impedance i 8 + j6ω. The voltae acro the load i 0. Find the real power and reactive power upplied by the enerator. Take the load voltae a the reference phaor for per unit analyi. Aume that the enerator ha zero impedance and that 500 A &. 0 olution F E The current i: A The impedance i: ( ) Ω 500 The per unit value are comted a follow: F

3 The enerator voltae in per unit i: E + F ( )( ) E E The complex power i: E ( )( ) j P Q Watt AR Three-Phae ytem n three- ytem, the power i the total three- power and the voltae i the lineto-line voltae. With the tar connection aumed, the line current i equal to the current. The impedance i per. The three- quantitie are related to the inle a follow: Bae Power, 3 3 φ φ Bae oltae, 3 Bae Current, total 3 Bae mpedance, ( ) ( ) total 33

4 The impedance characteritic of electrical equipment i uually expreed a a percentae d on it ratin. When uch a device i connected in a power ytem in which the elected value are different from the machine ratin, the per-unit quantitie have to be expreed in term of the ytem. The per unit value of any impedance may be converted to the new a follow: new old new old old new Example A three-, 60 Hz, 30 MA, 3.8k, wye connected ynchronou enerator ha an armature reitance R Ω per and a ynchronou reactance X 0Ω per. a a) Expre the machine impedance in per unit d on the machine ratin. b) Uin the reult in part (a), find the per unit impedance d on 50MA and 34. 5k. olution a) ( 3.8) 6. 35Ω 30 R + jx + j Ω a b) new j0. 4 new 34

5 NOTE n lare power ytem, the power i uually choen to be the ratin of one of the major equipment. The power i the ame for all part of the power ytem. The voltae are normally choen a the nominal voltae in the variou part of the ytem or are elected to be the voltae ratin of the tranformer windin. The current and impedance are comted d on the previouly elected value of the power and voltae. Example A tranformer of 30MA, 60Hz, 5/38k ha equivalent impedance referred to the primary ide of 0. 75Ω. Calculate the Pu impedance of the tranformer referred to both primary and econdary windin, takin the power a 30MA. olution From the primary ide 5k and 30MA 30,000, ( ) Ω A 35

6 From the econdary ide \ N Ω N 5 38k and 30MA 30, ( ) A Ω Example The one-line diaram of a three- power ytem i hown in the fiure below. elect a common of 00MA and k on the eneration ide and draw the impedance diaram with all impedance includin the load impedance in per unit. The manufacturer data for each device i iven in the table below. The three- load at bu #4 aborb 57MA, 0.6 power factor lain at 0.45k. ine and have reactance of 48.4 and ohm, repectively. Device Power (MA) oltae (k) Reactance (%) G T 50 /0 0.0 T 40 0/ 6.0 T 3 40 /0 6.4 T / 8.0 M

7 T ine # 3 T 4 0 k G M T3 ine # 5 0 k 6 T4 oad # olution ( voltaeof T ) k ow 0 Hih 0 ( voltaeof T ) k ( voltaeof T ) k Hih 0 3 ow 4 0 ( voltaeof T ) 0 k 0 0k 5 6 The new per unit value for the reactance of G, T, T, T 3 and T 4 are: (Note: there i no chane in the voltae ) 00 G: X T : T : T 3 : T 4 : 00 X X X X

8 The new per unit reactance for the motor M i: M: X ( ) Ω, ine #: X ( ) Ω, ine #: X The load impedance 3 φ Ω, but 3 3 φ φ 3 ( ) ( ine ) Ω 3 3 φ 3 φ 3 φ ince 57 co MA 3 φ ( 0.45) j. 5367Ω Ω j j The per unit impedance diaram can be drawn a hown in the fiure below: J0. J0. J0.5 J0. J0.6 J0.54 J0. J0.5 G 0.95 M J