EKT13 EECTRCA ENGNEERNG Chater 1 Three-Phase System 1
COURSE OUTCOME (CO) CO1: Ability to define and exlain the concet of single-hase and threehase system. 2
Revision A sinusoid is a signal that has the form of the sine or cosine function. A general exression for the sinusoid, v( t) sin( ω t + φ) m where m the amlitude of the sinusoid ω the angular frequency in radians/s Ф the hase 3
Revision A eriodic function is one that satisfies v(t) v(t + nt), for all t and for all integers n. T 2π ω f 1 Hz f T ω 2π Only two sinusoidal values with the same frequency can be comared by their amlitude and hase difference. f hase difference is zero, they are in hase; if hase difference is not zero, they are out of hase. 4
Revision Examle 1 o Given a sinusoid, 5 sin( 4πt 6 ), calculate its amlitude, hase, angular frequency, eriod, and frequency. 5
Revision Examle 1 o Given a sinusoid, 5 sin( 4πt 6 ), calculate its amlitude, hase, angular frequency, eriod, and frequency. Solution: Amlitude 5, hase 6 o, angular frequency 4π rad/s, Period.5 s, frequency 2 Hz. 6
Revision Examle 2 Find the hase angle between i 1 4sin(377t + and o i 5cos(377t 4 ), does i 1 lead or lag i 2? 2 25 o ) 7
Revision Examle 2 Find the hase angle between i 1 4sin(377t + and o i 5cos(377t 4 ), does i 1 lead or lag i 2? 2 25 o ) Solution: Since sin(ωt+9 o ) cosωt o o i 5sin(377t 4 + 9 ) 5sin(377t + 5 2 o o o i 4sin(377t + 25 ) 4sin(377t + 18 + 25 ) 4sin(377t + 1 therefore, i 1 leads i 2 155 o. o ) 25 o 8 )
Revision medance transformation Z 3Z Y Z Y 1 Z 3 9
Single-Phase Circuit A single hase circuit consists of a generator connected through a air of wires to a load Two wire system Three wired system same magnitude same hase
Two-Phase Circuit a A Three wired system Second source with 9 out of hase Three wired system same magnitude different hase
What is a Three-Phase Circuit? t is a system roduced by a generator consisting of three sources having the same amlitude and frequency but out of hase with each other by 12. Three sources with 12 out of hase Four wired system 12
Balance Three-Phase oltages A three-hase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator). A three-hase generator The generated voltages 13
Balance Three-Phase oltages Two ossible configurations: Three-hase voltage sources: (a) Y-connected ; (b) -connected 14
Balance Three-Phase oltages Phase sequences a) abc or ositive sequence b) acb or negative sequence 15
Balance Three-Phase oltages f the voltage source have the same amlitude and frequency ω and are out of hase with each other by 12 o, the voltage are said to be balanced. an + bn + cn an bn cn Balanced hase voltages are equal in magnitude and out of hase with each other by 12 o 16
Balance Three-Phase oltages abc sequence or ositive sequence: an bn cn 12 24 acb sequence or negative sequence: an cn bn 12 24 + 12 + 12 is the effective or rms value 17
Balance Three-Phase oltages Examle 1 Determine the hase sequence of the set of voltages. v v v an bn cn 2 cos( ωt + 1 ) 2 cos( ωt 23 ) 2 cos( ωt 11 ) 18
Solution: Balance Three-Phase oltages The voltages can be exressed in hasor form as an bn cn 2 1 2 23 2 11 We notice that an leads cn by 12 and cn in turn leads bn by 12. Hence, we have an acb sequence. 19
Balance Three-Phase oltages Two ossible three-hase load configurations: a) a wye-connected load b) a delta-connected load 2
Balance Three-Phase oltages A balanced load is one in which the hase imedances are equal in magnitude and in hase. For a balanced wye connected load: Z 1 Z2 Z3 ZY For a balanced delta connected load: Z a Z Z Z b c Z 3Z Y Z Y 1 Z 3 21
Balance Three-Phase Connection Four ossible connections 1. Y-Y connection (Y-connected source with a Y-connected load) 2. Y- connection (Y-connected source with a -connected load) 3. - connection 4. -Y connection 22
Balance Y-Y Y Y Connection A balanced Y-Y system is a three-hase system with a balanced y-connected source and a balanced y-connected load.
