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8/11/2019 Transistors Power Presentation
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1. Power and RMS Values
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Instantaneous power p(t) flowing into the box
)()()( titvtp Circuit in a box,
two wires
)(ti
)(tv
+
)(ti
)()()()()( titvtitvtp bbaa )(tvaCircuit in a box,
three wires
)(tia
+
)(tib
+
)(tvb
)()( titi ba Any wire can be thevoltage reference
Works for any circuit, as long as all N wires are accounted for. There must
be (N1) voltage measurements, and (N1) current measurements.
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Average value of
periodic instantaneous power p(t)
Tot
otavg dttpTP )(
1
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Two-wire sinusoidal case
)sin()sin()()()( tItVtitvtp oo
)cos(22)cos(2)(
1
IVVI
dttpTP
Tot
otavg
),sin()( tVtv o )sin()( tIti o
2
)2cos()cos()(
tVItp o
)cos( rmsrmsavg IVP Power factor
Average power
zero average
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Root-mean squared value of a
periodic waveform with period T
Tot
otavg dttpT
P )(1
R
VP rmsavg
2
Tot
otrms dttvT
V )(1 22
Apply v(t) to a resistor
Tot
ot
Tot
ot
Tot
otavg dttv
RTdt
R
tv
Tdttp
TP )(
1)(1)(
1 22
Compare to the average power
expression
rms is based on a power concept, describing the
equivalent voltage that will produce a given
average power to a resistor
The average value of the squared voltage
compare
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Root-mean squared value of a periodic
waveform with period T
Tot
ot orms dttV
TV )(sin1
222
Tot
oto
oTot
ot orms
ttT
Vdtt
T
VV
2
)(2sin
2)(2cos1
2
222
,2
22 V
Vrms
Tot
otrms dttvTV )(
1 22
For the sinusoidal case
2
VVrms
),sin()( tVtv o
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10-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
Voltage
Current
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
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11-100
-80
-60
-40
-20
0
20
40
60
80
100
0 30 60 90 120 150 180 210 240 270 300 330 360
Voltage
Current
Determine voltage and current magnitudes and phase angles
Voltage cosine has peak = 100V, phase angle = -90
Current cosine has peak = 50A, phase angle = -135
,902
100~VV
AI 1352
50~
Using a cosine reference,
Phasors
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The average power is
)cos(22
IVPavg
45cos2
50
2
100)135(90cos
2
50
2
100avgP
WPavg 1767
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VoltageCurrent Relationships
)(tiR )(tvR
R
tvti RR
)()(
)(tvL)(tiL
dt
tdiLtvL
)()(
)(tvC)(tiC
dt
tdvCtiC
)()(
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Thanks to Charles Steinmetz, Steady-State AC Problems
are Greatly Simplified with Phasor Analysis
(no differential equations are needed)
RI
VZ
R
RR ~
~
LjI
VZ
L
LL ~
~
CjI
VZ
C
CC
1~
~
R
tvti RR
)()(
dt
tdiLtvL
)()(
dt
tdvCtiC
)()(
Resistor
Inductor
Capacitor
Time Domain Frequency Domain
voltage leads current
current leads voltage
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V1 V2
Problem 10.17
1
20100
4
20100
~
~
2
11
2
1
2
1
2
1
2
1
3
1
4
1
2
1 j
V
V
j
j
2
1
2
1
2
11
2
1
2
1
3
1
4
1
jjD
D
j
j
V
2
11
2
1
1
20100
2
1
4
20100
~1
D
j
j
V
1
20100
2
11
2
1
4
20100
2
1
~2
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c EE411, Problem 10.17
implicit nonedimension v_phasor(2), i_injection_phasor(2), y(2,2)complex v_phasor, i_injection_phasor, y, determinant, i0_phasorreal pi
open(unit=6,file='EE411_Prob_10_17.txt')
