Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
UNIVERSITÄT DUISBURG – ESSEN
FACHGEBIET ELEKTRISCHE ANLAGEN UND NETZE
Solutions
Electric Power Systems I-III
Prof. Dr.-Ing. habil. Istvan Erlich
1-1
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1 THE THREE-PHASE SYSTEM
1-2
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.1 Working with complex numbers
Z 1 = 2.598 + j1.5 = 3∠30 0
Z 2 = 2.828 + j 2.828 = 4∠45 0
Z 3 = 2.5 + j 4.33= 5∠60 0
{
}
)
)
i( t ) = I ⋅ sin( ωt + φi ) = Im I ⋅ (cos( ωt + φi ) + j sin( ωt + φi ))
a. Evaluation of the expressions:
i.
Z 1 ⋅ Z *2 ⋅Z 3 = 3∠30 0 ⋅ 4∠ − 45 0 ⋅ 5∠60 0 = 60∠45 0
ii.
Z 1 + Z *2 ⋅Z 3 = 9∠60 0 + 4∠ − 45 0 ⋅ 5∠60 0 =
2
= 9∠60 0 + 20∠15 0 = 23.82 + j12.97 = 27.12∠28.57 0
iii.
1
−1
Z 1 ⋅ Im⋅ {Z 3 }= ∠ − 30 0 ⋅ 4.33 = 1.44Ω∠ − 30 0
3
b. The required complex number I:
{)
)
}
i. i( t ) = I ⋅ sin( ωt + φi ) = Im I ⋅ (cos( ωt + φi ) + j sin( ωt + φi )) =
{
}
= 2 ⋅ Im I ⋅ e jωt → I =
Î
2
∠φi
{)
)
}
ii. i( t ) = I ⋅ sin( ωt + φi ) = − j Re I ⋅ (cos( ωt + φi ) + j sin( ωt + φi )) =
)
π
 I)
j ( φi − + ωt ) 
 I

j ( φi + ωt ) 
2
⋅e
= − j 2 Re
⋅e

 = 2 Re
 2


 2
 Î

= 2 ⋅ Re
⋅ e j [φi −π / 2] ⋅ e jωt  =
 2

→I =
Î
2
∠φi − π / 2
1-3
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
c. The RMS (root mean square) or “effective” value of i(t):
i.
The radian frequency ω = 2π.50 s-1: I =
Î
ii.
The radian frequency ω = 2π.60 s-1: I =
Î
)
2
2
iii. A positive offset value I / 2 is added to i(t): I =
Î
2
+ offset value
1-4
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.2 Complex Calculation
a. Amplitude and root mean square (RMS) values
û = 141.4 V
û
U=
2
î
I=
2
=
=
141.4
2
11.31
2
î =11.31 A
Current and voltage amplitudes
= 100 V
RMS values
= 8A
b. Current and voltage phasors
)
)
u( t ) = u ⋅ sin( ωt + 30 0 ) = u ⋅ cos( ωt − 60 0 )
U=
141.4
2
V∠ − 60 0 = 100V∠ − 60 0
I=
11.31
2
A∠ − 30 0 = 8 A∠ − 30 0
c. Complex power and power factor
S = U ⋅ I * = 800VA∠ − 30 0 = 692.82W − j 400 var ;
Power factor = cos(-600 – (-300)) = cos(-300)= 0.866 leading
d. Values of R and X
i.
Series connection
U 100
=
Ω∠ − 30 0 = 12.5 Ω∠ − 30 0 = 10.825 − j 6.25 Ω
I
8
R = 10.825 Ω
Z=
X = −6.25 Ω
ii.
