Alternating Current Electricity Frequency and Period

Transcription

1 Frequency and Period 15-1 Frequency (linear and angular) Period = T

2 Average Value 15-2a Square wave, positive pulse = negative pulse: X ave = 0 Pulse pattern, positive, all the same: X ave = tx max T where t is the duration of the pulse and T is the period. 1 Sawtooth: X ave = 2 X max Symmetrical triangular wave: X ave = 0

3 Average Value 15-2b Example (FEIM): A plating tank with an effective resistance of 100 Ω is connected to the output of a full-wave rectifier. The applied voltage is sinusoidal with a maximum of 170 V. How long does it take to transfer faradays? V ave = 2V max " = (2)(170 V) " = V I ave = V ave R = V 100 " = A t = q I ave = # (0.005 F) 96,487 A " s & % ( $ F ' A = 446 s

4 Effective or Root-Mean-Squared (RMS) Value 15-3a Effective value of an alternating waveform: For a sinusoidal waveform, Pulse pattern, positive, all the same: X rms = t T X max where t is the duration of the pulse and T is the period. Symmetrical triangular: X rms = 1 3 X max For a half-wave sinusoidal waveform, Sawtooth: X rms = 1 3 X max

5 Effective or Root-Mean-Squared (RMS) Value 15-3b Example 1 (FEIM): A 170 V (maximum value) sinusoidal voltage is connected across a 4 Ω resistor. What is the power dissipated by the resistor? V rms = V max 2 = 170 V 2 = V ( ) 2 P = V 2 rms R = V 4 " = 3.61#10 3 W

6 Effective or Root-Mean-Squared (RMS) Value 15-3c1 Example 2 (FEIM): What is the I rms value for the following waveform? (A) (B) (C) (D) 2 4 I 3 4 I 10 4 I 3 2 I

7 Effective or Root-Mean-Squared (RMS) Value 15-3c2 I rms = 1 T T " (I(t)) 2 = 1 0 T " T 2 0 I 2 dt + 1 T " T T 2 2 # I & % ( dt $ 2' = I 2 T " T % $ # 2& ' + " 1 %" $ # T ' I 2 % $ ' T &# 4 & ( ) T 2 T I 2 = 2 + " I 2 %" $ # 4T & ' T % $ # 2 ' & = I I 2 8 = 5 8 I 2 = 10 4 I Therefore, the answer is (C).

8 Effective or Root-Mean-Squared (RMS) Value 15-3d Example 3 (FEIM): A sinusoidal AC voltage with a value of V rms = 60 V is applied to a purely resistive circuit. What steady-state voltage would dissipate the same power as the AC voltage? (A) 30 V (B) 42 V (C) 60 V (D) 85 V The answer is right in the problem statement. Since the voltage is given as an rms value, the equivalent DC voltage is simply 60 V. Therefore, the answer is (C).

9 Phasor Transforms 15-4 Phase angle = θ φ φ as the angle between the reference and voltage θ as the angle between the reference and current If the phase angle is positive, the signal is called leading or capacitive. If the phase angle is negative, the signal is called lagging or inductive. If the phase angle is zero, the signal is called in phase.

10 Impedance 15-5a Definitions Impedance: Resistor: Capacitor: Inductor:

11 Impedance 15-5b1 Example (FEIM): For the following circuit: (a) What is the impedance in polar form? (b) Is the current leading or lagging? (c) What is the voltage across the inductor?

12 Impedance 15-5b2 (a) # Z = 3 "+ j & % ((2.5 )10 *3 H) j $ s' "+40 # & % ((625 )10 *3 F) $ s' = 3 "+ j1"# j4 "+ (9 ")cos 40 + ( j9 ")sin40 = 3 " "+ j(1"# 4 " ") = "+ j2.785 " Z = (9.894") 2 + (2.785") 2 = 10.28" # imaginary& Phase angle = tan "1 % ( $ real ' $ 2.785# ' = tan "1 & ) = % 9.894# ( Z = 10.28"#15.72

13 Impedance 15-5b3 (b) I = V Z = $ 160V "0 ' & ) = A"*15.72 % 10.28#"15.72 ( Therefore, the current is lagging (ELI). (c) V L = IZ L = (15.56 A "#15.72 )(1$ "90 ) = V "74.28

14 Admittance 15-6 Multiplying by the complex conjugate,

15 Ohm s Law for AC Circuits 15-7 Example (FEIM): What is the current in the capacitor in the following circuits? (a) (b) Z = 2 "# j4 " = 4.47 " $# 63.4 I = V Z = 180 V"0 = 40.3 A " # "$ 63.4 Z c = 0.25 " #$ 90 I = V 180 V"0 = = 720 A "90 Z c 0.25 # "$ 90

