Transcript EE362L, Fall 2006 - Baylor University || School of

1. Power and RMS Values
1
2
3
4
Instantaneous power p(t) flowing into the box
i (t )
+
v(t )
p(t )  v(t )  i(t )
Circuit in a box,
two wires
−
i (t )
ia (t )
+
p(t )  va (t )  ia (t )  vb (t )  ib (t )
va (t )
−
Any wire can be the
voltage reference
ib (t )
+
vb (t )
Circuit in a box,
three wires
−
ia (t )  ib (t )
Works for any circuit, as long as all N wires are accounted for. There must
5
be (N – 1) voltage measurements, and (N – 1) current measurements.
Average value of
periodic instantaneous power p(t)
1 t o T
Pavg  
p(t )dt
T to
6
Two-wire sinusoidal case
v(t )  V sin(ot   ),
i(t )  I sin(ot   )
p(t )  v(t )  i(t )  V sin(ot   )  I sin(ot   )
zero average
 cos(   )  cos(2ot     ) 
p(t )  VI 

2


1 t o T
VI
V I
Pavg  
p(t )dt  cos(   ) 
cos(   )
T to
2
2 2
Pavg  VrmsI rms cos(  )
Average power
Power factor
7
Root-mean squared value of a
periodic waveform with period T
1 t o T 2
2
Vrms 
v (t )dt
T to

Compare to the average power
expression
1 t o T
Pavg  
p(t )dt
T to
The average value of the squared voltage
compare
Apply v(t) to a resistor
1 t o T
1 to T  v 2 (t ) 
1 t o T 2
Pavg  
p(t )dt  
v (t )dt

 dt 

t
t
t
T o
T o  R 
RT o
2
Vrms
Pavg 
R
rms is based on a power concept, describing the
equivalent voltage that will produce a given
average power to a resistor
8
Root-mean squared value of a periodic
waveform with period T
1 t o T 2
2
Vrms 
v (t )dt
T to

For the sinusoidal case v(t )  V sin(ot   ),
1 t o T 2 2
2
Vrms 
V sin (ot   )dt
T to

to T
2 t T
2

sin 2(ot   )
V
V
o
2
Vrms 
1  cos2(ot   ) dt 
t 

t
2T o
2T 
2o
 to



2
V
V
2
Vrms 
, Vrms 
2
2
9
Given single-phase v(t) and i(t) waveforms for a load
• Determine their magnitudes and phase angles
• Determine the average power
• Determine the impedance of the load
• Using a series RL or RC equivalent, determine the R
and L or C
100
80
60
40
20
Voltage
0
0
30
60
90
120
150
180
210
240
270
300
330
360
Current
-20
-40
-60
-80
-100
10
Determine voltage and current magnitudes and phase angles
Using a cosine reference,
Voltage cosine has peak = 100V, phase angle = -90º
Current cosine has peak = 50A, phase angle = -135º
~ 50
~ 100
  135 A
Phasors V 
  90 V , I 
2
2
100
80
60
40
20
Voltage
0
0
30
60
90
120
150
180
210
240
270
300
330
360
Current
-20
-40
-60
-80
-100
11
The average power is
V
I
Pavg 

cos(   )
2
2
100 50
100 50
Pavg 

cos 90  ( 135) 

cos45 
2
2
2
2
Pavg  1767W
12
Voltage – Current Relationships
i R (t )
i L (t )
iC (t )
 v R (t ) 
v R (t )
i R (t ) 
R
 v L (t ) 
v L (t )  L
 vC (t ) 
di( t )
dt
dv( t )
iC ( t )  C
dt
13
Thanks to Charles Steinmetz, Steady-State AC Problems
are Greatly Simplified with Phasor Analysis
(no differential equations are needed)
Resistor
Time Domain
Frequency Domain
v R (t )
i R (t ) 
R
~
V
Z R  ~R  R
IR
voltage leads current
Inductor
di( t )
v L (t )  L
dt
current leads voltage
Capacitor
dv( t )
iC ( t )  C
dt
~
VL
Z L  ~  jL
IL
~
V
1
ZC  ~C 
jC
IC
14
Problem 10.17
V1
1 1 1
 j4  3  2

1



2

1
2
V2
1
1 1 1

100 20
~
10020 
 j 4  3  2
V1   

2


~ 1  1  V~    j 4 
1



 V1   1 
j 4  2  10020 
  2~  
2   j 2 

~
1
1 V
 2  10020 
1
2
 j 2 
1 1  

 1 1 1 1
V1 
10020

j4

10020

1



1
1 
1
2
 j 2 
D

1
2
1  1
D        1
     

 j2   2   2 
 j4 3 2   2
 1 1 1 1
1 100201   1 1  
D        1
  j 
4
  2 2  
j
4
3
2
2

