AC POWER ANALYSIS

Tunku Muhammad Nizar Bin Tunku MansurPegawai Latihan Vokasional

Pusat Pengajian Kejuruteraan Sistem Elektrik

2

Content

Average Power Maximum Average Power Transfer Complex Power Power Factor Correction

3

AVERAGE POWER

4

Average Power

)cos(2

1ivmmIVP

Average Power, in watts (W), is the average of instantaneous power over one period

5

Average Power

RRIIVP mmm22 |I|

2

1

2

1

2

1

Resistive load (R) absorbs power all the time.

For a purely resistive circuit, the voltage and the current are in phase (v = i).

6

Average Power

090cos2

1 o mmIVP

Reactive load (L or C) absorbs zero average power.

For a purely reactive circuit, the voltage and the current are out of phase by 90o (v - i = ±90).

7

Exercise 11.3Find the average power supplied by the source and the average power absorb by the resistor

8

Solution

A56.571.118j24

305I o

o

2.5W)56.57cos(30(5)(1.118)2

1P oo

The current I is given by

The average power supplied by the voltage source is

9

Solution

2.5W118)(4.472)(1.2

1P

Notice that the average power supplied by the voltage source is same as the power absorbed by the resistor.

This result shows the capacitor absorbed zero average power.

The average power absorbed by the resistor is

V56.574.4724IV oRR

The current through the resistor is

A56.571.118II oR

The voltage across resistor is

10

Practice Problem 11.3Calculate the average power absorbed by the resistor and the inductor. Then find the average power supplied by the voltage source

11

Solution

A.576253.2j3

548I o

o

V26.577.593IV oR

The current I is given by

For the resistor

A26.572.53II oR

9.6W3)(7.59)(2.52

1IV

2

1P mmR

12

Solution

9.6W)26.57os(45(8)(2.53)c2

1P oo

V.5716153.2jIV oLL

The average power supplied by the voltage source is

For the inductor

A26.572.53II oL

0W)3)cos(90(2.53)(2.52

1P o

L

Notice that the average absorbed by the resistor is same as the power supplied by the voltage source.

This result shows the inductor also absorbed zero average power.

13

MAXIMUM AVERAGE POWER

TRANSFER

14

Maximum Power Transfer

*ZZ thL

For maximum power transfer, the load impedance ZL must equal to the complex conjugate of the Thevenin impedance Zth

ththLL jXRjXR

15

Maximum Average Power

The Maximum Average Power delivered to the load is

)jX(R)jX(R

V

ZZ

VI

LLthth

th

Lth

th

L2 R|I|

2

1P

2LTh

2LTh

L2

Th

)X(X)R(R

R|V|

2

1

The current through the load is

16

Maximum Average Power

In a situation in which the load is purely real, the load resistance must equal to the magnitude of the Thevenin impedance.

|Z|R thL

Th

2Th

Th

2Th

maxavg, 8R

|V|

4R

|V|

2

1P

By setting RL = Rth and XL = -Xth , the maximum average power is

17

Exercise 11.5Determine the load impedance ZL that maximize the power drawn and the maximum average power.

18

Solution

j6)(8||4j5ZTh

V10.37.454 o

j4.467)Ω(2.933

First we obtain the Thevenin equivalent

To find Zth, consider circuit (a)

)0(10j6)-(84

j6)-(8V o

Th

To find Vth, consider circuit (b)

19

Solution

j4.467)Ω(2.933*ZZ ThL

From the result obtained, the load impedance draws the maximum power from the circuit when

The maximum average power is

2.368W8(2.933)

(7.454)

8R

|V|P

2

Th

2Th

max

20

Practice Problem 11.5Determine the load impedance ZL that absorbs the maximum average power. Calculate the maximum average power.

21

Solution

j10)j4(8||5ZTh

A.3415249.1 o

j0.7317)Ω(3.415

First we obtain the Thevenin equivalentTo find Zth, consider circuit (a)

V51.346.255IV oTh

To find Vth, consider circuit (b)

By using current divider (2)j105j48

j48I

22

Solution

j0.7317)Ω(3.415*ZZ ThL

From the result obtained, the load impedance draws the maximum power from the circuit when

The maximum average power is

W429.18(3.415)

(6.25)

8R

|V|P

2

Th

2Th

max

23

Example 11.6Find the value of RL that will absorbs maximum average power. Then calculate that power.

