class notes of electrical & electronics.pptx

class notes of electrical & electronics.pptx

• 1.
  BEEE102L BASIC ELECTRICAL ANDELECTRONICS ENGINEERING Dr.S.ALBERT ALEXANDER SCHOOL OF ELECTRICAL ENGINEERING [email protected] 1 Dr.S.ALBERTALEXANDER- SELECT-VIT
• 2.
  Module 2 Dr.S.ALBERTALEXANDER-SELECT- VIT 2 Alternating voltages and currents  RMS, average, maximum values  Single Phase RL, RC, RLC series circuits  Power in AC circuits  Power Factor  Three phase balanced systems  Star and delta Connections  Electrical Safety, Fuses and Earthing
• 3.
  2.6 Three phasesystems Dr.S.ALBERTALEXANDER-SELECT- VIT 3  A three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 1200  All electric power is generated and distributed in three- phase, at the operating frequency of 60 Hz in the USA or 50 Hz in India  When one-phase or two-phase inputs are required, they are taken from the three-phase system rather than generated independently  The instantaneous power in a three-phase system can be constant (not pulsating) resulting in uniform power transmission and less vibration of three-phase machines
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  Three phase systems For the same amount of power, the three-phase system is more economical than the single phase  The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system Dr.S.ALBERTALEXANDER-SELECT- VIT 4
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  Three phase balancedsystems  Three-phase voltages are often produced with a three- phase ac generator (or alternator)  The generator basically consists of a rotating magnet (rotor) surrounded by a stationary winding (stator) Dr.S.ALBERTALEXANDER-SELECT- VIT 5
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  Three phase balancedsystems  Three separate windings or coils with terminals a-a’, b-b’ and c-c’ are physically placed 1200 apart around the stator  Terminals a and a’ for example, stand for one of the ends of coils going into and the other end coming out  As the rotor rotates, its magnetic field “cuts” the flux from the three coils and induces voltages in the coils  Because the coils are placed apart, the induced voltages in the coils are equal in magnitude but out of phase by 1200  Since each coil can be regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads Dr.S.ALBERTALEXANDER-SELECT- VIT 6
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  Three phase balancedsystems  A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines)  Three phase current sources are very scarce  A three-phase system is equivalent to three single-phase circuits  The voltage sources can be either wye (Y)-connected or delta ()-connected Dr.S.ALBERTALEXANDER-SELECT- VIT 7
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  Analysis  Let Van,Vbn and Vcn be the voltages between lines a, b, and c, and the neutral line n  “phase voltages”  If the voltage sources have the same amplitude and frequency and are out of phase with each other by the voltages are said to be balanced Van+Vbn+Vcn=0 │ V an│=│ Vbn│ =│ Vcn│  Balanced phase voltages are equal in magnitude and are out of phase with each other by 1200 Dr.S.ALBERTALEXANDER-SELECT- VIT 8
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  Analysis  Since thethree-phase voltages are out of phase with each other, there are two possible combinations Van= Vp00 Vbn= Vp-1200 Vcn= Vp-2400 or Vcn= Vp+1200 +ve sequence Rotor rotates counter clockwise Dr.S.ALBERTALEXANDER-SELECT- VIT 9 Van= Vp00 Vcn= Vp-1200 Vbn= Vp-2400 or Vcn= Vp+1200 -ve sequence Rotor rotates clockwise
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  Analysis Dr.S.ALBERTALEXANDER-SELECT- VIT 10  Thephase sequence is the time order in which the voltages pass through their respective maximum values  A balanced load is one in which the phase impedances are equal in magnitude and in phase  For a balanced wye-connected load, Z1=Z2=Z3=ZY  ZY is the load impedance per phase  For a balanced delta connected load, Za=Zb=Zc=Z  Z is the load impedance per phase  Wye-connected load can be transformed into a delta connected load, or vice versa, using Z=3ZY  Since both the 3 source and 3 load can be either wye or delta connected, we have four possible connections: Y-Y, Y-, - and -Y
• 11.
  Exercise 1 Dr.S.ALBERTALEXANDER-SELECT- VIT 11 Determinethe phase sequence of the set of voltages: van= 200 cos(t+100) vbn= 200 cos(t-2300) vcn= 200 cos(t-1100) SOLUTION  The voltages can be expressed in phasor form as, Van = 200100 Vbn = 200-2300 Vcn = 200-1100  Van leads Vcn by 1200 and in turn Vcn leads Vbn by 1200  Hence, the sequence is a-c-b
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  Power in abalanced system  For a Y-connected load, the phase voltages are vAN= 2 Vp cost vBN= 2 Vp cos (t-1200) vCN= 2 Vp cos (t+1200)  The factor 2 is necessary because Vp has been defined as the rms value of the phase voltage behind their  If ZY=Z, the phase currents lag corresponding phase voltages by . ia= 2 Ip cos (t-) ib= 2 Ip cos (t--1200) ic= 2 Ip cos (t-+1200)  Ip is the rms value of phase current Dr.S.ALBERTALEXANDER-SELECT- VIT 12
• 13.
