Transcript 幻灯片 1 - Sun Yat-sen University

Engineering Circuit Analysis
CH6 Polyphase Circuits
6.1 Polyphase System
6.2 Notations
6.3 Single-phase Three-wire Systems
6.4 Three Phase Y  Y connection
6.5 The Delta (  ) Connection
6.6 Power measurement
Ch6 Polyphsae Circuits
6.1 Polyphase System
Polyphase system : system with polyphase sources
Single source (Vs)
Notice the instantaneous voltage maybe zero
 The instantaneous power will be zero
V

T
V


3T
2T
t
They all have 120o phase differences
 The instantaneous power will
never be zero.
Poly sources ( Vs1 ,Vs2 ,Vs3 )
T
2T
3T
t
Ch6 Polyphsae Circuits
6.1 Polyphase System
V  Vs1  Vs2  Vs3
• The incident with a zero instantaneous power has been exempted.
• The source power can be delivered more stably.
• The polyphase systems can provide multiple output voltage levels.
• Polyphase systems in practice certain sources which maybe approximated
by ideal voltage sources, or ideal voltage sources in series with small
internal impedances.
Ch6 Polyphsae Circuits
6.2 Notations
a
b
8A
For note c :
5A  8A  I cd , I cd  3A
For note f :
I ef  4A  3A , I ef  7A
For note j :
Iij  3A  4A10A , Iij  7A
4A
e
d
c
I de  2A
I cd  ?
5A
I ef
 6A
g
h
f
Iij
j
i
2A
10 A
k
I fj  3A
l
Ch6 Polyphsae Circuits
6.2 Notations
c

Van  10000V
n

b
Vbn  100  1200V




a
Vcn  100  2400V

Vab  Van  Vnb
The voltage of
point a with
respect to point b
a +; b -;
 Van  Vbn
 10000V  100  1200V
 173.2300V
Similarly, Iab denotes the current from point a to b.
Test with graphical analysis ? (Using the phasor diagram)
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Function: allowing household electronics operating at two levels of
voltages to be applied.
a
n
1-phase
3-wire
Source
b
a
V1
V2
n
b
Voltage characteristics
Van  Vnb
Vab  2Van  2Vnb
Household electronics may either operate with
110 V or with 220 V
Phase characteristics
Van  Vnb
Van  Vbn  0
Van  Vbn
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
a
A
I Nn  IbB  I aA
Zp
V1
n
N
Zp
V1
B
Current characteristics
b
I bB
V1

Zp
I Aa 
V1
Zp
I Nn  0
This is no current in the neutral wire.
How if the two Z p are NOT equal, and all the wires have impedances ?
This is a more practical scenario.
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Example 9.1 (P242)
① Determine the power delivered to the
1
1150 V
rms
0
11500 V
rms
I1
3
I3
10
50 
20 
I2
100 
j10
50, 100  and the 20  j10 Loads.
② Determine the power lost in the three
lines represented by 1 3 and 1
respectively.
③ Determine the transmission efficiency?
total power absorbed by the loads
η=
total power generated by the sources
Hints: observe a structure with regular meshes and know the impedances, we can
determine the currents I1, I2 and I3 in order to find out the power being lost
and delivered!
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Apply KVL for the three meshes.
11500 V 1  I1  50  I1  I 2   3I1  I3   0
20  j10  I2  100  I2  I3   50I2  I1   0
11500 V  3  I3  I1   100I3  I 2  1  I3  0
Rearranging them in a matrix form as
 50
 3   I1  11500 
 54
 50 170  j10  100  I    0 


 2  
  3
 100
104   I 3  11500 
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
It can be calculated:
I1  11.24  19.830 A rms
I1 I 2  2.02∠2.27o A rms
I 2  9.389  24.470 A rms
I3
I 2  1.08∠2.12 o A rms
I 3  10.37  21.80 A rms
I3
I1  0.947∠2.3o A rms
0
Hence, the average power delivered to each of the loads are:
2
P50  I1  I 2  50  206W 

2
P100  I 3  I 2 100  117W  Total loaded power  2086 W
2
P20  j10  I 2  20  1763W 

Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Power lost in three wires are:


2
PbB  I 3 1  108W
Total lost power  237 W
2
2
PnN  I nN  3  I 3  I1  3  3W 

PaA  I1 1  126W
2
Transmission efficiencyη 
Powerdeliveredto theload
 100%
total power generated
Total power generated by the two voltage sources is:
Psources  11511.24cos19.830  11510.37cos21.800
 1216W  1107W  2323W
Transmission efficiency  2086 W  100 %  89 .8%
2323 W
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
a