pi = 4.0 * atan(1.0)
y(1,1) = 1.0 / cmplx(0.0,4.0)1 + 1.0 / 3.02 + 1.0 / 2.0y(1,2) = -1.0 / 2.0y(2,1) = y(1,2)y(2,2) = 1.0 / 2.0
1 + 1.02 + 1.0 / cmplx(0.0,-2.0)
i_injection_phasor(1) = 100.01 * cmplx(cos(20.0 * pi / 180.0),sin(20.0 * pi / 180.0))2 / cmplx(0.0,4.0)
i_injection_phasor(2) = 100.01 * 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 = ",determinantwrite(6,*) "determinant, polar = ", cabs(determinant),1 atan2(aimag(determinant),real(determinant)) * 180.0 / piwrite(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 / piwrite(6,*)
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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 / piwrite(6,*)
i0_phasor = (v_phasor(1) - v_phasor(2)) / 2.0
write(6,*) "i0_phasor, rectangular = ",i0_phasorwrite(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
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Active and Reactive Power Form a Power Triangle
),cos(22
IVPavg ),sin(22
IVQ
jQPIVS ~
~
)(
VV~
II~ P
Q
Projection of S
on the real axis
Projection of
S on the
imaginary
axis
Complex
power
S
)(
)cos(
is the power factor
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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
0~~~~
CBA IIIV
Consider a node, with voltage (to any reference), and three currents
IA
IB
IC
0~~~
CBA III
0~~~~ * CBA IIIV
0 CCBBAA jQPjQPjQP
0 CBA PPP
0 CBA QQQ
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Voltage and Current Phasors for Rs, Ls, Cs
RRR
RR IRVR
I
VZ
~~,~
~
LLL
LL ILjVLj
I
VZ
~~,~
~
Cj
IV
CjI
VZ CC
C
CC
~~
,1
~
~
Resistor
Inductor
Capacitor
Voltage and
Currentin phase Q = 0
Voltage leadsCurrentby 90
Q > 0
Currentleads
Voltage by 90 Q < 0
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VIIVIVjQPS **~~
cosVIP
sinVIQ
P
Q
Projection of S
on the real axis
Projection of
S on the
imaginaryaxis
Complex
power
S
)(
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RV
Z
V
Z
VVjQPS
2
*
2*~~
RIR
VP 2
2
0Q
RIZIIZIjQPS 22*~~
also
so
Resistor
, Use rms V, I
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LjV
LjV
LjV
ZVVjQPS
22
*
2*~
~
LIL
VQ
22
0P
LjILjIIZIjQPS 22*~~
also
so
Inductor
, Use rms V, I
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2
2
*
2*
11
~~
CVj
Cj
V
Cj
V
Z
V
VjQPS
C
ICVQ
22 0P
C
Ij
CjIIZIjQPS
22* 1~~
also
so
Capacitor
, Use rms V, I
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Active and Reactive Power for Rs, Ls, Cs(a positive value is consumed, a negative value is produced)
0
LIL
Vrms
rms
22
,
RI
R
Vrms
rms 22
,
0
0
Resistor
Inductor
Capacitor
Active Power P Reactive Power Q
,,2
2
C
ICV rmsrms
source of reactive power
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Now, demonstrate Excel spreadsheet
EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q
Vmag = 1
Vang = 0
Imag = 0.90 90
Iang = -30 150
Phase A Phase A Phase A P Q Phase B Phase B Phase B Phase C Phase C Phase C A+B+C Q
wt v(t) I(t) p(t) 0.389711 0.225 v(t) I(t) p(t) v(t) I(t) p(t) p(t) 0.675
0 1 0.779423 0.779423 0.389711 0.225 -0.5 -0.779423 0.389711 -0.5 5.51E-17 -2.76E-17 1.169134 0.675
2 0.999391 0.794653 0.794169 0.389711 0.225 -0.469472 -0.763243 0.358321 -0.529919 -0.03141 0.016645 1.169134 0.6754 0.997564 0.808915 0.806944 0.389711 0.225 -0.438371 -0.746134 0.327084 -0.559193 -0.062781 0.035107 1.169134 0.675
6 0.994522 0.822191 0.817687 0.389711 0.225 -0.406737 -0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134 0.675
8 0.990268 0.834465 0.826345 0.389711 0.225 -0.374607 -0.70921 0.265675 -0.615661 -0.125256 0.077115 1.169134 0.675
