Three-phase Systems Three-phase Voltages Phasor Diagram
EE230-Su05-Lecture 3
24 May 2005
1
Three-phase Systems Typical uses Commercial and industrial buildings More efficient Relative to singlesingle-phase systems Threephase motors provide constant torque Three SingleSingle-phase motors pulsate
2
Three-phase Voltages Three-Phase Voltages (V)
200 150 100 50 0 -50 0
0.005
0.01
0.015
0.02
0.025
Va Vb Vc
-100 -150 -200 time (s)
3
Phasor Diagram Vc 120° 120°
Va 120°
Vb
© 2003-2005 by Henry L. Welch
1
EE230-Su05-Lecture 3
24 May 2005
4
Single-phase Power Voltage (V), Current (A), and Power (W)
24 20 16 12
I V P
8 4 0 -4 0
0.005
0.01
0.015
0.02
0.025
-8 time (s)
5
Three Single-phase Powers 250
Power (W)
200 150
Pa Pb Pc
100 50 0 -50
0
0.005
0.01
0.015
0.02
0.025
time (s)
6
Total Power 350 300 Power (W)
250 Pa Pb Pc P total
200 150 100 50 0 -50 0
0.01
0.02
time (s)
© 2003-2005 by Henry L. Welch
2
EE230-Su05-Lecture 3
24 May 2005
7
Two Conventions
Positive phase convention Va = Vm/0o Vb = Vm/-120o Vc = Vm/120o Negative phase convention Va = Vm/0o Vb = Vm/120o Vc = Vm/-120o
8
Balanced Three-Phase System Ia
In
Ic
Ib
9
“Wye” Configuration The sources and loads look like a “Y” A, B, and C are called the “phases” “Neutral” is the common point on the “wye” connection VA-N is called the “Phase Voltage” VA-B is called the “Line Voltage”
© 2003-2005 by Henry L. Welch
3
EE230-Su05-Lecture 3
24 May 2005
10
Sizing the Neutral Wire KCL at Node N VN Van -VN Vbn -VN Vcn -VN = + + Zn Zs +Zl +ZL Zs +Zl +ZL Zs +Zl +ZL VN -3VN V +V + Vcn = + an bn Zn Zs +Zl +ZL Zs +Zl +ZL
⎛ 1 ⎞ Van +Vbn + Vcn 3 VN ⎜ + ⎟= Zs +Zl +ZL ⎝ Zn Zs +Zl +ZL ⎠
0
VN = 0! In = 0!
11
Single-Phase Equivalent Circuit
Subject to phase conventions for the other two legs
12
Line vs. Phase Voltages (1) Phase Voltages
Vc 120° 120°
Vca
Vbc Va 120°
Vb
© 2003-2005 by Henry L. Welch
Vab Line Voltages
4
EE230-Su05-Lecture 3
24 May 2005
13
Line vs. Phase Voltages (2)
Vab = Van – Vbn Similar for bc and ca
Vphase =
Vline 3
or Vline = 3 Vphase Vab = 3Vphase ∠30o
Including phase information
Vbc = 3Vphase ∠ − 90o Vca = 3Vphase ∠150o
14
Work Problem 1 Wye connected source Va = 240/0 240/0o V Wye connected load ZY = 80 + j60 Ω Compute all line/phase currents
15
Delta Load Configuration
© 2003-2005 by Henry L. Welch
5
EE230-Su05-Lecture 3
24 May 2005
16
Dealing with the Delta Load Use the line voltages IAB = VAB/Z∆ Equivalently Convert delta to wye (Sec. 2.7) Z∆ = 3ZY
17
Work Problem 2
Wye connected source Va = 120/0 120/0o V Source impedance Zs = 1 + j2 Ω Transmission line impedance Zl = 3 + j5 Ω Delta connected load Z ∆ = 80 + j60 Ω Compute all line/phase currents
18
Delta-* Systems DeltaDelta-Delta StraightStraight-forward Use line voltages and currents DeltaDelta-Wye Convert one end or the other Be very careful with signs and phases!