Van




n
b
A
B
Vbn
Balanced three-phase sources
(phasor voltages)
N
Van  Vbn  Vcn
Van  Vbn  Vcn  0


Voltage characteristics
Vcn
 C
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Positive phase sequence (abc) (clockwise rotation)
Vcn
Van  Vp 00
Vbn  Vp   1200
 240
0
Vcn  Vp   240
0
Vp
 120
0
Van
Vbn
Negative phase sequence (cba) (Anti-clockwise rotation)
Van  V p 0
Vbn
0
Vbn  V p 120
1200
0
Vcn  V p 2400
2400
Vcn
Van
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Line-to-line voltages (take the abc sequence as an example)
Vab  Van  Vnb  V p 00  V p 600  V p  V p cos60  jVp sin 60
3
3
 Vp  j
V p  3V p 300
2
2
Vca
Vbc  Vbn  Vnc  Vp   120  Vp   60
0
Vna
Vcn
0
1
3
1
3
  Vp  j
Vp  Vp  j
Vp  3V p   900
2
2
2
2
Vnb
Vnb
Vna
Van
Vca  Vcn  Vna  V p   2400  V p 1800
1
3
  Vp  j
V p  Vp  0  3V p   2100
2
2
Hence
Vab  Vbc  Vca  0
Vbn
Vnc
Vnc
Vbc
verifies KVL.
Vab
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Vab  3Vp 300
Vbc  3Vp   900
Voltage types
magnitude
Phase voltages ( Vp )
Vp
Line-to-line voltages (
VL )
3V p
Vca  3V p   2100
Phasor difference
1200
1200
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Current characteristics
I aA
I bB
a 




b
B
A
Zp
Zp
n
N


c
I cC
ZP
C
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Consider three impedances Z p are connected between each line and
the neutral line.
Vbn Van   1200
I bB

 I aA  1200
Zp
Zp
V
I aA an
Zp
I cC
Hence
Vcn

Zp
V p   2400
Zp
 I aA  2400
I aA + I bB + I cC = 0
When balanced impedances are applied to each of the three lines and
the neutral line carries no current.
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Example 9.2 (P247)
Phase voltages:
Van  20000 Vrms , Vbn  200  120 0 Vrms , Vcn  200  240 0 Vrms
line-to-line voltage:
Vab  200 3300 Vrms , Vbc  200 3  900 Vrms , Vca  200 3  210 0 Vrms
Line currents:
Van 20000
0
I aA


2


60
Arms
Z p 100600
I bB 2  180 0 Arms
I cC 2  300 0 Arms
Power absorbed by the three loads
P  3  200 2  cos60o  600W
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
Example 9.2 (P247)
How about the instantaneous power?
van t   200 2 cost V


iaA t   2 2 cos t  60 A
0
Note: Van = 200V rms


PaA t   van t  iaA t   200 2 cost V  2 2 cos t  600 A


 200 400cos 2t  600 W
Similarly , the instantaneous total power absorbed by the loads are :
Pt   PA t   PB t   PC t 
 600 400cos2t  60  400cos2t  300  400cos2t  180W
 600W
The total instantaneous power is NEVER ZERO.
Ch6 Polyphsae Circuits
6.4 Three Phase Y  Y connection
•
Example 9.3 (P249)
A balanced three-phase system with a line voltage of 300Vrms is supplying a
balanced Y-connected load with 1200W at a leading power factor (PF) of 0.8.
Determine line cuurent IL and per-phase load impedance Zp.
The phase voltage is: Vp = 300/ 3 Vrms.
The per-phase power is: 1200W/3 = 400W.
300
Therefore 400 =
( I L ) × 0.8 , and IL = 2.89Arms
3
V
300 3
The phase impedance is: | Z P |= P =
= 60Ω
IL
2.89