10 0.984808 0.845723 0.832875 0.389711 0.225 -0.34202 -0.68944 0.235802 -0.642788 -0.156283 0.100457 1.169134 0.675
12 0.978148 0.855951 0.837246 0.389711 0.225 -0.309017 -0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134 0.675
14 0.970296 0.865136 0.839437 0.389711 0.225 -0.275637 -0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134 0.675
16 0.961262 0.873266 0.839437 0.389711 0.225 -0.241922 -0.625193 0.151248 -0.71934 -0.248074 0.178449 1.169134 0.675
18 0.951057 0.880333 0.837246 0.389711 0.225 -0.207912 -0.602218 0.125208 -0.743145 -0.278115 0.20668 1.169134 0.675
20 0.939693 0.886327 0.832875 0.389711 0.225 -0.173648 -0.578509 0.100457 -0.766044 -0.307818 0.235802 1.169134 0.675
22 0.927184 0.891241 0.826345 0.389711 0.225 -0.139173 -0.554095 0.077115 -0.788011 -0.337146 0.265675 1.169134 0.675
24 0.913545 0.89507 0.817687 0.389711 0.225 -0.104528 -0.529007 0.055296 -0.809017 -0.366063 0.296151 1.169134 0.67526 0.898794 0.897808 0.806944 0.389711 0.225 -0.069756 -0.503274 0.035107 -0.829038 -0.394534 0.327084 1.169134 0.675
28 0.882948 0.899452 0.794169 0.389711 0.225 -0.034899 -0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134 0.675
30 0.866025 0.9 0.779423 0.389711 0.225 6.13E-17 -0.45 -2.76E-17 -0.866025 -0.45 0.389711 1.169134 0.675
32 0.848048 0.899452 0.762778 0.389711 0.225 0.034899 -0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134 0.675
34 0.829038 0.897808 0.744316 0.389711 0.225 0.069756 -0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134 0.675
36 0.809017 0.89507 0.724127 0.389711 0.225 0.104528 -0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134 0.675
38 0.788011 0.891241 0.702308 0.389711 0.225 0.139173 -0.337146 -0.046922 -0.927184 -0.554095 0.513748 1.169134 0.675
Instantaneous Power in Single-Phase Circuit
-1.5
0
1.5
0 90 180 270 360 450 540 630 720
va
ia
pa
P
Q
Instantaneous Power in Three-Phase Circuit
-1.5
0
1.5
0 90 180 270 360 450 540 630 720
va
ia
vb
ib
vc
icpa+pb+pc
Q
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A load consists of a 47 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 currentb.
Determine the load P, pf, Q, and S.
c.
Find an expression for instantaneous p(t)
A Single-Phase Power Example
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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.
0.05 + j0.15
pu ohms
j0.20 pu mhos
PL+ jQLVL= 1.020 /0 VR= 1.010 /-10
PR+ jQR
IS
IcapL IcapRj0.20 pu mhos
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implicit nonecomplex vl_phasor,sl,icapl_phasor,zcl,is_phasor,zlinecomplex vr_phasor,sr,icapr_phasor,zcr
real vlmag,vlang,vrmag,vrang,pi,qcapl,qcaprreal vl_mag,vl_ang,vr_mag,vr_angreal rline, xline, bcapreal pl,ql,pr,qr,is_mag,is_ang,icapl_mag,icapl_ang,icapr_mag,icapr_ang
real qline_loss
open(unit=6,file="EE411_Trans_Line.dat")pi = 4.0 * atan(1.0)
vl_mag = 1.02vl_ang = 0.0vr_mag = 1.01vr_ang = -10.0rline = 0.05xline = 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 = real(sl)ql = aimag(sl)
pr = real(sr)qr = aimag(sr)
write(6,*) "is_phasor (rectangular) = ",is_phasoris_mag = cabs(is_phasor)is_ang = atan2(aimag(is_phasor),real(is_phasor)) * 180.0 / pi
write(6,*) "is_phasor (polar) ",is_mag,is_angwrite(6,*)
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write(6,*) "icapl_phasor (rectangular) = ",icapl_phasoricapl_mag = cabs(icapl_phasor)icapl_ang = atan2(aimag(icapl_phasor),real(icapl_phasor)) * 180.0 / pi
write(6,*) "icapl_phasor (polar) ",icapl_mag,icapl_angwrite(6,*)