j
2
2

 
10020  1 1  1 
~ 
V1  
2
D
 j 2 
1
10020 



2
j4 


10020 
1 1 1
 2

 j2
1
~
V2 
D
15
c EE411, Problem 10.17
implicit none
dimension v_phasor(2), i_injection_phasor(2), y(2,2)
complex v_phasor, i_injection_phasor, y, determinant, i0_phasor
real pi
open(unit=6,file='EE411_Prob_10_17.txt')
pi = 4.0 * atan(1.0)
y(1,1)
1 + 1.0
2 + 1.0
y(1,2)
y(2,1)
y(2,2)
1 + 1.0
2 + 1.0
=
/
/
=
=
=
1.0 / cmplx(0.0,4.0)
3.0
2.0
-1.0 / 2.0
y(1,2)
1.0 / 2.0
/ cmplx(0.0,-2.0)
i_injection_phasor(1) = 100.0
1 * cmplx(cos(20.0 * pi / 180.0),sin(20.0 * pi / 180.0))
2 / cmplx(0.0,4.0)
i_injection_phasor(2) = 100.0
1 * cmplx(cos(20.0 * pi / 180.0),sin(20.0 * pi / 180.0))
determinant = y(1,1) * y(2,2) - y(1,2) * y(2,1)
write(6,*) "determinant, rectangular = ",determinant
write(6,*) "determinant, polar = ", cabs(determinant),
1 atan2(aimag(determinant),real(determinant)) * 180.0 / pi
write(6,*)
v_phasor(1) = (i_injection_phasor(1) * y(2,2)
1 - y(1,2) * i_injection_phasor(2)) / determinant
v_phasor(2) = (y(1,1) * i_injection_phasor(2)
1 - i_injection_phasor(1) * y(2,1)) / determinant
write(6,*) "v_phasor(1), rectangular = ",v_phasor(1)
write(6,*) "v_phasor(1), polar = ", cabs(v_phasor(1)),
1 atan2(aimag(v_phasor(1)),real(v_phasor(1))) * 180.0 / pi
write(6,*)
16
1 atan2(aimag(v_phasor(1)),real(v_phasor(1))) * 180.0 / pi
write(6,*)
write(6,*) "v_phasor(2), rectangular = ",v_phasor(2)
write(6,*) "v_phasor(2), polar = ", cabs(v_phasor(2)),
1 atan2(aimag(v_phasor(2)),real(v_phasor(2))) * 180.0 / pi
write(6,*)
i0_phasor = (v_phasor(1) - v_phasor(2)) / 2.0
write(6,*) "i0_phasor, rectangular = ",i0_phasor
write(6,*) "i0_phasor, polar = ", cabs(i0_phasor),
1 atan2(aimag(i0_phasor),real(i0_phasor)) * 180.0 / pi
write(6,*)
end
Program Results
determinant, rectangular = (1.125000,4.1666687E-02)
determinant, polar =
1.125771
2.121097
v_phasor(1), rectangular = (63.06294,-14.65763)
v_phasor(1), polar =
64.74397
-13.08485
v_phasor(2), rectangular = (80.67508,-8.976228)
v_phasor(2), polar =
81.17290
-6.348842
i0_phasor, rectangular = (-8.806068,-2.840703)
i0_phasor, polar =
9.252914
-162.1211
17
Active and Reactive Power Form a Power Triangle
V I
Pavg 
cos(   ),
2 2
V I
Q
sin(   ),
2 2
~ ~
S  V I  P  jQ
~
V  V
(   )
~
I  I
Complex
power
S
Q
(   )
Projection of
S on the
imaginary
axis
P
Projection of S
on the real axis
cos(   ) is the power factor
18
Question: Why is there conservation of P and Q in a circuit?
Answer: Because of KCL, power cannot simply vanish but must
be accounted for
Consider a node, with voltage (to any reference), and three currents
~ ~ ~
I A  I B  IC  0
IA
IB
IC




~~ ~ ~
V I A  I B  IC  0
~~ ~ ~ *
V I A  I B  IC  0
PA  jQA  PB  jQB  PC  jQC  0
PA  PB  PC  0
QA  QB  QC  0
19
Voltage and Current Phasors for R’s, L’s, C’s
Resistor
Inductor
~
VR
~
~
Z R  ~  R, VR  RI R
IR
~
VL
~
~
Z L  ~  jL, VL  jLI L
IL
~
~
VC
I
1
~
Capacitor ZC  ~ 
, VC  C
jC
jC
IC
Voltage and
Current in phase
Voltage leads
Current by 90°
Current leads
Voltage by 90°
Q=0
Q>0
Q<0
20
Complex
power
S
Q
(   )
Projection of
S on the
imaginary
axis
P
Projection of S
on the real axis

~ ~*
S  P  jQ  V  I  V  I *  VI    
P  VI cos   
Q  VI sin   
21
Resistor
~ *
~ V  V 2 V 2
S  P  jQ  V    

*
Z
R
Z
 

also
 
~
~*
2
2
S  P  jQ  I Z  I  I Z  I R
so
V2
P
 I 2R,
R
Q0
Use rms V, I
22
Inductor
~ *
V2
V2
jV 2
~ V 
S  P  jQ  V    