Parallel connection
I
8
=
S∠30 0 = 0.08 S∠30 0 = 0.06928 + j 0.04 S
U 100
R = 14.43Ω
Y=
X = 25 Ω
1-5
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.3 Circuit analysis using phasor representation
a. Voltage and current relationship
i. u ( t ) = i ⋅ R + L ⋅
ii. u ( t ) = i ⋅ R +
di
dt
1
⋅
C
∫ i ⋅ dt
b. Transformation into phasor form
Assume: u( t ) = û ⋅ cos( ωt + φu ) and
i( t ) = î ⋅ cos( ωt + φi )
u( t ) = û ⋅ cos( ωt + φu ) →
= Re{ û ⋅ cos( ωt + φu ) + jû ⋅ sin( ωt + φu )} = Re{ û ⋅ e j
φu
⋅ e jωt }
Definition of phasor U and I:
u(t) =
2 ⋅ Re{ U ⋅ e jωt }
with U =
û
2
⋅ e jφu = U∠φu
Similarly
i(t) = 2 ⋅ Re{ I ⋅ e jωt } ,
with I =
î
⋅ e jφi = I∠φi
2
di(t)
= 2 ⋅ Re{ I ⋅ jω ⋅ e jωt } ,
dt
∫ i(t) dt =
 I

2 ⋅ Re ⋅ e jωt 
 jω

1-6
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Now, substitute these relationships in the equations given under (a):
u( t ) = i ⋅ R + L ⋅
{
2 ⋅ Re U ⋅ e
jω t
di
=
dt
= 2 ⋅ Re I ⋅ R ⋅ e
}
{
jω t
}+
{
2 ⋅ Re I ⋅ j ω L ⋅ e
jω t
}→
U = I ⋅ ( R + jωL ) for ( i )
u( t ) = i ⋅ R +
{
1
⋅ i ⋅ dt =
C ∫
}
{
}


1
2 ⋅ Re U ⋅ e jωt = 2 ⋅ Re I ⋅ R ⋅ e jωt + 2 ⋅ Re I ⋅
⋅ e jωt  →
 jω C

1
U = I ⋅( R +
) for ( ii )
jωC
1-7
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.4 Impedance and power in RLC circuit
R
L
a
e
a) R in series with L
Magnetic field energy stored in the inductance L:
Wmf =
1
⋅ L ⋅i2
2
C
R
Electric field energy stored in the capacitance C:
a
e
Wef =
b) R in series with C
R
1
⋅C ⋅u2
2
L
e
a
C
c) R L in parallel with C
R = 10 Ω, L = 15.9 mH, C = 0.6366 mF. F = 50 Hz
⇒ ωL = 5 Ω ;
1
= 5Ω
ωC
a. Zae (in Ω) as well as the admittance Yae (in S) across the terminals a – e:
R –L : Z a −e = (10 + j 5)Ω = 11.18Ω∠26.56 0
R –C : Z a −e = (10 − j 5)Ω = 11.18Ω∠ − 26.56 0
R –L in parallel with C: Z a −e = (10 + j 5) // (− j 5) = 2.5 − j 5 = 5.56 Ω∠ − 63.430
1-8
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
b. Required: I, S, P and Q for all three connections.
Voltage across a – e: u( t ) = 2 ⋅ 220V ⋅ cos ωt .
R –L :
I=
R –C :
220V∠0 0
11.18∠26.56
= 19.678 A∠ − 26.56 0
0
S = U ⋅ I * = 4.32916 kVA∠26.56 0
I=
220V∠0 0
11.18∠ − 26.56
0
= 19.678 A∠26.56 0
S = U ⋅ I * = 4.32916 kVA∠ − 26.56 0
P = 3.872kW
P = 3.872kW
Q = 1.936k var
Q = −1.936k var
R –L in parallel with C:
I=
220V∠0 0
5.56∠ − 63.43
0
= 39.568 A∠63.43 0 →
S = U ⋅ I * = 220V∠0 0 ⋅ 39.568 A∠ − 63.430 = 8.704kVA∠ − 63.430 = 3.894kW − j 7.786k var ⇒
P = 3.872kW
Q = −7.786k var
c. A general expression for i ⋅ u L
uL = L ⋅
di
→
dt
uL ⋅ i = i ⋅ L ⋅
Wmf =
di
dt
∂Wmf
1
di
⋅ L ⋅i2 →
= L ⋅ i ⋅ = uL ⋅ i
2
dt
∂t
1-9
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
d. Re-work (c) for the circuit (b) to find an expression for i ⋅ u C
uC =
1
⋅ i ⋅ dt →
C ∫
i ⋅ uC = i ⋅
1
⋅ i ⋅ dt
C ∫
du 
du 1
du
du

substituting  i = C ⋅ C  → i ⋅ u C = C ⋅ C ⋅ ⋅ ∫ C ⋅ C ⋅ dt = C ⋅ u C ⋅ C
dt 
dt C
dt
dt

Wef =
∂Wef
du
1
⋅ C ⋅ uC2 →
= C ⋅ uC ⋅ C = uC ⋅ i
2
dt
∂t
e. What can you deduce from the solutions (c) and (d) with regard to the nature of Q
(the reactive power)?
