Transcript Introduction

Review
1
Review of Phasors
Goal of phasor analysis is to simplify the analysis of constant frequency ac
systems:
v(t) = Vmax cos(wt + qv),
i(t) = Imax cos(wt + qI),
where:
• v(t) and i(t) are the instantaneous voltage and current as a function of
time t,
• w is the angular frequency (2πf, with f the frequency in Hertz),
• Vmax and Imax are the magnitudes of voltage and current sinusoids,
• qv and qI are angular offsets of the peaks of sinusoids from a reference
waveform.
Root Mean Square (RMS) voltage of sinusoid:
T
Vmax
1
2
V 
v (t ) dt 
, so Vmax  2 V .

T0
2
2
Phasor Representation
The RMS, cosine-referenced voltage phasor is:
V
 V e jqV  V qV ,
v (t )
 Re 2 V e jwt e jqV ,
V
 V cosqV  j V sin qV ,
I
 I cosq I  j I sin q I .
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Phasor Analysis
Device
Resistor
Time Analysis
v (t )  Ri (t )
di (t )
v (t )  L
dt
Inductor
Phasor
V  RI
V  jw LI
t
1
v (t )   i (t )dt  v (0)
C0
Capacitor
1
V 
I
jwC
Z = Impedance  R  jX  Z  ,
(Note: Z is a
complex number but
not a phasor).
R = Resistance,
X = Reactance,
Z =
R X ,
2
2
X

 =arctan 
R
.


4
Complex Power
Instantaneous Power :
p (t )  v(t ) i (t ),
v(t ) = Vmax cos(w t  qV ),
i (t)
= I max cos(w t  q I ),
1
cos cos   [cos(   )  cos(   )],
2
1
p (t )  Vmax I max [cos(qV  q I ) 
2
cos(2w t  qV  q I )].
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Complex Power, cont’d
Instantaneous Power is sum of average and varying terms :
1
p (t )  Vmax I max [cos(qV  q I )  cos(2w t  qV  q I )],
2
T
Pavg
1

p (t )dt ,

T0
1
 Vmax I max cos(qV  q I ),
2
 V I cos(qV  q I ),
Power Factor Angle =  =qV  q I .
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Complex Power
S  V I  cos(qV  q I )  j sin(qV  q I )  ,
 P  jQ,
 V I *,
(Note: S is a complex number but not a phasor.)
P = Real Power (W, kW, MW),
Q = Reactive Power (VAr, kVAr, MVAr),
= magnitude of power into electric and magnetic fields,
S = Complex power (VA, kVA, MVA),
Power Factor (pf) = cos ,
If current leads voltage then pf is leading,
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If current lags voltage then pf is lagging.
Complex Power, cont’d
Relationships between real, reactive, and complex power:
P  S cos  ,
Q  S sin 
  S 1  pf 2 ,
Example: A load draws 100 kW with a leading pf of 0.85.
What are  (power factor angle), Q and S ?
   cos1 0.85  31.8,
100kW
S 
 117.6 kVA,
0.85
Q  117.6sin( 31.8)  62.0 kVAr.
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example
ZL=jwL=j*1000*1*10^-3 =j
9
10
Example
Power flowing from
source to load at bus
Earlier we found
I = 20-6.9 amps
S  V I *  10030  206.9  200036.9 VA,
= 1600W + j1200VAr
  36.9 pf = 0.8 lagging,
S R  VR I  ( RI ) I  4  20  6.9  206.9,
*
*
2
PR  1600W  I R
(QR  0),
S L  VL I *  ( jXI ) I *  3 j  20  6.9  206.9,
QL  1200VA r
2
 I X,
(PL  0).
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Power Consumption in Devices
Resistors only consume real power:
2
PResistor  I Resistor R,
Inductors only "consume" reactive power:
2
QInductor  I Inductor X L ,
Capacitors only "generate" reactive power:
2
QCapacitor   I Capacitor X C
QCapacitor  
VCapacitor
XC
1
XC 
.
wC
2
. (Note-some define X C negative.)
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Example
I
First solve
basic circuit
400000 V
I 
 4000 Amps
1000 
V  400000  (5  j 40) 4000
 42000  j16000  44.920.8 kV
S  V I *  44.9k20.8 4000
 17.9820.8 MVA  16.8  j 6.4 MVA
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Example, cont’d
Now add additional
reactive power load
and re-solve, assuming
that load voltage is
maintained at 40 kV.
Z Load  70.7
pf  0.7 lagging
I  564  45 Amps
V  59.713.6 kV
S  33.758.6 MVA  17.6  j 28.8 MVA
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Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn.
17.6 MW
16.0 MW
28.8 MVR
-16.0 MVR
59.7 kV
17.6 MW
28.8 MVR
Generators are
shown as circles
40.0 kV
16.0 MW
16.0 MVR
Transmission lines are shown as
a single line
Arrows are
used to
show loads
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Reactive Compensation
Key idea of reactive compensation is to supply reactive
power locally. In the previous example this can
be done by adding a 16 MVAr capacitor at the load.
16.8 MW
16.0 MW
6.4 MVR
0.0 MVR
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.0 MVR
Compensated circuit is identical to first example with just real power load.
Supply voltage magnitude and line current is lower with compensation.
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Reactive Compensation, cont’d
 Reactive compensation decreased the line flow
from 564 Amps to 400 Amps. This has
advantages:
– Lines losses, which are equal to I2 R, decrease,
– Lower current allows use of smaller wires, or
alternatively, supply more load over the same wires,
– Voltage drop on the line is less.
 Reactive compensation is used extensively
throughout transmission and distribution
systems.
 Capacitors can be used to “correct” a load’s
power factor to an arbitrary value.
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Power Factor Correction Example
Assume we have 100 kVA load with pf=0.8 lagging,
and would like to correct the pf to 0.95 lagging
S  80  j 60 kVA
  cos 1 0.8  36.9
PF of 0.95 requires desired
 cos 1 0.95  18.2
Snew  80  j (60  Qcap )
60 - Qcap
80
Qcap
 tan18.2  60  Qcap  26.3 kVAr
 33.7 kVAr
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Distribution System Capacitors
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Balanced 3 Phase () Systems
 A balanced 3 phase () system has:
– three voltage sources with equal magnitude, but
with an angle shift of 120,
– equal loads on each phase,
– equal impedance on the lines connecting the
generators to the loads.
 Bulk power systems are almost exclusively 3.
 Single phase is used primarily only in low
voltage, low power settings, such as residential
and some commercial.
 Single phase transmission used for electric
trains in Europe.
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Balanced 3 -- Zero Neutral Current
In  Ia  Ib  Ic
V
In 
(10  1   1  
Z
S  Van I a*  Vbn I b*  Vcn I c*  3 Van I a*
Note: Vxy means voltage at point x with respect to point y.
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Advantages of 3 Power
Can transmit more power for same amount of
wire (twice as much as single phase).
Total torque produced by 3 machines is
constant, so less vibration.
Three phase machines start more easily than
single phase machines.
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Three Phase - Wye Connection
 There are two ways to connect 3 systems:
– Wye (Y), and
– Delta ().
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
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Wye Connection Line Voltages
Vcn
Vab
Vca
-Vbn
Van
Vbn
(α = 0 in this case)
Vbc
Vab
Vbc
Vca
 Van  Vbn  V (1  1  120

