Download Balanced poly phase circuits

Balanced poly phase circuits
Two and four phase systems
 A two phase system is an electrical system in which the voltages of the
phases are 90 degree out of time phase.
 The emf’s Eac and Ebd are 90 degree out of time phase.
 A two phase system is the equivalent of two separate single-phase systems
that are separated 90 degree in time phase.
 If the connection is made between the two windings at n and n`, the system
would be called a four-phase system.
 The voltages Eda, Eab, Ebc and Ecd are called the line voltages, while
voltages E0a, E0b, E0c and E0d are called the phase voltages or voltages to
neutral.
 Eda=Ed0+E0a. Thus, the line voltages= 2 times the phase voltages in the
four phase star.
Two and four phase systems
 The line voltages are 45 degree or 135 degree out of time phase from the
phase voltages.
 For four-phase mesh, the aa` line current is Iaa`=Ida+Iba.
 Thus, the line currents= 2 times the phase currents and 45 degree or 135
degree degree out of time phase in the four phase mesh
Three phase four wire system
Vcn
- Vbn
Vab
30 o
Van
Vbn
Vab = Van - Vbn
Vab = 3 Van 30o
N is the neutral wire.
Lighting loads are placed from line to neutral. Motors and other three phase power
loads are connected between the three lines.
Three phase three wire system
Three phase four wire system
Vab  Van  Vbn
| V p | 0 | V p | 120
| V p | 1  (cos 120  j sin 120) 
 1
3

| V | p 1   j

2
2


3
3

| V | p   j

2
2


 3 | V p | 30
VL  3 | V p |  Line Voltage
Three phase four wire system
The current in any line is the same as the current in the corresponding phase.
I L  I
Current in the neutral wire is obtained through the application of KCL.
I nn  I na  I nb  I nc
These currents are equal in magnitude and displaced from one another in time
phase by 120 degree.
Thus the current in the neutral wire is 0 since
I nn  I na  I nb  I nc  0
Three phase three wire system
VL  V
Three phase three wire system
Balanced Wye loads
Given the line voltages as 220 volts balanced three phase and R and X of each phase 6
ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase
and total power. Draw the vector diagram
VL
 127volts
3
127
 Ip 
 12.7 amperes
2
2
6 8
VP 
IL
power
per
phase  I P RP  968watts
2
total power  3  968  2904 watts
Vna  127  j 0volts


Vnb  127  120  127 cos 120  j sin 120  63.5  j110volts
Vnc  127  120  63.5  j110volts
Vba  Vbn V na 190.5  j110volts
I na 
Vna
 12.7  53.13
Z na
I nb  12.7  173.13
I nc  12.766.87 
Pna  vi  vi
Balanced Delta loads
Given the line voltages as 220 volts balanced three phase and R and X of each phase 6
ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase
and total power. Draw the vector diagram
self
Power calculation in balanced systems
Pp  V p I P cos  P
Pt  nPP  nV p I P cos  P
VL
Pt  3V p I P cos  P  3
I L cos  P  3VL I L cos  P
3
IL
Pt  3V p I P cos  P  3VL
cos  P  3VL I L cos  P
3
Wye connection
Dealta connection
Volt-Amperes
vat  3vap  3V p I P
The sine of the angle
between
phase
voltage and phase
Wye connection current is called the
reactive factor of a
balanced system.
VL
3
I L  3VL I L
3
IL
 3VL
 3VL I L
Dealta connection
3
Reactive Volt-Amperes P  3V I sin 
X
p P
P
Wye connection
Dealta connection
VL
PX  3
I L sin  P  3VL I L sin  P
3
IL
PX  3VL
sin  P  3VL I L sin  P
3
Single phase and balanced three phase power
Pa  Vm I m sin t sin t   

 

sin t  240 sin t  240   
Pb  Vm I m sin t  120 sin t  120  
Pa  Vm I m

P3  Pa  Pb  Pa


 


 
 Vm I m sin t sin t     Vm I m sin t  120 sin t  120    Vm I m sin t  240 sin t  240  
 1.5Vm I m cos 
For any phase a
P1  Vm I m sin t sin t   
Vm I m
Vm I m

cos  
cos2t   
2
2
Balanced three phase power
under steady state conditions is
constant from instant to instant.
But it is a double frequency
variation with respect to time
for single phase power..

Three wattmeter method
It
is not feasible to break
into the phases of a delta
connected load.
For the Y load it is
necessary to connect to the
neutral point. This point is not
always accessible.
Two wattmeter method
Two wattmeter method
Watt ratio curve
For each value of ϴ, there is a definite
ratio of Wa/Wb.
If the ratio of the smaller to the larger
reading is always taken and plotted
against the corresponding cosϴ, a curve
called the watt-ratio curve results.
At 0.5 power factor, one wattmeter
reads 0.
At 0, each wattmeter has the same
deflection but the readings are of
opposite signs.
Fig:46. watt-ratio curve for two wattmeter
method of measuring power
Sign of wattmeter
Method 1
Open line a, all power must be transferred to the load over lines b and c.
If wattmeter Wb reads upscale, the power is going to the load.
Reconnect line a and open line b. then connect Wa so that it reads upscale.
Reconnect line b.
If at any time after this wither wattmeter needle goes backward against down-scale,
power through this wattmeter channel is being transferred to the generator and this
power must be of opposite sign to that registered by the other.
Either the potential or current coil will have to be reversed to secure an up-scale
reading.
Method 2
disconnect the potential coil of Wb from line c and connect to line a.
It the needle goes against the down-scale stop, the wattmeter reading was negative.
Copper required to transmit power under fixed
conditions
P1  VI1 cos 
P3  3VI 3 cos 
P1  P3
VI1 cos   3VI 3 cos 
I1  3 I 3
I1 R1  2  I 3 R3  3
2
2
2
R1 3I 3
1


2
R3 2 I1
2
copper 3 number of

copper1
number of
wires 3
wires 1
R1 3


R3 4
Copper required to transmit power under fixed
conditions
P1  VI1 cos 
P3  3VI 3 cos 
P1  P3
VI1 cos   3VI 3 cos 
I1  3 I 3
I1 R1  2  I 3 R3  3
2
2
2
R1 3I 3
1


2
R3 2 I1
2
copper 3 number of

copper1
number of
wires 3
wires 1
R1 3


R3 4
Unbalanced system
 An unbalanced system is due to unbalanced voltage sources or unbalanced load.
In a unbalanced system the neutral current is NOT zero.
Line currents DO NOT add up to zero.
In= -(Ia+ Ib+ Ic) ≠ 0