− Reactive power is the change of stored magnetic/electric field energy with
respect to time.
1-10
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.5 Calculation of Equivalent Circuit Parameters
a. Voltage and current sign convention
Current and voltage directions: (in accordance with the generation-oriented sign
convention)
IA
A
XAE
UA
IE
E
UE
RE
XE
b. Load admittance
YE =
1
1
1
=
−j
= G E − jB E = (15.75 − j9.09 ) mS
XE
ZE RE
Y E = (15.75 − j9.09) 10 −3 S
Y E = 18.18 ⋅ 10 −3 S ∠ − 30°
c. Load current
IE = Y E UE
IE = (18.18 ⋅ 10 −3 S ∠ − 30°) (220 V ∠ − 30°)
IE = 4.00 A ∠ − 60°
1-11
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
d. complex power
∗
S E = UE ⋅ IE
SE = (220 V ∠ − 30°) ⋅ ( 4.00 A ∠60°)
SE = 880 VA ∠ 30°
=
PE + jQ E
e. Voltage drop and the sending-end voltage
∆U AE = jX AE ⋅ I AE = jX AE ⋅ IE
∆U AE = j 10 Ω ⋅ 4.00 A ∠ − 60°
∆U AE = 40.0 V ∠ 30°
U A = ∆U AE + UE
U A = 40.0 V (
1
1
1
1
3 + j ) + 220 V(
3−j )
2
2
2
2
U A = (130 3 − j90 ) V
U A = 242.5 V ∠ − 21.8°
f.
The current at the receiving-end
IE = Re{IE } + j Im{IE }
Real component:
Re{IE } = 4.00 A ⋅ cos( −60°) = 2A
Imaginary component:
Im{IE } = j 4.00 A ⋅ sin( −60°) = −3.46 A
Active component:
I wE = IE cos(ϕuE − ϕiE ) = 3.46 A
Reactive component:
IbE = IE sin(ϕ uE − ϕ iE ) = 2.0 A
1-12
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
g. Phasor diagram
IERe
IEIm
1-13
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.6 Alternative schemes for power transmission
a. DC two-wire scheme
IL
Transmission line
IL
Generator
Load
Load current:
IL = J ⋅
A
,
2
(where A = cross-sectional area of the available conductor, J = allowable current density).
Note: only half of the available conductor can be used to carry the load current as the other
half is needed for the return current.
Maximum power:
P = J⋅
P
A
⋅ U0 = 0
2
2
with P0 = J ⋅ A ⋅ U0
b. DC three wire scheme
IL
Transmission line
Transmission line
IL
Generator
Load
1-14
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Load current:
IL = J ⋅
A
2
Maximum power:
P = 2⋅J⋅
A
⋅ U0 = P0
2
(By arranging the transmission scheme in such a way that voltage sources of opposing
polarity use the same return conductor, the need for conductor to carry the return current can
be obviated.)
c. AC single-phase
IL
Transmission line
IL
Generator
Load
Load current:
IL = J ⋅
A
2
Maximum power:
P = J⋅
A U0
⋅
= 0.35 ⋅ P0
2
2
Note: the conductor is insulated for a maximum voltage of U0. For power calculation, the
effective value of the voltage (
U0
) needs to be used.
2
1-15
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
d. AC single-phase three wire
IL
Transmission line
IR
Transmission line
IL
Generator
P = 2⋅J⋅
Load
A U0
⋅
= 0.7 ⋅ P0
2
2
e. AC two-phase scheme
Ix
Transmission line
IR = Ix + Iy
Transmission line
Iy
Generator
Load
1-16
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Assuming:
u x = u m cos ωt
and with a simplifying assumption of unity power factor:
u y = u m sin ωt
i x = im cos ωt
i y = im sin ωt
The maximum current in the return conductor can be obtained by setting the derivative of the
return current zero.
