1 1. power and rms values. 2 instantaneous power p(t) flowing into the box circuit in a box, two...
TRANSCRIPT
1
1. Power and RMS Values
2
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 the voltage reference
Works for any circuit, as long as all N wires are accounted for. There must be (N – 1) voltage measurements, and (N – 1) current measurements.
3
Average value ofperiodic instantaneous power p(t)
Tot
otavg dttp
TP )(
1
4
Two-wire sinusoidal case
)sin()sin()()()( tItVtitvtp oo
)cos(22
)cos(2
)(1
IVVIdttp
TP
Tot
otavg
),sin()( tVtv o )sin()( tIti o
2
)2cos()cos()(
tVItp o
)cos( rmsrmsavg IVP Power factor
Average power
zero average
5
Root-mean squared value of a periodic waveform with period T
Tot
otavg dttp
TP )(
1
R
VP rmsavg
2
Tot
otrms dttv
TV )(
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
6
Root-mean squared value of a periodic waveform with period T
Tot
otorms dttV
TV )(sin
1 222
Tot
oto
oTot
otorms
tt
T
Vdtt
T
VV
2
)(2sin
2)(2cos1
2
222
,2
22 V
Vrms
Tot
otrms dttv
TV )(
1 22
For the sinusoidal case
2
VVrms
),sin()( tVtv o
7-100
-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
8-100
-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 = -90º
Current cosine has peak = 50A, phase angle = -135º
, 902
100~VV AI 135
2
50~
Using a cosine reference,
Phasors
9
The average power is
)cos(22
IV
Pavg
45cos2
50
2
100)135(90cos
2
50
2
100avgP
WPavg 1767
10
Voltage – Current Relationships
)(tiR )(tvR
R
tvti R
R)(
)(
)(tvL)(tiL
dt
tdiLtvL
)()(
)(tvC)(tiC
dt
tdvCtiC
)()(
11
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 R
R)(
)(
dt
tdiLtvL
)()(
dt
tdvCtiC
)()(
Resistor
Inductor
Capacitor
Time Domain Frequency Domain
voltage leads current
current leads voltage
12
201004
20100
~
~
2
11
2
1
2
12
1
2
1
3
1
4
1
2
1j
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
120100
2
1
4
20100
~1
D
j
j
V
20100
2
11
2
14
20100
2
1
~2
V1 V2
Problem 10.17
1
201004
20100
~
~
2
11
2
1
2
12
1
2
1
3
1
4
1
2
1 jV
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
201002
1
4
20100
~1
D
j
j
V
1
20100
2
11
2
14
20100
2
1
~2
13
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) = 1.0 / cmplx(0.0,4.0) 1 + 1.0 / 3.0 2 + 1.0 / 2.0 y(1,2) = -1.0 / 2.0 y(2,1) = y(1,2) y(2,2) = 1.0 / 2.0 1 + 1.0 2 + 1.0 / 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,*) 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
14
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) = 1.0 / cmplx(0.0,4.0) 1 + 1.0 / 3.0 2 + 1.0 / 2.0 y(1,2) = -1.0 / 2.0 y(2,1) = y(1,2) y(2,2) = 1.0 / 2.0 1 + 1.0 2 + 1.0 / 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,*) 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
15
Active and Reactive Power Form a Power Triangle
),cos(22
IV
Pavg ),sin(22
IV
Q
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
16
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
17
Voltage and Current Phasors for R’s, L’s, C’s
RRR
RR IRVR
I
VZ
~~ ,~
~
LLL
LL ILjVLj
I
VZ
~~ ,~
~
Cj
IV
CjI
VZ C
CC
CC
~~
,1
~
~
Resistor
Inductor
Capacitor
Voltage and Current in phase Q = 0
Voltage leads Current by 90°
Q > 0
Current leads Voltage by 90° Q < 0
18
VIIVIVjQPS **~~
cosVIP
sinVIQ
P
Q
Projection of S on the real axis
Projection of S on the
imaginary axis
Complex power
S
)(
19
R
V
Z
V
Z
VVjQPS
2
*
2*~~
RIR
VP 2
2 0Q
RIZIIZIjQPS 22*~~
also
so
Resistor
, Use rms V, I
20
L
jV
Lj
V
Lj
V
Z
VVjQPS
22
*
2*~~
LIL
VQ
2
20P
LjILjIIZIjQPS 22*~~
also
so
Inductor
, Use rms V, I
21
22
*
2*
11
~~
CVj
Cj
V
Cj
V
Z
VVjQPS
C
ICVQ
22 0P
C
Ij
CjIIZIjQPS
22* 1~~
also
so
Capacitor
, Use rms V, I
22
Active and Reactive Power for R’s, L’s, C’s(a positive value is consumed, a negative value is produced)
0
LIL
Vrms
rms
22
,
RIR
Vrms
rms 22
,
0
0
Resistor
Inductor
Capacitor
Active Power P Reactive Power Q
, , 2
2
C
ICV rmsrms
source of reactive power
23
Now, demonstrate Excel spreadsheet
