Chapter 12: Three‐Phase Circuits 12.1 What Is a Three‐Phase Circuit?
12.2 Balance Three‐Phase Voltages
12.3 Balance Three‐Phase Y‐to‐Y Connection
12.4 Other Balance Three‐Phase Connections
12.5 Power in a Balanced System
Nikola Tesla
10/07/1856 – 7/01/1943 12.6 Unbalanced Three‐Phase Systems
was a Serbian American inventor, electrical 12.7 Design & Applications – Three‐Phase engineer, mechanical engineer, physicist, and Power Measurement & Residential futurist best known for his contributions to the Wiring
design of the modern ac electricity supply system.
12.8 Summary
A three‐phase system with rotating magnetic fields
1
12.1 What is a Three‐Phase Circuit? (1)
Single‐Phase Systems
Two‐wire type Three‐wire type Two‐Phase Three‐Wire System
Normal household system is a single‐
phase three‐wire system: • The same magnitude and the same phase of the terminal voltages
• The connection of both 120 V and 240 V appliances.
2
12.1 What is a Three‐Phase Circuit? (2)
• Three‐Phase System: produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°.
Three sources with 120° out of phase
Four wired system
Advantages:
1. Most of the electric power is generated and distributed in three‐phase.
2. The instantaneous power in a three‐phase system can be constant.
3. The amount of power, the three‐phase system is more economical that the single‐phase. 4. In fact, the amount of wire required for a three‐phase system is less 3
than that required for an equivalent single‐phase system.
12.2 Balance Three‐Phase Voltages (1)
• A three‐phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator).
Three‐Phase Generator
Generated Voltages
• Three‐phase voltage sources: (l) Y‐connected ; (r) Δ‐connected
4
12.2 Balance Three‐Phase Voltages (2)
• Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°.
• 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.
Balanced Y‐connected load, Z1 = Z2 = Z3 = ZY
Balanced Δ‐connected load, Za = Zb = Zc = ZΔ
→ ZΔ = 3ZY
5
12.3 Balance Three‐Phase Y‐to‐Y Connection (1)
• Four possible connections
1. Y‐Y connection (Y‐connected source with a Y‐connected load)
2. Y‐Δ connection (Y‐connected source with a Δ‐connected load)
3. Δ‐Δ connection
4. Δ‐Y connection
• A balanced Y‐Y system is a three‐phase system with a balanced y‐
connected source and a balanced y‐connected load.
‐ Phase (line‐to‐neutral) voltages:
‐ Line (line‐to‐line) voltages:
6
12.3 Balance Three‐Phase Y‐to‐Y Connection (2)
ZY  Z s + Z l + Z L
‐ Line currents: (Phase currents: same as)
Hint: The neutral line can be removed w/o affecting the system.
7
12.3 Balance Three‐Phase Y‐to‐Y Connection (3)
Example: Calculate the line currents in the three‐wire Y‐Y system. 8
12.4 Other Balance Three‐Phase Connections (1)
• A balanced Y‐Δ system is a (most practice) three‐phase system with a balanced y‐connected source and a balanced Δ‐connected load.
‐ Phase currents:
‐ Line currents:
lag by 30o
I L  3I p , where
I L  Ia  Ib  Ic
I p  I AB  I BC  ICA
9
12.4 Other Balance Three‐Phase Connections (2)
• A balanced Δ‐Δ system is a three‐phase system with a balanced Δ ‐connected source and a balanced Δ ‐connected load.
‐ Phase voltages: (Line voltages: same as)
‐ Phase currents:
‐ Line currents:
Each line current lags the corresponding phase current by 30o,
the magnitude of the line current is √3 times the magnitude Ip of
the phase current,
I L  3I p
10
12.4 Other Balance Three‐Phase Connections (3)
• A balanced Δ‐Y system is a three‐phase system with a balanced y‐connected source and a balanced y‐connected load.
‐ Phase voltages: (Line voltages: same as)
‐ Line currents: (Phase currents: same as)
11
12
12.5 Power in a Balanced System (1)
• For a Y‐connected load, the phase voltages and currents are
Note: ZY  Z 
• Total real power in the load:
The total instantaneous power in a balanced three‐
phase system is constant.
• Total complex power: 13
12.5 Power in a Balanced System (2)
Example (Line losses): Figure shows a balanced three‐wire Y‐to‐Y
circuit. Determine average power delivered by the three‐phase
source, delivered to the three‐phase load, and absorbed by the
