Power Analysis
Electrical power is the time rate of receiving or delivering electrical energy that depends on the voltage and current quantities. In an ac circuit, the current and voltage quantities vary with time, and so the electrical power. Every electrical device su
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Power Analysis
1.1
Introduction
Electrical power is the time rate of receiving or delivering electrical energy that depends on the voltage and current quantities. In an ac circuit, the current and voltage quantities vary with time, and so the electrical power. Every electrical device such as ceiling fan, bulb, television, iron, micro-wave oven, DVD player, water-heater, refrigerator, etc. has a power rating that specifies the amount of power that device requires to operate. An electrical equipment with high power rating generally draws large amount of current from the energy source (e.g., voltage source) which increases the energy consumption. Nowadays, Scientists and Engineers are jointly working on the design issues of the electrical equipment to reduce the energy consumption. Since in the design stage, power analysis plays a vital role, a clear understanding on power analysis fundamentals becomes a most important prep work for any electrical engineer. This chapter reviews these fundamental concepts that include instantaneous power, average power, complex power, power factor, power factor correction and three-phase power.
1.2
Instantaneous Power
The instantaneous power (in watt) pðtÞ is defined as the product of time varying voltage vðtÞ and current iðtÞ, and it is written as [1], pðtÞ ¼ vðtÞ iðtÞ
© Springer Science+Business Media Singapore 2016 Md.A. Salam and Q.M. Rahman, Power Systems Grounding, Power Systems, DOI 10.1007/978-981-10-0446-9_1
ð1:1Þ
1
2
1 Power Analysis
Consider that the time varying excitation voltage vðtÞ for an ac circuit is given by, vðtÞ ¼ Vm sinðxt þ /Þ
ð1:2Þ
where, x is the angular frequency and / is the phase angle associated to the voltage source. In this case, the expression of the resulting current iðtÞ in an ac circuit as shown in Fig. 1.1 can be written as, vðtÞ iðtÞ ¼ Z /
ð1:3Þ
where, Z / is the impedance of the circuit in polar form, in which Z is the magnitude of the circuit impedance. Substituting Eq. (1.2) into Eq. (1.3) yields, Vm sinðxt þ /Þ Vm / Vm iðtÞ ¼ ¼ ¼ j0 Z Z / Z / iðtÞ ¼ Im sin xt
ð1:4Þ ð1:5Þ
where, the maximum current is, Im ¼
Vm Z
ð1:6Þ
Substituting Eqs. (1.2) and (1.5) into Eq. (1.1) yields, pðtÞ ¼ Vm sinðxt þ /Þ Im sin xt
ð1:7Þ
V m Im 2 sinðxt þ /Þ sin xt 2
ð1:8Þ
pðtÞ ¼
Fig. 1.1 A simple ac circuit
1.2 Instantaneous Power
3
Vm Im ½cos / cosð2xt þ /Þ 2
ð1:9Þ
V m Im Vm Im cos / cosð2xt þ /Þ 2 2
ð1:10Þ
pðtÞ ¼ pðtÞ ¼
Equation (1.10) presents the resultant instantaneous power. Example 1.1 An ac series circuit is having the excitation voltage and the impedance vðtÞ ¼ 5 sinðxt 25 Þ V and Z ¼ 2j15 X, respectively. Determine the instantaneous power. Solution The value of the series current is, 5 25 I¼ ¼ 2:540 A 2j 15 iðtÞ ¼ 2:5 sinðxt 40 Þ A The instantaneous power can be determined as, 5 2:5 2 sinðxt 25 Þ sinðxt 40 Þ 2 5 2:5 ½cos 15 cosð2xt 65 Þ pðtÞ ¼ 2 pðtÞ ¼ 6:04 6:25 cosð2xt 65 Þ W pðtÞ ¼
Practice problem 1.1 The impedance and the current in an ac circ
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