24

Solution

j30)(40||20jZTh

V13476.72 o

j22.35)Ω(9.412

First we obtain the Thevenin equivalent

Find Zth

Find Vth

By using voltage divider

)30(150j3040j20

j20V o

Th

25

Solution

22ThL (22.35)(9.412)|Z|R

24.25j22.35)(9.412

13472.76

RZ

VI

o

LTh

Th

25.24

The value of RL that will absorb the maximum average power is

The maximum average power is

39.29W24.25(1.8)2

1R|I|

2

1P 2

L2

max

A100.421.8 o

The current through the load is

26

Practice Problem 11.6Find the value of RL that will absorbs maximize average power, Then calculate the power.

27

Solution

j24.57)Ω(17.181Z||ZZ 21Th

V91.3198.35 o

j6008Z1

First we obtain the Thevenin equivalentTo find Zth let

To find Vth

By using voltage divider

)60(120ZZ

ZV o

21

2Th

j27)9(j30)(||90Z2 and

Then

28

Solution

30|Z|R ThL

03j24.57)(17.181

91.31-98.53

RZ

VI

o

LTh

Th

The value of RL that will absorb the maximum average power is

The maximum average power is

6.863W)0(3(0.6764)2

1R|I|

2

1P 2

L2

max

A4.40.6764 o

The current through the load is

29

COMPLEX POWER

30

Complex Power

Apparent Power, S (VA) Real Power, P (Watts) Reactive Power, Q (VAR) Power Factor, cos

31

Complex Power Complex power is the product of the

rms voltage phasor and the complex conjugate of the rms current phasor.

Measured in volt-amperes or VA As a complex quantity

Its real part is real power, P Its imaginary part is reactive power, Q

32

Complex Power (Derivation)

*VI2

1S

irmsrms θI2

II

ivrmsrms θθIVS

vrmsrms θV2

VV

)θsin(θIjV)θcos(θIV ivrmsrmsivrmsrms

*IVS rmsrms

33

Complex Power (Derivation)

jX)(RIS rms2

XjIRI rms2

rms2

ZIS rms2

34

Complex Power (Derivation)

)θcos(θIVP ivrmsrms

)θθsin(IVQ ivrmsrms

XIQ rms2

RIP rms2

From derivation, we notice that the real power is

and also the reactive power

or

or

35

Real or Average Power

The real power is the average power delivered to a load.

Measured in watts (W) The only useful power The actual power dissipated by the

load

36

Reactive Power The reactive power, Q is the imaginary

parts of complex power. The unit of Q is volt-ampere reactive

(VAR). It represents a lossless interchange

between the load and the source Q = 0 for resistive load (unity pf) Q < 0 for capacitive load (leading pf) Q > 0 for inductive load (lagging pf)

37

Apparent Power The apparent power is the product of

rms values of voltage and current Measured in volt-amperes or VA Magnitude of the complex power

22|S| QPIV rmsrms

38

Power Factor Power factor is the cosine of the

phase difference between voltage and current.

It is also cosine of the angle of the load impedance.

)cos(pf ivS

P

39

Power Factor The range of pf is between zero and

unity. For a purely resistive load, the voltage

and current are in phase so that v- i = 0 and pf = 1, the apparent power is equal to average power.

For a purely reactive load, v- i = 90 and pf = 0, the average power is zero.

40

Power Triangular

Comparison between the power triangular (a) and the impedance triangular (b).

41

Problem 11.46

For the following voltage and current phasors, calculate the complex power, apparent power, real power and reactive power. Specify whether the pf is leading or lagging.

a) V = 22030o Vrms, I = 0.560o Arms.

b) V = 250-10o Vrms, I = 6.2-25o Arms.

c) V = 1200o Vrms, I = 2.4-15o Arms.

d) V = 16045o Vrms, I = 8.590o Arms.