Thus,
d
π
(im cos ωt + im sin ωt ) = 0 → ωt =
dt
4
Maximum current in the return conductor:
iR max = im (cos
π
π
+ sin ) = 2 ⋅ im
4
4
The available conductor cross-section for each of the two phase conductors is thus:
x + x + 2 ⋅x = A → x =
A
2+ 2
= 0.293 ⋅ A
Maximum power:
P = 2⋅J⋅
A
2+ 2
⋅
U0
2
= 0.41 ⋅ P0
1-17
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
f. Three-phase scheme
Ia
Transmission line
Ib
Transmission line
Ic
Generator
Load
Maximum power:
P = 3⋅J⋅
A U0
⋅
= 0.7 ⋅ P0
3
2
(return current zero→ no conductor required)
Summary:
Scheme
Instantaneous
power
Maximum
power
1.6.1 DC two-wire
constant
0.5 P0
DC three-wire
constant
P0
Single-phase two-wire
pulsating
0.35 P0
Single-phase three-wire pulsating
0.7 P0
Two-phase
constant
0.41 P0
Three-phase
constant
0.7 P0
1-18
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Discussion of the results:
The DC scheme has the maximum transmission capability, but cannot be chosen due to the
limitations of DC generators for large-scale power generation. Single-phase AC is
unacceptable due to the pulsating instantaneous power. Two-phase system has a small
transmission capability. The three-phase system is, therefore, the simplest of the poly-phase
systems with constant output power and the highest transmission capability (compared to
other AC options).
1-19
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.7 The complex operator a
a. 1 + a + a2 = 0
b. 1 – a 2 + a = 1 + a – a 2 = -2 a 2 = 1 + j 3 = 2∠600
c. a 2 + a + j = -1+j =
2 ∠1350
d. j a + a 2 = -1.366 – j1.366 = 1.932∠2250
e.
f.
ea = e-0.5-j0.866 = e-0.5(cos 0.866 rad +j sin 0.866 rad)=0.39+j0.46 = 0.6∠49.70
a 10 = (a 3)3. a = a
g. (1- a 2)3 = ( 3 ∠300)3 = 3. 3 ∠900 = j3. 3
h. a -1 = -1.5-j0.866 =
3 ∠2100
1-20
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.8 Symmetrical three-phase load I
Z1
~
L1
UL1-L2
L2
Z2
~
L3
Z3
~
Z 1 = Z 2 = Z 3 = 30Ω∠31.8 0
U L1− L 2 = 3 ⋅ 220V∠30 0 (symmetrical)
a. The line to line voltages U L 2− L 3 ,U L 3− L1
2
U L 2− L3 = a ⋅ U L1− L 2
U L3− L1 = a ⋅ U L1− L 2
b. The phase voltages U L1 ,U L 2 ,U L 3
U L1 =
U L1− L 2
3
∠0 0
2
U L 2 = a ⋅ U L1
U L3 = a ⋅ U L1
c.
The line currents I L1 , I L 2 , I L 3
I L1 =
U L1
220V
=
= 7.33 A∠ − 31.8 0
0
Z1
30Ω∠31.8
I L 2 = a 2 ⋅ I L1
I L3 = a ⋅ I L1
d. The complex powers in each phase S L1 , S L 2 , S L 3
*
S L1 = S L 2 = S L3 = U L1 ⋅ I L1 = 1.613kVA∠31.8 0
e. The total complex three phase power absorbed by the load S 3φ
S 3φ = 3 ⋅ S L1 = 4.84kVA∠31.8 0
1-21
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.9 Symmetrical three-phase load II
Ia
a
Zab
Ib
b
Ic
c
Zca
Zbc
Z ab = Z bc = Z ca = (30 + j 27 )Ω = 40.36Ω∠41.99 0 → Z Y = (10 + j 9)Ω = 13.45Ω∠41.99 0
2
U a = 220V∠0 0
U b = a ⋅U a
U c = a ⋅U a
a. The currents I a , I b , I c .
Ia =
Ua
220V∠0 0
=
= 16.35 A∠ − 41.99 0
0
Z Y 13.45Ω∠41.99
2
Ib = a ⋅Ia
Ic = a⋅Ia
b. The total complex power and the corresponding power factor (cosϕ)?