write(6,*) "icapr_phasor (rectangular) = ",icapr_phasoricapr_mag = cabs(icapr_phasor)icapr_ang = atan2(aimag(icapr_phasor),real(icapr_phasor)) * 180.0 / pi
write(6,*) "icapr_phasor (polar) ",icapr_mag,icapr_angwrite(6,*)
qcapl = cabs(vl_phasor) * cabs(vl_phasor) * (-bcap)qcapr = cabs(vr_phasor) * cabs(vr_phasor) * (-bcap)
write(6,*) "pl = ",plwrite(6,*) "ql = ",qlwrite(6,*)
write(6,*) "pr = ",prwrite(6,*) "qr = ",qrwrite(6,*)
write(6,*) "qcapl = ",qcaplwrite(6,*) "qcapr = ",qcaprwrite(6,*)
write(6,*) "pl + pr = ",(pl + pr)write(6,*) "ql + qr = ",(ql + qr)write(6,*)
write(6,*) "pline_loss = ",cabs(is_phasor) * cabs(is_phasor) * rlineqline_loss = cabs(is_phasor) * cabs(is_phasor) * xline
write(6,*) "qline_loss = ",qline_losswrite(6,*) "qline_loss + qcapl + qcapr = ",(qline_loss + qcapl + qcapr)write(6,*)
end
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-----------------------------------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 = 1.125056ql = -0.4107586
pr = -1.062252qr = 0.1870712
qcapl = -0.2080800qcapr = -0.2040200
pl + pr = 6.2804222E-02ql + qr = -0.2236874
pline_loss = 6.2804200E-02qline_loss = 0.1884126
qline_loss + qcapl + qcapr = -0.2236874
0.05 + j0.15
pu ohms
j0.20 pu mhos
PL+ jQL
VL= 1.020 /0 VR= 1.010 /-10
PR+ jQR
IS
IcapL IcapRj0.20 pu mhos
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RMS of some common periodic waveforms
22
0
2
0
22 1)(1
DVDT
T
VdtV
T
dttv
T
V
DTT
rms
DVVrms
Duty cycle controller
DTT
V
0
0 < D < 1
By inspection, this is
the average value of
the squaredwaveform
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RMS of common periodic waveforms, cont.
TTT
rms t
T
Vdtt
T
Vdtt
T
V
T
V
0
3
3
2
0
2
3
2
0
22
3
1
T
V
0
3
VVrms
Sawtooth
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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
0
V
0
V
0
0
-V
V
0
3
VVrms
V
0
V
0
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2. Three-Phase Circuits
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Three Important Properties of Three-Phase
Balanced Systems
Because they form a balanced set, the a-b-ccurrents sum to zero. Thus, there is no returncurrent through the neutral or ground, whichreduces wiring losses.
A N-wire system needs (N1) meters. A three-phase, four-wire system needs three meters. Athree-phase, three-wire system needs only twometers.
The instantaneous power is constant
Three-phase,
four wire system
a
b
c
n
Reference
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Observe Constant Three-Phase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls
Vmag = 1
Vang = 0
Imag = 0.90 90
Iang = -30 150
Phase A Phase A Phase A P Q Phase B Phase B Phase B Phase C Phase C Phase C A+B+C Q
wt v(t) I(t) p(t) 0.389711 0.225 v(t) I(t) p(t) v(t) I(t) p(t) p(t) 0.675
0 1 0.779423 0.779423 0.389711 0.225 -0.5 -0.779423 0.389711 -0.5 5.51E-17 -2.76E-17 1.169134 0.675
2 0.999391 0.794653 0.794169 0.389711 0.225 -0.469472 -0.763243 0.358321 -0.529919 -0.03141 0.016645 1.169134 0.6754 0.997564 0.808915 0.806944 0.389711 0.225 -0.438371 -0.746134 0.327084 -0.559193 -0.062781 0.035107 1.169134 0.675
6 0.994522 0.822191 0.817687 0.389711 0.225 -0.406737 -0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134 0.675
8 0.990268 0.834465 0.826345 0.389711 0.225 -0.374607 -0.70921 0.265675 -0.615661 -0.125256 0.077115 1.169134 0.675
10 0.984808 0.845723 0.832875 0.389711 0.225 -0.34202 -0.68944 0.235802 -0.642788 -0.156283 0.100457 1.169134 0.675
