Chapter 2: Three-Phase Circuits
2-1.
Three impedances of 4 + j3 Ω are Δ-connected and tied to a three-phase 208-V power line. Find I φ , I L ,
P, Q, S, and the power factor of this load.
SOLUTION
IL
+
240 V
Iφ
Zφ
-
Zφ = 3 + j4 Ω
Zφ
Zφ
Here, VL = Vφ = 208 V , and Z φ = 4 + j 3 Ω = 5∠36.87° Ω , so
Iφ =
Vφ
Zφ
=
208 V
= 41.6 A
5Ω
I L = 3I φ = 3 ( 41.6 A ) = 72.05 A
P=3
Vφ
2
2
208 V )
(
cos θ = 3
cos 36.87° = 20.77 kW
5Ω
2
208 V )
(
Q=3
sin θ = 3
sin 36.87° = 15.58 kvar
Z
5Ω
Z
2
Vφ
S = P 2 + Q 2 = 25.96 kVA
PF = cos θ = 0.8 lagging
2-2.
Figure P2-1 shows a three-phase power system with two loads. The Δ-connected generator is producing a
line voltage of 480 V, and the line impedance is 0.09 + j0.16 Ω. Load 1 is Y-connected, with a phase
impedance of 2.5∠36.87° Ω and load 2 is Δ-connected, with a phase impedance of 5∠-20° Ω.
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(a) What is the line voltage of the two loads?
(b) What is the voltage drop on the transmission lines?
(c) Find the real and reactive powers supplied to each load.
(d) Find the real and reactive power losses in the transmission line.
(e) Find the real power, reactive power, and power factor supplied by the generator.
SOLUTION To solve this problem, first convert the two deltas to equivalent wyes, and get the per-phase
equivalent circuit.
0.090 Ω
j0.16 Ω
+
Line
277∠0° V
+
-
Zφ 2
Zφ 1
Vφ ,load
Z φ1 = 2.5∠36.87° Ω
Zφ 2 = 1.67∠ − 20° Ω
(a)
The phase voltage of the equivalent Y-loads can be found by nodal analysis.
Vφ ,load − 277∠0° V
+
Vφ ,load
+
Vφ ,load
=0
0.09 + j 0.16 Ω
2.5∠36.87° Ω 1.67∠ − 20° Ω
(5.443∠ − 60.6°) (Vφ ,load − 277∠0° V ) + (0.4∠ − 36.87°)Vφ ,load + (0.6∠20°)Vφ ,load = 0
(5.955∠ − 53.34°) Vφ ,load = 1508∠ − 60.6°
Vφ ,load = 253.2∠ − 7.3° V
20
Therefore, the line voltage at the loads is VL 3 Vφ = 439 V .
(b)
The voltage drop in the transmission lines is
ΔVline = Vφ ,gen − Vφ ,load = 277∠0° V - 253.2∠ - 7.3° = 41.3∠52° V
(c)
The real and reactive power of each load is
P1 = 3
Vφ
2
cos θ = 3
(253.2 V )2 cos 36.87° = 61.6 kW
2.5 Ω
( 253.2 V ) 2 sin 36.87° = 46.2 kvar
Q1 = 3
sin θ = 3
Z
2.5 Ω
2
2
Vφ
(
253.2 V )
cos (- 20°) = 108.4 kW
P2 = 3
cos θ = 3
1.67 Ω
Z
2
2
Vφ
253.2 V )
(
Q2 = 3
sin θ = 3
sin ( -20°) = −39.5 kvar
Z
1.67 Ω
(d)
Z
2
Vφ
The line current is
I line =
ΔVline
41.3∠52° V
= 225∠ − 8.6°A
=
0.09 + j 0.16 Ω
Z line
Therefore, the loses in the transmission line are
Pline = 3I line Rline = 3 (225 A ) (0.09 Ω ) = 13.7 kW
2
2
Qline = 3I line X line = 3 ( 225 A ) ( 0.16 Ω ) = 24.3 kvar
2
(e)
2
The real and reactive power supplied by the generator is
Pgen = Pline + P1 + P2 = 13.7 kW + 61.6 kW + 108.4 kW = 183.7 kW
Qgen = Qline + Q1 + Q2 = 24.3 kvar + 46.2 kvar − 39.5 kvar = 31 kvar
The power factor of the generator is
ª
Q º
31 kvar º
ª
PF = cos « tan -1 gen » = cos « tan −1
= 0.986 lagging
Pgen ¼»
183.7 kW »¼
¬
¬«
2-3.
