EECE310_F05_Lecture_Notes__wk_of_11_7_05_

EECE310_F05_Lecture_Notes__wk_of_11_7_05_ - College of...

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Unformatted text preview: College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Up until this point, the schematics studied have all been single-phase. Now the analysis techniques will be extended to include three-phase (3-) circuits. The three-phase networks contain 3 independent voltage sources that are one-third of a cycle apart in time. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis If the three sinusoidal voltages have the same magnitude and frequency and each voltage is 120 out of phase with the other two, and if the loads are such that the currents produced by the voltages also have the the same magnitude and frequency and each current is 120 out of phase with the other two then the entire circuit is referred to as a balanced three-phase network. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis A balanced set of three-phase (3-) voltages can be represented in phasor form: r Vbn = 120 - 120 V rms = 120 + 240 V rms r Vcn = 120 - 240 V rms = 120 + 120 V rms r Van = 120 0 V rms EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The three-phase (3-) voltages may also be written in the time domain: v an ( t ) = 120 2 cos t V 2 v bn ( t ) = 120 2 cos( t - )V 3 4 v cn ( t ) = 120 2 cos( t - )V 3 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The instantaneous power produced by the balanced three-phase system is equated as: p (t ) = p a (t ) + p b (t ) + p c (t ) The expression for instantaneous power is: VM I M p (t ) = 3 cos W 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Important property of the balanced voltage set r r r is: Van + Vbn + Vcn = 0 Proof: 120 + (-60- j103.9) + (-60 + j103.9) = 0 The voltage set may be described as a positive voltage sequence, if the voltages reach their peak values in the order abc (Van leads Vbn and Vbn leads Vcn). EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Three-phase (3-) networks are always designed as three different types of configurations (designated by the sources and the loads): 1) Wye-Wye (Y-Y) configuration 2) Wye-Delta (Y-) configuration 3) Delta-Delta (-) configuration EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Given any balanced 3- Y-Y configuration, the phase voltage is the magnitude of the phasor voltage from the neutral to any line. The phase voltages with positive phase sequence are: r Van = V p 0 V rms r Vbn = V P - 120 V rms r Vcn = V P + 120 V rms EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis For a balanced 3- Y-Y configuration, the line voltage is the difference in voltage between each line. r 1 3 Vab = V P 0 - V P - 120 = V P - V P [ - - j ] 2 2 r 1 3 3 3 Vab = V P [1 + + j ] = VP [ + j ] = 3V P 30 2 2 2 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP The line-to-line voltages can be calculated as r r r follows: V = V -V ab an bn College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The set of line-to-line voltages are: r Vbc = 3V P - 90 = 3V P + 270 r Vca = 3V P - 210 = 3V P + 150 r Vab = 3V P 30 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The currents flowing through the transmission lines (the line currents) can be computed as r follows: r Van V P 0 r Ib r Ic Ia = r = r ZY r ZY Vbn V P - 120 r = r = ZY ZY r Vcn V P + 120 r = r = ZY ZY EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Important Note about balanced 3- Y-Y configuration: Using the phase sequence, only one phase of the network needs to be analyze to obtain the voltages and currents of the other two phases. Also: V = 3V L P IL = IP r r r r I n = ( I a + Ib + I c ) = 0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The second configuration is a balanced 3- Y- connection. Here the phase voltage is computed the same way as in a balanced 3- Y-Y connection. The phase voltages with positive phase sequence are: r Van = V p 0 V rms r Vbn = V P - 120 V rms r Vcn = V P + 120 V rms EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis For a balanced 3- Y- configuration, the line voltage at the Y source is the difference in voltage between each line. r 1 3 Vab = V P 0 - V P - 120 = V P - V P [ - - j ] 2 2 r 1 3 3 3 Vab = V P [1 + + j ] = VP [ + j ] = 3V P 30 2 2 2 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP The line-to-line voltages can be calculated as r r r follows: V = V -V ab an bn College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The set of line-to-line voltages for the Y source r are: r Vbc = 3V P - 90 = 3V P + 270 r Vca = 3V P - 210 = 3V P + 150 Vab = 3V P 30 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Since the configuration has a load, then the phase currents at the can be easily discovered. r VP 0 = I P (0 - z ) I AB = r Z r V P - 120 r = I P ( - 120 - Z ) I BC = Z r V P + 120 r = I P ( + 120 - Z ) ICA = Z EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The line