ECE 421/521
Electric Energy Systems
Power Systems Analysis I
2 – Basic Principles
Instructor:
Kai Sun
Fall 2013
1
Outline
• Power in a 1-phase AC circuit
• Complex power
• Balanced 3-phase circuit
2
Single Phase AC System
=
i (t ) I m cos(ωt + θi )
+
=
v (t ) Vm cos(ωt + θ v )
-
Z
3
Load
Phasor Representation
=
v(t ) Vm cos(ωt +=
θv )
=
+ θi )
i (t ) I m cos(ωt =
• V=|V|∠θv
2 | V | cos(ωt + θ v )
2 | I | cos(ωt + θi )
I=|I|∠θi
– RMS phasors of v(t) and i(t)
Reference
• A phasor is a complex number that carries the amplitude and phase
angle information of a sinusoidal function
• Phasors represent sinusoidal signals of a common frequency (ω) as
vectors in the complex plane w.r.t. a chosen reference signal.
• Phasor representation is a mapping from the time domain to complex
number domain.
4
• Instantaneous power delivered to the load:
p (t ) =
v (t )i (t ) =
Vm I m cos(ωt + θ v ) cos(ωt + θi )
1
Vm I m [cos(θ v − θi ) + cos(2ωt + θ v + θi )]
1
cos A cos
B
=
[
]
2
2
1
= Vm I m {cos(θ v − θi ) + cos[2(ωt + θ v ) − (θ v − θi )]}
2
θ= θ v − θi
1
Impedance angle >0 for inductive load=
and
Vm I m {cos θ + cos[2(ωt + θ v ) − θ ]}
<0 for capacitive load
2
1
= Vm I m [cos θ + cos 2(ωt + θ v ) cos θ + sin 2(ωt + θ v ) sin θ ]
2
Using trigonometric identity
=
cos( A - B ) + cos( A + B )
| V | V=
2, | I | I m / 2
m /
p(t ) = V ⋅ I cos θ [1 + cos 2(ωt + θ v )] + V ⋅ I sin θ sin 2(ωt + θ v )
 
pR ( t )
Root-mean-square (RMS) values
5
pX ( t )
Example 2.1
v (t ) = 100 cos ωt
Z= 1.25∠60° Ω
1 100∠0°
Calculate the RMS current phasor:=
I
=
1.25
60°
∠
2
=
i (t ) 80 cos(ωt − 60o )
Assume
1
80∠ − 60°
2
v(t)=V cos ωt, i(t)=I cos(ωt -60)
m
m
p(t)=v(t) i(t)
100
6000
50
4000
0
2000
-50
0
-100
0
90
180
ωt, degree
270
-2000
360
0
90
180
ωt, degree
pX(t) Eq. 2.8
270
360
0
90
180
ωt, degree
270
360
pR(t) Eq. 2.6
4000
4000
3000
2000
2000
0
1000
0
-2000
0
90
180
ωt, degree
270
360
-4000
6
p (t ) = V ⋅ I cos θ [1 + cos 2(ωt + θ v )] + V ⋅ I sin θ sin 2(ωt + θ v )
 
pR ( t )
pX ( t )
• Observations on pR(t):
– Oscillating at frequency 2*ω (twice of the source frequency)
– Changing between 0 and 2|V|⋅|I|cosθ (always positive)
– Average value is |V|⋅|I|cosθ
– It is the energy flow into the circuit
• Observations on pX(t):
– Oscillating at frequency 2*ω (twice of the source frequency)
– Changing between -|V|⋅|I|sinθ and |V|⋅|I|sinθ (either positive or negative)
– Average value is zero
– It is the energy borrowed & returned by the circuit. It does no useful
work in the load but takes some line capacity
7
Real and Reactive Powers
p(t ) = V ⋅ I cos θ [1 + cos 2(ωt + θ v )] + V ⋅ I sin θ sin 2(ωt + θ v )
 
pR ( t )
pX ( t )
|V | ⋅ | I |
• Apparent power
• Unit ~ volt ampere or VA (kVA or MVA)
def
Q
P | V | ⋅ | I | cos θ
=
• Real power or active power (average power)
• Unit ~ watt or W
• Power factor (PF): cosθ= cos(θv-θi)
– Lagging or leading when θ =θv-θi >0 or <0
def
=
Q | V | ⋅ | I | sin θ
• Reactive power
• Unit ~ var (volt-ampere reactive). Some people use Var, VAR or VAr
• Q>0 or <0 when θ=θv-θi >0 or <0
8
P
Characteristics of instantaneous power p(t)
p (t ) = V ⋅ I cos θ [1 + cos 2(ωt + θ v )] + V ⋅ I sin θ sin 2(ωt + θ v )

