chapter 12 polyphase circuits - kocwelearning.kocw.net/kocw/document/2016/pusan/yimoonsuk/1.pdf ·...
TRANSCRIPT
회로이론-ІI 2016년1학기 이문석 1
Chapter 12Polyphase Circuits
12.1 Polyphase Systems
12.2 Single-Phase Three-Wire Systems
12.3 Three-Phase Y-Y Connection
12.4 The Delta(∆) Connection
12.5 Power Measurement in Three-Phase Systems
회로이론-ІI 2016년1학기 이문석 2
12.1 Polyphase Systems
• A three-phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator).
A three-phase generator The generated voltages
• Three separate windings or coils with terminals a-a’, b-b’, c-c’ are physically placed 120° apart around the stator.
회로이론-ІI 2016년1학기 이문석 3
12.1 Polyphase Systems
120° out of phase with each other.
Advantages of Polyphase systems (mostly three-phase system :
• Most of the electric power is generated and distributed in three-phase.• The instantaneous power in a three-phase system can be constant.• The amount of power, the three-phase system is more economical than the single-phase. • In fact, the amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.
sdfadsf
회로이론-ІI 2016년1학기 이문석 4
12.1 Polyphase Systems
• The notation Vab indicates the voltage between point a (+) and point b (−).
• Indicating the + and − terminals is redundant if double-subscript voltages are used.
• KVL equations can be written correctly without reference to a circuit diagram:
V V V , V V V V , V Vad ab bd ad ab bc cd ab ba= + = + + = −
Double-subscript Notation
The double subscript notation Iab means the current from a to b by the direct path
When the direct path is not obvious (e.g. between c and d) we need to use a different notation.
회로이론-ІI 2016년1학기 이문석 5
12.1 Polyphase Systems
This is an example three-phase source with a neutral.
V V VV V 100 0 100 120100 ( 50 86.6) 173.2 30
100 3 30
ab an nb
an bn
j
= += − = ∠ °− ∠− °= − − − = ∠ °
= ∠ °
One possible representation of a three-phase system of voltages
V 100 0V 100 120V 100 240
an
bn
cn
VVV
= ∠ °= ∠− °= ∠− °
회로이론-ІI 2016년1학기 이문석 6
12.2 Single-Phase Three-Wire Systems
Three output terminals a, n, and b, and Van=Vbn
1V V V V V , V 2V 2Van nb an bn ab an nb= = → = − = =
Single-phase three-wire system:
The typical North American household is provided single-phase 3-wire power, where V1≅110 V and Vab=2V1.
When the A and B loads are balanced, the neutral wire from n to N carries no current:
V VI , I I IZ Z
an nbaA Bb aA Bb
p p
= = → =
0nN Bb Aa Bb aAΙ = Ι + Ι = Ι − Ι =
회로이론-ІI 2016년1학기 이문석 7
12.2 Single-Phase Three-Wire Systems
Example 12.1 Determine the power delivered to each of the three loads as well as the power lost in the neutral wire and each of the two lines.
1 1 2 1 3
2 2 3 2 1
3 1 3 2 3
1
2
3
115 0 50( ) 3( ) 0(20 10) 100( ) 50( ) 0
115 0 3( ) 100( ) 011.24 19.839.398 24.4710.37 21.80
j− ∠ °+ Ι + Ι − Ι + Ι − Ι =
+ Ι + Ι − Ι + Ι − Ι =− ∠ °+ Ι − Ι + Ι − Ι + Ι =
Ι = ∠− ° Α→ Ι = ∠− ° ΑΙ = ∠− ° Α
1
3
3 1
11.24 19.8310.37 158.20
0.9459 177.70
aA
bB
nN
aA bB nN
Ι = Ι = ∠− ° Α→ Ι = −Ι = ∠ ° ΑΙ = Ι − Ι = ∠− ° Α
→ Ι + Ι + Ι =
250 1 2
2100 3 2
220 10 2
(50) 206
(100) 117
(20) 1763j
P W
P W
P W+
= Ι − Ι = = Ι − Ι =
= Ι =
21
23
2
(1) 126
(1) 108
(3) 3
aA
bB
nN nN
P W
P W
P W
= Ι = = Ι =
= Ι =
Load power Line loss
115 11.24 cos(19.83 ) 1216115 10.37 cos(21.80 ) 1107
an
bn
P WP W
= × × ° = = × × ° =
Delivered PowerTransmission efficiency
2086 89.8%2086 237
= =+
회로이론-ІI 2016년1학기 이문석 8
12.3 Three-Phase Y-Y Connection
Balanced three-phase sources
V V Van bn cn= = V V V 0an bn cn+ + =
V V 0 , V 120 , V 240
V 0 , V 120 , V 240
an p bn p cn p
an p bn p cn p
V Vor
V V V
= ∠ ° = ∠− ° = ∠− °
= ∠ ° = ∠ ° = ∠ °
The abc (positive) and cba (negative) sequences:
Positive sequences:Van leads Vbn by 120°Vbn leads Vcn by 120°
Negative sequences:Van leads Vcn by 120°Vcn leads Vbn by 120°
회로이론-ІI 2016년1학기 이문석 9
12.3 Three-Phase Y-Y Connection
Line-to-Line Voltages
For a balanced source and balanced loads, the neutral wire current INn=0 and so the neutral wire could have any impedance, including ∞.