*
S = 3 ⋅ U a ⋅ I a = 10.791kVA∠41.99 0 = 8.02kW + j 7.219k var = P + jQ
cos φ = cos 41.99 0 = 0.74ind .
c. The power factor at, when C= 25 µF/phase capacitor connected:
i. in star,
Qc =
(
tan φ =
3 ⋅ 220V
)
2
⋅ ωC = 1.14k var → Qnet = 7.219k var − 1.14k var = 6.079k var
6.079k var
→ φ = 37.16 0 → cos φ = 0.797ind .
8.02kW
ii. in delta.
Qc = 3 ⋅
(
tan φ =
3.8k var
→ φ = 25.35 0 → cos φ = 0.9ind .
8.02kW
3 ⋅ 220V
)
2
⋅ ωC = 3.42k var → Qnet = 7.219k var − 3.42k var = 3.8k var
1-22
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.10 Unsymmetrical Load I
R
a
Xc
b
V
c
a. Positive-sequence
Ia = −Ib =
3 ⋅ 220 V∠30 0
= 2.69 A∠75 0
100 − j100 Ω
Voltage measured by the voltmeter:
U V = 220 V − Ia ⋅ 100 Ω − Uc = 220 V − 2.69 A∠75 0 ⋅ 100 Ω − 220 V∠120 0
= 520 V∠ − 60 0 → 520 V
b. Negative-sequence
Ia = −Ib =
3 ⋅ 220 V∠ − 30 0
= 2.69 A∠15 0
100 − j100 Ω
Voltage measured by the voltmeter:
U V = 220 V − Ia ⋅ 100 Ω − Uc = 220 V − 2.69 A∠15 0 ⋅ 100 Ω − 220 V∠ − 120 0
= 139.85 V∠60 0 → 139.85 V
1-23
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.11 Unsymmetrical load II
a. Elements of the load impedance
10 MW =
38 kV 2
→ R = 144.4 Ω
R
X L = X C = 3 ⋅ R = 250.1Ω
b. The complex power
S = P + j(QL − Q C ) = 10 MW
c. Generator output currents
I ab
38 kV∠0 0
= 263.2 A∠0 0
=
144.4 Ω
I bc =
38 kV∠ − 120 0
= 152.35 A∠ − 30 0
− j 250.1Ω
I ca ==
38 kV∠120 0
= 152.35 A∠30 0
j 250.1Ω
I a = I ab − I ca = 263.2∠0 0 − 152.35∠30 0 = −131.58 − j 75.97 = 151.93 A∠ − 30 0
I b = I bc − I ab = 152.35∠ − 30 0 − 263.2∠0 0 = −131.58 − j 75.97 = 151.93 A∠210 0
I c = I ca − I bc = 152.35∠30 0 − 152.35∠ − 30 0 = 0 + j151.93 = 151.93 A∠90 0
Equivalent resistance (per phase):
R = 144.4 Ω
1-24
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.12 Reactive power compensation
a. Size of the capacitor
I=
S
3 ⋅U
=
20 kVA
3 ⋅ 380 V
= 30.39 A
I = 30.39 A∠ − 45.57 0 = 21.27 − j21.7 A
After compensation:
I = 25 A = 21.27 2 + (21.7 − IC ) 2 → IC = 8.56 A
i. Star connection
IC = ω ⋅ C Y ⋅
380 V
3
→ C Y = 0.124 µF
ii. Delta connection
C∆ =
C Y 0.124 µF
=
= 0.041µF
3
3
b. Power factor after compensation
i. In the input line
 21.7 − 8.56 
0
= tan −1 
 = 31.7
 21.27 
cos φ = 0.85 lagging
ii. At the motor terminals
Cos ϕ = 0.7 lagging (not affected by the compensation).
1-25
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.13 Reactive power compensation
Z∆
Zb
IA
A
IB
B
IC
C
~
IAB∆∆
~
Voltage source
Line
IAY
ZY
~
~
~
Load
a.
The load impedance
ZY =
(0.4kV )2
20kW
= 8Ω
3 ⋅ (0.4kV )
= 3 ⋅ Z Y = 24Ω
20kW
2
ZΔ =
b.
