Review of Electric Circuits and Introduction of 3-phase system

1. Review of Electric Circuits and Introduction of 3-phase system Ahmad Firdaus Bin Ahmad Zaidi

2. Sinusoidal voltage • AC voltage generated by commercial alternators is nearly a perfect sine wave and is expressed by equation: e = Em.sin(ωt + θ), where: • e = instantaneous voltage (V) • Em = peak value of e (V) • ω = angular frequency (rad/sec) • t = time (sec) • θ = phase angle (rad) • Another way to express the equation in term of angle in degree unit is e = Em.sin(360ft + θ), where: • f= frequency (Hz) • θ = phase angle (degree)

3. Example 2-3 • The sine wave in the figure represents voltage Eab across terminal a and b of an ac motor that operates at 50Hz. Given Em = 100V, calculate voltage at t = 0 and at t = 27.144s

4. Effective value of ac voltage • Effective value of an ac voltage is sometimes called RMS (root mean square) value of the voltage. • RMS value of voltage and current are given as follows: • Erms = Em/√2 • Irms = Im/√2 • This measures heating effect of the ac voltage compared to that of equivalent dc voltage. • For example, ac voltage having rms value of 135 produces the same heating effect in a resistor as does dc voltage of 135V.

5. Example 2-4 • A 60Hz source having effective voltage of 240V delivers effective current of 10A to a circuit. The current lags the voltage by 30°. Draw the waveshape of e and i.

6. V or E = potential difference, [V] I = current, [A] R = resistance, [] In alternating current: XL = Inductive reactance () Is the opposition to ac current due to inductance in the circuit XL = 2fL, f = frequency of the circuit (Hz), L = inductance (H) Xc = Capacitive reactance () Is the opposition to the flow of ac current due to the capacitance in the circuit Xc = 1/2fC, f = frequency of the circuit (Hz), C = capacitance (F) IMPEDANCE Ohm’s law: V=IR Impedance = a measurement of total opposition of to the flow of an AC current H=henry = 1Wb/A F=farad = 1coulomb/volt (Xc is inversely proportional to the signal frequency f and the capacitance C)

7. The instantaneous voltage across a pure resistor, VR is “in-phase” with the current. • The instantaneous voltage across a pure inductor, VL “leads” the current by 90o • The instantaneous voltage across a pure capacitor, VC “lags” the current by 90o • Therefore, VL and VC are 180o “out-of-phase” and in opposition to each other. • For the series RLC circuit, this can be shown as:

9. Phasor Representation in Sinusoidal Steady State • Linear circuits with sinusoidal voltage and current are at frequency f=w/2pi • To analyze such circuits, calculation is simplified by means of phasor-domain analysis. • In phasor domain, time-domain variables v(t) and i(t) are transformed into phasors which are represented by the complex variables and Ī. • In a complex plane, these phasors can be drawn with a magnitude and an angle. • v(t) = cos wt=∠0 • i(t)= Î cos (wt-ϕ) Ī= Î ∠-ϕ

10. Using phasors, differential equations can be converted into easily solved algebraic equations containing complex variables. • In order to calculate the current in the circuit, following differential equation would have to be solved: • In the phasor-domain circuit, impedance Z of the series-connected elements is obtained by the impedance triangle where:

11. In the figure, a)time domain circuit b)phasor domain circuit and c) impedance triangle • Impedance can be expressed as: • , where:

12. Example • Calculate the impedance seen from the terminals of the circuit under sinusoidal steady state at frequency f=60Hz • Z = j0.1+(-j5 II 2.0) = j0.1+(-j10/(2j-5)) • = 1.72-j0.59 = 1.82∠-18.9 • Using impedance, current can be obtained as: • Current can also be expressed in time domain as:

13. Example • Calculate I1 and i1(t) in the circuit if the applied voltage is 120Vrms and frequency 60Hz. Assume V1 to be reference phasor.

14. Power, Reactive Power and Power Factor • Consider generic circuit in a sinusoidal steady state. • Each sub circuit may consist of passive R-L-C elements and active voltage and current sources. • Instantenous power p(t)=v(t)i(t) is delivered by subcircuit1 and absorbed by subcircuit2. • A negative value of p(t) reverses the roles of subcircuit1 and subcircuit2.

15. Black Boxes Distinction between sources and load A definition: Asources delivers electrical power whereas a load absorb it Distinction between a source and a load A variable voltage exists across the terminal A1,A2 and B1,B2 and its magnitude and polarity also continually changing Under such highly variable conditions, how we can tell whether A or B is sources or load?