*
 jL  jL L
Z 

also
 
~
~*
S  P  jQ  I Z  I  I 2  jL   jI 2L
so
P 0, Q 
V
2
L
 I 2L
Use rms V, I
23
Capacitor
~ *
V2
V2
~ V 
S  P  jQ  V    

  jCV 2
*
1
Z 
 1 


 jC
j

C


also

 
2


1
I
~
~*
   j
S  P  jQ  I Z  I  I 2 
C
 jC 
so
2
I
P  0 , Q  CV 2  
C
Use rms V, I
24
Active and Reactive Power for R’s, L’s, C’s
(a positive value is consumed, a negative value is produced)
Active Power P
Resistor
Inductor
Capacitor
Reactive Power Q
2
Vrms
2
, I rms
R
R
0
0
2
Vrms
2
, I rms
L
L
0
2

I
2
 CVrms
, rms ,
C
source of reactive power
25
Now, demonstrate Excel spreadsheet
EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q
Vmag =
Vang =
Imag =
Iang =
wt
0
2
4
6
1.5
8
10
12
14
16
18
0 20
22
24
26
28
30
32
-1.5
34
0
36
38
1
0
0.90
90
-30
150
Phase A Phase A Phase A
v(t)
I(t)
p(t)
1 0.779423 0.779423
0.999391
0.794653 Power
0.794169
Instantaneous
in
0.997564 0.808915 0.806944
0.994522 0.822191 0.817687
0.990268 0.834465 0.826345
0.984808 0.845723 0.832875
0.978148 0.855951 0.837246
0.970296 0.865136 0.839437
0.961262 0.873266 0.839437
0.951057 0.880333 0.837246
0.939693 0.886327 0.832875
0.927184 0.891241 0.826345
0.913545 0.89507 0.817687
0.898794 0.897808 0.806944
0.882948 0.899452 0.794169
0.866025
0.9 0.779423
0.848048 0.899452 0.762778
0.829038 0.897808 0.744316
90
180
270
360
0.809017 0.89507 0.724127
0.788011 0.891241 0.702308
P
Q
Phase B
0.389711
0.225
v(t)
0.389711
0.225
-0.5
0.389711
0.225
-0.469472
Single-Phase
Circuit
0.389711
0.225 -0.438371
0.389711
0.225 -0.406737
0.389711
0.225 -0.374607
0.389711
0.225 -0.34202
0.389711
0.225 -0.309017
0.389711
0.225 -0.275637
0.389711
0.225 -0.241922
0.389711
0.225 -0.207912
0.389711
0.225 -0.173648
0.389711
0.225 -0.139173
0.389711
0.225 -0.104528
0.389711
0.225 -0.069756
0.389711
0.225 -0.034899
0.389711
0.225 6.13E-17
0.389711
0.225 0.034899
0.389711
0.225 0.069756
450
540
630
720
0.389711
0.225 0.104528
0.389711
0.225 0.139173
Phase B Phase B Phase C Phase C Phase C
A+B+C
I(t)
p(t)
v(t)
I(t)
p(t)
p(t)
-0.779423 0.389711
-0.5 5.51E-17 -2.76E-17 1.169134
-0.763243 0.358321 Instantaneous
-0.529919 -0.03141
0.016645
1.169134
Power
in Three-Phase
-0.746134 0.327084 -0.559193 -0.062781 0.035107 1.169134
-0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134
1.5
-0.70921 0.265675 -0.615661 -0.125256 0.077115 1.169134
-0.68944 0.235802 -0.642788 -0.156283 0.100457 1.169134
-0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134
-0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134
va
-0.625193
0.151248 -0.71934 -0.248074 0.178449 1.169134
ia
-0.602218
0.125208 -0.743145 -0.278115 0.20668 1.169134
0
-0.578509
0.100457
-0.766044 -0.307818 0.235802 1.169134
pa
-0.554095
0.077115 -0.788011 -0.337146 0.265675 1.169134
P
-0.529007 0.055296 -0.809017 -0.366063 0.296151 1.169134
Q
-0.503274 0.035107 -0.829038 -0.394534 0.327084 1.169134
-0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134
-0.45 -2.76E-17 -0.866025
-0.45 0.389711 1.169134
-0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134
-1.5
-0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134
0
90
180
270
360
450
540
630
-0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134
-0.337146 -0.046922 -0.927184 -0.554095 0.513748 1.169134
Q
0.675
0.675
0.675
Circuit
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
720
0.675
0.675
va
ia
vb
ib
vc
ic
pa+pb+pc
Q
26
A Single-Phase Power Example
A load consists of a 47Ω resistor and 10mH inductor in
series. The load is energized by a 120V, 60Hz voltage
source. The phase angle of the voltage source is zero.
a. Determine the phasor current
b. Determine the load P, pf, Q, and S.
c. Find an expression for instantaneous p(t)
27
A Transmission Line Example
