transistors power presentation

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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9

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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15

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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22

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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23

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


38/65

38

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


39/65

39

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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40

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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41

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+


42/65

42

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


43/65

43

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


44/65

44

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?


45/65

45

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.


46/65

46

3. Transformers


47/65

47

Transformer Core Types


48/65

48

High-Voltage Grid Transformers, 100s of MW


49/65

49

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)


50/65

50

Short Circuit Test

Rs jXsIdeal


Rm jXm

7200V 240V




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


51/65

51

Open Circuit Test

Rs jXsIdeal


Rm jXm

7200V 240V




+

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,


52/65

52

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.


53/65

53

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%


54/65

54

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


55/65

55

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


56/65

56

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


57/65

57

-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


58/65

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


59/65

59

Single-Phase Transformer

Impedance Reflection from High-Side (H) to Low-Side (L) by the

Square of the Turns Ratio

Rs jXsIdeal


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


60/65

60

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


Rm jXm

7200V 240V


(30 : 1)



A three phase transformer can be three separate


61/65

61

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


62/65

62

-

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


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


63/65

63

- Y

For Delta WyeConnection Model, Convert the


N1:N2

N1:N2

N1:N2

IdealTransformer

3

Rs

2:3

1N

N

3

jXs

3

Rm

3

jXm

A

N


Reflect side 1 delta ohms to side 2 wye ohms


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


64/65

64

Y -

For Wye-DeltaConnection Model, Convert the


N1:N2

N1:N2

N1:N2

IdealTransformer

3

2:1 N

N

jXs

Rm jXm

A

N

Rs


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


Convert side 2 impedances from delta ohms

to wye ohms by dividing by 3

Above circuit results in the proper reflection


65/65

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

transistors power presentation

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