three‐phase line. (a) A balanced three‐wire Y‐to‐Y circuit and (b)
the per‐phase equivalent circuit.
The line current is calculated:
I aA ( ) 
100
 1.894  18.7A
50  j (377)(0.045)
The phase voltage at the load is
VAN ( )  (40  j (377)(0.04))I aA ( )  812V
The power delivered by the source is calculated as
I aA ( )  1.894-18.7A and Van ( )  1000V
So
Pa 
(100)(1.894)
cos(18.7)  89.7W
2
The power delivered to the load is calculated as
I aA ( )  1.894  18.7A and
1.8942
RA  40, so PA 
40  71.7W
2
The power lost in the line is calculated as
1.8942
10  17.9W
I aA ( )  1.894  18.7A and RaA  10, so PaA 
2
The three‐phase load receives 3PA = 215.1 W, and 3PaA= 53.7 W is
lost in the line. A total of 80 percent of the power supplied by the
source is delivered to the load. The other 20 percent is lost in the
line. The three‐phase source delivers 3Pa = 269.1 W.
12.5 Power in a Balanced System (4)
Example (Reducing line losses): 80% of the power supplied by the
source is delivered to the load, and the other 20% is lost in the line
in the last example. The loss in the line can be reduced by reducing
the current in the line. Reducing the current in the load would
reduce the power delivered to the load. Transformers provide a
way of reducing the line current without reducing the load current.
In this example, two three‐phase transformers are added to the
three‐phase circuit. A transformer at the source steps up the
voltage and steps down the current. Conversely, a transformer at
the load steps down the voltage and steps up the current. Because
the turns ratios of these transformers are reciprocals of each other,
the voltage and current at the load are unchanged. The current in
the line will be reduced to reduce the power lost in line. The line
voltage will increase. The higher line voltage will require increased
insulation and increased attention to safety.
Figure shows the per‐phase equivalent circuit of the balanced three‐
wire Y‐to‐Y circuit that includes the two transformers. Determine
the average power delivered by the three‐phase source, delivered
to the three‐phase load, and absorbed by the three‐phase line. (a) A
per‐phase equivalent circuit for a balanced Y‐to‐Y circuit with step‐
up and step‐down transformers and (b) the corresponding
frequency‐domain circuit used to calculate the line current.
To analyze the per‐phase equivalent circuit in (a), notice that:
‐ The secondary voltage of the left‐hand transformer is 10 times the primary
voltage, that is, 1000 cos (377t).
‐ The impedance connected to the secondary of the right‐hand transformer can
be reflected to the primary of this transformer by multiplying by 100. The result
is a 4000‐Ω resistor in series with a 4‐H inductor.
These observations lead to the one‐mesh circuit shown in (b). The mesh current in
this circuit is the line current of the three‐phase circuit. This line current is
calculated as
1000
I aA ( ) 
4010  j (377)(4.005)
 0.2334  20.6A
The current into the dotted end of the secondary of the left‐hand transformer in
(a) is IaA, so the current into the dotted end of the primary of this transformer is
I a ( )  10(I aA ( ))  2.334  20.6A
The current into the dotted end of the primary of the right‐hand transformer is
IaA(ω), so the current into the dotted end of the secondary is
I A ( )  (10I aA ( ))  2.334  20.6A
The phase voltage at the load is
VAN ( )  (40  j (377)(0.04))I A ( )  99.70V
The power delivered by the source is calculated as
I a ( )  2.334  20.6A and
Van ( )  1000A so Pa 
(100)(2.334)
cos(20.6)  109.2W
2
The power delivered to the load is calculated as
I A ( )  2.334  20.6A and
2.3342
RA  40, so PA 
40  108.95W
2
The power lost in the line is calculated as
0.23342
I aA ( )  0.2334  20.6A and RaA  10, so PA 
10  0.27W
2
Now 98% of the power supplied by the source is delivered to the load. Only 2%
lost in the line.
12.5 Power in a Balanced System (8)
• Comparing the power loss:
Ploss
PL2
 2I R  2R 2
VL
2
L
PL2
PL2
  3( I L ) R  3R 2  R 2
Ploss
VL
3VL
2
Ploss 2 R 2 r2
material for single-phase 2( r 2l ) 2r 2 4