Calculate the P and Q flows (in per unit) for the loadflow situation shown below,
and also check conservation of P and Q.
PL + jQL
0.05 + j0.15
pu ohms
VL = 1.020 /0°
PR + jQR
VR = 1.010 /-10°
IS
j0.20 pu mhos
IcapL
IcapR
j0.20 pu mhos
28
implicit none
complex vl_phasor,sl,icapl_phasor,zcl,is_phasor,zline
complex vr_phasor,sr,icapr_phasor,zcr
real
real
real
real
real
vlmag,vlang,vrmag,vrang,pi,qcapl,qcapr
vl_mag,vl_ang,vr_mag,vr_ang
rline, xline, bcap
pl,ql,pr,qr,is_mag,is_ang,icapl_mag,icapl_ang,icapr_mag,icapr_ang
qline_loss
open(unit=6,file="EE411_Trans_Line.dat")
pi = 4.0 * atan(1.0)
vl_mag = 1.02
vl_ang = 0.0
vr_mag = 1.01
vr_ang = -10.0
rline = 0.05
xline = 0.15
bcap = 0.20
vl_phasor = vl_mag * cmplx(cos(vl_ang * pi / 180.0),sin(vl_ang * pi / 180.0))
vr_phasor = vr_mag * cmplx(cos(vr_ang * pi / 180.0),sin(vr_ang * pi / 180.0))
is_phasor = (vl_phasor - vr_phasor) / cmplx(rline,xline)
icapl_phasor = vl_phasor * cmplx(0.0,bcap)
icapr_phasor = vr_phasor * cmplx(0.0,bcap)
sl = vl_phasor * conjg(is_phasor + icapl_phasor)
sr = vr_phasor * conjg(-is_phasor + icapr_phasor)
pl
ql
pr
qr
=
=
=
=
real(sl)
aimag(sl)
real(sr)
aimag(sr)
write(6,*) "is_phasor (rectangular) = ",is_phasor
is_mag = cabs(is_phasor)
is_ang = atan2(aimag(is_phasor),real(is_phasor)) * 180.0 / pi
write(6,*) "is_phasor (polar) ",is_mag,is_ang
write(6,*)
write(6,*) "icapl_phasor (rectangular) = ",icapl_phasor
icapl_mag = cabs(icapl_phasor)
icapl_ang = atan2(aimag(icapl_phasor),real(icapl_phasor)) * 180.0 / pi
write(6,*) "icapl_phasor (polar) ",icapl_mag,icapl_ang
write(6,*)
29
write(6,*) "is_phasor (rectangular) = ",is_phasor
is_mag = cabs(is_phasor)
is_ang = atan2(aimag(is_phasor),real(is_phasor)) * 180.0 / pi
write(6,*) "is_phasor (polar) ",is_mag,is_ang
write(6,*)
write(6,*) "icapl_phasor (rectangular) = ",icapl_phasor
icapl_mag = cabs(icapl_phasor)
icapl_ang = atan2(aimag(icapl_phasor),real(icapl_phasor)) * 180.0 / pi
write(6,*) "icapl_phasor (polar) ",icapl_mag,icapl_ang
write(6,*)
write(6,*) "icapr_phasor (rectangular) = ",icapr_phasor
icapr_mag = cabs(icapr_phasor)
icapr_ang = atan2(aimag(icapr_phasor),real(icapr_phasor)) * 180.0 / pi
write(6,*) "icapr_phasor (polar) ",icapr_mag,icapr_ang
write(6,*)
qcapl = cabs(vl_phasor) * cabs(vl_phasor) * (-bcap)
qcapr = cabs(vr_phasor) * cabs(vr_phasor) * (-bcap)
write(6,*) "pl = ",pl
write(6,*) "ql = ",ql
write(6,*)
write(6,*) "pr = ",pr
write(6,*) "qr = ",qr
write(6,*)
write(6,*) "qcapl = ",qcapl
write(6,*) "qcapr = ",qcapr
write(6,*)
write(6,*) "pl + pr = ",(pl + pr)
write(6,*) "ql + qr = ",(ql + qr)
write(6,*)
write(6,*)
qline_loss
write(6,*)
write(6,*)
write(6,*)
"pline_loss = ",cabs(is_phasor) * cabs(is_phasor) * rline
= cabs(is_phasor) * cabs(is_phasor) * xline
"qline_loss = ",qline_loss
"qline_loss + qcapl + qcapr = ",(qline_loss + qcapl + qcapr)
end
30
PL + jQL
0.05 + j0.15
pu ohms
PR + jQR
VL = 1.020 /0°
VR = 1.010 /-10°
IS
j0.20 pu mhos
IcapL
IcapR
j0.20 pu mhos
----------------------------------Results
is_phasor (rectangular) = (1.102996,0.1987045)
is_phasor (polar)
1.120752
10.21229
icapl_phasor (rectangular) = (0.0000000E+00,0.2040000)
icapl_phasor (polar)
0.2040000
90.00000
icapr_phasor (rectangular) = (3.5076931E-02,0.1989312)
icapr_phasor (polar)
0.2020000
80.00000
pl =
ql =
1.125056
-0.4107586
pr =
qr =
-1.062252
0.1870712
qcapl =
qcapr =
-0.2080800
-0.2040200
pl + pr =
ql + qr =
6.2804222E-02
-0.2236874
pline_loss =
6.2804200E-02
qline_loss =
0.1884126
qline_loss + qcapl + qcapr =
-0.2236874
31
RMS of some common periodic waveforms
Duty cycle controller
V
0<D<1
By inspection, this is
the average value of
the squared
waveform
0
DT
T
T
DT
0
0
1 2
1
2
Vrms   v (t )dt 
T
T
2
V
2
2
V
dt