currents for the load can be calculated from the rload phase currents as: r r r r IaA = I AB + I AC = I AB - ICA r r r r r IbB = I BC + I BA = I BC - I AB r r r r r IcC = ICA + ICB = ICA - I BC EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The line currents for the load are formulated as r follows: r r r IaA = I AB + I AC = I AB r - ICA r VP VP VP VP 1 3 IaA = 0 - + 120 = - [- + j ] Z Z Z Z 2 2 r 3 1 3 VP VP 3 VP ]= 3 - 30 ]= [ - j IaA = [1 + - j 2 Z 2 2 Z 2 Z r IaA = 3 I P - 30 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The set of line currents for the load are: r IbB = 3 I P - 150 = 3 I P + 210 r IcC = 3 I P - 270 = 3 I P + 90 r IaA = 3 I P - 30 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Important Note about balanced 3- Y- configuration: Using the phase sequence, only one phase of the network needs to be analyze to obtain the voltages and currents of the other two phases. Also the following equations correspond to the load: VL = VP IL = 3I P EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The third and final configuration is a balanced 3- - connection. In this connection the sources are connected line-to-line. The line voltages (at the source) with positive r phase sequence are: r r Vab = Va - Vb = V L 0 V rms r r r Vbc = Vb - Vc = V L - 120 V rms r r r Vca = Vc - Va = V L + 120 V rms EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Since the configuration has a source, then the phase voltages translated for a Y-source would be: V 0 r V 3 30 3 r V L - 120 V L Vbn = = - 150 3 30 3 r V L + 120 V L Vcn = = + 90 ( = - 270 ) 3 30 3 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP Van = L = L - 30 College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Since the configuration has a load, then the phase currents at the load are: r VL 0 = I P (0 - z ) I AB = r Z r V L - 120 r = I P ( - 120 - Z ) I BC = Z r V L + 120 r = I P ( + 120 - Z ) ICA = Z EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis For a balanced 3- - configuration, the line current at the load is the difference in current between each phase. The line-to-line currents can be calculated as r r r follows: IaA = I AB - ICA r 1 3 IaA = I P 0 - I P 120 = I P - I P [ - + j ] 2 2 r 1 3 3 3 IaA = I P [1 + - j ] = IP[ - j ] = 3 I P - 30 2 2 2 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The set of line currents for the load are: The line-phase current relationship at the load is: r IbB = 3 I P - 150 = 3 I P + 210 r IcC = 3 I P - 270 = 3 I P + 90 r IaA = 3 I P - 30 IL = 3I P EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Important Note about balanced 3- - configuration: Given this configuration, the sources can simply be converted to a Y-source (as to the load a to Y-load) to comply with the preceding two connections. Also the following: r Vab Z = 3 ZY r r + Vbc + Vca = 0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis Complex power (S) for a single-phase system for a Y-connected or -connected load: The total complex power for a balanced 3- network is: r VL I L S1 = V P I P ( v - i ) = ( v - i ) 3 r S 3 = 3V L I L Z LOAD EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 5-Steady-State Power Analysis The rectangular relationships for apparent power (S) in a balanced 3- system: r r PT = Re( S T ) = r QT = Im( S T ) = ST S 3 = S T = PT jQ T r 3V L I L cos( z LOAD ) r 2 r 2 = (Re( S T )) + (Im(rS T )) -1 Im( S T ) r ) z LOAD = tan ( Re( S T ) EECE 310 Preston D. Frazier, Ph.D., P.E., PMP 3V L I L sin( z LOAD ) College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 7 Network Frequency Characteristics EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 7-Network Frequency Characteristics This chapter explores the frequency characteristics of an electrical network. Particularly how the frequencies at the source affect the circuit. The frequency response (i.e., the sinusoidal frequency response) defines the behavior of the network as a function of the frequency. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 7-Network Frequency Characteristics Elucidating the frequency response of a electrical circuit will provide a better understanding of its performance. Electrical networks are designed to operate at a single frequency. In general, schematics will respond differently to sinusoidal signals of various frequencies. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 7-Network Frequency Characteristics Achieving the proper frequency response in certain circuits is extremely important in the circuit's functionality. To examine the sinusoidal frequency response in a network, we need to examine the ratio of the output to the input (ergo the system's transfer function). EECE 310 Preston D. Frazier, Ph.D., P.E., PMP ...
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This note was uploaded on 10/15/2008 for the course EECE 310 taught by Professor Frazier during the Fall '08 term at Howard.

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