42

Solution

a) S = VI* = (22030o)( 0.5-60o) = 110-30o VA = 95.26 – j55 VA

Apparent power = 110 VAReal Power = 95.26 WReactive Power = -55 VARpf is leading because current leads voltage

b) S = VI* = (250-10o)(6.225o) = 155015o VA = 1497.2 + j401.2 VA

Apparent power = 1550 VAReal Power = 1497.2 WReactive Power = 401.2 VARpf is lagging because current lags voltage

c) S = VI* = (1200o)( 2.415o) = 28815o VA = 278.2 + j74.54 VA

Apparent power = 288 VAReal Power = 278.2 WReactive Power = 74.54 VARpf is lagging because current lags voltage

d) S = VI* = (16045o)(8.5-90o) = 1360-45o VA = 961.7 – j961.7 VA

Apparent power = 1360 VAReal Power = 961.7 WReactive Power = -961.7 VARpf is leading because current leads voltage

43

Problem 11.48

Determine the complex power for the following cases:

a) P = 269 W, Q = 150 VAR (capacitive)

b) Q = 2000 VAR, pf = 0.9 (leading)

c) S = 600 VA, Q = 450 VAR (inductive)

d) Vrms = 220 V, P = 1 kW, |Z| = 40 (inductive)

44

Solutiona) Given P = 269W, Q = 150VAR

(capacitive)

Complex power,

b) Given Q = 2000VAR, pf = 0.9 (leading)

j150)VA269(j QPS

j2000)VA4129(j QPS

opf 84.259.0cos

VA25.84-4588 o

31.4588)84.25sin(

2000

sinsin

o

QSSQ

48.4129cos SP

VA29.14308 o

Complex power,

45

Solution

c) Given S = 600VA, Q = 450VAR (inductive)

j450)VA9.396(j QPS

6614.0cos pf

VA59.84600 o

75.0600

450sinsin

S

QSQ

86.396cos SP

Complex power,

o59.48

46

Solution

d) Given Vrms = 220V, P = 1kW, |Z| = 40 (inductive)

j681.2)VA1000(j QPS

121040

220

|Z|

|V|S

22

VA34.261210 o

8264.01210

1000coscos

S

PSP

25.681sin SQ

Complex power,

o26.34

47

Problem 11.42

A 110Vrms, 60Hz source is applied to a load impedance Z. The apparent power entering the load is 120VA at a power factor of 0.707 lagging. Calculate

a) The complex power

b) The rms current supplied to the load.

c) Determine Z

d) Assuming that Z = R + j L, find the value of R and L.

48

Solution

j84.84VA84.48sinjcos SSS

rmsrms IVS

Given S = 120VA, pf = 0.707 = cos = 45o

a) the complex power

A091.1110

120

rmsrms V

SI

b) the rms current supplied to the load

49

Solution

ZIS rms2

71.278ωL

d) value of R and L

If Z = R + jL then Z = 71.278 + j 71.278

)278.71278.71(2

jI

SZ

rms

0.1891H2

278.71L

f

71.278ΩR

c) the impedance Z

50

Problem 11.83

Oscilloscope measurement indicate that the voltage across a load and the current through is are 21060o

V and 825o A respectively. Determine

a) The real power

b) The apparent power

c) The reactive power

d) The power factor

51

Solution

688.1W)(35cos840)(35cos oo SP

VA840S

a) the real power

d) the power factor

ging)0.8191(lag)35cos( o S

Ppf

c) the reactive power

j481.8)VA(688.1)VA35(840 o

)25)(860(2102

1VI*

2

1 oo S

.8VAR814)(35sin840)(35sin oo SQ

b) the apparent power

52

POWER FACTOR CORRECTION

53

Power Factor Correction

The process of increasing the power factor without altering the voltage or current to the original load.

It may be viewed as the addition of a reactive element (usually capacitor) in parallel with the load in order to make the power factor closer to unity.

54

Power Factor Correction

Normally, most loads are inductive. Thus power factor is improved or corrected by installing a capacitor in parallel with the load.

In circuit analysis, an inductive load is modeled as a series combination of an inductor and a resistor.

