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BEEE102L BASIC ELECTRICAL
AND ELECTRONICS
ENGINEERING
Dr.S.ALBERT ALEXANDER
SCHOOL OF ELECTRICAL ENGINEERING
albert.alexander@vit.ac.in
1
Dr.S.ALBERTALEXANDER-
SELECT-VIT
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
2.6 Three phase systems
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
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
Three phase balanced systems
 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
Three phase balanced systems
 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
Three phase balanced systems
 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
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
Analysis
 Since the three-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
Analysis
Dr.S.ALBERTALEXANDER-SELECT-
VIT 10
 The phase 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
Exercise 1
Dr.S.ALBERTALEXANDER-SELECT-
VIT 11
Determine the 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
Power in a balanced 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
Power in a balanced 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
Power in a balanced 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
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
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
Exercise 3
Determine the total 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
Exercise 3 (Contd..)
Dr.S.ALBERTALEXANDER-SELECT-
VIT 18
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
Exercise 4
A three-phase motor 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
Dr.S.ALBERTALEXANDER-SELECT-
VIT 20

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class notes of electrical & electronics.pptx

  • 1. BEEE102L BASIC ELECTRICAL AND ELECTRONICS ENGINEERING Dr.S.ALBERT ALEXANDER SCHOOL OF ELECTRICAL ENGINEERING albert.alexander@vit.ac.in 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 phase systems 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
  • 4. 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
  • 5. Three phase balanced systems  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
  • 6. Three phase balanced systems  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
  • 7. Three phase balanced systems  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
  • 8. 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
  • 9. Analysis  Since the three-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
  • 10. Analysis Dr.S.ALBERTALEXANDER-SELECT- VIT 10  The phase 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 Determine the 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
  • 12. Power in a balanced 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 a balanced 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
  • 14. Power in a balanced 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
  • 15. 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
  • 16. 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
  • 17. Exercise 3 Determine the total 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
  • 18. Exercise 3 (Contd..) Dr.S.ALBERTALEXANDER-SELECT- VIT 18 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
  • 19. Exercise 4 A three-phase motor 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