EE411_Voltage_Current_Power.xls
to show the relationship between v(t), i(t), p(t), P, and Q
Vmag = 1Vang = 0Imag = 0.90 90Iang = -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 Qwt 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 0.6752 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 0.6756 0.994522 0.822191 0.817687 0.389711 0.225 -0.406737 -0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134 0.6758 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 0.67512 0.978148 0.855951 0.837246 0.389711 0.225 -0.309017 -0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134 0.67514 0.970296 0.865136 0.839437 0.389711 0.225 -0.275637 -0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134 0.67516 0.961262 0.873266 0.839437 0.389711 0.225 -0.241922 -0.625193 0.151248 -0.71934 -0.248074 0.178449 1.169134 0.67518 0.951057 0.880333 0.837246 0.389711 0.225 -0.207912 -0.602218 0.125208 -0.743145 -0.278115 0.20668 1.169134 0.67520 0.939693 0.886327 0.832875 0.389711 0.225 -0.173648 -0.578509 0.100457 -0.766044 -0.307818 0.235802 1.169134 0.67522 0.927184 0.891241 0.826345 0.389711 0.225 -0.139173 -0.554095 0.077115 -0.788011 -0.337146 0.265675 1.169134 0.67524 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 0.67528 0.882948 0.899452 0.794169 0.389711 0.225 -0.034899 -0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134 0.67530 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 0.67532 0.848048 0.899452 0.762778 0.389711 0.225 0.034899 -0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134 0.67534 0.829038 0.897808 0.744316 0.389711 0.225 0.069756 -0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134 0.67536 0.809017 0.89507 0.724127 0.389711 0.225 0.104528 -0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134 0.67538 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
ic
pa+pb+pc
Q
24
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)
A Single-Phase Power Example
25
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 + jQL
VL = 1.020 /0° VR = 1.010 /-10°
PR + jQR
IS
IcapL IcapRj0.20 pu mhos
26
implicit none complex vl_phasor,sl,icapl_phasor,zcl,is_phasor,zline complex vr_phasor,sr,icapr_phasor,zcr real vlmag,vlang,vrmag,vrang,pi,qcapl,qcapr real vl_mag,vl_ang,vr_mag,vr_ang real rline, xline, bcap real 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 = 1.02 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 = real(sl) ql = aimag(sl) pr = real(sr) qr = 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,*) 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
27
implicit none complex vl_phasor,sl,icapl_phasor,zcl,is_phasor,zline complex vr_phasor,sr,icapr_phasor,zcr real vlmag,vlang,vrmag,vrang,pi,qcapl,qcapr real vl_mag,vl_ang,vr_mag,vr_ang real rline, xline, bcap real 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 = 1.02 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 = real(sl) ql = aimag(sl) pr = real(sr) qr = 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,*) 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,*) "pline_loss = ",cabs(is_phasor) * cabs(is_phasor) * rline qline_loss = cabs(is_phasor) * cabs(is_phasor) * xline write(6,*) "qline_loss = ",qline_loss write(6,*) "qline_loss + qcapl + qcapr = ",(qline_loss + qcapl + qcapr) write(6,*) end
28
----------------------------------- 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 = 1.125056 ql = -0.4107586 pr = -1.062252 qr = 0.1870712 qcapl = -0.2080800 qcapr = -0.2040200 pl + pr = 6.2804222E-02 ql + qr = -0.2236874 pline_loss = 6.2804200E-02 qline_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
29
RMS of some common periodic waveforms
22
0
2
0
22 1)(
1DVDT
T
VdtV
Tdttv
TV
DTT
rms
DVVrms
Duty cycle controller
DT
T
V
0
0 < D < 1
By inspection, this is the average value of
the squared waveform
30
RMS of common periodic waveforms, cont.
