1 1. Power and RMS Values. 2 Instantaneous power p(t) flowing into the box Circuit in a box, two...

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