© 2003-2005 by Henry L. Welch
6
EE230-Su05-Lecture 3
24 May 2005
19
Wye and Delta Voltages
Phase
Van Line
Vab
Y- Y
Y- ∆
Vp ∠0°
Vp ∠0°
3 Vp∠30°
3 Vp ∠30°
∆-∆
Vp ∠0° Same as phase
20
Wye and Delta Currents
Phase
Ia , IAB Line
Ia
Y- Y
Y- ∆
∆-∆
Same as line
VAB Z∆
Vab Z∆
Van ZY
3 IAB ∠ − 30°
3 IAB ∠ − 30°
21
Per-Phase Analysis If necessary, convert all the loads and sources into wye configurations All delta configurations is another option Calculate the currents in one phase Use KCL to total the currents Use vectors, not just the magnitudes
© 2003-2005 by Henry L. Welch
7
EE230-Su05-Lecture 3
24 May 2005
22
Combining Loads
All wyewye-connected Add up the impedances in parallel All deltadelta-connected Add up the impedances in parallel Not all of one type? Convert them! Also, you can add the currents
23
Work Problem 3
208208-V/120V/120-V source Load 1 Wye connected: 2 + j4 Ω Load 2 Wye connected: 3 - j5 Ω Find the total impedance per phase Find the total current from the source to the loads
24
Work Problem 4
208208-V/120V/120-V source Load 1 Wye connected: 2 + j4 Ω Load 2 Delta connected: 3 - j6 Ω Find the total impedance per phase Find the total current from the source to the loads
© 2003-2005 by Henry L. Welch
8
24 May 2005
1
Three-phase Systems Typical uses Commercial and industrial buildings More efficient Relative to singlesingle-phase systems Threephase motors provide constant torque Three SingleSingle-phase motors pulsate
2
Three-phase Voltages Three-Phase Voltages (V)
200 150 100 50 0 -50 0
0.005
0.01
0.015
0.02
0.025
Va Vb Vc
-100 -150 -200 time (s)
3
Phasor Diagram Vc 120° 120°
Va 120°
Vb
© 2003-2005 by Henry L. Welch
1
EE230-Su05-Lecture 3
24 May 2005
4
Single-phase Power Voltage (V), Current (A), and Power (W)
24 20 16 12
I V P
8 4 0 -4 0
0.005
0.01
0.015
0.02
0.025
-8 time (s)
5
Three Single-phase Powers 250
Power (W)
200 150
Pa Pb Pc
100 50 0 -50
0
0.005
0.01
0.015
0.02
0.025
time (s)
6
Total Power 350 300 Power (W)
250 Pa Pb Pc P total
200 150 100 50 0 -50 0
0.01
0.02
time (s)
© 2003-2005 by Henry L. Welch
2
EE230-Su05-Lecture 3
24 May 2005
7
Two Conventions
Positive phase convention Va = Vm/0o Vb = Vm/-120o Vc = Vm/120o Negative phase convention Va = Vm/0o Vb = Vm/120o Vc = Vm/-120o
8
Balanced Three-Phase System Ia
In
Ic
Ib
9
“Wye” Configuration The sources and loads look like a “Y” A, B, and C are called the “phases” “Neutral” is the common point on the “wye” connection VA-N is called the “Phase Voltage” VA-B is called the “Line Voltage”
© 2003-2005 by Henry L. Welch
3
EE230-Su05-Lecture 3
24 May 2005
10
Sizing the Neutral Wire KCL at Node N VN Van -VN Vbn -VN Vcn -VN = + + Zn Zs +Zl +ZL Zs +Zl +ZL Zs +Zl +ZL VN -3VN V +V + Vcn = + an bn Zn Zs +Zl +ZL Zs +Zl +ZL
⎛ 1 ⎞ Van +Vbn + Vcn 3 VN ⎜ + ⎟= Zs +Zl +ZL ⎝ Zn Zs +Zl +ZL ⎠
0
VN = 0! In = 0!
11
Single-Phase Equivalent Circuit
Subject to phase conventions for the other two legs
12
Line vs. Phase Voltages (1) Phase Voltages
Vc 120° 120°
Vca
Vbc Va 120°
Vb
© 2003-2005 by Henry L. Welch
Vab Line Voltages
4
EE230-Su05-Lecture 3
24 May 2005
13
Line vs. Phase Voltages (2)
Vab = Van – Vbn Similar for bc and ca
Vphase =
Vline 3
or Vline = 3 Vphase Vab = 3Vphase ∠30o
Including phase information
Vbc = 3Vphase ∠ − 90o Vca = 3Vphase ∠150o
14
Work Problem 1 Wye connected source Va = 240/0 240/0o V Wye connected load ZY = 80 + j60 Ω Compute all line/phase currents
15
Delta Load Configuration
© 2003-2005 by Henry L. Welch
5
EE230-Su05-Lecture 3
24 May 2005
16
Dealing with the Delta Load Use the line voltages IAB = VAB/Z∆ Equivalently Convert delta to wye (Sec. 2.7) Z∆ = 3ZY
17
Work Problem 2
Wye connected source Va = 120/0 120/0o V Source impedance Zs = 1 + j2 Ω Transmission line impedance Zl = 3 + j5 Ω Delta connected load Z ∆ = 80 + j60 Ω Compute all line/phase currents
18
Delta-* Systems DeltaDelta-Delta StraightStraight-forward Use line voltages and currents DeltaDelta-Wye Convert one end or the other Be very careful with signs and phases!
© 2003-2005 by Henry L. Welch
6
EE230-Su05-Lecture 3
24 May 2005
19
Wye and Delta Voltages
Phase
Van Line
Vab
Y- Y
Y- ∆
Vp ∠0°
Vp ∠0°
3 Vp∠30°
3 Vp ∠30°
∆-∆
Vp ∠0° Same as phase
20
Wye and Delta Currents
Phase
Ia , IAB Line
Ia
Y- Y
Y- ∆
∆-∆
Same as line
VAB Z∆
Vab Z∆
Van ZY
3 IAB ∠ − 30°
3 IAB ∠ − 30°
21
Per-Phase Analysis If necessary, convert all the loads and sources into wye configurations All delta configurations is another option Calculate the currents in one phase Use KCL to total the currents Use vectors, not just the magnitudes
© 2003-2005 by Henry L. Welch
7
EE230-Su05-Lecture 3
24 May 2005
22
Combining Loads
All wyewye-connected Add up the impedances in parallel All deltadelta-connected Add up the impedances in parallel Not all of one type? Convert them! Also, you can add the currents
23
Work Problem 3
208208-V/120V/120-V source Load 1 Wye connected: 2 + j4 Ω Load 2 Wye connected: 3 - j5 Ω Find the total impedance per phase Find the total current from the source to the loads
24
Work Problem 4
208208-V/120V/120-V source Load 1 Wye connected: 2 + j4 Ω Load 2 Delta connected: 3 - j6 Ω Find the total impedance per phase Find the total current from the source to the loads
© 2003-2005 by Henry L. Welch
8