 

pR ( t )
pX (t )
= P[1 + cos 2(ωt + θ v )] + Q sin 2(ωt + θ v )
For a pure resistor load (thermal load),
• θ=θv-θi =0, PF=1 (unity PF)
• P=|V||I|, so all electric energy becomes thermal energy
For a pure inductive load,
• θ=θv-θi =90, PF=0
• P=0, so electric energy = constant (no transformation to other forms)
• p(t) oscillates between the source and the magnetic field associated with the
inductive load
For a pure capacitive load,
• θ=θv-θi =-90, PF=0
• P=0, so electric energy = constant (no transformation to other forms)
• p(t) oscillates between the source and the electric field associated with the
capacitive load
9
Example 2.1
Z= 1.25∠60° Ω
RMS phasors:
v (t ) = 100 cos ωt =
i (t ) 80 cos(ωt − 60o )
=
V
100
∠0°, =
I
2
80
∠ − 60°
2
o
4000
cos 60
[1 + cos 2ω
sin
60o sin 2
p (t ) =
V ⋅ I cos θ [1 + cos 2ωt ] + V ⋅ I sin θ sin 2ωt =
t ] + 4000
ωt








 


p (t )
pR ( t )
pX ( t )
P
pR ( t )
Q
v(t)=Vm cos ωt, i(t)=Im cos(ωt -60)
X
p(t)=v(t) i(t)
100
6000
50
4000
0
2000
-50
0
P
-100
0
90
180
ωt, degree
pR(t)
270
-2000
360
4000
3000
2000
2000
P
0
180
ωt, degree
pX(t) Eq. 2.8
270
Borrowing
Returning
Q
-2000
90
180
ωt, degree
270
360
-4000
10
360
Q
0
1000
90
Eq. 2.6
4000
0
0
0
90
180
ωt, degree
270
360
What does “absorbing” or “generating” reactive
power mean?
• “Reactive power” with a circuit element does not
lead to real energy consumption
• Starting from the time of v(t)=Vm,
– An inductive element (lagging PF) first absorbs and
then returns the same amount of electric energy.
Such a pattern is considered “absorbing” reactive
power, and it helps reduce oscillations of power
– A capacitive element (leading PF) first generates
and then absorbs the same amount of electric
energy. Such a pattern is considered “generating”
reactive power, and it helps maintain voltage
• For a real-world power system with long-distance
transmission, low voltage at receiving ends is
usually a big reliability concern. Shunt/series
capacitors can be added as “sources” of reactive
power or var
11
Complex Power
def
S = P + jQ
V I cos θ + j V I sin θ =
V I ∠θ
=
*
= V I ∠θ v − θ=
VI
i
= | S | ∠θ =
P 2 + Q 2 ∠θ
Reference
• |S| is the apparent power
• θ=tan-1(Q/P)
– When θ>0 (θi <θv, i.e. lagging PF), Q>0 i.e.
absorbing Q (inductive load)
– When θ<0 (θi >θv, i.e. leading PF), Q<0 i.e.
generating Q (capacitive load)
12
PF=cosθ=P/|S| (lagging/leading is told by +/- sign of Q)
P=|S|⋅ cosθ=|S|⋅ PF
 | S | ⋅ sin θ = | S | ⋅ 1 − PF 2
, if lagging
Q= 
− | S | ⋅ 1 − PF 2 , if leading
 − | S | ⋅ sin θ =
If the load impedance is Z, i.e .V=ZI, then
|Z|
X
S = VI*= ZII*= Z|I|2= R|I|2+jX|I|2 = P + jQ
P=R|I|2
R
Q=X|I|2
• The load impedance angle θ is also called power angle
(ϕ in some literature)
• Apparent power |S|∝|I|2 indicates heating and is used
as a rating unit of power equipment
13
• Some useful equations:
*
2
2
V
VV
2 *
*
=
S VI
=
=
=
V Y
*
*
Z
Z
V
Z= *
S
V
P=
R
• If Z is purely resistive
2
2
2
2
V
V
V
Q =
or
• If Z is purely reactive=
X
−1/ ωC
ωL
14
Y=
S*
V
2
Complex Power Balance
