BEF 23803 - Lesson 3 - Balanced Delta Load Thre-Phase Systems.ppt

Balanced Delta Load Thre-Phase Systems

1. Lecture 3:
Balanced Delta-Load
Systems

2. •Know the basic delta three-phase connections
•Know how to calculate voltage and currents in balanced,
three-phase circuits with delta-connected loads.
•Know how to calculate complex power in balanced, three-
phase circuits with delta-connected loads.
THE LEARNING GOALS FOR THIS LESSON ARE:

3. Balanced Wye-Delta Connection
•Three phase sources are usually Wye connected and three
phase loads are Delta connected.
•There is no neutral connection for the Y-∆ system.

4. Balanced Delta-delta Connection
•Both the source and load are delta-connected and balanced.

5. THREE-PHASE DELTA-CONNECTED BALANCED LOADS
For delta connection, voltages across elements equal line voltages.
For phase currentsABZ
V
I
AB
AB BCZ
V
I
CB
BC CAZ
V
I
CA
CA
These phase currents also form a balanced set of phasors as they
did in the star connection. They have equal magnitudes and are out
of phase with each other by 120
o
.

6. THREE-PHASE DELTA-CONNECTED BALANCED LOADSo
I
III
303 

AB
CAABA
ABBCB III  BCCAC III 
The line currents are obtained by using Kirchhoff’s current laws at
each node of the delta.
At node A
At node B
At node C

7. For a balanced delta load,
I
A+ I
B+ I
C= 0
Then
I
A+ I
B+ I
C= (I
AB–I
CA) + (I
BC–I
AB) + (I
CA–I
BC) = 0
To find the relationship betweenI
Aand I
AB, that is the relationship
between the line current and the phase current in a delta load,
consider the following.
If we let
I
AB= Iamperes
then in ABC sequence
I
BC= I-120amperes and I
CA= I+120amperes
These are phase current in the load.
THREE-PHASE DELTA-CONNECTED BALANCED LOADS

8. Phasor diagram:
Sinceo
I
III
303 

AB
CAABA
ABA II 3
Then and lags I
ABby 30
o
.
Then in general the expression isphL
II 3
THREE-PHASE DELTA-CONNECTED BALANCED LOADS

9. *
phphT3IVS V
AB
V
BC
V
CA
V
AN
120
o
120
o
120
o
30
o
V
AB
I
BC
I
CA
I
AB120
o
120
o
120
o

o
Total complex power
THREE-PHASE DELTA-CONNECTED BALANCED LOADS

10. A three-phase three-wire 415 V ABC system supplies a delta-
connected load whose phase impedance is 6045
o
.
Find the phase currents and the line currents in this system and
draw the phasor diagrams.
Example
Firstly, let us draw the connection diagram.
Solution

11. For ABC sequence with line-to-line voltage of 415 V, we can writeV
AB  30415V V
BC  90415V V
CA  150415V
To calculate the phase currents I
AB, I
BC, and I
CA. A
AB
AB 


 159.6
4560
30415
Z
V
I
Solution (continued)

12. A
CB
BC 


 1359.6
4560
90415
Z
V
I A
CA
CA 


 1059.6
4560
150415
Z
V
I Solution (continued)

13. To find the line currents I
A, I
B, and I
C. A



CAABA





4594.11
44.844.8
66.678.178.166.6
1059.6159.6
j
jj
III A



ABBCB





16595.11
1.354.11
78.166.688.488.4
159.61359.6
j
jj
III
Solution (continued)

14. To find the line currents I
A, I
B, and I
C. A



BCCAC





7594.11
54.111.3
88.488.466.678.1
159.61059.6
j
jj
III
Notes:The currents I
A, I
B, and I
Care of equal magnitude and are
spaced at 120
o
. They also lag their respective phase currents by
30
o
.
Solution (continued)

15. The phasor diagram for all currents is shown in the figure.
Solution (continued)

16. In this example it would be possible, after calculating I
ABto find I
BC
and I
CAby following the phase sequence and separating them from
I
ABby 120
o
. AAB  159.6I A
ABC


13596
120159.6
-.
I A
ACA


10596
120159.6
.
I
Solution (continued)

17. Consequently once finding I
Ayou could follow the same sequence,
giving: AA  4595.11I A
AB


16595.11
1204595.11
-
I A
AC


7595.11
1204595.11I
Solution (continued)

18. For the same reasons as when we calculated power factor in the
star-connected load, we must consider the voltage which causes the
current in the load. With a delta load there is only one voltage, the
phase-to-phase voltage, but there are two currents.
If we consider the impedance in A phase then we have V
ABas the
voltage which causes I
ABto flow and you will recall the formula:
Power FactorZ
V
I
AB
AB
So the angle of the power factor is the angle between VAB and
IAB. Thus the power factor for a delta circuit can be determined by
considering the impedance angle and applying the power factor
formula
PF = cos 
z

19. A three-phase three-wire 300 V ABC sequence system supplies a
balanced delta load. Each phase impedance is made up of a
resistance of 35 and an inductance of 0.25 henries is series. The
system frequency is 50 Hz.
Draw the diagram of the load connections.
Determine the phase currents, line currents and draw the complete
phasor diagram.
Exercise

20. A three-phase star-connected generator having phase sequence
ABC and a phase-to-neutral voltage V
ANof 200 volts supplies a
balanced delta load. The phase impedance of the delta load is 35-
29.
Determine the phase currents, line currents and power factor and
draw the complete phasor diagram.
Exercise