TTT
rms tT
Vdtt
T
Vdtt
T
V
TV
0
33
2
0
23
2
0
22
3
1
T
V
0
3
VVrms
Sawtooth
31
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
32
2. Three-Phase Circuits
33
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 three-phase, four-wire system needs three meters. A three-phase, three-wire system needs only two meters.
• The instantaneous power is constant
Three-phase, four wire system
abcn
Reference
34
Observe Constant Three-Phase P and Q in Excel spreadsheet
1_Single_Phase_Three_Phase_Instantaneous_Power.xls
Vmag = 1Vang = 0Imag = 0.90 90Iang = -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 Qwt 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 0.6752 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 0.6756 0.994522 0.822191 0.817687 0.389711 0.225 -0.406737 -0.728115 0.296151 -0.587785 -0.094076 0.055296 1.169134 0.6758 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 0.67512 0.978148 0.855951 0.837246 0.389711 0.225 -0.309017 -0.66883 0.20668 -0.669131 -0.187121 0.125208 1.169134 0.67514 0.970296 0.865136 0.839437 0.389711 0.225 -0.275637 -0.647406 0.178449 -0.694658 -0.21773 0.151248 1.169134 0.67516 0.961262 0.873266 0.839437 0.389711 0.225 -0.241922 -0.625193 0.151248 -0.71934 -0.248074 0.178449 1.169134 0.67518 0.951057 0.880333 0.837246 0.389711 0.225 -0.207912 -0.602218 0.125208 -0.743145 -0.278115 0.20668 1.169134 0.67520 0.939693 0.886327 0.832875 0.389711 0.225 -0.173648 -0.578509 0.100457 -0.766044 -0.307818 0.235802 1.169134 0.67522 0.927184 0.891241 0.826345 0.389711 0.225 -0.139173 -0.554095 0.077115 -0.788011 -0.337146 0.265675 1.169134 0.67524 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 0.67528 0.882948 0.899452 0.794169 0.389711 0.225 -0.034899 -0.476927 0.016645 -0.848048 -0.422524 0.358321 1.169134 0.67530 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 0.67532 0.848048 0.899452 0.762778 0.389711 0.225 0.034899 -0.422524 -0.014746 -0.882948 -0.476927 0.421102 1.169134 0.67534 0.829038 0.897808 0.744316 0.389711 0.225 0.069756 -0.394534 -0.027521 -0.898794 -0.503274 0.452339 1.169134 0.67536 0.809017 0.89507 0.724127 0.389711 0.225 0.104528 -0.366063 -0.038264 -0.913545 -0.529007 0.483272 1.169134 0.67538 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
ic
pa+pb+pc
Q
35
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.
Vbn
Vab = Van – Vbn
Vbc =
Vbn – Vcn
Van
Vcn
30°
120°
Imaginary
Real
Vca = Vcn – Van
36
Conservation of power requires that the magnitudes of delta currents Iab, Ica, and Ibc are 3
1
times the magnitude of line currents Ia, Ib, Ic.
Van
Vbn
Vcn
Real
Imaginary
Vab = Van – Vbn
Vbc =
Vbn – Vcn
30°
Vca = Vcn – Van
Ia
Ib
Ic
Iab
Ibc
Ica
Ib
Ic
Ibc
Ia
c
b
– Vab +
Balanced Sets Add to Zero in Both Time 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
37
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)
3Z
3Z 3Z
a
c
b
– Vab +
Ia
Ib
Ic
Z
Z Z
c
b
– Vab +
Ia
Ib
Ic
n
38
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)
a
c
b
– Vab +
Ia
Ib
Ic
Van
c
b
– Vab +
Ia
Ib
Ic
n +
–
39
Z
Z Z
a b
– Vab +
Ia
Ib
Ic
c
n In
KCL: In = Ia + Ib + Ic But for a 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 In 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 balanced 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.
