10-three Phase
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ELECTRIC CIRCUITS THEORY 1 These lecture slides have been compiled by Mohammed LECTURE 10 SalahUdDin Ayubi. Three Phase Power Systems 25 August 2005
Engineer M S Ayubi
1
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
2
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
3
Three Phase Voltages •3 AC voltage of equal amplitude, out of phase by 120° Va = Vm sin ( ωt ) = Vm ∠0° = Vm (1 + j 0) 1 ( ) Vb = Vm sin ωt − 120° = Vm ∠ − 120° = Vm (− − j
3 ) 2 2 1 3 Vc = Vm sin ( ωt − 240°) = Vm ∠ − 240° = Vm (− + j ) 2 2 25 August 2005
Engineer M S Ayubi
4
Phase Sequence Vc=V∠ -240°
Vb=V∠ -240° Va=V∠ 0°
Vb=V∠ -120° Phase sequence: abc 25 August 2005
Va=V∠ 0°
Vc=V∠ -120° Reverse sequence: acb Engineer M S Ayubi
5
Three Phase Voltages • Sum of phasor voltages is zero.
25 August 2005
Engineer M S Ayubi
6
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
7
Wye (Y) Connection C
neutral
O B
4 wire system
5 wire system
A ground
25 August 2005
Engineer M S Ayubi
8
Y Phase-Line Relationships C
neutral
O B A
ground
Va − Vb − Vab = 0 Vab = Va − Vb 1 3 Vab = Vm (1 + j 0) − Vm (− − j ) 2 2 3 3 = Vm ( + j ) 2 2 Vab = 3Vm ∠30° Vbc = 3Vm ∠90° Vca = 3Vm ∠150°
25 August 2005
Engineer M S Ayubi
9
Y Phase-Line Relationships Vc=V∠ -240°
Vab =V∠ 30°
Va=V∠ 0°
Vca =V∠ 150°
Vb=V∠ -120° Vbc =V∠ -90°
25 August 2005
Vline = V phase * 3∠30° I line = I phase Engineer M S Ayubi
10
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
11
Delta (∆ ) Connection A B
4 wire system
C ground
25 August 2005
Engineer M S Ayubi
12
∆ Phase-Line Relationships IC=I∠ -270° Ic=I∠ -240°
IB=I∠ -150°
Ia=I∠ 0°
IA=I∠ -30°
I line = I phase * 3∠ − 30° Vline = V phase
Ib=I∠ -120°
25 August 2005
Engineer M S Ayubi
13
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
14
Three Phase Power p (t ) = 3V prms I prms cosθ Z Constant in time! P = 3V prms I prms cosθ Z P = 3Vlrms I lrms cosθ Z
25 August 2005
Engineer M S Ayubi
15
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
16
Three Phase Load Conversion Zd=3Zw
25 August 2005
Engineer M S Ayubi
17
Example • A 220V three phase source drives a load with impedance of 2 + j5Ω per phase. For a wye and a delta connected load, find the magnitude of VP, IP, VL, IL and find the total power absorbed in the load.
25 August 2005
Engineer M S Ayubi
18
Example Delta
Wye
VP IP VL IL P 25 August 2005
Engineer M S Ayubi
19
Example VP
Delta 220
Wye 127
IP
40.9
23.6
VL
220
220
IL
70.8
23.6
P
10 KW
3.3 KW
25 August 2005
Engineer M S Ayubi
20
Engineer M S Ayubi
1
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
2
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
3
Three Phase Voltages •3 AC voltage of equal amplitude, out of phase by 120° Va = Vm sin ( ωt ) = Vm ∠0° = Vm (1 + j 0) 1 ( ) Vb = Vm sin ωt − 120° = Vm ∠ − 120° = Vm (− − j
3 ) 2 2 1 3 Vc = Vm sin ( ωt − 240°) = Vm ∠ − 240° = Vm (− + j ) 2 2 25 August 2005
Engineer M S Ayubi
4
Phase Sequence Vc=V∠ -240°
Vb=V∠ -240° Va=V∠ 0°
Vb=V∠ -120° Phase sequence: abc 25 August 2005
Va=V∠ 0°
Vc=V∠ -120° Reverse sequence: acb Engineer M S Ayubi
5
Three Phase Voltages • Sum of phasor voltages is zero.
25 August 2005
Engineer M S Ayubi
6
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
7
Wye (Y) Connection C
neutral
O B
4 wire system
5 wire system
A ground
25 August 2005
Engineer M S Ayubi
8
Y Phase-Line Relationships C
neutral
O B A
ground
Va − Vb − Vab = 0 Vab = Va − Vb 1 3 Vab = Vm (1 + j 0) − Vm (− − j ) 2 2 3 3 = Vm ( + j ) 2 2 Vab = 3Vm ∠30° Vbc = 3Vm ∠90° Vca = 3Vm ∠150°
25 August 2005
Engineer M S Ayubi
9
Y Phase-Line Relationships Vc=V∠ -240°
Vab =V∠ 30°
Va=V∠ 0°
Vca =V∠ 150°
Vb=V∠ -120° Vbc =V∠ -90°
25 August 2005
Vline = V phase * 3∠30° I line = I phase Engineer M S Ayubi
10
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
11
Delta (∆ ) Connection A B
4 wire system
C ground
25 August 2005
Engineer M S Ayubi
12
∆ Phase-Line Relationships IC=I∠ -270° Ic=I∠ -240°
IB=I∠ -150°
Ia=I∠ 0°
IA=I∠ -30°
I line = I phase * 3∠ − 30° Vline = V phase
Ib=I∠ -120°
25 August 2005
Engineer M S Ayubi
13
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
14
Three Phase Power p (t ) = 3V prms I prms cosθ Z Constant in time! P = 3V prms I prms cosθ Z P = 3Vlrms I lrms cosθ Z
25 August 2005
Engineer M S Ayubi
15
Outline • Three Phase Voltages • Three Phase Connections – Wye (Y) – Delta (∆ )
• Three Phase Power • Three Phase Load Conversion
25 August 2005
Engineer M S Ayubi
16
Three Phase Load Conversion Zd=3Zw
25 August 2005
Engineer M S Ayubi
17
Example • A 220V three phase source drives a load with impedance of 2 + j5Ω per phase. For a wye and a delta connected load, find the magnitude of VP, IP, VL, IL and find the total power absorbed in the load.
25 August 2005
Engineer M S Ayubi
18
Example Delta
Wye
VP IP VL IL P 25 August 2005
Engineer M S Ayubi
19
Example VP
Delta 220
Wye 127
IP
40.9
23.6
VL
220
220
IL
70.8
23.6
P
10 KW
3.3 KW
25 August 2005
Engineer M S Ayubi
20
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