Lecture7

KF1063 Introduction to Electrical Engineering, ®mbi, bb

KF 1063

LECTURE 7


OUTLINE

Power in A.C. Networks

Active Power Reactive Power Apparent Power Power Factor (p.f.) Power Factor Correction


INTRODUCTION Instantaneous power, p(t) = v(t)i(t)

Power, p(t) value positive – power transmit/dissipate by load

negative – power return from the load

Since p(t) is power transmits by load, then it is the average power, P at load

Sometimes P is also known as active power, real power or true power measured in unit of Watts.


Z = R (purely resistive)

ACTIVE POWER

P = VI = I2R = V2/R (Watt)


Z = jXL (inductive)

REACTIVE POWER

Instantaneous power

p(t) = v(t)i(t) = VI sin 2ωt

Average power is zero

The product of VI is called reactive power (QL or PQ) with unit Volt-Amp Reactive (VAR)

Reactive power (inductive)

QL = VI = I2XL = V2/XL (VAR)


Z = – jXC (capacitive)

KUASA REGANGAN

Reactive power (capacitive)

QC = VI = I2XC = V2/XC (VAR)

Note:

To distinguish between inductive reactive power (QL) and capacitive reactive power (QC), we use two different signs (+ or –) depending on our reference (i or v), for example jQL and – jQC or otherwise.


ACTIVE/REACTIVE POWER - Example

• I = 100 V/25 Ω = 4 A, P = VI = (100 V)(4 A) = 400 W, Q = 0 VAR

• I = 100 V/20 Ω = 5 A, P = 0, QL = VI = (100 V)(5 A) = 500 VAR (inductive)

• I = 100 V/40 Ω = 2.5 A, P = 0, QC = VI = (100 V)(2.5) = 250 VAR (capacitive)

= – 250 VAR

Note: use the magnitude of I and V


Determine the total PT and QT for the circuit. Sketch the series equivalent circuit.

R = PT/I2 = 1200/202 = 3 Ω

Xeq = XL = QT/I2 = 1600/202 = 4 Ω

ACTIVE/REACTIVE POWER - Example


For load consisting of series resistance and reactance, Z = R ± jX = Z /±θ , the power produced is called Apparent Power or Complex Power), S or PS with unit Volt-Amp (VA)

APPARENT POWER

S = VI

S = VI

θ positive, inductive load θ negative, capacitive load

S = VI (VA) P = VI kos θ = I2R = VR

2/R (W) = S kos θ (W)Q = VI sin θ = I2X = Vx

2/X (VAR) = S sin θS = √(P2 + Q2) = VI*

V/0°

Power Triangle

S = P + jQL S = P – jQC


POWER TRIANGLE - Example

Sketch the power triangle.


S (Z) Q (X) θ P (R)

POWER FACTOR Power factor, p.f. = cos θ = P/S = R/Z p.f. depends on the load type:

Purely resistive load, R , p.f. = 1 Inductive load, RL, p.f. <1 (lagging) and Capacitive load, RC, p.f. < 1 (leading)

Most of the loads are inductive (lagging p.f.) and must be corrected until p.f. approximately become unity (p.f. = 1) using capacitor.


Leading p.f. (final) = cos θJ; QJ = P tan θJ

QC = Q – QJ

QC = V2/XC ; XC = 1/ jωC = V2/ QC

R C jX V S QC

Q θ θJ QJ P

POWER FACTOR - Correction


Figure 7.18, 7.19, 7.20

Given: Vs = 117∠0 V, RL = 50 Ω, jXL = 86.7 Ω ω = 377 rad/s

ZL = 50 + j86.7 = 100∠1.047 Ω

IL = VL/ZL = (117∠0 )/(100∠1.047) = 1.17∠–1.047 A

S = VLIL* = 137 ∠1.047) = 68.4 + j118.5 VA

QC = –118.5 VAR XC = VL2/118.5 = – j115 Ω

C = 1/ωXc = 23.1 µF

Find the complex power for the circuit. Correct the circuit power factor to p.f. = 1 using parallel reactance.

POWER FACTOR - Example


POWER MEASUREMENT

Wattmeter only reads active power on the load side


POWER MEASUREMENT - ExampleWhat is the reading of wattmeter, W?

PT = 10 + 40 + 700 = 750 W


P.F. MEASUREMENT - Example

S = VI = (9.615)(240) = 2.308 kVA

p.f. = 1.5 kW/2.308 kVA = 0.65

Q = 1.754 kVAR

If p.f. = 1 (correction)

Xc = V2/Q then C = 80.761 µF

What happen if C = 80 µF?

Lecture7

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