12 0.978148 0.855951 0.837246 0.389711 0.225 -0.309017 -0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134 0.675
14 0.970296 0.865136 0.839437 0.389711 0.225 -0.275637 -0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134 0.675
16 0.961262 0.873266 0.839437 0.389711 0.225 -0.241922 -0.625193 0.151248 -0.71934 -0.248074 0.178449 1.169134 0.675
18 0.951057 0.880333 0.837246 0.389711 0.225 -0.207912 -0.602218 0.125208 -0.743145 -0.278115 0.20668 1.169134 0.675
20 0.939693 0.886327 0.832875 0.389711 0.225 -0.173648 -0.578509 0.100457 -0.766044 -0.307818 0.235802 1.169134 0.675
22 0.927184 0.891241 0.826345 0.389711 0.225 -0.139173 -0.554095 0.077115 -0.788011 -0.337146 0.265675 1.169134 0.675
24 0.913545 0.89507 0.817687 0.389711 0.225 -0.104528 -0.529007 0.055296 -0.809017 -0.366063 0.296151 1.169134 0.67526 0.898794 0.897808 0.806944 0.389711 0.225 -0.069756 -0.503274 0.035107 -0.829038 -0.394534 0.327084 1.169134 0.675
28 0.882948 0.899452 0.794169 0.389711 0.225 -0.034899 -0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134 0.675
30 0.866025 0.9 0.779423 0.389711 0.225 6.13E-17 -0.45 -2.76E-17 -0.866025 -0.45 0.389711 1.169134 0.675
32 0.848048 0.899452 0.762778 0.389711 0.225 0.034899 -0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134 0.675
34 0.829038 0.897808 0.744316 0.389711 0.225 0.069756 -0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134 0.675
36 0.809017 0.89507 0.724127 0.389711 0.225 0.104528 -0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134 0.675
38 0.788011 0.891241 0.702308 0.389711 0.225 0.139173 -0.337146 -0.046922 -0.927184 -0.554095 0.513748 1.169134 0.675
Instantaneous Power in Single-Phase Circuit
-1.5
0
1.5
0 90 180 270 360 450 540 630 720
va
ia
pa
P
Q
Instantaneous Power in Three-Phase Circuit
-1.5
0
1.5
0 90 180 270 360 450 540 630 720
va
ia
vb
ib
vc
icpa+pb+pc
Q
Imaginary
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The phasorsare rotating counter-clockwise.
The magnitude of line-to-line voltage phasors is 3 times the magnitude of line-to-neutral voltage phasors.
Vbn
Vab= VanVbn
Vbc=
VbnVcn
Van
Vcn
30
120
Real
Vca= VcnVan
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Conservation of power requires that the magnitudes of deltacurrents Iab, Ica, and Ibcare3
1
times the magnitude of line currents Ia, Ib, Ic.
Van
Vbn
Vcn
Real
Imaginary
Vab= VanVbn
Vbc=
VbnVcn
30
Vca= VcnVan
Ia
Ib
Ic
Iab
Ibc
Ica
Ib
Ic
Iab
Ica
Ibc
Ia
a
c
b
Vab +
Balanced Sets Add to Zero in BothTime and Phasor Domains
Ia+ Ib+ Ic= 0
Van+ Vbn+ Vcn= 0
Vab+ Vbc+ Vca= 0
Line currents Ia, Ib, and Ic
Delta currents Iab, Ibc, and Ica
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The Two Above Loads are Equivalent in Balanced Systems
(i.e., same line currents Ia, Ib, Icand phase-to-phase voltages Vab, Vbc, Vcain both cases)
3Z
3Z3Z
a
c
b
Vab +
Ia
Ib
Ic
Z
ZZ
c
b
Vab
Ia
Ib
Ic
n
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The Two Above Sourcesare Equivalent in Balanced Systems(i.e., same line currents Ia, Ib, Icand phase-to-phase voltages Vab, Vbc, Vcain both cases)
a
c
b
Vab +
Ia
Ib
Ic
Van
a
c
b
Vab +
Ia
Ib
Ic
n+
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Z
ZZ
ab
Vab +
Ia
Ib
Ic
c
nIn
KCL: In= Ia+ Ib+ Ic
But fora balanced set,
Ia
+ Ib
+ Ic
= 0, so In
= 0
Ground (i.e., V = 0)
The Experiment: Opening and closing the switch has no effect because I nisalready zero for a three-phase
balanced set. Since no current flows, even if there is a resistance in the grounding path, we must conclude thatVn= 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.