S = VI * = V [ I1 + I 2 + I 3 ]∗ = VI1* + VI 2* + VI 3* = S1 + S2 + S3
P + jQ = P1 + jQ1 + P2 + jQ2 + P3 + jQ3
• Due to the conservation of energy, the total complex power
delivered to the loads in parallel is the sum of complex
powers delivered to each (by KCL).
• In other words, a balance must be maintained all the time
for both real power and reactive power
15
Example 2.2
V= 1200∠0° V,
Z1= 60 + j 0Ω,
Z 2 =6 + j12Ω,
Z 3 =30 − j 30Ω
Find the power absorbed by each
load and the total complex power
I=
1
V 1200∠0° 1200
=
=
= 20 + j 0 A
Z1
60 + j 0
60
I=
2
V 1200∠0°
200
200(1 − j 2)
=
=
=
= 40 − j80 A
Z2
6 + j12 1 + j 2
5
I=
3
V 1200∠0° 40
40(1 + j )
=
=
=
= 20 + j 20 A
Z 3 30 − j 30 1 − j
2
*
S=
VI=
1200∠0°(20 − j 0)
= 24, 000 W+j 0 var
1
1
*
S=
VI=
1200∠0°(40 + j80)
= 48, 000 W+j 96, 000 var
2
2
*
S=
VI=
1200∠0°(20 − j 20)
= 24, 000 W − j 24, 000 var
3
3
S = S1 + S2 + S3 = 96, 000 W+j 72, 000 var
16
• Other approaches
I =I1 + I 2 + I 3 =(20 + j 0) + (40 − j80) + (20 + j 20)
= 80 − j 60= 100∠ − 36.87° A
S =VI * =(1200∠0°)(100∠36.87°) =120, 000∠36.87° VA
96, 000 W + j 72, 000 var
2
V
(1200)2
=
= 24, 000 W + j 0 var
S1 =
Z1*
60
2
V
(1200)2
=
= 48, 000 W + j 96, 000 var
S2 =
Z 2* 6 − j12
2
V
(1200)2
=
S3 =
= 24, 000 W − j 24, 000 var
Z 3* 30 + j 30
17
Theorem of Conservation of Complex Power
For a network supplied by independent sources all at the
same frequency (all voltages and currents are assumed to
be sinusoids), the sum of the complex power supplied by
the independent sources equals the sum of the complex
power received by all the other branches of the network
• For a single source with elements in series or parallel, it is proved by
Kirchhoff’s voltage or current law (KVL/KCL)
• For a general case, it is proved by Tellegen’s theorem.
• Application of the theorem: a part of the network can be replaced by
an equivalent independent source
18
• Some examples on Bergen and Vittal’s book
19
20
21
Homework #2
• Read through Saadat’s Chapter 2.1~2.4
• ECE521: 2.1~2.6
• ECE421: 2.1, 2.3, 2.4, 2.5
• Due date: 9/11 (Wednesday) submitted in class or by
email
22
Power Factor Correction
P=|V||I|⋅ cosθ=|S|⋅ PF
• When PF<1, current I needs to increase by 1/PF to deliver
the same P as that with PF=1.
• Thus, the major loads of the system are expected to have
close to unity PFs. That is not a problem for residential
and small commercial customers.
• However, industrial loads are inductive with low lagging
PFs, so capacitors may be installed to improve PFs
23
P+jQ
Example 2.3
(a) Find P, Q, I and PF at the source
(b) Find capacitance C of the shunt capacitor
to improve the overall PF to 0.8 lagging
C
Solution: (a)
(b)
200∠0°
= 2∠0° A
100
200∠0°
= 4 − j8 A
I 2=
10 + j 20
I1 =
*
S=
VI1=
200∠0°(2 − j 0)= 400 W+j 0 var
1
*
S=
VI =
200∠0°(4 + j8)= 800 W + j1600 var
2
2
New power angle
=
θ ′ cos −1 (0.8)
= 36.87°
Q ′ P tan
θ ′ 1200 tan(36.87
=
=
=
°) 900 var
Qc = 1600 − 900 = 700 var
S =P + jQ =1200 + j1600 =2000∠53.13° VA
S * 2000∠ − 53.13°
I=
=
= 10∠ − 53.13° A
V*
200∠0°
PF
2
V
(200) 2
Zc =
= =
− j57.14 Ω
Sc*
j 700