The figure shown below shows a one-line diagram of a simple power system containing a single 480 V
generator and three loads. Assume that the transmission lines in this power system are lossless, and
answer the following questions.
(a) Assume that Load 1 is Y-connected. What are the phase voltage and currents in that load?
(b) Assume that Load 2 is Δ-connected. What are the phase voltage and currents in that load?
(c) What real, reactive, and apparent power does the generator supply when the switch is open?
(d) What is the total line current I L when the switch is open?
(e) What real, reactive, and apparent power does the generator supply when the switch is closed?
(f) What is the total line current I L when the switch is closed?
(g) How does the total line current I L compare to the sum of the three individual currents I1 + I 2 + I 3 ?
If they are not equal, why not?
21
SOLUTION Since the transmission lines are lossless in this power system, the full voltage generated by G1
will be present at each of the loads.
(a)
Since this load is Y-connected, the phase voltage is
Vφ1 =
480 V
= 277 V
3
The phase current can be derived from the equation P = 3Vφ I φ cos θ as follows:
I φ1 =
(b)
P
100 kW
=
= 133.7 A
3Vφ cos θ 3 ( 277 V )( 0.9 )
Since this load is Δ-connected, the phase voltage is
Vφ 2 = 480 V
The phase current can be derived from the equation S = 3Vφ I φ as follows:
Iφ 2 =
S
80 kVA
=
= 55.56 A
3Vφ 3 ( 480 V )
(c) The real and reactive power supplied by the generator when the switch is open is just the sum of the
real and reactive powers of Loads 1 and 2.
P1 = 100 kW
(
)
Q1 = P tan θ = P tan cos −1 PF = (100 kW )( tan 25.84°) = 48.4 kvar
P2 = S cos θ = ( 80 kVA )( 0.8) = 64 kW
Q2 = S sin θ = ( 80 kVA )( 0.6) = 48 kvar
PG = P1 + P2 = 100 kW + 64 kW = 164 kW
QG = Q1 + Q2 = 48.4 kvar + 48 kvar = 96.4 kvar
(d)
The line current when the switch is open is given by I L =
QG
96.4 kvar
= tan −1
= 30.45°
PG
164 kW
P
164 kW
=
= 228.8 A
IL =
3 VL cosθ
3 ( 480 V ) cos ( 30.45°)
θ = tan −1
22
P
, where θ = tan −1 QG .
PG
3 VL cos θ
(e) The real and reactive power supplied by the generator when the switch is closed is just the sum of
the real and reactive powers of Loads 1, 2, and 3. The powers of Loads 1 and 2 have already been
calculated. The real and reactive power of Load 3 are:
P3 = 80 kW
(
)
Q3 = P tan θ = P tan cos−1 PF = ( 80 kW ) ª¬ tan ( −31.79°) ¼º = −49.6 kvar
PG = P1 + P2 + P3 = 100 kW + 64 kW + 80 kW = 244 kW
QG = Q1 + Q2 + Q3 = 48.4 kvar + 48 kvar − 49.6 kvar = 46.8 kvar
(f)
The line current when the switch is closed is given by I L =
P
, where θ = tan −1 QG .
PG
3 VL cos θ
QG
46.8 kvar
= tan −1
= 10.86°
PG
244 kW
P
244 kW
IL =
=
= 298.8 A
3 VL cosθ
3 ( 480 V ) cos (10.86°)
θ = tan −1
(g)
The total line current from the generator is 298.8 A. The line currents to each individual load are:
P1
100 kW
=
= 133.6 A
I L1 =
3 VL cosθ1
3 ( 480 V )( 0.9 )
S2
80 kVA
I L2 =
=
= 96.2 A
3 VL
3 ( 480 V )
P3
80 kW
I L3 =
=
= 113.2 A
3 VL cosθ 3
3 ( 480 V )( 0.85)
The sum of the three individual line currents is 343 A, while the current supplied by the generator is 298.8
A. These values are not the same, because the three loads have different impedance angles. Essentially,
Load 3 is supplying some of the reactive power being consumed by Loads 1 and 2, so that it does not
have to come from the generator.
2-4.
Prove that the line voltage of a Y-connected generator with an acb phase sequence lags the corresponding
phase voltage by 30°. Draw a phasor diagram showing the phase and line voltages for this generator.