55

Implementation of Power Factor Correction

56

Calculation

If the original inductive load has apparent power S1, then

P = S1 cos 1 and Q1 = S1 sin 1 = P tan 1

If we desired to increased the power factor from cos1 to cos2 without altering the real power, then the new reactive power is

Q2 = P tan 2

The reduction in the reactive power is caused by the shunt capacitor is given by

QC = Q1 – Q2 = P (tan 1 - tan 2)

57

Calculation

rms2

21

rms2C

ωV

)tanθP(tanθ

ωV

QC

The value of the required shunt capacitance is determined by the formula

Notice that the real power, P dissipated by the load is not affected by the power factor correction because the average power due to the capacitor is zero

58

Example 11.15

When connected to a 120V (rms), 60Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

59

Solution

VA50008.0

4000

cos 11

P

S

If the pf = 0.8 then,

cos1 = 0.8 1 = 36.87o

where 1 is the phase difference between the voltage and current.

We obtained the apparent power from the real power and the pf as shown below.

The reactive power is

VAR300087.36sin5000sin 111 SQ

60

Solution

VA5.421095.0

4000

cos 22

P

S

When the pf raised to 0.95,

cos2 = 0.95 2 = 18.19o

The real power P has not changed. But the apparent power has changed. The new value is

The new reactive power is

VAR4.1314sin 222 SQ

61

Solution

VAR6.16854.1314300021 QQQC

The difference between the new and the old reactive power is due to the parallel addition of the capacitor to the load.

The reactive power due to the capacitor is

The value of capacitance added is

μF5.310)120)(60(2

6.168522

rms

C

V

QC

62

Practice Problem 11.15

Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume the load is supplied by a 110V (rms) 60Hz power line.

63

Solution

265.8kVA)79.31sin(

140kVAR

sin 1

11

Q

S

If the pf = 0.85 then,

cos1 = 0.85 1 = 31.79o

where 1 is the phase difference between the voltage and current.

We obtained the apparent power from the reactive power and the pf as shown below.

The real power is

kW93.225)79.31cos(8.265cos 11 SP

64

Solution

kVA93.2251

kW93.225

cos 22

P

S

When the pf raised to 1 (unity),

cos2 = 1 2 = 0o

The real power P has not changed. But the apparent power has changed. The new value is

The new reactive power is

0sin 222 SQ

65

Solution

kVAR140014000021 QQQC

The difference between the new and the old reactive power due to the parallel addition of the capacitor to the load.

The reactive power due to the capacitor is

The value of capacitance is

mF69.30)110)(60(2

140kVAR22

rms

C

V

QC

66

Problem 11.82

A 240Vrms, 60Hz source supplies a parallel combination of a 5 kW heater and a 30 kVA induction motor whose power factor is 0.82. Determine

a) The system apparent powerb) The system reactive powerc) The kVA rating of a capacitor required to

adjust the system power factor to 0.9 laggingd) The value of capacitance required

67

Solution

W4600282.0x30000)91.34cos(22 SP

For the heater

P1 = 5000 Q1 = 0

For the 30kVA induction motor, the pf = 0.82 then,

cos1 = 0.82 1 = 34.92o

The real and the reactive power for the induction motor

kVAR17171)91.34sin(22 SQ

68

Solution

The total system complex power

Stotal = S1 + S2 = (P1 + P2) + j (Q1 + Q2) = 29600 + j17171

The system apparent power

S = |Stotal| = 34.33kVA

The system reactive power

Q = 17171 kVAR

The system power factor

865.034220

29600

S

Ppf

69

Solution

The system pf = 0.865 then,

cos1 = 0.865 1 = 30.12o

The new system pf = 0.9 then, cos2 = 0.9 2 = 25.84o

The rating for the capacitance required to adjust the power factor to 0.9

QC = P (tan 1 + tan 2) = 29600 (tan 30.12 + tan 25.84) = 2833kVAR

70

Solution

The value of capacitance is

μF46.130)240)(60(2

283322

rms

C

V

QC

71

Problem 11.91

The nameplate of an electric motor has the following information.

Line voltage: 220 Vrms

Line current: 15 Arms

Line frequency: 60 Hz Power: 2700 W

Determine the power factor (lagging) of the motor.

Find the value of the capacitance, C that must be connected across the motor to raise the pf to unity.

72

Additional Problem

The nameplate of an electric motor has the following information.

Line voltage: 230 Vrms

Line current: 13 Arms

Line frequency: 60 Hz Reactive Power: 1500 VAR

Determine the power factor (lagging) of the motor.

Find the reactive power required to increase the pf to 0.95.

Find the value of the capacitance, C.