V V V 3 30
V V V 3 90
V V V 3 210
ab an bn p
bc bn cn p
ca cn an p
V
V
V
= − = ∠ °
= − = ∠− °
= − = ∠− °
V 3 , V V V 0L p ab bc caV and= + + =
( )VI , I I IZ
V V 120I I 120Z Z
I I 240I I I I 0
anaA aA bB cC L p
p
bn anbB aA
p p
cC aA
Nn aA bB cC
I I= = = = =
∠− °= = = ∠− °
= ∠− °→ = + + =
회로이론-ІI 2016년1학기 이문석 10
12.3 Three-Phase Y-Y Connection
Example 12.2 Find the total power delivered to the loads.
V 200 0 , V 200 120 , V 200 240an bn cn= ∠ ° = ∠− ° = ∠− °
V 200 0 2 60100 60
V 200 120 2 180100 60
V 200 240 2 300100 60
anaA
p
bnbB
p
cncC
p
Z
Z
Z
∠ °Ι = = = ∠− °
∠ °
∠− °Ι = = = ∠− °
∠ °
∠− °Ι = = = ∠− °
∠ °
* of phase A Re V
200 2 cos(0 60 ) 200
Total average power is 600
av an aAP
W
W
= Ι
= × × °+ ° =
회로이론-ІI 2016년1학기 이문석 11
12.3 Three-Phase Y-Y Connection
Example 12.3 A balanced three-phase system with a line voltage of 300V is supplying a balanced Y-connected load with 1200W at a leading PF of 0.8.Find the line current and the per-phase load impedance.
300Single phase(per-phase) power is 400W, and the phase voltage is 3
300400 0.83
2.89 ( )
p p p
p L
V I PF I
I I A rms
= × × = × ×
→ = =
1cos (0.8)=36.9 , leading PF : current leads voltage
V 300 3Z 60 36.9I 2.89 36.9
anp
aA
− °
= = = ∠− ° Ω∠ °
회로이론-ІI 2016년1학기 이문석 12
12.3 Three-Phase Y-Y Connection
Example 12.4 A balanced 600W lighting load is added (in parallel) to the system of Example 12.3. Determine the new line current.
1 1
2 2
300200 I cos(0 ) I 1.1553
300400 I 0.8 I 2.89 ( )3
A
A rms
= × × ° → =
= × × → =
1 2
1 2
If assume the angle of phase voltage is 0
I 1.155 0 , I 2.89 36.9I I I 3.87 26.6
300 3.87 cos(26.6 ) 6003
L
pP W
°
= ∠ ° = ∠ °→ = + = ∠ °
= × × ° =
회로이론-ІI 2016년1학기 이문석 13
12.4 The Delta (∆) Connection
A delta-connected load is also commonly used (note the absence of the neutral wire).
