ELECTRICAL TECHNOLOGY ET 201

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Explain and calculate average power, apparent power, reactive power

Calculate the total P, Q and S and sketch the power triangle.

ELECTRICAL TECHNOLOGY ET 201

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• Power is distributed into the resistance and reactance in AC circuit.

• The power delivered to a load at any instant is defined by the product of the applied voltage and the resulting current:

• Since v and i are sinusoidal quantities

Power in AC Circuit

vip

tVv m sin

tIi m sin

3

The chosen v and i include all possibilities because

• If the load is purely resistive: = 0°

• If the load is purely inductive: = 90°

• If the load is purely capacitive: = - 90°

• For a network that is primarily inductive, is positive (v leads i or i lags v)

• For a network that is primarily capacitive, is negative (i leads v)

Power in AC Circuit tVv m sin tIi m sin

4

• Positive power means that power has been distributed from supply into circuit.

• Negative power means that the power has been distributed from circuit into supply.

• There are three type of power in AC circuit:i) Average power, Pii) Apparent Power, S iii) Reactive Power, Q

Power in AC Circuit

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Average Power, P

• The average power (real power) is the power delivered to the load and dissipated by the load.

[watt, W]

Where, θ : phase angle between Vrms and Irms

OR ; [watt, W]

Power in AC Circuit

cosrmsrms IVP

RIP rms2

R

VP rms

2

6

14.5 Power Factor (Review)Average Power, P

• For a purely resistive load;

Hence;

• For purely inductive or purely capacitive load;

Hence;

0T

1cos PFrmsrmsIVP

90T0P

0cos PF

cosrmsrms IVP

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Apparent Power, S

• From analysis of DC networks (and resistive elements above), it would seem apparent that the power delivered to the load is simply determined by P = VI, with no concern for the components of the load.

• However, in Chapter 14 (Lecture 10) the power factor (cos θ) of the load has a pronounced effect on the power dissipated, less pronounced for more reactive loads.

• Therefore P = VI is called apparent power, S.

[volt-amperes, VA]

Power in AC Circuit

rmsrmsIVS

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Apparent Power, S

OR ; [volt-amperes, VA]

• The average power to the load is;

However;

Therefore

• The power factor of a system Fp is

Power in AC Circuit

ZIS rms2

cosVIP

VIS

cosSP

S

PFp cos

Z

VS rms

2

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Reactive Power, Q

• In general, the reactive power associated with any circuit is defined to be

[volt-ampere reactive, VAR]

Where, θ : phase angle between Vrms and Irms

OR [volt-ampere reactive, VAR]

• For the resistor,

Power in AC Circuit

sinrmsrms IVQ

XIQ rms2

VAR 0Q

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Reactive Power, Q

• For a purely inductive circuit,

OR ;

Since the apparent power, S = VI , and the average power for inductor, P = 0

Power in AC Circuit

rmsrmsL IVQ

LrmsL XIQ 2

00

cos VIS

PFp

L

rmsL X

VQ

2

sinrmsrms IVQ

vL leads iL by 90°

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Reactive Power, Q

• For a purely capacitive circuit,

OR ;

Since the apparent power, S = VI , and the average power for capacitor, P = 0

Power in AC Circuit

rmsrmsC IVQ

CrmsC XIQ 2

00

cos VIS

PFp

C

rmsC X

VQ

2

sinrmsrms IVQ

iC leads vC by 90°

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19.7 Power Triangle

• The three quantities average power P, apparent power S and reactive power Q can be related in the vector domain by

with

90

90

0

CC

LL

Q

Q

P

Q

Q

P

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19.7 Power Triangle

• For an inductive load, the phasor power S, as it is often called, is defined by

S = P + jQL

• For a capacitive load, the phasor power S is defined by

S = P - jQC

22 QPS

14

• If a network has both capacitive and inductive elements, the reactive component of the power triangle will be determined by the difference between the reactive power delivered to each.

• If QL QC, the resultant power triangle will be similar to the inductive load power diagram.

• If QC QL, the resultant power triangle will be similar to the capacitive load power diagram.

• That the total reactive power is the difference between the reactive powers of the inductive and capacitive elements.

19.7 Power Triangle

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19.7 The Total P, Q, and SThe total number of watts PT, volt-amperes reactive QT, and volt-amperes ST, and the power factor Fp of any system can be found using the following procedure:

1. Find the real (average) power and reactive power for each branch of the circuit.

2. The total real power of the system (PT) is the sum of the average power delivered to each branch

3. The total reactive power (QT ) is the difference between the reactive power of the inductive loads and that of the capacitive loads.

4. The total apparent power is ST2 = PT

2 + QT2.

5. The total power factor is PT / ST. 22 QPS

16

There are two important points in the previous slide:

• First, the total apparent power, ST must be determined from the total average PT and total reactive powers QT and cannot be determined from the apparent powers of each branch.

• Second, and more important, it is not necessary to consider the series-parallel arrangement of branches. In other words, the total real PT , total reactive QT , or total apparent power ST is independent of whether the loads are in series, parallel, or series-parallel.

19.7 The Total P, Q, and S

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Example 19.3

1. Find the total number of watts PT, volt-amperes reactive QT, and volt-amperes ST and draw the power triangle.2. Find the power factor Fp

3. Find the current in phasor form.


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Example 19.3 – Solution

1.

Load 1

VAR 0

W;100

1

1

Q

P

VA 100

01001

jS

22 QPS


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Example 19.3 – solution (cont’d)

Load 2

VAR 700

W;200

2

2

Q

P

VA 728

7002002

jS


22 QPS

20

Load 3

VAR 1500

W;300

3

3

Q

P

VA 71.1529

15003003

jS



22 QPS

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Total

300200100321

PPPPT



W600TP

22

Total

VAR 800

15007000321

QQQQT



VAR(C) 800TQ

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Total

800600 j

jQPS TTT

The power triangle;



W600TP

VAR(C) 800TQ

VA 13.531000 TS

Note: ST is NOT equal to sum of each branch!!

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2. The power factor FP

VA 1000

W600

T

Tp S

PF



)( leading 6.0 CFp

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3. The current

A 10V 100

VA 1000

VA 1000

VA 1000

I

I

VI

ST



)( leading 6.0 CFp

Since Fp is leading I leads E, predominantly capacitive circuit.

53.13 A 10I

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