power system protection

# power system protection - Per-unit Quantities(1 Networks...

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Unformatted text preview: Per-unit Quantities (1) Networks can be analyzed using Volts, Amperes, Ohms, and Voltamperes Referring these quantities to a per-unit (pu) or percent value of base voltage, current, impedance, or VA can make calculations much easier Per-unit Quantities (2) The per-unit method expresses a quantity as a decimal fraction of a chosen base quantity For a chosen base of 10,000V, a voltage of 9,000V would have a pu value of 0.9pu, while 11,000V would be 1.1pu h.ﬁ%‘pi.,glygui:pwﬂhbﬁ it a Per-unit Quantities (3) as w Per-unit quantities are determined by dividing the actual value by the base value: villailulll Per-unit value = actual value/base value [Ea a: 59 ha is l! W Per-unit Quantities (4) All relationships between voltage, current, impedance, and Voltamperes must be maintained when using pu analysis 3., If base values of any two of these quantities are speciﬁed, then base values for the other two are ﬁxed Usually, base Volts and base Voltamps are the quantities chosen to be speciﬁed 1 i awtaaaaaaw it» is Per-unit Quantities (5) For a single-phase system the base current is then base voltamps divided by base volts Base 1 = Base VA/Base V The base impedance is also ﬁxed as base voltage divided by base current Base Z = Base V/Base l = (Base V)2/Base VA 5' as a e m in a; twill 551 :A" 3"1 gr. Per—unit Quantities (6) Most power systems networks are composed of balanced three-phase circuits and are commonly solved as single-phase circuits with a neutral return as a '3 ms '7 a .2. I However, system data are usually given and results desired in total 3-phase VA and line- to-line voltages Per-unit Quantities (7) actual V, _1 VP“ : J 4 J3 (base VL_N ) actual VA?"25 VA u = W p 3(base VA1 #5) tumhalasuwésmar-muuuuté Per-unit Quantities (8) V = actual VL_L base ‘5» actual VA” A. VA u = —————— p base V14345 Per—unit Quantities (9) ' ‘ Note that these pu quantities will still be used in a i single-phase representation of a three—phase circuits H! In order for the pu line currents and pu impedances to have the same values with the L-L and 3-phase bases as they had with the L-N and single-phase bases the base current and base __ __ impedance calculated from each set of voltage and ii; voltampere bases must be the same I : Per—unlt Quantltles (10) met—a 3 t base VAW ﬂ base VA1 #5 ‘ 3 base VA3 ¢ BaSQII—w—Iw:_ _ . ' .3 base VL—N base VInC 6([7059 VII-L) s ~/§ m ﬂ was "' " if Base Z 2 (base VH.)2 : (base VH /\/§)2 2 (base VHY . é ase VA1 a, base VA“ / 3 base VAM ‘5 ._.. .4: «Wu; Per—unit Quant1t1es (11) m P .a'myv‘; d These bases can be used to convert L-L voltages, 5‘ 3-phase Voltamps, line currents, and per-phase impedances to pu quantities without ﬁrst "3 converting to L-N and Single-Phase vaiues W ,5,- The L-L and 3-phase set of bases is the most _ frequently used for 3-phase circuits a" When a voltage and VA are given as bases for 3- ’3 phase circuits, they are assumed to be L-L and 3- w...w_. phase quantities unless otherwise stated "in. is: Per-unit Quantities (12) VIII The impedance of a device is usually given in pu in terms of that device’s voltage and VA ratings it tt If the device is to be incorporated in a system analysis, all of its given pu values must be changed to values in the system pu bases .uii it -:lna %' a : can be converted to a pu impedance on impedance base 2 in the following manner: _ (baseVl)2 Actual Z : Z W} base VA1 2 base VI base VA2 3W m p112 = Z pu] . w j base V2 base VA1 in at: Per-unit Quantities (15) a? .. '3 {I bﬁuﬂﬁﬂﬁ Next, convert the actual