Subido por Mauricio Zapata Quintana

# 3-Per Unit System

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Lecture 3:
Per Unit Systems
Basic Units
 The 4 basic electrical quantities are:

Voltage V
(volt)
Current I
(amp)
Impedance Z (ohm)
P
Power
S
(VA)
For single-phase circuits,
V(volt) = Z(ohm) &times; I(amp);
S (VA) = V(volt) &times; I(amp)
I(amp)*
Per unit notation
 In per unit notation, the physical quantity is
expressed as a fraction of the reference
value,
va
ue, i.e.
.e.
per unit value = actual value/base value in
the same unit.
unit
e.g. V(in per unit) = V(in kV)/V base (in kV)
where the base value is a reference value for
magnitude.
Base Quantities
Q
 In per unit notation we would like to keep
the basic relations:
Vpu = Zpu Ipu; Spu = Vpu Ipu*
 Hence the base quantities should be chosen
such that
Base voltage (VB) =
b
base
impedance
i
d
(ZB) &times; base
b
currentt (IB)
Base power (SB) =
b
base
voltage
l
(VB) &times; base
b
current(I
(IB)
Base Quantities
Q
 Thus only two of the base quantities can be
arbitrarily chosen, the other two will follow
d ect y.
directly.
 It is common practice to specify
b
base
pwer (SB) andd base
b
voltage
lt
(VB)
 Then it follows
base current
IB = SB/VB
base impedance
p
ZB = VB/IB =VB2/SB
Percentage
g Values
 An equivalent way to express the per unit
value is the percentage value where
Percentage value = per unit value &times; 100%
 However, percentage values are not so
convenient to use since
Vpercent ≠ Zpercent &times; Ipercent
Example
p 1
 Given
 Find
Fi d
V = 100∠30o
Z = 3 + j4 = 5∠53.1o Ω
currentt
active, reactive, &amp; apparent power
power factor
I
Z
V
Solution 1
Take ((for example)
p )
Base power
Base voltage
SB = 1 kVA
VB = 100 V
Base current
Base impedance
IB = SB/VB = 10 A
ZB = VB/IB = 10 Ω
Then
Given
V = 100 ∠30o V = 1.0 ∠30o p.u.
Z = 5∠53.1
5∠53 1o Ω = 0.5∠53.1
0 5∠53 1o p.u.
Solution 1 (cont)
(
)
 Current
&raquo; I = V/Z = 2.0 ∠-23.1o p.u

= 2.0&times;10∠
2 0&times;10∠-23
23.11o A = 20∠
20∠-23
23.1
1o A
Complex power
&raquo; S = VI* = 2.0
2 0 ∠53.1
∠53 1o p.u
p u = 1.2
1 2 + j1.6
j1 6 p.u.
pu
&raquo; Apparent power
S = 2.0&times;1 = 2 kVA
&raquo; Active
Acti e power
po er
P = 11.2&times;1
2&times;1 = 1.2
1 2 kW
&raquo; Reactive power
Q = 1.6&times;1 = 1.6 kVAr
&raquo; Power
P
ffactor
t
p.f.
f = P/S = 11.2/2.0
2/2 0 = 0.6
06
Base Value for 3-phase
p
systems
y
 For 3-phase
3 phase systems it is common practice
to describe system operation with:
total 3-phase power S = S3-Ԅ
line voltage V = Vline
line current I = Iline
equivalent impedance/phase Z = Zph
p
with (in magnitude)
V = √3ZI;
S = √3VI.
√3VI
Base Value for 3-phase
p
systems
y
 Hence if base values are chosen for:
total 3-phase power SB
li voltage
line
lt
VB
Then
base line current
√3V
VB
IB = SB/ √
base impedance
ZB = VB/ √3IB = VB2/SB
Example
p 2
 Supply: 400 V, 50 Hz, 33-phase
phase
 Load: 3 identical coils with Z = 20+j15 Ω
in star connection.
connection
 Find:
line current
power supplied
power factor.
Solution 2
Take ((for example)
p )
Base power (total 3-phase)
Base voltage (line-to-line)
SB = 10 kVA
VB = 400 V
Then
Base current
Base impedance
Given
IB = SB/√3VB = 14.43
14 43 A
ZB = VB2/SB = 16 Ω
V = 400 V = 1.0 p.u.
Z = 25∠36.9
25∠36 9o Ω = 1.5625∠36.9
1 5625∠36 9o p.u.
Solution 2 (cont)
(
)
 Current
I = V/Z
/ = 1.0
1 0 /1.5625
/1 62 p.u. = 0.64
0 64 p.u.