cos(53.13
=
) 0.6 lagging
106
C
=
= 46.42 μF
2π (60)(57.14)
24
New apparent power and current
S ′ = 1200 + j 900 = 1500∠36.87°
2000→1500
S ′* 1500∠ − 36.87°
=
= 7.5∠ − 36.87° 10→7.5
I=′
*
V
200∠0°
High-voltage capacitor bank (150kV - 75MVAR)
Learn Example 2.4 (high-voltage)
(Source: wikipedia.org)
25
P12+jQ12
Complex Power Flow
V=
V1 ∠δ1
1
I12 =
=
P21+jQ21
V=
V2 ∠δ 2
2
V1 ∠δ1 − V2 ∠δ 2
Z ∠γ
V1
V
∠δ1 − γ − 2 ∠δ 2 − γ
Z
Z
S12 = V1 I12* = V1 ∠δ1 [
V1
V
∠γ − δ1 − 2 ∠γ − δ 2 ]
Z
Z
2
V
V V
= 1 ∠γ − 1 2 ∠γ + δ1 − δ 2
Z
Z
2
V1
V V
=
P12
cos γ − 1 2 cos(γ + δ1 − δ 2 )
Z
Z
=
P12
If R=0, i.e.
|Z|=X and γ=90o
2
V1 V2
sin(δ1 − δ 2 )
X
V
Q12 =1 [ V1 − V2 cos(δ1 − δ 2 )]
X
V1
V V
Q12
sin γ − 1 2 sin(γ + δ1 − δ 2 )
=
Z
Z
26
P12
V1 V2
sin(δ1 − δ 2 )
X
V
Q12 =1 [ V1 − V2 cos(δ1 − δ 2 )]
X
• Since R=0, there are no transmission line losses, and P12=-P21
• In a normal power systems, |Vi|≈100% and |δ1-δ2| is small (e.g. 5~30o).
• P12∝sin(δ1-δ2)≈δ1-δ2
– Real power flow is governed mainly by the angle difference of the
terminal voltages.
– For example, if V1 leads V2, i.e. δ1-δ2>0, the real power flows from
node 1 to node 2.
– Theoretical maximum power transfer (i.e. static transmission
capacity):
V1 V2
V1 V2