  Power in abalanced system  The total instantaneous power in the load is the sum of the instantaneous powers in the three phases p=pa+pb+pc= vANia+vBNib+vCNic =3VpIp cos  The total instantaneous power is independent of time  The average power per phase Pp for -connected load or Y-connected load is p/3 or Pp =VpIp cos  The reactive power per phase is: Qp =VpIp sin  The apparent power per phase is: Sp= VpIp  The complex power per phase is: Sp= Pp +jQp =VpIp*  Vp and Ip are the phase voltage and phase current with magnitudes Vp and Ip respectively  The total =3VpIp cos= average power is: P=Pa+Pb+Pc=3Pp 3 VLIL cos Dr.S.ALBERTALEXANDER-SELECT- VIT 13
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  Power in abalanced system  For a Y-connected load, IL=IP, VL= 3 Vp  For a -connected load, IL= 3 Ip, VL=Vp  For both Y and -connected loads: Total Average Power: P= 3 VLIL cos Total Reactive Power: Q= 3 VLIL sin Total Complex Power: S= 3 VLIL   Vp, Ip, VL and IL are rms values and  is the angle of the load impedance or the angle between the phase voltage and the phase current Dr.S.ALBERTALEXANDER-SELECT- VIT 14
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  Exercise 2 A balanced∆-connected load having an impedance 20−j15 Ω is connected to a ∆- connected, positive-sequence generator having Vab = 330∠0° V. Calculate a) Phase currents of the load, b) Line currents, c) Total real power d) Total reactive power. SOLUTION  Vab = 330∠0°, Vbc= 330∠-120° and Vca = 330∠120°  Z1=Z2=Z3=Z = 20− j15 Ω= 25∠-36.87° a) Phase currents of the load: Iab 1 Z 25∠−36.87° =Vab = 330∠0° = 13.2∠36.87°A bc Z2 25∠−36.87° I =Vbc = 330∠− 120°= 13.2∠-83.13° A ca Z3 25∠−36.87° I =Vca = 330∠120° = 13.2∠156.87° A In  connection, line currents are 300 lagging with phase current and magnitude is 3 times the phase current. Dr.S.ALBERTALEXANDER-SELECT- VIT 15
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  Exercise 2 (Contd..) b)Line currents:  IL= 3 Ip (-300)  Ia= 3 × 13.2(36.870-300)= 22.866.80 A  Ib= 3 × 13.2(-83.130-300)= 22.86-113.130 A  Ic= 3 × 13.2(156.870-300)= 22.86126.870 A c) Total real power: P= 3 VLIL cos = 3 × 330 × 22.86 ×cos (-36.87)= 10.452 kW d) Total reactive power. P= 3 VLIL sin = 3 × 330 × 22.86 × sin (-36.87)= -7840 VAR Dr.S.ALBERTALEXANDER-SELECT- VIT 16
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  Exercise 3 Determine thetotal average power, reactive complex power at the source and at the load. power, and SOLUTION  It is sufficient to consider one phase, as the system is balanced. For phase a, Van  Ia= ZY = 11000 11000 11000 (5− j2)+(10+j8)= 15+j6 = 16.15521.80 0 = 6.81− 21.8 A Dr.S.ALBERTALEXANDER-SELECT- VIT 17
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  Exercise 3 (Contd..) Dr.S.ALBERTALEXANDER-SELECT- VIT18 At the source, the complex power absorbed is,  SS=-3VpIp*= -3(11000)(6.81−21.80) =-224721.80 =-(2087+j834.6) VA  The real or average power absorbed is -2087 W  The reactive power is -834.6 VAR At the load, the complex power absorbed is, SL=3│Ip│2 Zp= 3(6.81)2(10+j8)= 3(6.81)2(12.8138.660) =1782 38.660 = (1392+j1113.3) VA  The real power absorbed is 1392 W  The reactive power absorbed is 1113.3 VAR
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  Exercise 4 A three-phasemotor can be regarded as a balanced Y-load. A three phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor. SOLUTION  The apparent power is, S= 3 VLIL 3 (220)(18.2)= 6935.13 VA  Since the real power is, P = 5600 W  Power factor is, cos  = P/S = 5600/ 6935.13 =0.8075 Dr.S.ALBERTALEXANDER-SELECT- VIT 19
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  Dr.S.ALBERTALEXANDER-SELECT- VIT 20