impedance to a pu value on base 2: actual Z W : (bagel/2f /base VA2 its a; Z ts: set a Li M <43} Example T2 \ - .n= v ' _ Il‘tQ ' Y r .. A T3 wanna" It: ﬁn. is m m m n m “#1,... a: '2; as. :ﬂ- Overview of Per—unit Analysis (1) Per-unit = actual value /base value (1) - The actual value can be a complex number or a ‘ ‘ phasor (magnitude and angle) in units, whether the base value is simply a real number in units - The key factor of a per-unit normalization procedure is the selection of the base values of the electrical quantities. In practice, the base values of 3-phase apparent power in MVA (i.e. base MVA) and line-to-lme voltage in kV (i.e. base kV) are assigned on each side of every 3-phase power transformer in accordance with the following simple rules, 3 Q m 1:: la: a is; it i;- iﬁ it as} ﬁi its I l i 'i ‘9 Overview of Per-unit Analysis (2) First, a convenient base MVA is chosen (e.g., base MVA = l, 10, 100, 10000, etc.), and is common throughout the entire power system. Second, base IN on any side of a 3-phase power transformer is designated to be equal to the nominal line-to-line nameplate kV rating of the transformer ! ‘l Winnie-ﬁlial» Overview of Per-unit Analysis (3) e E: 2v- ‘E The base values of per-phase impedance in Ohms g (i.e. base Z) and line current in kA (Le. base I) are then derived from base MVA and base kV, : according to the following equations: (base kV)2 (2) 1' base Z = ——-—~—-—-— base M VA a ba M VA ,3 base KA = -—————S€ (3) y - ; \E - basekV ' «r Overv1ew of Per-umt AnalySIS (4) m, J .3 a a To adjust the per—unit impedance of power 3 systems apparatus whenever the 3-phase 3 nameplate ratings are diﬂerent than the a power system’s 3-phase quantities: M. A: W a 2 . ) / ' ' ‘aajusted Z M : unadjusted Z W (5%] (20:12:?) (4) ' 1 s In 3! a 10 a Overview of Per-unit Analysis (5) f Utility j: z" = base MVA/(SCA MVA) (5) Here SCA MVA represents the available 9m, ,3, short-circuit apparent power delivered by the utility from all sources outside the plant a Overview of Per-unit Analysis (6) ﬂash-{ts Transformers The per-unit equivalent circuit model of any 3-phase ’ transformer connection is simply a per-unit series impedance. The unaaﬁ‘usied impedance is provided on its nameplate and is expressed as a percentage of rated impedance. In other words, you simply divide it by 100% to arrive at its unadjusted per-unit value. If is given in pa, than this step is not necessary. A? This per-unit impedance is adjusted with respect to utilise J's; in r 341 m .. m ‘ It: w . 13%? . .ﬂf m w Jim. "W r r, ._ ..‘K A the system base quantities per formula (4) ll _-. i. Overview of Per—unit Analysis (7) .3. a ‘3‘ Generators . 3 g In general, the per-phase equivalent circuit * model of a rotating machine is a voltage a 112 Wk 3. source in series with impedance. It should be adjusted to the system base quantities by Li} .5 using formula (4) '% m Overview of Per-unit Analysis (8) :“3‘KW r Transmission Lines _ The per-phase equivalent circuit model of the line is simply a pu series impedance. To ﬁnd the line’s pu reactance XL and resistance RL, the following equations are used: (Xin QperlOOOﬁ)-(fengthafruninﬁ) I _—_..__— (6) (numberof paraﬂel conductors per phase)-(baseZ m (2) I, _ (Rm QperlOOOft) - (length of runinft) ‘ — (numberqf parallel conductors per phase)-(baseZ in Q) (7) 12 ...
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• Spring '10
• Dr.IlyaGrinberg
• Electric power transmission, Electricity distribution, Three-phase electric power, per-unit quantities

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power system protection - Per-unit Quantities(1 Networks...

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