= 0.64&times;14.43 A = 9.235 A
Apparent power
S = VI = 1.0&times;0.64 = 0.64 p.u. = 6.4 kVA
Power factor
pp.f. = cos 36.9o = 0.8
Active power
P = VI &times; ppf = 0.64 &times; 0.8 pp.u = 0.512 pp.u.
= 0.512 &times; 10 = 5.12 kW
Choice of Base values
 For a connected circuit,, it is obvious that the same
bases should be used for the whole network such
that the normal circuit theorems would also apply
to per unit values, e.g.
Kirchhoff laws
At a given node,
∑i(Ipu)i = 0
Around a mesh
∑i(ΔVpu)i = 0
Impedance in series
Zpu = (Zpu)1+ (Zpu)2
Ypu = (Ypu)1+ (Ypu)2
Base values for a transformer
In a transformer, two circuits are not directly
connected but magnetically coupled. The
voltages
vo
tages of
o the
t e windings
w d gs are
a e in the
t e ratio
at o of
o
turns and currents in inverse ratio.
For the coupled circuit
circuit, we should then choose
&raquo; The same base power
&raquo; Base voltages in the ratio of turns.
turns
This will ensure Spu, Vpu, Ipu, to remain unchanged
when passing through an ideal transformer
Base values for a transformer
Let
n1,n2 be the number of turns in primary and
secondary winding.
winding
Z1, Z2 be the primary and secondary winding
impedance.
Then total impedance referred to primary
ZT11 = Z1 + (n1/n2)2Z2
and total impedance referred to secondary
ZT2 = Z2 + (n2/n1)2Z1 = (n2/n1)2ZT1
Base values for a transformer
If base values were chosen for the transformer:
SB1 = SB2; VB1 = (n1/n2)VB2
Then
Th
IB1 = (n2/n1)IB2 ; ZB1 = (n1/n2)2ZB2
Thus per unit impedance of transformer
Zpu = ZT1/ZB1 = ZT2/ZB2
is the same whether we use the total
impedance referred to primary or secondary.
secondary
Equivalent
q
circuit for transformer
 In the per unit representation, the equivalent
circuit of a transformer is a simple winding
peda ce Zpu (w
(with
t excitation
e c tat o branch
ba c
impedance
ignored)
(I1)pu
(V1)pu
(I2)pu
(Z)pu
(V2)pu
Base Conversion
 If the per unit values are given based on SB1
and VB1 which are different from the chosen
base SB2 aandd VB2 for
o analysis,
a a ys s, tthee given
g ve per
pe
unit values must be modified before they
ccan be used. Thus
us
(Vpu)2 = V/VB2 = (Vpu)1&times;VB1/VB2
(Spu)2 = S/SB2 = (Spu)1&times;SB1/SB2
Base Conversion
 Similarly
(Ipu)2 = I/IB2 = (Ipu)1&times;IB1/IB2
= (Ipu)1&times;VB2/VB1 &times;SB1/SB2
(Zpu)2 = Z/ZB2 = (Zpu)1&times;ZB1/ZB2
= (Zpu)1&times;(VB1/VB2)2 &times;SB2/SB1
Example
p
 Given a 50 MVA, 3.3 kV generator has a
synchronous impedance of 10%.
 Find the per unit impedance on a base of
100 MVA and 5 kV.
Zp.u. = 0.1 x (3.3/5.0)2 x (100/50)
= 0.087 p.u.
Example
p
 Choose base power
p
Sbase = 20 MVA
 Base
ase vo
voltages
tages
Vbase1 = 11 kV
Vbase2 = 33 kV
 Then
XG1 = 0.15pu
XG2 = 0.20x20/40=0.10pu
XT = 0.12x20/60=0.04pu
20MVA
X=15%
40MVA
X=20%
G1
G2
11 kV
T
33 kV
11/33kV
60MVA
X=12%
g of Per Unit System
y
 Normally we are dealing with numerics
near unity rather than over a wide range.
 Provides a more meaningful comparison of
parameters of machines with different
ratings.
ratings
 As the per unit values of parameters of a
machine
hi off a given
i
ddesign
i normally
ll falls
f ll
within a certain range, a typical value can
be used
sed if such
s ch parameters are not provided.
pro ided
```