DT

DV

T
Vrms  V D
32
RMS of common periodic waveforms, cont.
Sawtooth
V
0
T
T
2
2T
2
1
V
V
V


2
2
3T
Vrms    t  dt 
t dt 
t

3
3
0
T T 
T
3
T
0
0
V
Vrms 
3
33
RMS of common periodic waveforms, cont.
Using the power concept, it is easy to reason that the following waveforms
would all produce the same average power to a resistor, and thus their rms
values are identical and equal to the previous example
V
V
0
0
0
-V
V
V
V
0
0
0
V
0
V
Vrms 
3
34
2. Three-Phase Circuits
35
Three Important Properties of Three-Phase
Balanced Systems
•
•
Because they form a balanced set, the a-b-c
currents sum to zero. Thus, there is no return
current through the neutral or ground, which
reduces wiring losses.
A N-wire system needs (N – 1) meters. A threephase, four-wire system needs three meters. A
three-phase, three-wire system needs only two
meters.
a
b
c
n
Three-phase,
four wire system
Reference
•
The instantaneous power is constant
36
Observe Constant Three-Phase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls
Vmag =
Vang =
Imag =
Iang =
wt
0
2
4
6
1.5
8
10
12
14
16
18
0 20
22
24
26
28
30
32
-1.5
34
0
36
38
1
0
0.90
90
-30
150
Phase A Phase A Phase A
v(t)
I(t)
p(t)
1 0.779423 0.779423
0.999391
0.794653 Power
0.794169
Instantaneous
in
0.997564 0.808915 0.806944
0.994522 0.822191 0.817687
0.990268 0.834465 0.826345
0.984808 0.845723 0.832875
0.978148 0.855951 0.837246
0.970296 0.865136 0.839437
0.961262 0.873266 0.839437
0.951057 0.880333 0.837246
0.939693 0.886327 0.832875
0.927184 0.891241 0.826345
0.913545 0.89507 0.817687
0.898794 0.897808 0.806944
0.882948 0.899452 0.794169
0.866025
0.9 0.779423
0.848048 0.899452 0.762778
0.829038 0.897808 0.744316
90
180
270
360
0.809017 0.89507 0.724127
0.788011 0.891241 0.702308
P
Q
Phase B
0.389711
0.225
v(t)
0.389711
0.225
-0.5
0.389711
0.225
-0.469472
Single-Phase
Circuit
0.389711
0.225 -0.438371
0.389711
0.225 -0.406737
0.389711
0.225 -0.374607
0.389711
0.225 -0.34202
0.389711
0.225 -0.309017
0.389711
0.225 -0.275637
0.389711
0.225 -0.241922
0.389711
0.225 -0.207912
0.389711
0.225 -0.173648
0.389711
0.225 -0.139173
0.389711
0.225 -0.104528
0.389711
0.225 -0.069756
0.389711
0.225 -0.034899
0.389711
0.225 6.13E-17
0.389711
0.225 0.034899
0.389711
0.225 0.069756
450
540
630
720
0.389711
0.225 0.104528
0.389711
0.225 0.139173
Phase B Phase B Phase C Phase C Phase C
A+B+C
I(t)
p(t)
v(t)
I(t)
p(t)
p(t)
-0.779423 0.389711
-0.5 5.51E-17 -2.76E-17 1.169134
-0.763243 0.358321 Instantaneous
-0.529919 -0.03141
0.016645
1.169134
Power
in Three-Phase
-0.746134 0.327084 -0.559193 -0.062781 0.035107 1.169134
-0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134
1.5
-0.70921 0.265675 -0.615661 -0.125256 0.077115 1.169134
-0.68944 0.235802 -0.642788 -0.156283 0.100457 1.169134
-0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134
-0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134
va
-0.625193
0.151248 -0.71934 -0.248074 0.178449 1.169134
ia
-0.602218
0.125208 -0.743145 -0.278115 0.20668 1.169134
0
-0.578509
0.100457
-0.766044 -0.307818 0.235802 1.169134
pa
-0.554095
0.077115 -0.788011 -0.337146 0.265675 1.169134
P
-0.529007 0.055296 -0.809017 -0.366063 0.296151 1.169134
Q
-0.503274 0.035107 -0.829038 -0.394534 0.327084 1.169134
-0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134
-0.45 -2.76E-17 -0.866025
-0.45 0.389711 1.169134
-0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134
-1.5
-0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134
0
90
180
270
360
450
540
630
-0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134
-0.337146 -0.046922 -0.927184 -0.554095 0.513748 1.169134
Q
0.675
0.675
0.675
Circuit
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
0.675
720
0.675
0.675
va
ia
vb
ib
vc
ic
pa+pb+pc
Q
37
Imaginary
Vab = Van – Vbn
Vcn
Vca = Vcn – Van
30°
Real
Van
120°
Vbn
Vbc =
Vbn – Vcn
The phasors are rotating counter-clockwise.
The magnitude of line-to-line voltage phasors is 3 times the magnitude of line-to-neutral voltage phasors.
38
Imaginary
Vab = Van – Vbn