V V V
3 3 V 3 V 3 VL ab bc ca
p an bn cn
V
V
= = =
= = = =
V V VI , I , IZ Z Z
ab bc caAB BC CA
p p p
= = =
Line currentI I I , I I I , I I I
I I I 3aA AB CA bB BC AB cC CA BC
L aA bB cC L pI I I
= − = − = −
= = = → =
Phase currentI I Ip AB BC CAI = = =
회로이론-ІI 2016년1학기 이문석 14
12.4 The Delta (∆) Connection
회로이론-ІI 2016년1학기 이문석 15
12.4 The Delta (∆) Connection
Example 12.5 A balanced three-phase system with a line voltage of 300 V is supplying a balanced Y-connected load with 1200 W at a lagging PF of 0.8. Find the line current and the per-phase load impedance
Single phase power is 400W400 300 0.8
1.667 3 1.667 2.89p p p
p L
V I PF I
I A I A
= × × = × ×
→ = → = × =
1cos (0.8)=36.9V V 300Z 180 36.9I I 1.667 36.9
ab ABp
AB AB
− °
= = = = ∠ ° Ω∠− °
Average power :1 cos( ) cos( ) cos( )2 m m eff eff rms rmsP V I V I V Iθ φ θ φ θ φ= − = − = −
회로이론-ІI 2016년1학기 이문석 16
12.5 Power Measurement in Three-phase systems
Wattmeter in a single-phase system
The wattmeter is a four-terminal device that measures power delivered to the network if connected as shown: the torque applied to the moving system is proportional to the instantaneous product of the currents flowing in the two coils.
• Current coil : very low resistance, Icurrent-coil is flowing into network
• Potential coil : relatively high resistance (voltage coil), Vpotential-coil is Vnetwork
• For an upscale reading, a positive current is flowing into the (+) end of the
current coil while (+) terminal of the potential coil is positive with respect to
the unmarked end.
회로이론-ІI 2016년1학기 이문석 17
12.5 Power Measurement in Three-phase systems
Practice 12.9 Determine the wattmeter readings when (+) terminal of the wattmeter is connected to x, y, and z points.
100 2I (150 130) 4I (6 12)II=15.81 108.4
j j j= + + + + −→ ∠− °
PI
(a) Probe at point , we measure the potential,(6 12)I=(5.367 26.57 )(15.81 108.4 )
84.85 135 (84.85)(15.81)cos( 135 108.4 ) 1199.5This power is absorbed by the 6 resistor.
xj
P W− ∠− ° ∠− °
= ∠− ° → = − °+ ° =Ω
P
(b) Probe at point , we measure the potential,84.85 135 4I=84.85 135 63.24 108.4 144.2 123.7
(144.2)(15.81)cos( 123.7 108.4 ) 2199 W: absorbed by the 4 and 6 resistors.
y
P∠− °+ ∠− °+ ∠− ° = ∠− °
→ = − °+ ° = Ω Ω
(c) Probe at point , we measure the potential,100 2I=70.70 8.115
(70.70)(15.81)cos(8.115 108.4 ) 499 W: absorbed by the 100V source
zj
P− ∠ °
→ = °+ ° = −
회로이론-ІI 2016년1학기 이문석 18
12.5 Power Measurement in Three-phase systems
Wattmeter in a three-phase system
• Theoretically correct configuration, but useless in practice because the neutral of the Y is not always accessible and the phases of the ∆ are not available.
The sum of the powers measured by the wattmeters is the total power delivered.
The common node x is arbitrary.
회로이론-ІI 2016년1학기 이문석 19
12.5 Power Measurement in Three-phase systems
One wattmeter can be eliminated if the point x is moved to a line (as shown, B).
회로이론-ІI 2016년1학기 이문석 20
12.5 Power Measurement in Three-phase systems
Example 12.7 The balanced load is fed by a balanced three-phase system having Vab=230∠0° V rms and positive phase sequence. Find the reading of each wattmeter and the total power drawn by the load.
V 230 0 , V 230 120 ,V 230 120 , V 230 60
ab bc
ca ac
= ∠ ° = ∠− °= ∠ ° = ∠− °
1
2
1 2
V 230 3 304 15 4 15
8.554 105.1
V cos( V )(230)(8.554)cos( 60 105.1 )1389
V cos( V )(230)(8.554)cos( 120 134.9 )
512.5
876.5
anaA
ac aA ac aA
bc bB bc bB
j j
P ang ang
WP ang ang
W
P P P W
∠− °Ι = =
+ += ∠− °
= Ι − Ι
= − °+ °=
= Ι − Ι
= − °− °= −
= + =
Check ! Total power is 3 V cos( V )
2303 8.554 cos( 30 105.1 )3
876.3
an aA an aAang ang
W
Ι − Ι
= × × × − °+ °
=