40
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 wye with 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.
a
n
a
n
Zload +
Van –
Zline
Ia
a
c
b
– Vab + 3Zload
a
c
b
Ib
Ia
Ic
Zline
Zline
Zline
3Zload 3Zload
The “One-Line” Diagram
41
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 supply voltage, what will be the new after the kVArs are added?
42
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. 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 anV~
at the source has phase angle zero, find ''~baV at the load.
Z
Z Z
43
3. Transformers
44
Single-Phase Transformer
Rs jXs Ideal
Transformer 7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding)
Φ
45
Short Circuit Test
Rs jXs Ideal
Transformer 7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(but approx. same amount of copper in each winding)
Φ
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. Calculate
sc
scss I
VjXR ~
~
+Vsc
-
Isc
46
Open Circuit Test
Rs jXs Ideal
Transformer 7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(but approx. same amount of copper in each winding)
Φ
+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. Calculate
oc
ocmm I
VjXR ~
~||
Ioc
47
Single-Phase TransformerImpedance Reflection by the Square of the Turns Ratio
Rs jXs Ideal
Transformer 7200:240V
Rm jXm
7200V 240V
Ideal Transformer 7200:240V
7200V 240V
2
7200
240
sjX
2
7200
240
sR
2
7200
240
mjX2
7200
240
mR
2
~/
~
~/
~
~/
~
~/
~ ,~
~ ,~
~
LS
HS
LS
HSHS
HS
LSHS
HSHS
LSLS
HSHS
LS
HS
HS
LS
LS
HS
LS
HS
LS
HS
N
N
N
NI
N
NV
IV
IV
IV
Z
Z
N
N
I
I
N
N
V
V
48
Now Work a Single-Phase Transformer Example
Open circuit and short circuit tests are performed on a single-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, Qoc = 250VAr. Determine the four impedance values (in ohms) for the transformer model shown.
Rs jXs Ideal
Transformer 7200:240V
Rm jXm
7200V 240V
Turns ratio 7200:240
(30 : 1)
(but approx. same amount of copper in each winding)
Φ
See Grady 2007, pp. 76-77
49
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 Ideal
Transformer N1 : N2
Rm jXm
Wye-Equivalent One-Line Model
A
N
• Values for one of the transformer windings, on side 1
• Can reflect to side 2 using either individual transformer turns ratio N1:N2, or three-phase bank line-to-line turns ratio (which are identical ratios)
23 : 13 NN
50
Δ - Δ
For Delta-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
Ideal Transformer
3
Rs
3
2 :
3
1 NN
3
jXs
3
Rm
3
jXm
A
N
Wye-Equivalent One-Line Model
• Converting side 1 impedances from delta to equivalent wye
• Can reflect to side 2 using either individual transformer turns ratio N1:N2, or three-phase
bank line-to-line turns ratio (which are identical ratios) 3
2 :
3
1 NN
51
Δ - Y
For Delta-Wye Connection Model, Convert the Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
Ideal Transformer
3
Rs
2 : 3
1N
N
3
jXs
3
Rm
3
jXm
A
N
Wye-Equivalent One-Line Model
• Converting side 1 impedances from delta to wye
• Can then reflect to side 2 using three-phase bank line-to-line turns ratio 23 : 1 NN
52
Y - Δ
For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye
N1:N2
N1:N2
N1:N2
Ideal Transformer
3
2 : 1N
N
jXs
Rm jXm
A
N
Rs
Wye-Equivalent One-Line Model
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 turns ratio
Can then reflect to side 2 using three-phase bank line-to-line turns ratio 2 : 13 NN
53
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 one side are in phase with line-to-neutral voltages on the other side
• Thus, and phase shift in line-to-line voltages is unavoidable, but it can be managed to avoid problems