16. Under sinusoidal steady state condition at frequency f, the complex power S, reactive power Q and power factor express how effectively the real power P is transferred from one sub-circuit to the other. • If v(t) and i(t) are in phase, p(t)=v(t)i(t) pulsates at twice the frequency. • At all times, p(t)>0 and therefore power always flow in one direction: from subcircuit1 to subcircuit2. • Consider waveforms where i(t) lags behind v(t) by phase angle ϕ(t). • Now p(t) becomes negative during a time interval of (ϕ/ω) during each half cycle. • A negative instantenous power implies power flow in opposite direction. • This back and forth flow of power indicates that real power is not optimally transferred from one subcircuit to the other.

17. The circuit is redrawn in phasor domain. The voltage and current phasors are defined by their magnitudes and phase angles as: • It is normally assumed that and that has negative value. • To express real, reactive and complex powers, it is convenient to use rms voltage, V and current, I instead of peak amplitude. • Complex power S is defined as: • , therefore: • , where • Therefore: • , where: • , is real power, and • , is reactive power • S is also called apparent power where: • and

18. The cost of most electrical equipment such as generators, transformers and transmission lines is proportional to |S|, since electrical insulation level and magnetic core size depend on the voltage V and conductor size depend on current I. • Real power P has physical significance since it represents useful work being performed plus the losses. • In most situations, it is desirable to have reactive power, Q to be zero.

19. Power Factor • Power factor is a measure of how effectively a load draws real power: • Power factor is dimensionless quantity. • Ideally, power factor should be 1.0 (Q=0) in order to draw real power with a minimum current magnitude and hence minimize losses in electrical equipment and transmission lines. • An inductive load draws power at a lagging power factor where current lags behind voltage. • A capacitive load draws power at leading power factor where load current leads load voltage.

20. Example • Calculate P, Q, S and power factor for circuit below:

21. Three-Phase Circuits

22. Three-Phase Circuits • Nearly all electricity is generated by means of three-phase ac generators. • Generated voltage (22kV to 69kV) are stepped up by means of transformers to 230kV to 500kV level for transferring power over transmission lines from the generation site to load centers. • Most motor loads above few kW operate from three phase voltages. • The most common configurations of three-phase ac circuits are wye and delta connections. • We will assume a balanced conditions, implies that all three voltages are equal in magnitude and displaced by 120deg (2pi/3rad) with respect to each other.

23. Three-Phase Circuits • The phase sequence is commonly assumed to be a-b-c which is considered positive sequence. • In these sequence, phase a voltage leads phase b voltage by 120deg, and phase b voltage leads phase c by 120deg.

24. Three-Phase Circuits • These voltages are represented in phasor form as: • , , • Where is phase voltage amplitude and for a balanced set of voltages, at any instant the sum of these phase voltages equals zero:

25. Per-Phase Analysis • A three-phase circuit can be analyzed on a per phase basis, provided that it has a balanced set of source voltages and equal impedances in each of the phases. • In such circuit, the source neutral n a and the load neutral N are at the same potential. • Therefore, hypothetically connecting these with a zero impedance wire, does not change the original three-phase circuit which can now be analyzed on a per phase basis.

26. Per-Phase Analysis • If , using the fact that in a balanced three phase circuit, phase quantities are displaced by with respect to each other, we find that • Total real and reactive powers can be obtained by multiplying per phase value by factor of 3. Power factor is the same as per phase value.

27. Example • In balanced star-connected source and load circuit, the rms phase voltages equal 120V and the load impedance . Calculate power factor of operation and total real and reactive power consumed by three phase load.

28. Line to Line Voltages • In the balanced wye-connected circuit, it is often necessary to consider line to line voltages, such as those between phases a and b. • Therefore:

29. Line to Line Voltages • It can be shown that: • Comparing to phase equations, line to line voltages have amplitude of times phase voltage amplitude and leads by

30. Delta-Connected Loads • In ac motor drives, the three motor phases may be connected in a delta configuration.

31. Delta-Connected Loads • Under a totally balanced condition, it is possible to replace delta-connected loads with equivalent wye connected loads where per phase analysis can be done after that. • Consider the delta connected load impedances in a three phase circuit. In terms of current drawn, these are equivalent to the wye connected impedance where:

32. Conversion