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Unformatted text preview: Perunit Quantities (1) Networks can be analyzed using Volts,
Amperes, Ohms, and Voltamperes Referring these quantities to a perunit (pu)
or percent value of base voltage, current,
impedance, or VA can make calculations
much easier Perunit Quantities (2) The perunit 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 Perunit Quantities (3) as w Perunit quantities are determined by
dividing the actual value by the base value: villailulll Perunit value = actual value/base value [Ea a: 59 ha is l! W Perunit 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 Perunit Quantities (5) For a singlephase 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 threephase circuits
and are commonly solved as singlephase
circuits with a neutral return as a
'3
ms
'7 a
.2. I However, system data are usually given and
results desired in total 3phase VA and line
toline voltages Perunit Quantities (7) actual V, _1
VP“ : J 4
J3 (base VL_N ) actual VA?"25
VA u = W
p 3(base VA1 #5) tumhalasuwésmarmuuuuté Perunit 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 singlephase representation of a three—phase
circuits H! In order for the pu line currents and pu impedances to have the same values with the LL and 3phase bases as they had with the LN and
singlephase 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 VIIL) 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 LL voltages, 5‘ 3phase Voltamps, line currents, and perphase impedances to pu quantities without ﬁrst "3 converting to LN and SinglePhase vaiues W ,5, The LL and 3phase set of bases is the most
_ frequently used for 3phase circuits a" When a voltage and VA are given as bases for 3 ’3 phase circuits, they are assumed to be LL and 3 w...w_. phase quantities unless otherwise stated "in. is: Perunit 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: Perunit 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) Perunit = 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 perunit normalization procedure is the
selection of the base values of the electrical quantities. In
practice, the base values of 3phase apparent power in MVA
(i.e. base MVA) and linetolme voltage in kV (i.e. base kV)
are assigned on each side of every 3phase 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 Perunit 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
3phase power transformer is designated
to be equal to the nominal linetoline
nameplate kV rating of the transformer !
‘l
Winnieﬁlial» Overview of Perunit Analysis (3) e
E: 2v ‘E The base values of perphase 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 Perumt AnalySIS (4)
m, J .3
a a To adjust the per—unit impedance of power 3 systems apparatus whenever the 3phase 3 nameplate ratings are diﬂerent than the a power system’s 3phase 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 Perunit Analysis (5) f Utility j: z" = base MVA/(SCA MVA) (5) Here SCA MVA represents the available 9m, ,3, shortcircuit apparent power delivered by the utility from all sources outside the plant a Overview of Perunit Analysis (6) ﬂash{ts Transformers
The perunit equivalent circuit model of any 3phase
’ transformer connection is simply a perunit 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 perunit value.
If is given in pa, than this step is not necessary.
A? This perunit 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 perphase 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 Perunit Analysis (8) :“3‘KW r Transmission Lines _ The perphase 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

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