16 Complex Power 15-8a P: real power (W) Q: reactive power (VAR) S: apparent power (VA) Power Factor: Real Power: Reactive Power:

17 Complex Power 15-8b1 Example (FEIM): For the following circuit, find the (a) real power, and (b) reactive power. (c) Draw the power triangle. (a) The real power is: P = V 2 R R = (120 V)2 10 " = 1440 W (b) The reactive power is: Q = V 2 L = X L (120 V)2 4 " = 3600 VAR

18 Complex Power 15-8b2 (c) The power triangle is: In this example, the impedance of the inductor has a lagging Current so the current has a negative phase angle. The complex conjugate of the current has a positive phase angle, so the reactive power, Q, is positive and the power triangle is in the first quadrant. For a leading current (which has a positive phase angle compared to the voltage) the power triangle has a negative imaginary part and a negative power angle, so it is in the fourth quadrant.

19 Resonance 15-9a Series Resonance Quality Factor: Bandwidth: Half-power point is when R = X: " 1/2power = " 0 ± 1 2 BW Parallel Resonance Quality Factor: Bandwidth: Half-power point is when R = X: " 1/2power = " 0 ± 1 2 Q

20 Resonance 15-9b1 Example (FEIM): For the following circuit, find (a) the resonance frequency (rad/s) (b) the half-power points (rad/s) (c) the peak current (at resonance) (d) the peak voltage across each component at resonance (a) " 0 = 1 LC = 1 (200 #10 $6 H)(200 #10 $12 F) = 5 rad #106 s

21 Resonance 15-9b2 (b) " 1 2 power = " 0 ± 1 2 BW = " 0 ± R 2L = 5 #106 rad s ± 50 $ (2)(200 #10 %6 H) = "10 6, "10 6 rad s (c) At resonance, Z = R, I(resonance) = V R = 20 V "0 50 # = 0.4 A"0 (d) V R = I 0 R = ( 0.4 A "0 )( 50 #) = 20 V "0 % rad( V L = I 0 X L = I 0 j"l = (0.4 A #0 )' 5 $10 6 *(200 $10 +6 H)#90 & s ) = 400 "90 V C = "V L = 400 V #" 90

22 Resonance 15-9c Example (FEIM): A parallel resonance circuit with a 10 Ω resistor has a resonance frequency of 1 MHz and bandwidth of 10 khz. What resistor value will increase the BW to 20 khz? BW = " 0 Q = " 0 " 0 RC = 1 RC C remains the same, so to double the BW. R new = 1 2 R old = 5 "

23 Transformers 15-10a

24 Transformers 15-10b Example 1 (FEIM): What is the voltage V 3? Disregard losses. (A) 45 V (B) 65 V (C) 75 V (D) 90 V Therefore, the answer is (D). V 2 = N 2 N V V 3 = N 3 N A V 2 " V 3 = N %" 3 $ ' N % 2 $ ' 120 V # &# & N A N 1 ( ) = $ 300 " %" ' 50 % $ ' 120 V # 100& # 200& ( ) = 90 V

25 Transformers 15-10c Example 2 (FEIM): What is the total impedance in the primary circuit? Assume an ideal transformer. Z total = Z p + Z = 50 2 " % $ ' (12 (+ j4 () + 3 (+ j ( # 100& = 4"+ 3"+ j(1"+1") = 7"+ j2"

26 Rotating Machines 15-11a Synchronous Machines 1. Synchronous motors 2. Synchronous generators

27 Rotating Machines 15-11b Example (FEIM): A six-pole, three-phase synchronous generator supplies current with a frequency of 60 Hz. What is the angular velocity of the rotor in the generator? (A) 10" rad s (B) 40" rad s (C) 60" rad s (D) 80" rad s n s = 120f p = 60" 2# (Note: Here, Ω does not mean ohms. It is the angular velocity.) " = 4#f p = # 4" rad &# % ( 60 cycle & % ( $ cycle' $ s ' 6 = 40" rad/s Therefore, the answer is (B).