Load current
Star load:
I AY =
0.4kV / 3
= 28.87 A
8Ω
Delta load:
0.4kV∠30 0
= 16.67 A∠30 0
24Ω
0.4kV∠150 0
I CAΔ =
= 16.67 A∠150 0
24Ω
I AΔ = I ABΔ − I CAΔ = 16.67 A∠30 0 − 16.67 A∠150 0 = (28.87 − j 0 )A
I ABΔ =
Total current:
I A = 28.87 A + 28.87 A = 57.74 A
c.
Complex power
UA =
0.4kV∠0 0
3
+ 57.74 A ⋅ j 2Ω = 230.94 + j115.48V = 258.2V∠26.56 0
S A = 3 ⋅ U A ⋅ I A = 3 ⋅ 258.2 ⋅ 57.74∠26.56 0 VA = 44.725 KVA∠26.56 0
= 40kW + j 20k var
1-26
Solutions
Electric Power Systems I - III
d.
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Unsymmetrical loading
Zb
IA
A
IB
B
IC
~
C
~
Voltage source
Line
i. Line currents and power absorbed by the load
Currents:
IA
U A 258.2V∠26.56 0
=
=
= 129.1A∠ − 63.44 0
Zb
j2
IB =
IC
U A 258.2V∠ − 93.44 0
= 31.31A∠ − 107.44 0
=
Zb
8 + j2
U A 258.2V∠146.56 0
= 31.31A∠132.56 0
=
=
Zb
8 + j2
The power absorbed by the load:
*
*
*
S = U AI A +U B I B +UC IC =
= 258.2V∠26.56 0 ⋅ 129.1A∠63.44 0 + 258.2V∠ − 93.44 0 ⋅ 31.31A∠107.44 0 +
258.2V∠146.56 0 ⋅ 31.31A∠ − 132.56 0 = 33.334∠90 0 + 8.084∠14 0 + 8.084∠14 0 =
(15.69 + j37.245) kVA
ii. Current in the neutral
I n = I A + I B + I C = 129.1A∠ − 63.44 0 + 31.31A∠ − 107.44 0 + 31.31A∠132.56 0
= 27.16 − j122.28 A = 125.26 A∠ − 77.48 0
I A = 129.1A∠ − 63.44 0
I B = 31.31A∠ − 107.44 0
I C = 31.31A∠132.56 0
I n = 125.26 A∠ − 77.48 0
iii. Phase currents without a neutral wire
1-27
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
I A + IB + IC = 0
U AB = I A ⋅ Z b − I B ⋅ (Z b + Z Y ) = I A ⋅ Z b + ( I A + I C ) ⋅ (Z b + Z Y )
= I A ⋅ (2 ⋅ Z b + Z Y ) + I C ⋅ (Z b + Z Y )
U CA = − I A ⋅ Z b + I C ⋅ (Z b + Z Y ) →
U AB − U CA = I A ⋅ (3 ⋅ Z b + Z Y ) → I A
U AB − U CA
3⋅U A
3 ⋅ 258.2V∠26.56 0
=
=
=
(3 ⋅ Z b + Z Y ) (3 ⋅ Z b + Z Y )
(8 + j 6)Ω
I A = 77.46 A∠ − 10.3 0
U CA = − I A ⋅ Z b + I C ⋅ (Z b + Z Y ) → I C =
IC =
U CA + I A ⋅ Z b 447.22V∠176.56 0 + 154.92∠79.7 0
→
=
(Z b + Z Y )
(8 + j 2)Ω
− 418.71 + j179.26 455.45∠156.82 0
= 55.2 A∠142.78 0
=
0
(8 + j 2)Ω
8.25∠14.04
I B = −(I A + I C ) = −32.25 − j19.54 A = 37.07 A∠ − 148.79 0
I A = 77.46 A∠ − 10.30
I B =37.07 A∠ − 148.79 0
I C = 55.2 A∠142.78 0
iv. The size of the neutral wire
For symmetrical loading, the neutral current carries no or small current, but for an
unsymmetrical load, the current in the neutral wire can exceed the current in the phase
conductors.
e. Unsymmetrical delta load
Zb
Voltage source
IA
A
IB
B
IC
C
IAB∆∆
Z∆
Line
Z∆/2
1-28
Solutions
Electric Power Systems I - III
i.