=
P12 max
=
sin 90
X
X
• Q12 ∝|V1|-|V2|cos(δ1-δ2) ≈|V1|-|V2|
– Reactive power flow is determined by the magnitude difference of
terminal voltages
27
S12=P12+jQ12
Example 2.5
S21=P21+jQ21
I21
Determine the real and reactive powers
supplied or received by each source and
the power loss in the line
V=
120∠ − 5° V，V=
100∠0° V, Z
= 1 + j7 Ω
1
2
120∠ − 5° − 100∠0°
= 3.135∠ − 110.02° A
1 + j7
100∠0° − 120∠ − 5°
= 3.135∠69.98° A
I 21 =
1 + j7
=
I12
1000
800
P
600
21
400
S12 = V1 I12* = 376.2∠105.02° = −97.5 W + j 363.3 var
P, Watts
200
=
S21 V1=
I 21* 313.5∠ − 69.98
=
° 107.3 W − j 294.5 var
PL
0
-200
S L = S12 + S21 = 9.8 W + j 68.8 var =PL + jQL
-400
PL
P
R=
I12
(1)(3.135)
= 9.8 W
2
2
-800
=
QL X=
I12
(7)(3.135)
= 68.8 var
2
12
-600
2
-1000
-40
28
-30
10
0
-10
-20
Source #1 Voltage Phase Angle, δ
20
1
30
Balanced Three-Phase Circuits
• A balanced source: three sinusoidal voltages are
generated having the same amplitude but displaced in
phase by 120o
– Instantaneous power delivered to the external loads is
constant
Positive phase sequence
Negative phase sequence
29
Balanced Y-connected Loads
E=
E p ∠α A
An
EBn = E p ∠α A − 120°
ECn = E p ∠α A − 240°
V=
E An − Z G I a
An
V=
VAn − Z L I a
an
•
Let Van be the reference:
Phase voltages (line-to-neutral voltages):
Van= V p ∠0°
Vbn
= V p ∠ − 120°
Vcn
= V p ∠ − 240°
Line voltages (line-to-line voltages):
Vab
= Van − Vbn
= V p (1∠0° − 1∠ − 120°=
)
3 V p ∠30°
Vbc
= Vbn − Vcn
= V p (1∠ − 120° − 1∠ − 240°=
)
Vca= Vcn − Van= V p (1∠ − 240° − 1∠0°=
)
3 V p ∠ − 90°
3 V p ∠150°
30
Vbc
| VL |= 3 V p
I=
a
Van | Van | ∠0
=
=
Z p | Z p | ∠θ
I=
b
Vbn
=
Zp
I p ∠ − 120° − θ
I=
c
Vcn
=
Zp
I p ∠ − 240° − θ
Ip ∠ −θ
IL = I p
• On the neutral line (return line)
In=Ia+Ib+Ic=0
• The neutral line may not actually exist
(one line per phase)
• The balanced 3-phase power system
problems can be solved on a “per-phase”
basis, e.g. for Phase A only. The other
two phases carry identical currents
except for phase shifts.
31
Balanced △-connected Loads
VL = V p
•
Let Iab be the reference:
I ab=
I p ∠0°
I bc=
I p ∠ − 120°
I ca=
I p ∠ − 240°
I=
I ab − I ca=
a
I p (1∠0° − 1∠ − 240°=
)
I=
I bc − I ab=
b
I p (1∠ − 120° − 1∠0°=
)
I=
I ca − I bc=
c
3 I p ∠ − 30°
3 I p ∠ − 150°
I p (1∠ − 240° − 1∠ − 120°=
)
3 I p ∠90°
| I L |= 3 I p
32
△-Y Transformation
• Idea: compare Ia=f∆(Van, Z∆)
and Ia=fY(Van, ZY)
Z∆
Z∆
n
Z∆
– Find Ia=f∆(Van, Z∆):
Ia =
Vab Vac Vab + Vac
+
=
Z∆ Z∆
Z∆
Ia =
Vca
3Van
Z∆
– Find Ia=fY(Van, Z∆):
Van
Ia =
ZY
ZY =
Vab=
+ Vac
3 Van ∠30° + 3 Van ∠ − 30°
=
3 Van ( 3 / 2 + j / 2 + 3 / 2=
− j / 2) 3Van
33
Z∆
3
Balanced Three-Phase Power
van
2 V p cos(ωt + θ v )
vbn
2 V p cos(ωt + θ v − 120°)
vcn
2 V p cos(ωt + θ v − 240°)
For balanced loads
ian
2 I p cos(ωt + θi )
ibn
2 I p cos(ωt + θi − 120°)
+ V p I p [cos(θ v − θi ) + cos(2ωt + θ v + θi − 240°)]
icn
2 I p cos(ωt + θi − 240°)
+ V p I p [cos(θ v − θi ) + cos(2ωt + θ v + θi − 480°)]
=
p3φ V p I p [cos(θ v − θi ) + cos(2ωt + θ v + θi )]
Total instantaneous power:
p3φ = vania + vbnib + vcnic
Zero
Total instantaneous power is constant:
= 2 V p I p cos(ωt + θ v ) cos(ωt + θi )
p=
P=
3 V p I p cos θ
3φ
3φ
+2 V p I p cos(ωt + θ v − 120°) cos(ωt + θi − 120°)
+2 V p I p cos(ωt + θ v − 240°) cos(ωt + θi − 240°)
Where is reactive power?
34
• Extend the concept of complex power to 3-phase systems
Q3φ = 3 V p I p sin θ
S3φ =P3φ + jQ3φ =3V p I *p
• Expressed in terms of line voltage VL and line current IL
– Y-connection: V = V / 3 I p = I L
I = I / 3
– ∆-connection: Vp = VL
p
L
p
L
P3φ = 3 VL I L cos θ
Q3φ = 3 VL I L cos θ
θ is the angle between the phase voltage and the phase current for the
same phase (e.g. Phase A)
35
Example 2.7
(a) The current, real power and reactive
powers drawn from the supply
(b) The line voltage at the combined loads
(c) The current per phase in each load
(d) The total real and reactive powers in
each load and the line
36
V2=110-j20
37
Homework #3
• Read through Saadat’s Chapter 2.5~2.12
• ECE521: 2.7~2.10, 2.14~2.16
• ECE421: 2.7, 2.8, 2.10, 2.15, 2.16
• Due date: 9/20 (Friday) submitted in class or by email
38