~
IL

Vp  300 3 Vrms
Zp
A leading PF of 0.8 implies the current leads the voltage, and the impedance angle
is: -argcos(0.8) = -36.9o
and Zp = 60∠-36.9o Ω
Note: the apparent power of a Y-Y connected load is P = Van × IAN
(phase voltage × line current)
Ch6 Polyphsae Circuits
6.5 The Delta (  ) Connection
The neural line dose not exist. Balanced impedances are connected
between each pair of lines.
a





b
A
Zp
Zp
n


c
B
ZP
C
Ch6 Polyphsae Circuits
6.5 The Delta (  ) Connection
Voltage characteristics
Phase voltages
Vp  Van  Vbn  Vcn
Line voltages
VL  Vab  Vbc  Vca
VL  3Vp
﹠
Vab  3Vp 300
Current characteristics
Phase currents
I p  I AB  I BC  ICA
Line currents
I L  I aA  I bB  I cC  3I p
Ch6 Polyphsae Circuits
6.5 The Delta (  ) Connection
Y connections
Phase voltages
Vp
Line voltages
VL  3V p
Phase currents
Ip
Line currents
√
IL  I p √
 connections
Vp
VL  3V p √
Ip
I L  3I p
√
Ch6 Polyphsae Circuits
6.5 The Delta (  ) Connection
•
Example 9.5 (p251)
Determine the amplitude of line current in a three-phase system with a line voltage
of 300Vrms that supplies 1200W to a Δ-connected load at a lagging PF of 0.8.
The per-phase average power is: 1200W/3 = 400W
Therefore, 400W = VL ∙ IP ∙ 0.8 = 300V ∙ IP ∙ 0.8, and IP = 1.667Arms
The line current is: IL = 3 IP = 3 1.667A = 2.89Arms
Moreover, a lagging PF implies the voltage leads the current by argcos(0.8) = 36.9o

The impedance is: Z  VP  300 36.9o  180∠36.9o 
P
IP
1.667
Note: the apparent power of a Δ connected load is P = Vab × IAB
(line voltage × phase current)
Ch6 Polyphsae Circuits
6.6 Power measurement
P  I V
Wattmeter
measured by
current coil
I
current coil


potential coil
E.g.
V
measured by
potential coil
Passive
Network
I  11.18153.4Arms
V  1000Vrms
P  V  I cosangV  angI 
 10011.18 cos0  153.4  1000W
Ch6 Polyphsae Circuits
6.6 Power measurement
A

a
IaA


ZP 
IAB
B
IBC
ZP 
ZP 
1
IbB
b 
ICA
c 
IcC

2
C

Validate the power meter reads the actual
power absorbed/delivered by the three
impedances.
Ch6 Polyphsae Circuits
6.6 Power measurement


P1  VAB  I aA cosangVAB  angI aA   VL I L cos 300    




 VL I L cos 300  


P2  VCB  I cC cosangVCB  angI cC   VL I L cos 900  120   
 VL I L cos 300  
3 1
 tg
P1 cos 30  
cos30 cos  sin 30 sin 
3  tg
2
2




P2 cos 300  
cos300 cos  sin 300 sin 
3 1
3  tg
 tg
P2  P1 2 2
P2  P1
tg  3
  arctg 3
P2  P1
P2  P1


0


reactive (PF=0)

   , tg  
2
P1   P2
0
0
capacitive / inductive (0<PF<1)


    , tg   , tg  
2
2
π
P1  P2 ,     0, capacitive
2 
P1  P2 , 0    , inductive
2
resistive (PF=1)
  0 , tg  0
P1  P2
Ch6 Polyphsae Circuits
6.6 Power measurement
•
a
.

A 4

Example 9.7 (p256)
1

Vab  2300Vrms with positive phase sequence. b
(1) Find the reading of each wattmeter.
2
(2) The total power absorbed by the loads.
With positive phase sequence , we know :
Vab  2300Vrms
Vbc  230 120Vrms

c
Vca  230 120Vrms
Wattmeter 1 reads IaA and Vac :
V  V  230  60Vrms
ac
.
B

j15
ca
 230   30



V
3

an
IaA 

 8.554  105.1A
4  j15
4  j15
.
C
.
N
Ch6 Polyphsae Circuits
6.6 Power measurement
•
a
.

A 4

Example 9.7 (p256)
Wattmeter 1 reads :
P1  Vac IaA cosangVac  angIaA 
1
j15

b
.
B


 230 8.554 cos 60  105.1  1389W
c
Wattmeter 2 reads IbB and Vbc :
 230   150



V
3


bn
IaB 

 8.554134.9A
4  j15
4  j15
2
.
C
P2  Vbc IbB cosangVbc  angIbB 
 230 8.554 cos 120  134.9  512.5W
Hence , P  P1  P2  876.5W
Q: Please try to prove the two wattmeters read the power associated with the
three impedances.
.
N