Zline
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Balanced three-phase systems, no matter if they are deltaconnected, wye connected, or a mix, are easy to solve if youfollow these steps:1. Convert the entire circuit to an equivalent wyewith a
groundedneutral.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 usingthe 120 relationships.5. If needed, compute line-to-line voltages and deltacurrents
usingthe 3 and 30relationships.
a
n
a
n
Zload+
Van
Zline
Ia
a
c
b
Vab +
3Zload
a
c
b
Ib
Ia
Ic
Zline
Zline
3Zload3Zload
The One-LineDiagram
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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 Vanhas phase angle zero,
Find phasor currents Iaand Iaband (noteIabis 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 Iaafter the kVArs are added?
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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 5.
Part a. If each Z is (90 + j45), 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 anV~
at the source has phase angle zero, find ''~baV at the load.
Z
ZZ
Part c. Draw a phasor diagram that shows line currents Ia, Ib, and Ic, and
load currents Iab, Ibc, and Ica.
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3. Transformers
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Transformer Core Types
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High-Voltage Grid Transformers, 100s of MW
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Single-Phase Transformer
Rs jXs
Ideal
Transformer7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of
copper in each winding)
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Short Circuit Test
Rs jXsIdeal
Transformer7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(but approx. same amount of
copper in each winding)
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.
sc
scss
I
VjXR ~
~
+
Vsc
-
Isc
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Open Circuit Test
Rs jXsIdeal
Transformer7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(but approx. same amount of
copper in each winding)
+
Voc
-
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.
oc
ocmm
I
VjXR ~
~
||
Ioc
1. Given the standard percentage values below for a 125kVA transformer,
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Single Phase Transformer.
Percent values are given
on transformer base.
Winding 1
kv = 7.2, kVA = 125
Winding 2
kv = 0.24, kVA = 125
%imag = 0.5
%loadloss = 0.9
%noloadloss = 0.2
%Xs = 2.2
Rs jXs
Ideal
Transformer
7200:240V
Rm jXm
7200V 240V
Magnetizing
current
No
load
loss
XsLoad
loss
3. If standard open circuit and short circuit
tests are performed on this transformer, what
will be the Ps and Qs (Watts and VArs)
measured in those tests?
1. Given the standard percentage values below for a 125kVA transformer,
determine the Rs and Xs in the diagram, in .
2. If the Rs and Xs are moved to the 240V side, compute the new values.
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Rs jXs
Ideal
TransformerRm jXm
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%
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Linear Scale Log10 Scale
Saturationrelative 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
T f C S t t
i
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Transformer Core Saturation
-6
-4
-2
0
2
4
6
Am
peres
Magnetizing Currentfor Single-Phase
25 kVA. 12.5kV/240V Transformer.
THDi = 76.1%, Mostly 3rdHarmonic.