Vcn
Vca = Vcn – Van
Ic
Ica
30°
Iab
Real
Ia
Van
Line currents I a, Ib, and I c
Ib
Ibc
Delta currents I ab, Ibc, and I ca
Ic
c
Vbn
Ica
Balanced Sets Add to Zero in Both
Time and Phasor Domains
Ibc
Ia + Ib + Ic = 0
Van + Vbn + Vcn = 0
Vab + Vbc + Vca = 0
Ib
Vbc =
Vbn – Vcn
Iab
b
a
–
Vab
+
Ia
Conservation of power requires that the magnitudes of delta currents I ab, Ica, and I bc are
times the magnitude of line currents I a, Ib, Ic.
1
3
39
c
c
Ic
Ic
Z
3Z
3Z
n
Z
a
b
Ib
–
Ia
3Z
Vab
Z
a
b
Ib
–
Vab
+
+
Ia
The Two Above Loads are Equivalent in Balanced Systems
(i.e., same line currents Ia, Ib, Ic and phase-to-phase voltages Vab, Vbc, Vca in both cases)
40
c
c
Ic
Ic
n
–
Van
+
a
b
Ib
Ib
–
Ia
Vab
a
b
–
Vab
+
+
Ia
The Two Above Sources are Equivalent in Balanced Systems
(i.e., same line currents Ia, Ib, Ic and phase-to-phase voltages Vab, Vbc, Vca in both cases)
41
c
KCL: I n = I a + I b + I c
Ic
But for a balanced set,
Ia + Ib + I c = 0, so I n = 0
Z
In
n
Z
Z
a
b
Ib
–
V ab
+
Ia
Ground (i.e., V = 0)
The Experiment: Opening and closing the switch has no effect because I n is already zero for a three -phase
balanced set. Since no current flows, even if there is a resistance in the grounding path, we must conclude that
Vn = 0 at the neutral point (or equivalent neutral point) of any balanced three phase load or source in a bala nced
system. This allows us to draw a “one -line” diagram (typically for phase a) and solve a single -phase problem.
Solutions for phases b and c follow from the phase shifts that must exist.
42
Zline
c
c
Ic
3Zload
3Zload
a
b
Zline
–
Vab
a
b
Ia
3Zload
+
Zline
Ib
Zline
a
a
Ia
+
Van
–
The “One-Line”
Diagram
n
Zload
n
Balanced three-phase systems, no matter if they are delta
connected, wye connected, or a mix, are easy to solve if you
follow these steps:
1. Convert the entire circuit to an equivalent wyewith a
grounded neutral.
2. Draw the one-line diagram for phase a, recognizing that
phase a has one third of the P and Q.
3. Solve the one-line diagram for line-to-neutral voltages and
line currents.
4. If needed, compute line-to-neutral voltages and line currents
for phases b and c using the ±120° relationships.
5. If needed, compute line-to-line voltages and delta currents
using the 3 and ±30° relationships.
43
Now Work a Three-Phase Motor Power Factor
Correction Example
A three-phase, 460V motor draws 5kW with a power factor of 0.80
lagging. Assuming that phasor voltage Van has phase angle zero,
• Find phasor currents Ia and Iab and (note – Iab is inside
the motor delta windings)
• Find the three phase motor Q and S
• How much capacitive kVAr (three-phase) should be connected in
parallel with the motor to improve the net power factor to 0.95?
• Assuming no change in motor voltage magnitude, what will be the
new phasor current Ia after the kVArs are added?
44
Now Work a Delta-Wye Conversion Example
The 60Hz system shown below is balanced. The line-to-line voltage of the source is 460V.
Resistors R are each 5Ω.
Z
Z
Z
Part a. If each Z is (90 + j45)Ω, determine the three-phase complex power delivered by the
source, and the three-phase complex power absorbed by the delta-connected Z loads.
~
~
Part b. If Van at the source has phase angle zero, find Va'b' at the load.
Part c. Draw a phasor diagram that shows line currents Ia, Ib, and Ic, and
load currents Iab, Ibc, and Ica.
45
3. Transformers
46
Transformer Core Types
47
High-Voltage Grid Transformers, 100’s of MW
48
Single-Phase Transformer
Φ
Rs
jXs
Rm
jXm
Ideal
Transformer
7200:240V
Turns ratio 7200:240
7200V
240V
(30 : 1)
(but approx. same amount of
copper in each winding)
49
Short Circuit Test
Isc
Rs
+
Vsc
Short circuit test: Short circuit
the 240V-side, and raise the
7200V-side voltage to a few
percent of 7200, until rated
current flows. There is almost
no core flux so the
magnetizing terms are
negligible.
jXs
Rm
jXm
Ideal
Transformer
7200:240V
7200V
240V
Φ
~
Vsc
Rs  jX s  ~
I sc
Turns ratio 7200:240
(but approx. same amount of
copper in each winding)
50