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Phase and line currents
I A + IB + IC = 0
U AB = I A ⋅ Z b − I B ⋅ Z b
= I A ⋅2⋅ Zb + IC ⋅ Zb
U CA = − I A ⋅ Z b + I C ⋅ (Z b + Z Δ / 2 ) →
U AB + 2 ⋅ U CA = I C ⋅ (3 ⋅ Z b + Z Δ ) → I C =
3⋅U C
U AB + 2 ⋅ U CA
3 ⋅ 258.2V∠146.56 0
=
=
(3 ⋅ Z b + Z Δ ) (3 ⋅ Z b + Z Δ )
(24 + j 6)Ω
I C = 31.31A∠132.52 0
U AB = I A ⋅ 2 ⋅ Z b + I C ⋅ Z b →
IA
U AB − I C ⋅ Z b
3 ⋅ 258.2V∠26.56 0
=
= 111.8 A∠ − 63.44 0
=
2⋅ Zb
j4
(
)
I B = −(I A + I C ) = − 111.8 A∠ − 63.44 0 + 31.31A∠132.52 0 = −28.82 + j 76.92 A = 82.14 A∠110.5 0
I A = 111.8 A∠ − 63.44 0
I B = 82.14 A∠110.5 0
I C = 31.31A∠132.520
ii.
Level of asymmetry
The asymmetry is greater for the delta load.
1-29
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.14 Voltage drop, transmission losses with/without compensation
a. Draw the three-phase and single-phase equivalent circuit
~
2.5+j3Ω
∼
20kV/ 3
110+j50Ω
b. Calculate:
− I=
20kV / 3
11547
=
= 92.85 A∠ − 25.22 0
2.5 + 110 + j (3 + 50 )Ω 112.5 + j 53
− P = 3⋅ I
2
⋅ R = 3 ⋅ 92.85 2 ⋅ 110W = 2844.97kW
− Q = 3⋅ I
2
⋅ X = 3 ⋅ 92.85 2 ⋅ 50W = 1293.168k var
− S=
P 2 + Q 2 = 3125.08kVA
− φ = tan −1 (Q / P ) = 24.44 0 → S = 3125.08kVA∠24.44 0
− cos φ = cos 24.44 0 = 0.91 lag
c. Determine the output power and the power factor at the generator terminals
2
Ploss = 3 ⋅ I ⋅ Rb = 3 ⋅ 92.85 2 ⋅ 2.5 = 64.6584kW
2
−
Q loss = 3 ⋅ I ⋅ X b = 3 ⋅ 92.85 2 ⋅ 3 = 77.59k var
S G = (P + Ploss ) + j (Q + Q loss ) = (2844.97 + 64.658) + j (1293.18 + 77.59 ) kVA
= 2909.598 + j1370.77 kVA = 3216.329kVA∠25.230
d. The power factor at the load is to be increased by connecting capacitors in parallel to
the load. Determine the value of the capacitors per phase needed to improve the
power factor to 0.98, if the capacitors are to be connected in:
− Delta
− Star.
1-30
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Z
CY
Determination of C
Z = 110 + j 50Ω → Y =
− jX C =
1
= 7.53 − j 3.42 mS
Z
1
→ jB = jωCY (unknown initially )
jωCY
Overall admittance after compensation
Y Ges = Y + jB = 7.53 − j (3.42 − B ) mS
New power factor=0.98→
cos φ = 0.98 → φ = 11.478 → tan φ = 0.203 =
C Y = 6.02 μF ⇒ C Δ =
3.42 − B
→ B = ωC Y = 1.89 mS →
7.53
CY
= 2 μF
3
Y Ges = 7.53 − j1.53 mS → Z neu =
1
= 127.54 + j 25.91 Ω
Y Ges
e. Compute the line losses with and without compensation.
Transmission loss without compensation:
2
PV = 3 ⋅ I ⋅ R
20 / 3kV
→ I
I=
112.5 + j 53 Ω
(20 / 3 )
(20 / 3 )
2
2
=
112.5 + 53
2
2
⋅ 10 6 A 2 →
2
PV = 3 ⋅
112.5 + 53
2
2
⋅ 10 6 ⋅ 2.5 = 64.65 kW
Transmission loss after compensation:
2
PV = 3 ⋅ I ⋅ R
20 / 3kV
→ I
I=
130.04 + j 28.91 Ω
(20 / 3 )
(20 / 3 )
2
2
=
130.04 + 28.91
2
2
⋅ 10 6 A 2 →
2
PV = 3 ⋅
130.04 + 28.91
2
2
⋅ 10 6 ⋅ 2.5 = 56.35 kW
1-31
Solutions
Electric Power Systems I - III
f.
