1. EMLAB 1 Chapter 9. Steady-state power analysis

2. EMLAB 2 Contents 1.Instantaneous Power: For the special case of steady state sinusoidal signals 2.Average Power : Power absorbed or supplied during one cycle 3.Maximum Average Power Transfer :When the circuit is in sinusoidal steady state 4.Effective or RMS Values : For the case of sinusoidal signals 5.Power Factor : A measure of the angle between current and voltage phasors 6.Power Factor Correction : How to improve power transfer to a load by “aligning” phasors 7.Complex Power : Measure of power using phasors 8.Single Phase Three-Wire Circuits : Typical distribution method for households and small loads

3. EMLAB 3 1. Instantaneous power Net power transfer is positive! Net power transfer is zero!

4. EMLAB 4 Power vs. voltage

5. EMLAB 5 2. Average power If voltage and current are in phase If voltage and current are in quadrature Purely inductive or capacitive

6. EMLAB 6 Example 9.2 Find the average power absorbed by impedance  15,60,53.3,10 ivMM IV 

7. EMLAB 7 Example 9.3 Determine the average power absorbed by each resistor, the total average power absorbed and the average power supplied by the source

8. EMLAB 8 3. Effective or rms Values Definition is valid for ANY periodic signal with period T root mean square

9. EMLAB 9 Compute rms value of the voltage waveform Example

10. EMLAB 10 4. Complex power ; Average power ; Complex power ; Average power

11. EMLAB 11 Example Evaluate the average power consumed by the following circuit.

12. EMLAB 12 5. Maximum average power transfer

13. EMLAB 13 Maximum average power transfer condition

14. EMLAB 14 Example Find Z L for maximum average power transfer. Compute the maximum average power supplied to the load.

15. EMLAB 15 6. Power factor Instantaneous power in an ac circuit. Positive p represents power to the load; negative p represents power returned from the load. Power to a purely resistive load. The peak value of p is V m I m.

16. EMLAB 16 Power to a purely capacitive load. Average power is zero. Power to a purely inductive load. Energy stored during each quarter cycle is returned during the next quarter-cycle. Average power is zero.

17. EMLAB 17 Power factor ☆ The reference phase angle is that of voltage. A leading or lagging phase angle is the state that the phase angle of current is leading or lagging behind that of voltage. ( 전압의 위상을 기준으로 하여 전류의 위상이 lead 인지 lag 인지 결정됨.) Power factorPhase difference (θ v - θ i ) Q (Complex power) leading -- Capacitive in phase 00 Resistive lagging ++ Inductive

18. EMLAB 18 Transmission line

19. EMLAB 19

20. EMLAB 20 Electric company 전기 회사 Electric company 전기 회사 House 가정 +-+- 계량기 Power meter Z 전선에 의한 손실을 줄이기 위해서 가능한 전류를 적게 흘려야 한다. To save power losses due to resistance of transmission lines, current should be kept as small as possible. Transmission line loss 220V

21. EMLAB 21 +-+- 220W 220V +-+- 220W 110V Example

22. EMLAB 22 An industrial load consumes 88 kW at a pf of 0.707 lagging from a 480-V rms line. The transmission line resistance from the power company’s transformer to the plant is 0.08. Let us determine the power that must be supplied by the power company (a) under present conditions and (b) if the pf is somehow changed to 0.90 lagging. (It is economically advantageous to have a power factor as close to one as possible.) Example 9.10 (a) (b)

23. EMLAB 23 Example A load operates at 20 kW, 0.8 pf lagging. The load voltage is at 60 Hz. The impedance of the line is 0.09 + j0.3. We wish to determine the voltage and power factor at the input to the line.

24. EMLAB 24 7. Power factor correction Simple approach to power factor correction Industrial load with lagging pf Electrical source

25. EMLAB 25 Example

26. EMLAB 26 8. Single-phase three wire circuits

27. EMLAB 27 +-+- 1kW 100V (rms) Disadvantage of single phase two wire system ( 단상 2 선 식의 단점 ) 1.If the load increases to 2kW, P loss becomes 40W. 2.If each wire can sustain heat only to 5W power loss, the cross-section area should increase 4 times, whereby cost for copper wires increase 4 times. 100V (rms) +-+- 2kW 100V (rms)

28. EMLAB 28 +-+- 1kW 100V (rms) +-+- 1kW 100V (rms) 1.Although the power load increases to 2kW, P loss remains 10W. 2.With 2 times power load, only one copper wire is added. Single-phase 3 wire circuits

29. EMLAB 29

30. EMLAB 30 Example 9.27 A light-duty commercial single-phase three-wire 60-Hz circuit serves lighting, heating, and motor loads, as shown below. Lighting and heating loads are essentially pure resistance and, hence, unity power factor (pf), whereas motor loads have lagging pf. Unbalanced connection

31. EMLAB 31 Balanced connection

32. EMLAB 32 이행선 Safety considerations Average effect of 60Hz current from hand to hand and passing the heart Required voltage depends on contact, person and other factors Typical residential circuit with ground and neutral Ground conductor is not needed for normal operation

33. EMLAB 33 Grounding is needed to protect from lightning. Grounding for safety considerations

34. EMLAB 34 기기 고장에 의한 short

35. EMLAB 35 Increased safety due to grounding When switched on the tool case is energized without the ground connector the user can be exposed to the full supply voltage! Conducting due to wet floor If case is grounded then the supply is shorted and the fuse acts to open the circuit More detailed numbers in a related case study Example

36. EMLAB 36 Ground prong removed Suggested resistances for human body Wet skin Can cause ventricular fibrillation