Log10 Scale
Linear Scale
No DC
No DC
With DC
A l DC V lt t T f d W t h It S t t
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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
-20
0
20
40
60
80
100
120
140
160
-0.005 0.000 0.005 0.010 0.015 0.020 0.025
Time - Seconds
TransformerCurrentin120V
W
inding
-20
0
20
40
60
80
100
120
140
160
TransformerVoltageAcross
120V
Winding
\
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
VoltageCurrent
Saturates
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-0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
-150 -100 -50 0 50 100 150 200 250 300
Est. Magnetizing Amps
Volt-Seconds
B-H Curve Constructed from V-I Measurements Shows Linear
Region, Saturation, Hysteresis, and Residual Magnetism
Shape of normal
hysteresis path
Severe hysteresis
Residual
magnetism
Residual
magnetism
Distribution Feeder Loss Secondary Lines21%Annual Loss
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Example Annual energy loss = 2.40%
Largest component is transformer no-load loss (45% of the 2.40%)
Transformer No-
Load
45%
Transformer Load
8%
Primary Lines
26%
21%
Demand values for the peak hour of (load + loss) Total kW % of Consump Total kWh % of Consumpt
Consumption/Demand 5665 22222498
Total Loss 173 3.06% 534293 2.40%
Line Loss (Wires) 123 2.18% 250568 1.13%
Transformer Loss (load plus no-load) 50 0.88% 283726 1.28%Load Loss (Wires and transformers) 144 2.54% 291879 1.31%
No-Load Loss (Transformer magnetizing) 29 0.52% 242414 1.09%
Primary Loss (Includes transformers) 116 2.05% 421316 1.90%
Secondary Loss (No transformers) 57 1.01% 112978 0.51%
Primary Lines (Wires) 66 1.17% 137590 0.62%
Secondary Lines (Wires) 57 1.01% 112978 0.51%
No-Load Loss (Transformer magnetizing) 29 0.52% 242414 1.09%
Transformer Load Loss 21 0.36% 41312 0.19%
Annual EnergyAt Peak Hour
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
Si l Ph T f
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Single-Phase Transformer
Impedance Reflection from High-Side (H) to Low-Side (L) by the
Square of the Turns Ratio
Rs jXsIdeal
Transformer7200:240V
Rm jXm
7200V 240V
IdealTransformer7200:240V
7200V 240V
2
7200
240
sjX 2
7200
240
sR
2
7200
240
mjX
2
7200
240
mR
2
~/
~
~/
~
~/
~
~/
~
,~
~
~
~
so,~~~~
,~
~
L
H
L
HH
H
LH
HH
LL
HH
L
H
H
L
H
L
L
HLLHH
L
H
L
H
N
N
N
NI
N
NV
IV
IV
IV
Z
Z
N
N
V
V
I
IIVIV
N
N
V
V
Faradays law Conservation ofpower
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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 thatrated current flows, and measure Pscand Qsc). MeasuredPsc= 400W, Qsc= 200VAr.
Open circuit test (open circuit the high-voltage side, apply rated voltage to the low-voltageside, and measure Pocand Qoc). MeasuredPoc= 100W, Qoc= 250VAr.
Determine the four impedancevalues(in ohms) for the transformer model shown.
Rs jXs
Ideal
Transformer7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of
copper in each winding)
A three phase transformer can be three separate
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Y - Y
A three-phase transformer can be three separate
single-phase transformers, or one large
transformer with three sets of windings
N1:N2
N1:N2
N1:N2
Rs jXs
IdealTransformer
N1 : N2
Rm jXm
Wye-Equivalent One-Line Model
A
N
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 winding to provide
circulating 3rdharmonic current
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-
For Delta-DeltaConnection Model, Convert the
Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
IdealTransformer
3
Rs
3
2:
3
1 NN
3
jXs
3
Rm
3
jXm
A
N
Wye-Equivalent One-Line Model
Reflect side 1 delta ohms to side 2 delta
ohms by multiplying by [N2 / N1]^2
Convert side 2 delta ohms to wye ohms bydividing by 3
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
same as N2 divided by N1
For Delta-Wye Connection Model Convert the
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- Y
For Delta WyeConnection Model, Convert the
Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
IdealTransformer
3
Rs
2:3
1N
N
3
jXs
3
Rm
3
jXm
A
N
Wye-Equivalent One-Line Model
Reflect side 1 delta ohms to side 2 wye ohms
by multiplying by [N2 / N1]^2
Convert side 1 delta ohms to wye ohms bydividing by 3
Above circuit results in the proper reflection
Standard building entrance
and substation transformers.
high side/ GYlow side
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Y -
For Wye-DeltaConnection Model, Convert the
Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
IdealTransformer
3
2:1 N
N
jXs
Rm jXm
A
N
Rs
Wye-Equivalent One-Line Model
So, for all configurations, the equivalent wye-wye transformer ohmscan be
reflected from one side to the other using the three-phase bank line-to-line
turns ratio
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
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For wye-delta and delta-wye configurations, there is a
phase shift in line-to-linevoltages because
the individual transformer windings on one sideare 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