Open Circuit Test
Ioc
Rs
jXs
Rm
jXm
7200V
Φ
+
Voc
-
Ideal
Transformer
7200:240V
240V
Open circuit test: Open circuit
the 7200V-side, and apply
240V to the 240V-side. The
winding currents are small, so
the series terms are negligible.
~
Voc
Rm || jX m  ~
I oc
Turns ratio 7200:240
(but approx. same amount of
copper in each winding)
51
1. Given the standard percentage values below for a 125kVA transformer,
determine the R’s and X’s in the diagram, in Ω.
2. If the R’s and X’s are moved to the 240V side, compute the new Ω values.
Single Phase Transformer.
Percent values are given
on transformer base.
Winding 1
kv = 7.2, kVA = 125
Winding 2
kv = 0.24, kVA = 125
Load
loss
Rs
Xs
No
load
loss
Magnetizing
current
jXs
Ideal
Rm
jXm
Transformer
7200:240V
%imag = 0.5
%loadloss = 0.9
%noloadloss = 0.2
%Xs = 2.2
7200V
240V
3. If standard open circuit and short circuit
tests are performed on this transformer, what
will be the P’s and Q’s (Watts and VArs)
measured in those tests?
52
X / R Ratios for Three-Phase Transformers
• 345kV to 138kV, X/R = 10
• Substation transformers (e.g., 138kV to 25kV or 12.5kV, X/R = 2, X = 12%
• 25kV or 12.5kV to 480V, X/R = 1, X = 5%
• 480V class, X/R = 0.1, X = 1.5% to 4.5%
Rs
jXs
Ideal
Rm
jXm
Transformer
53
Saturation – relative permeability decreases rapidly after 1.7 Tesla
Relative permeability drops from about 2000 to about 1 (becomes air core)
Magnetizing inductance of the core decreases, yielding a highly peaked
magnetizing current
Linear Scale
Log10 Scale
54
Transformer Core Saturation
6
Amperes
4
No DC
Log10 Scale
2
0
-2
-4
-6
Magnetizing Current for Single-Phase
25 kVA. 12.5kV/240V Transformer.
THDi = 76.1%, Mostly 3rd Harmonic.
No DC
With DC
Linear Scale
55
Apply a DC Voltage to a Transformer and Watch It Saturate
Where there is a DC current, there is a DC voltage, and vice-versa
Cold Core Test on 1kVA Transformer
(120V Winding Excited, 480V Winding Connected to 25 Ohm Resistor, Vdc = 150V on 6500 uF)
MG, March 12, 2009
Transformer Current in 120V
Winding
140
160
Voltage
140
Current
120
120
100
100
Saturates
80
80
\
60
60
40
40
20
20
0
0
-20
-0.005
0.000
0.005
0.010
0.015
0.020
Transformer Voltage Across
120V Winding
160
-20
0.025
Time - Seconds
56
B-H Curve Constructed from V-I Measurements Shows Linear
Region, Saturation, Hysteresis, and Residual Magnetism
0.9
Severe hysteresis
0.8
Residual
magnetism
0.7
Volt-Seconds
0.6
0.5
0.4
0.3
Shape of normal
hysteresis path
0.2
0.1
0.0
Residual
magnetism
-0.1
-150
-100
-50
0
50
100
150
200
250
300
Est. Magnetizing Amps
57
Distribution Feeder Loss
Example
Annual Loss
Secondary Lines
21%
Transformer NoLoad
45%
• Annual energy loss = 2.40%
• Largest component is transformer noload loss (45% of the 2.40%)
Primary Lines
26%
Demand values for the peak hour of (load + loss)
Consumption/Demand
Total Loss
Line Loss (Wires)
Transformer Loss (load plus no-load)
Load Loss (Wires and transformers)
No-Load Loss (Transformer magnetizing)
Primary Loss (Includes transformers)
Secondary Loss (No transformers)
Primary Lines (Wires)
Secondary Lines (Wires)
No-Load Loss (Transformer magnetizing)
Transformer Load Loss
At Peak Hour
Total kW
% of Consump
5665
173
3.06%
123
2.18%
50
0.88%
144
2.54%
29
0.52%
116
2.05%
57
1.01%
66
1.17%
57
1.01%
29
0.52%
21
0.36%
Annual Energy
Total kWh % of Consumpt
22222498
534293
2.40%
250568
1.13%
283726
1.28%
291879
1.31%
242414
1.09%
421316
1.90%
112978
0.51%
137590
0.62%
112978
0.51%
242414
1.09%
41312
0.19%
Transformer Load
8%
Modern Distribution Transformer:
• Load loss at rated load (I2R in conductors) = 0.75% of rated transformer kW.
• No load loss at rated voltage (magnetizing, core steel) = 0.2% of rated
transformer kW.
• Magnetizing current = 0.5% of rated transformer amperes
Single-Phase Transformer
Impedance Reflection from High-Side (H) to Low-Side (L) by the
Square of the Turns Ratio
jXs
Rs
Rm
Faraday’s law
Conservation of
power
Ideal
Transformer
7200:240V
jXm
7200V
240V
2
~
~
~
~ ~
~ ~
 NH 
~~
VH N H ~ ~
I H VL
N L Z H VH / I H
VH / I H