21. Effect of Phase Reversal on Delta Load Currents
To determine the effect of phase reversal on the currents we must
consider the effect of the angular change of the voltage.
If the reversed phase sequence is CBA and V
ANis taken as
reference, then the phase-to-phase voltages are nowvolts
AB  30VV volts
AC 90VV volts
CA  150VV

22. Here we want to find the relationship between I
Aand I
ABwith the
reverse phase sequence in the delta load.
If we let I
AB= Iamperes
Then in CBA sequence
I
BC= I+120amperes
and
I
CA= I-120
These are the phase currents in the load.
To find the line current IA, consider the figure shown.CAABA III 
thenABA II 3
and leads I
ABby 30
o
.
Effect of Phase Reversal on Delta Load Currents

23. A three-phase three wire 415 volt CBA system supplies a delta-
connected load whose phase impedance is 6045.
Find the phase currents and line currents in the system and draw the
phasor diagram.
Worked Example

24. Solution
The voltage phase diagram is as shown in the figure.
The phase-to-phase voltages areV AB  30415V V BC  90415V V CA  150415V
Now the phase currents are A
AB
AB 


 7596
4560
30415
-.
Z
V
I

25. A
BC
BC 


 4596
4560
12075415
.
Z
V
I A
CA
CA 


 19596
4560
12075415
.
Z
V
I To calculate the line current I
A, I
Band I
C.




4595.11
44.844.8
78.166.666.678.1
1959.6759.6




CAA
j
jj
III AB
Solution (continued)

26. 

7595.11
1204595.11

BI 

16595.11
1204595.11

CI and
Note that the line current I
Ais now leading the phase current I
ABby
30. All other magnitudes are the same as the ABC case.
Solution C(ontinued)

27. Calculation of power in delta-connected balanced loads
With delta-connected load, as it was with star-connected loads, the
average power dissipated in each phase is the same and is called
P
ph.
The total power dissipated by the three-phase balanced load is
P
T=3 P
ph
The wattmeter is connected as shown in Figure x to measure the
single-phase power.

28. If 
zis the impedance angle of the load, thenwatts
zL
L
zphphph
 cos
3
cos V
I
VIP 
Then the power in the three-phase load iswatts
zLLzL
L
T
 cos3cos
3
3 VIV
I
P 
This formula is identical to the formula used to calculate the three-
phase balanced star total load power.
So, providing we know the line current and phase-to-phase (line)
voltage and the impedance angle, we can calculate the power of the
load without having to know how it is connected.

29. Three identical coils, each having a resistance of 20and an
inductance of 0.5 H connected in delta to a three phase supply of 400
V;50 Hz. Calculate the line current and the total power absorbed by
the load.
Worked Example20
400V400V
400V
0.5H
0.5H
0.5H
20
20
Firstly, let us draw the connection diagram.
SolutionV 400
ph
V
Phase voltage

30. H

5.0
20


coil
coil
L
R 
22
coilcoilcoilcoilcoil
XRjXRZ Calculate the impedance of the coils.
















83
20
157
tantan
11
coil
coil
R
X

Solution (continued)
Given:
Impedance of coil,
where
and 1575.05022 fLX
coil

31. Solution (continued)
83158
20
157
tan15720
122








coil
Z 1264.083coscos 


Therefore,
and
Since it is a balanced loadA 38.4
158
400

ph
ph
Lph
Z
V
II
Total power absorbed by loadcos3
LLT IVP W3831264.038.44003 

32. The Y-TransformationAZ BZ CZ 1Z 3Z 2Z 1
2
3
A B B C A C
B
A B B C A C
A
A B B C A C
C







Z Z Z Z Z Z
Z
Z Z Z Z Z Z
Z
Z
Z
Z
Z Z Z Z Z Z
Z

33. AZ BZ CZ 1Z 3Z 2Z 13
1 2 3
23
1 2 3
12
1 2 3
A
B
C






ZZ
Z Z Z
ZZ
Z Z Z
ZZ
ZZ
Z
Z
Z
Z The -YTransformation
For balanced delta-connected
load,
Z
1= Z
2= Z
3= Z

Z
A= Z
B= Z
C= Z
Y3


Z
Z
Y

34. •Presence of line impedance Z
Lin each line makes it difficult to
solve the circuit directly.
•Need to transform delta-connected load into a star-connected
load.
BALANCED Y-SOURCE DELTA-LOADCIRCUIT

35. Circuit after delta-star transformation:
BALANCED Y-SOURCE DELTA-LOADCIRCUIT

36. Given the following circuit find the phase currents flowing in the
delta-connected load.rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V 10 5
75 225
L j
j

  
  
Z
Z
Worked Example

37. 25 75
3
Y j

   
Z
Z Solution

38. rms rms1.26 186 A and 1.26 54 A
bB cC     II rms
rms
rms
99.6 5 V
99.6 115 V
99.6 125 V
AN aA Y
BN
CN
   
   
  
V I Z
V
V The voltages in the per-phase equivalent circuit arerms1.26 66 A
a
aA
LY
    

V
I
ZZ
Solution

39. rms
rms
rms
172 35 V
172 85 V
172 155 V
AB AN BN
BC BN CN
CA CN AN
    
     
    
V V V
V V V
V V V The line-to-line voltages arerms
rms
rms
0.727 36 A
0.727 156 A
0.727 84 A
AB
AB
BC
BC
CA
CA



    
    
   
V
I
Z
V
I
Z
V
I
Z
Solution

40. rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V 10 5
75 225
L j
j

  
  
Z
Z Given the following circuit. Compute the total power dissipated in the
load.
Exercise
[Answer: 122.6 W]

41. END