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
Compare the voltage drop on the overhead line with and without compensation.
Voltage at the load without compensation:
UL
20
110 + j 50
→ U L = U G ⋅ 0.972∠ − 0.78 0 =
⋅ 0.972∠ − 0.78 0 = 11.22∠ − 0.78 0 kV
=
U G (110 + j 50) + (2.5 + j 3)
3
Voltage at the load after compensation:
UL
20
127.54 + j 25.91
→ U L = U G ⋅ 0.98∠ − 1.02 0 =
⋅ 0.98∠ − 1.02 0 = 11.32∠ − 0.78 0 kV
=
U G (127.54 + j 25.91) + (2.5 + j 3)
3
1-32
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
1.15 Working with per-unit (p.u.) quantities
1 Ω j8 Ω
IL
39 Ω
∼
220 V ∠00
UL
26 Ω
In the circuit shown above, a load having an impedance of 39 + j26 Ω is fed from a voltage
source through a line having an impedance of 1 + j8Ω. The effective (RMS) value of the
source voltage is 220 V.
Z load = (39 + j 26)Ω
Z line = (1 + j8)Ω
U = 220V∠0 0
a. Calculate the load current IL and the voltage at the load terminal UL.
IL =
220V∠0 0
= (3.193 − j 2.714 )A = 4.1906 A∠ − 40.364 0
(40 + j34 )Ω
U L = I L ⋅ Z load = (39 + j 26 )Ω ⋅ 4.1906 A∠ − 40.364 0 = (195.09 − j 22.83)V = 196.424V∠ − 6.674 0 V
b. Calculate the active and reactive power delivered to the load.
S = U L ⋅ I L = (195.09 − j 22.83) A ⋅ (3.193 + j 2.714 )V = (684.889 + j 46.92 )VA
*
c.
Repeat the above calculation in per-unit choosing a base of 220 V for the voltage and
1500 VA for the apparent power S.
Voltage Base = 220 V, Voltampere Base = 1500 VA →
1500VA
= 6.8181A
220V
220V
= 32.267 Ω
Im pedance Base = Z Base =
6.8181A
Current Base = I Base =
1-33
Solutions
Electric Power Systems I - III
UNIVERSITÄT DUISBURG - ESSEN
FG Elektrische Anlagen und Netze
Univ. Prof. Dr.-Ing. habil. I. Erlich
220V∠0 0
= 1 p .u .∠0 0
220V
(1 + j8)Ω = (0.3099 + j 0.2479) p.u .
Z line− p .u . =
32.267Ω
(39 + j 26)Ω = (1.20867 + j 0.805785) p.u .
Z load − p .u . =
32.267Ω
U p .u . =
The circuit diagram given above in physical units can now be represented in per-unit
by the following circuit diagram in which the per-unit values of the various quantities
are given.
0.3099
j0.2479
IL
1.20867
∼
IL =
1 ∠00
UL
j0.805785
1 p.u .∠0 0
= 0.614642 p.u .∠ − 40.364 0
(1.23966 + j1.053685) p.u .
U L − p .u . = I L − p .u . ⋅ Z load − p .u . =
= (1.20867 + j 0.805785) ⋅ 0.614642∠ − 40.364 0 = (0.886805 − j 0.103761) p .u . = 0.8928557∠ − 6.674 0 p.u .
*
S p .u . = U L − p .u . ⋅ I L − p .u . = 0.456618 + j 0.3044 = 0.54879∠33.69 0
d. Compare the values obtained in c) with those obtained in a) and b).
I L = I L − p .u . ⋅ I Base = 0.614642∠ − 40.364 0 ⋅ 6.8181A = 4.1906 A∠ − 40.364 0
U L = U L − p .u . ⋅ U Base = 0.8928557∠ − 6.674 0 ⋅ 220V = 196.42V∠ − 6.674 0
S = S p .u . ⋅ S Base = (0.456618 + j 0.3044 ) ⋅ 1500VA = (684.899 + j 456.592 )VA = 823.185VA∠33.69 0
The values obtained in c) are the same as those obtained in a) and b).
1-34