 2 

 2 
2

Ploss
R
r
material for three-phase 3( r l ) 3r
3
If same power loss is tolerated in both system, three‐phase system use only 75% of materials of a single‐phase system.
20
12.5 Power in a Balanced System (9)
Example:
21
12.6 Unbalanced Three‐Phase Systems (1)
• An unbalanced system is due to unbalanced voltage sources or an unbalanced load.
Ia 
VAN
V
V
, I b  BN , I c  CN ,
ZA
ZB
ZC
I n  (I a  I b  I c )
• To calculate power in an unbalanced three-phase system requires
that we find the power in each phase.
• The total power is not simply three times the power in one phase
but the sum of the powers in the three phases.
22
12.6 Unbalanced Three‐Phase Systems (2)
Example: For the unbalanced circuit, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.
23
12.7 Design & Applications (1)
Design (Power factor correction): Figure shows a three‐phase circuit.
The capacitors are added to improve the power factor of the load.
We need to determine the value of the capacitance, C, required to
obtain a power factor of 0.9 lagging.
 The circuit is excited by sinusoidal sources all having the same frequency, 60 Hz or 377 rad/s. The
circuit is at steady state. The circuit is a linear circuit. Phasors can be used to analyze this circuit.
 The circuit is a balanced three‐phase circuit. A per‐phase equivalent circuit can be used to analyze
this circuit.
 The load consists of two parts. The part comprising resistors and inductors is connected as a Y. The
part comprising capacitors is connected as a Δ. A Δ‐to‐Y transformation can be used to simplify the
load.
The per‐phase equivalent circuit is shown as State the Goal: Determine the value of C required to correct the power factor to
0.9 lagging.
Generate a Plan:
A formula was provided for calculating the reactance, X1, needed to correct the
power factor of a load
2
2
X1 
R X
R tan(cos 1 pfc)  X
where R and X are the real and imaginary parts of the load impedance before the
power factor is corrected and pfc is the corrected power factor. After this equation
is used to calculate X1, the capacitance, C, can be calculated from X1. Notice that
X1 will be the reactance of the equivalent Y‐connected capacitors. We will need to
calculate the Δ‐connected capacitor equivalent of the Y‐connected capacitor.
Act on the Plan:
We note that: Z = R + jX = 20 + j75.4 Ω. Therefore, the reactance, X1, needed to
correct the power factor is
202  75.42
X1 
20 tan(cos 1 0.9)  75.4
 92.6
The Y‐connected capacitor equivalent to the Δ‐connected capacitor can be
calculated from ZY = ZΔ/3. Therefore, the capacitance of the equivalent Y‐
connected capacitor is 3C. Finally, because X1 = 1 /(3Cω), we have
C
1
1

 9.548 F
377  3(92.6)
  3  X1
Verify the Proposed Solution:
When C = 9.548 μF, the impedance of one phase of the equivalent Y‐connected 1
load will be
(20  j 75.4)
ZY 
j 377  3  C
 246.45  j119.4
1
 (20  j 75.4)
j 377  3  C
The value of the power factor is

 119.4  
pf  cos  tan 1 
   0.90
 246.45  

so the specifications have been satisfied.
12.7 Design & Applications (4)
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12.7 Design & Applications (5)
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12.8 Summary (1)
•
The generation & transmission of electrical power are more
efficient in three‐phase systems employing 3 voltages of the same
magnitude & frequency and differing in phase by 120° from each
other.
•
The three‐phase source consists of either three Y‐connected
sinusoidal voltage sources or three Δ‐connected sinusoidal voltage
sources. Similarly, the circuit elements that comprise the load are
connected to form either a Y or a Δ. The transmission line connects
the source to the load and consists of either three or four wires.
•
Analysis of three‐phase circuits using phasors and impedances will
determine the steady‐state response of the three‐phase circuit.
We are particularly interested in the power the three‐phase source
delivers to the three‐phase load.
12.8 Summary (2)
•
The current in the neutral wire of a balanced Y‐to‐Y connection is
zero; thus, the wire may be removed if desired. The key to the
analysis of the Y‐to‐Y circuit is the calculation of the line currents.
When the circuit is not balanced, the first step in the analysis of
this circuit is to calculate VnN, the voltage at the neutral node of
the three‐phase load with respect to the voltage at the neutral
node of the three‐phase source.
•
For a Δ load, we converted the Δ load to a Y‐connected load by
using the relation Δ‐to‐Y transformation. Then we proceeded with
the Y‐to‐Y analysis.
•
The line current for a balanced Δ load is √3 times the phase
current and is displaced ‐30° in phase. The line‐to‐line voltage of a
Δ load is equal to the phase voltage.