SOLUTION If the generator has an acb phase sequence, then the three phase voltages will be
Van = Vφ ∠0°
Vbn = Vφ ∠ − 240°
Vcn = Vφ ∠ − 120°
The relationship between line voltage and phase voltage is derived below. By Kirchhoff’s voltage law,
the line-to-line voltage Vab is given by
Vab = Va − Vb
Vab = Vφ ∠0° − Vφ ∠ − 240°
§ 1
3 · 3
3
Vab = Vφ − ¨ − Vφ + j
Vφ ¸ = Vφ − j
Vφ
2 ¹ 2
2
© 2
§ 3
1·
−j ¸
Vab = 3Vφ ¨
2¹
© 2
Vab = 3Vφ ∠ − 30°
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Thus the line voltage lags the corresponding phase voltage by 30°.
connection is shown below.
The phasor diagram for this
Vbc
Vbn
Van
Vab
Vcn
2-5.
Find the magnitudes and angles of each line and phase voltage and current on the load shown in Figure
P2-3.
SOLUTION Note that because this load is Δ-connected, the line and phase voltages are identical.
Vab = Van − Vbn = 120∠0° V - 120∠ - 120° V = 208∠30° V
Vbc = Vbn − Vcn = 120∠ − 120° V - 120∠ - 240° V = 208∠ - 90° V
Vca = Vcn − Van = 120∠ − 240° V - 120∠0° V = 208∠150° V
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I ab =
Vab 208∠30° V
=
= 20.8∠10° A
10∠20° Ω
Zφ
I bc =
Vbc 208∠ − 90° V
=
= 20.8∠ − 110° A
10∠20° Ω
Zφ
I ca =
Vca 208∠150° V
=
= 20.8∠130° A
10∠20° Ω
Zφ
I a = I ab − I ca = 20.8∠10° A - 20.8∠130° A = 36∠ - 20° A
Ib = Ibc − I ab = 20.8∠ − 110° A - 20.8∠10° A = 36∠ - 140° A
I c = I ca − Ibc = 20.8∠130° A - 20.8∠-110° A = 36∠100° A
2-6.
Figure P2-4 shows a small 480-V distribution system. Assume that the lines in the system have zero
impedance.
(a) If the switch shown is open, find the real, reactive, and apparent powers in the system. Find the total
current supplied to the distribution system by the utility.
(b) Repeat part (a) with the switch closed. What happened to the total current supplied? Why?
SOLUTION
(a)
With the switch open, the power supplied to each load is
2
Vφ
(480 V )2 cos 30° = 59.86 kW
cos θ = 3
P1 = 3
10 Ω
Z
2
Vφ
( 480 V ) 2 sin 30° = 34.56 kvar
sin θ = 3
Q1 = 3
10 Ω
Z
2
Vφ
(277 V )2 cos 36.87° = 46.04 kW
cos θ = 3
P2 = 3
Z
4Ω
2
2
Vφ
277 V )
(
sin θ = 3
sin 36.87° = 34.53 kvar
Q2 = 3
4Ω
Z
PTOT = P1 + P2 = 59.86 kW + 46.04 kW = 105.9 kW
QTOT = Q1 + Q2 = 34.56 kvar + 34.53 kvar = 69.09 kvar
The apparent power supplied by the utility is
STOT = PTOT + QTOT = 126.4 kVA
The power factor supplied by the utility is
2
2
25
ª
Q º
69.09 kvar º
ª
= 0.838 lagging
PF = cos « tan -1 TOT » = cos « tan −1
PTOT ¼
105.9 kW »¼
¬
¬
The current supplied by the utility is
IL =
(b)
PTOT
=
3 VT PF
105.9 kW
= 152 A
3 ( 480 V ) ( 0.838)
With the switch closed, P3 is added to the circuit. The real and reactive power of P3 is
P3 = 3
Vφ
2
cos θ = 3
(277 V )2 cos (-90°) = 0 kW
5Ω
( 277 V ) 2 sin -90° = −46.06 kvar
P3 = 3
sin θ = 3
( )
Z
5Ω
PTOT = P1 + P2 + P3 = 59.86 kW + 46.04 kW + 0 kW = 105.9 kW
QTOT = Q1 + Q2 + Q3 = 34.56 kvar + 34.53 kvar − 46.06 kvar = 23.03 kvar
The apparent power supplied by the utility is
Z
2
Vφ
STOT = PTOT + QTOT = 108.4 kVA
The power factor supplied by the utility is
2
2
ª
Q º
23.03 kVAR º
ª
= 0.977 lagging
PF = cos « tan -1 TOT » = cos « tan −1
P
105.9 kW »¼
¬
TOT ¼
¬
The current supplied by the utility is
IL =
PTOT
=
3 VT PF
105.9 kW
= 130.4 A
3 ( 480 V ) ( 0.977 )
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