28 Rotating Machines 15-11c Induction Motors Slip: A constant-speed device that receives power though induction without using brushes or slip rings

29 Rotating Machines 15-11d Example (FEIM): A four-pole induction motor operates on a three-phase, 240 V rms line-toline supply. The slip is 5%. The operating speed is 1600 rpm. What is most nearly the operating frequency? (A) 56 Hz (B) 60 Hz (C) 64 Hz (D) 102 Hz n s = n(1+ s) = ( 1600 rpm)( ) = 1680 rpm Solving the synchronous speed equation for the operating frequency yields Therefore, the answer is (A). f = pn s 120 ( )( 1680 rpm) = = 56 Hz

30 Rotating Machines 15-11e DC Machines Have a constant magnetic field in the stator; the magnetic field of rotor responds to the stator field The magnetic flux produced is described by the equation

31 Rotating Machines 15-11f DC Generators A device that produces DC potential

32 Rotating Machines 15-11g DC Generators For the DC machine equivalent circuit Terminal Voltage: For the simplified DC machine equivalent circuit Terminal Voltage:

33 Rotating Machines 15-11h1 Example (FEIM): A DC generator provides a current of 12 A for a resistive load when operating at 1800 rpm. The speed of the armature is reduced, and the new steady-state current is 10 A. What is most nearly the operating speed when the generator reaches steady-state conditions? Ignore armature resistance. (A) 1500 rpm (B) 1700 rpm (C) 1800 rpm (D) 2200 rpm

34 Rotating Machines 15-11h2 The current is proportional to the voltage by Ohm s law. From the equation for the armature voltage, ignoring the armature resistance, the armature voltage is proportional to the speed. V a = K a n" Therefore, the armature current is proportional to the speed. I 2 I 1 = V 2 V 1 = n 2 n 1 Rearranging yields the new operating speed. n 2 = I 2 I 1 n 1 " = 10 A % $ '(1800 rpm) # 12 A& = 1500 rpm Therefore, the answer is (A).

35 Rotating Machines 15-11i DC Motors Very similar to a DC generator, only the direction of current changes Power (for the motor model in the DC machine equivalent circuit): Ignoring the power dissipated as heat for the DC machine equivalent circuit, Torque: Mechanical Power:

36 Rotating Machines Example (FEIM): A DC motor is operating on 100 V with a magnetic flux of 0.02 Wb produces an output torque of 20 N m. The magnetic flux is changed to 0.03 Wb. What is most nearly the new steady-state output torque? Ignore armature resistance. (A) = 19 N"m (B) = 30 N"m (C) = 32 N"m (D) = 51 N"m The motor voltage is not important to the solution because the torque equation is proportional to the magnetic flux. T 1 = " 1 T 2 " j T 2 = " 2 T 1 " 1 " 0.03 Wb% = $ ' 20 N(m # 0.02 Wb& = 30 N"m The answer is (B). ( )

37 Additional Examples 15-12a Example 1 (FEIM): For the circuit elements shown, draw the conventional impedance diagram.

38 Additional Examples 15-12b Example 2 (FEIM): What is the steady-state magnitude of the rms voltage across the capacitor? (A) 15 V (B) 30 V (C) 45 V (D) 60 V Z = 10 "+ j(15 "#15 ") = 10 " I rms = V rms R = 40 V 10 " = 4 A V c rms = I rms X c = (4 A)(15 ") = 60 V Therefore, the answer is (D).

39 Additional Examples 15-12c1 Example 3 (FEIM): For the following circuit, the rms steady-state currents are I 1 = 14.4"#36.9 and I 2 = 6"90. The impedance seen by the voltage source is most nearly (A) 9.5 " #18.4 (B) 10.0 " #36.9 (C) 10.7 " #16.0 (D) 11.0 " #7.1

40 Additional Examples 15-12c2 Z = R = R + Z 1Z 2 Z 1 + Z 2 = 5 "+ Z Z 2 (4 + j3)(# j12) 4 + j3 # j12 $ 36 # j48' $ " = 5 "+ & ) 4 + j9 ' & )" % 4 # j9 (% 4 + j9( Note: The preceding calculation is an example of rationalizing the denominator of a complex number without converting it to polar form. 144 "+ 432 "+ j(324 "#192 ") Z = 5 "+ = 5 "+ 576 " 16 "+ 81" 97 " + j $ 132 " ' & ) % 97 " ( = 5 " "+ j1.361" = "+ j1.361" Z = ( ") 2 + (1.361") 2 = 11.0 " tan " # # = 7.1 Z = 11"#7.1 Therefore, the answer is (D).

41 Additional Examples 15-12d Example 4 (FEIM): The phasor form of I 3 is most nearly (A) 10.3 A "#32.5 (B) 11.8 A "#12.9 (C) 18.6 A "51.9 (D) 18.6 A "#51.9 I 3 = I 1 + I 2 = 14.4 A"# A"90 = (14.4 A) cos(#36.9 ) + j(14.4 A) sin# j6 A = A # j8.646 A + j6 A = A # j2.646 A I 3 = ( A) 2 + (2.646 A) 2 = 11.8 A tan -1 "2.646 A A = "12.9 I 3 = 11.8 A#"12.9 Therefore, the answer is (B).