3 V   30

3 V   90

Line to line
voltages are
also balanced.
3 V   150
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Wye Connection, cont’d
We call the voltage across each element of a
wye connected device the “phase” voltage.
We call the current through each element of a
wye connected device the “phase” current.
Call the voltage across lines the “line-to-line” or
just the “line” voltage.
Call the current through lines the “line” current.
VLine  3 VPhase 130  3 VPhase e
j
6
I Line  I Phase
S3
*
 3 VPhase I Phase
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Delta Connection
For Delta connection,
voltages across elements
equals line voltages
Ica
For currents
I a  I ab  I ca
Ic

Ib
Ibc
Iab
Ia
3 I ab   
I b  I bc  I ab
I c  I ca  I bc
*
S3  3 VPhase I Phase
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Three Phase Example
Assume a -connected load, with each leg Z =
10020, is supplied from a 3 13.8 kV (L-L) source
Vab  13.80 kV
Vbc  13.8 0 kV
Vca  13.80 kV
13.80 kV
I ab 
 138  20 amps
 
I bc  138  140 amps
I ca  1380 amps
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Three Phase Example, cont’d
I a  I ab  I ca  138  20  1380
 239  50 amps
I b  239  170 amps I c  2390 amps
*
S  3  Vab I ab
 3  13.80kV  138 amps
 5.7 MVA
 5.37  j1.95 MVA
pf  cos 20   lagging
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Delta-Wye Transformation
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with Z Y  Z 
3
2) Δ-connected sources can be replaced by
VLine
Y-connected sources with Vphase 
330
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Delta-Wye Transformation Proof
-
+
Suppose the two sides have identical terminal behavior.
For the  side we get
Vab Vca
Vab  Vca
Ia 


Z Z
Z
Hence
Z
Vab  Vca

Ia
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Delta-Wye Transformation, cont’d
For the Y side we get
Vab
 ZY ( I a  I b )
Vca  ZY ( I c  I a )
Vab  Vca  ZY (2 I a  I b  I c )
Ia  Ib  Ic  0  Ia   Ib  Ic
Since
Vab  Vca  3 ZY I a
Hence
3 ZY
Vab  Vca

 Z
Ia
Therefore
ZY
1
 Z
3
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3 phase power calculation
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