Balance Y-Y Y Y Connection Z Z Z Z s l Y Source imedance ine imedance oad imedance Total imedance er hase Since all imedance are in series, Thus Z Y Z s + Zl + Z Z Y Z 24
Balance Y-Y Y Y Connection a Z an Y
Balance Y-Y Y Y Connection Alying K to each hase: a Z an Y 12 bn an b a 12 ZY ZY 24 cn an c a 24 ZY ZY nn Znn a + b + c n 26
Balance Y-Y Y Y Connection ine to line voltages or line voltages: ab bc ca 3 3 3 3 9 21 Magnitude of line voltages: 3 an ab bn bc cn ca
Balance Y-Y Y Y Connection Examle 2 Calculate the line currents in the three-wire Y-Y system shown below: 28
Balance Y-Y Y Y Connection Examle 2 Calculate the line currents in the three-wire Y-Y system shown below: Ans a b c 6.81 21.8 A 6.81 141.8 A 6.81 98.2 A 29
Balance Y- Connection A balanced Y- system is a three-hase system with a balanced y-connected source and a balanced -connected load. 3
Balance Y- Connection A single hase equivalent circuit a Z an Y Z an / 3 Z Y Z 3
Balance Y- Connection A single hase equivalent circuit ine voltages: ab bc ca 3 3 3 9 3 21 AB BC CA
Balance Y- Connection A single-hase equivalent circuit of a balanced Y- circuit ine currents: a b c AB BC CA CA AB BC Phase currents: 3 3 3 AB AB AB AB BC CA 3 15 9 Z Z Z AB BC CA
Balance Y- Connection A single-hase equivalent circuit of a balanced Y- circuit CA a AB AB 24 CA AB (1 1 24 ) a AB 3 3 Magnitude line currents: 3 a AB b BC c CA
Examle 3 A balanced abc-sequence Y-connected source with ( an 1 1 ) is connected to a -connected load (8+j4)Ω er hase. Calculate the hase and line currents. Solution Balance Y- Connection Using single-hase analysis, 1 1 an 33.54 16.57 Z / 3 2.981 26.57 a A Other line currents are obtained using the abc hase sequence 35
Balance - Connection A balanced - system is a three-hase system with a balanced -connected source and a balanced -connected load. 36
ine voltages: Balance - Connection ine currents: Phase currents: ab bc ca AB BC CA Magnitude line currents: 3 a b c AB BC CA CA AB BC Total imedance: Z Y 3 3 3 AB AB AB Z 3 3 15 9 AB BC CA Z Z Z AB BC CA 37
Examle 4 A balanced -connected load having an imedance 2-j15 Ω isconnected to a -connected ositive-sequence generator having ( ab 33 ). Calculate the hase currents of the load and the line currents. Ans: The hase currents 13.2 36.87 A; 13.2 81.13 A; 13.2 156.87 A AB BC AB The line currents 22.86 6.87 A; 22.86 113.13 A; 22.86 126.87 A a Balance - Connection b c 38
Balance -Y Connection A balanced -Y system is a three-hase system with a balanced y-connected source and a balanced y-connected load. 39
Balance -Y Connection Alying K to loo aanbba: From: a b Z b a 12 a b a 3 3 Y ine currents: a 3 Z Y 3 4
Balance -Y Connection Relace connected source to equivalent Y connected source. Phase voltages: an bn cn 3 3 15 3 + 9 3 41
Balance -Y Connection A single hase equivalent circuit a Z an Y 3 Z Y 3 42
Balance -Y Connection Examle 5 A balanced Y-connected load with a hase imedance 4+j25 Ω is sulied by a balanced, ositive-sequence -connected source with a line voltage of 21. Calculate the hase currents. Use ab as reference. Answer The hase currents AN BN CN 2.57 62 A; 2.57 178 A; 2.57 58 A; 43
Power in a Balanced System Comaring the ower loss in (a) a single-hase system, and (b) a three-hase system 2 P P ' loss 2R 2, single - hase 2 ( ) R 2R, single- hase P loss P 2 2 2 2 P P loss 3 2 3 2 ( ' ) R' 3R' R', three - hase ' 2 P 2 f same ower loss is tolerated in both system, three-hase system use only 75% of materials of a single-hase system 44
Power in a Balanced System For Y connected load, the hase voltage: v AN 2 cos ωt v BN 2 cos( ωt 12 ) v CN 2 cos( ωt + 12 ) 45
Power in a Balanced System Phase current lag hase voltage by θ. f Z Y Z θ The hase current: i a 2 cos( ω t θ ) i b 2 cos( ωt θ 12 ) i c 2 cos( ωt θ + 12 ) 46
Power in a Balanced System Total instantaneous ower: + + v i + v i + a b c AN 3 cos a θ BN b v CN i c Average ower er hase: P cos θ Reactive ower er hase: Q sin θ Aarent ower er hase: Comlex ower er hase: S S P + jq * 47
Power in a Balanced System Total average ower: P 3P 3 cos θ 3 cos θ Total reactive ower: Q 3Q 3 sin θ 3 sin θ Total comlex ower: S 3S 3 3 Z * 2 3 Z 2 * S P + jq 3 θ 48
Power in a Balanced System Power loss in two wires: P 2 R loss 2 2R P 2 2 Power loss in three wires: P P loss P 2 2 2 3 R 3R R 2 2 P 3 : ower absorbed by the load : magnitude of line current : line voltage R : line resistance 49
Examle 6 A three-hase motor can be regarded as a balanced Y-load. A three-hase motor draws 5.6 kw when the line voltage is 22 and the line current is 18.2 A. Determine the ower factor of the motor. 5
Examle 6 A three-hase motor can be regarded as a balanced Y-load. A three-hase motor draws 5.6 kw when the line voltage is 22 and the line current is 18.2 A. Determine the ower factor of the motor. The aarent ower is S 3 3(22)(18.2) 6935. 13A The real ower is P S cos θ 56W The ower factor is f P cos θ.875 S 51
Exercise 6 Calculate the line current required for a 3-kW three-hase motor having a ower factor of.85 lagging if it is connected to a balanced source with a line voltage of 44. 52
Exercise 6 Calculate the line current required for a 3-kW three-hase motor having a ower factor of.85 lagging if it is connected to a balanced source with a line voltage of 44. Answer : 46. 31 53