, VH I H  VL I L , so ~  ~ 
,
 ~ ~ 
 
~ 

~
~
N
N
N
N
Z
N
VL
I L VH
VL / I L V
L /I
H
L
H
L
L 

H
H
NH
NL
Ideal
Transformer
7200:240V
7200V
 240 
jX m 

 7200 
240V
2
 240 
Rm 

 7200 
2
 240 
jX s 

7200


2
 240 
Rs 

7200


2
59
Now Work a Single-Phase Transformer Example
Open circuit and short circuit tests are performed on asingle-phase, 7200:240V, 25kVA, 60Hz
distribution transformer. The results are:

Short circuit test (short circuit the low-voltage side, energize the high-voltage side so that
rated current flows, and measure Psc and Qsc). Measured Psc = 400W, Qsc = 200VAr.

Open circuit test (open circuit the high-voltage side, apply rated voltage to the low-voltage
side, and measure Poc and Qoc). Measured Poc = 100W, Q oc = 250VAr.
Determine the four impedance values (in ohms) for the transformer model shown.
Rs
Φ
jXs
Rm
jXm
Ideal
Transformer
7200:240V
Turns ratio 7200:240
7200V
240V
(30 : 1)
(but approx. same amount of
copper in each winding) 60
A three-phase transformer can be three separate
single-phase transformers, or one large
transformer with three sets of windings
Wye-Equivalent One-Line Model
A
Rs
jXs
N1:N2
Rm
jXm
Ideal
Transformer
N1 : N2
N
N1:N2
N1:N2
Y-Y
• Reflect side 1 wye ohms to side 2 wye ohms
by multiplying by [N2 / N1]^2
Standard 345/138kV autotransformers, GY - GY,
with a tertiary 12.5kV Δ winding to provide
circulating 3rd harmonic current
61
For Delta-Delta Connection Model, Convert the
Transformer to Equivalent Wye-Wye
Wye-Equivalent One-Line Model
A
N1:N2
jXs
3
Rm
3
N
N1:N2
Rs
3
jXm
3
Ideal
Transformer
N1 N 2
:
3
3
• Reflect side 1 delta ohms to side 2 delta
ohms by multiplying by [N2 / N1]^2
• Convert side 2 delta ohms to wye ohms by
dividing by 3
N1:N2
Δ-Δ
• Convert side 1 delta ohms to wye ohms by
dividing by 3
• Above circuit results in the proper reflection.
Note that N2/Sqrt3 divided by N1/Sqrt3 is the
62
same as N2 divided by N1
For Delta-Wye Connection Model, Convert the
Transformer to Equivalent Wye-Wye
Standard building entrance
and substation transformers.
Δ high side/ GY low side
Wye-Equivalent One-Line Model
A
N1:N2
N1:N2
Rs
3
jXs
3
Rm
3
N
jXm
3
Ideal
Transformer
N1
: N2
3
• Reflect side 1 delta ohms to side 2 wye ohms
by multiplying by [N2 / N1]^2
• Convert side 1 delta ohms to wye ohms by
dividing by 3
• Above circuit results in the proper reflection
N1:N2
Δ-Y
63
For Wye-Delta Connection Model, Convert the
Transformer to Equivalent Wye-Wye
Wye-Equivalent One-Line Model
A
N1:N2
Rs
jXs
Ideal
Transformer
Rm
jXm
N1 :
N
N1:N2
N2
3
• Reflect side 1 wye ohms to side 2 delta ohms
by multiplying by [N2 / N1]^2
• Convert side 2 impedances from delta ohms
to wye ohms by dividing by 3
• Above circuit results in the proper reflection
N1:N2
Y-Δ
So, for all configurations, the equivalent wye-wye transformer ohms can be
reflected from one side to the other using the three-phase bank line-to-line
64
turns ratio
For wye-delta and delta-wye configurations, there is a
phase shift in line-to-line voltages because
• the individual transformer windings on one side
are connected line-to-neutral, and on the other
side are connected line-to-line
• But there is no phase shift in any of the
individual transformers
• This means that line-to-line voltages on the
delta side are in phase with line-to-neutral
voltages on the wye side
• Thus, phase shift in line-to-line voltages from
one side to the other is unavoidable, but it can
be managed by standard labeling to avoid
problems caused by paralleling transformers
65