LAB SHEET
TRIMESTER II (2010-2011)
PTD1 - Performance of Transmission Line under Different Loading Conditions
PTD2 - Parameters which affect Real and Reactive Power Flow
*Note: On-the-spot evaluation may be carried out during or at the end of the experiment.
Students are advised to read through this lab sheet before doing experiment. Your performance, teamwork effort, and learning attitude will count towards the marks.

1.
Before coming to the laboratory read the lab sheet carefully and understands the procedure of performing the experiments.
2. Do not switch-on the power supply unless permitted by the lab supervisor.
3.
Do not make or break any connection with the power supply on.
4.
Handle the equipments with care.
5.
Do the necessary calculation, draw the graphs and submit the report within the specified time of the lab session.
Marking Scheme
Lab Report Writing: (The report should consists of Results and Answers for all the questions, Discussion and Conclusion) ----------------------------------7 marks
Spot evaluation (Oral assessment at the end of lab) ------------------- 3 marks
Experiment # 1

To analyze the voltage regulation under resistive, inductive and capacitive loading conditions of the transmission line

To compare the voltage drop in the transmission line when the sending-end and receiving-end voltages have the same magnitude

To perform regulation of the receiving-end voltage
A short transmission line is modeled by a single reactance as shown in Fig. 1. A good understanding of the behaviour of most of the transmission lines can be obtained by the short line model. It is this model which will be used in this experiment.
Depending upon the loading condition the phase angle difference between the sending-end and receiving-end voltages and the voltage drop along the line will vary. These effects can be easily understood from the phasor diagram shown in Fig. 1. It may also be observed that a significant voltage drop will exist across the line even when the sending -end voltage, E
1
and the receiving-end voltage, E
2
are equal in magnitude.
I X
L E
1
E
1
E
2
I
Fig. 1 (a) Transmission line (b) Phasor diagram
E
2
IX
L
2

We have studied that the voltage drop along the transmission line and the receiving-end voltage vary widely for inductive loads. In order to regulate the voltage at the receiving-end of the line in some way so as to keep it at as constant as possible we should adopt some type of compensation. One method commonly used is to connect shunt capacitors at the end of the line. These capacitors produce a significant voltage rise thus compensating for the voltage drop. Static capacitors are switched in and out in a practical system and their value is adjusted depending on the loads. For purely inductive loads, the capacitor should deliver reactive power equal to that consumed by the inductive load. For resistive loads, the reactive power, which the capacitor must supply to regulate the voltage, is not easy to calculate. In this experiment, we shall determine the reactive power (the value of capacitor) by trial and error, adjusting the capacitors until the receiving-end voltage is approximately equal to the sendingend voltage. For loads, which draw both real and reactive power, the same trial and error method is adopted.
Note that for a short transmission line having a line reactance of X

/phase and resistance neglected. The following formulas will be useful.
Sending-end voltage (L-L) = E
1

1
; Receiving end voltage (L-L) = E
2

2
Three-phase sending-end power = Three-phase receiving end Power
= P
1
= P
2
=
E
1
E
2
Sin
X
(

1
 
2
)
Three-phase sending-end reactive power = Q
1
=
E
1
2
X
-
E
1
E
2
Cos (
 
1
X

2
)
Three-phase receiving-end reactive power = Q
2
=
E
1
E
2
Cos (
 
1

2
X
)
-
E
2
2
X
Apparent power at sending end = S
1
Apparent power at receiving end = S
2



P
1
2

P
2
2

Q
1
2

Q
2
2
;


;
Three-phase transmission line (8329)
Resistive load (8311)
Inductive load (8321)
Capacitive load (8331)
AC voltmeter (8426)
Phase meter (8451)
Three-phase wattmeter/varmeter (8446)
Power supply (8821)
Connection leads (9128)
1.
Set the impedance of the transmission line to 200

and connect the meters as shown in
Fig. 2. The circuit should be connected to the three-phase variable supply. Note that watt/var meters and phase meter need 24V AC supply provided in the power supply unit.
Connect all the loads in star. Verify your connections with the lab supervisor before switching on the power supply.
3

0-500V
E
2
0-500V
E
1
4
5
6
1
2
3
P
1
Q
1
4
5
6
1
2
3
P
2
Q
2
4
5
6
Yconnected
LOAD
8821
8329
0-415V
3-phase
8446 8446 8311
8321
8331
Fig. 2 Connection diagram for steps 2, 3 and 4
2.
Adjust the sending-end voltage E
1
to 300 V and keep it constant for the reminder part of the experiment. Use a three-phase resistive load and increase the load in steps making sure that the loads are balanced. Take readings of E
1
, Q
1
, P
1
, E
2
, Q
2
, and P
2.
Record your results in Table 1.
3.
Switch off the power supply and connect a three-phase balanced inductive load in parallel with the balanced resistive load. Don’t remove any other connections shown in Fig.2.
Increase the load in steps making sure that the loads are balanced. Take readings of E
1
,
Q
1
, P
1
, E
2
, Q
2
, and P
2.
Record your results in Table 2.
4.
Switch off the power supply, remove the inductive load and connect a three-phase balanced capacitive load in parallel with the balanced resistive load. Take readings of E
1
,
Q
1
, P
1
, E
2
, Q
2
, and P
2 for different loadings
.
Record your results in Table 3.
5.
Draw three graphs of E
2
(obtained from steps 2, 3, and 4) on the same graph paper as a function of the receiving-end power P
2 and discuss your results.
6.
Switch off the power supply and connect a phase meter to measure the phase angle difference between E
1
and E
2
and a voltmeter to measure the voltage across the transmission line as shown in Fig. 3. Note that the load consists of resistances in parallel with capacitances. Now for each resistive load, adjust the capacitive load so that the load voltage E
2
is as close as possible to 300 V. Take readings of X
C
, E
1
, P
1
, Q
1
, E
2
, P
2
, Q
2
, and the phase angle for different loadings
.
Record your results in Table 4.
7.
Draw the graphs of E
2
and the phase angle difference between E
1
and E
2
as a function of
P
2
from the results in Table 4. Note that the addition of static capacitors has yielded a much more constant voltage, and further more, the power P
2
which can be delivered has increased. On this curve, indicate the phase angle between E
2
and E
1
as well as the reactive power Q
2
used for individual resistive load settings.
8.
In this part of the experiment, we shall observe a significant voltage drop along the transmission line even when the voltages E
1
and E
2
are equal in magnitude. This voltage drop is due to the phase angle difference between the two voltages. Switch off the supply and insert an ammeter in series with the transmission line as shown in Fig. 3 to measure
4

the line current without removing any other connection. Using the circuit shown in Fig. 3, set the load resistance per phase at 686

and E
1
= 300 V, adjust the capacitive reactance until the load voltage is as close as possible to 300 V. Measure and record E
1
, Q
1
, P
1
, E
2
,
Q
2
, P
2
, E
3
, the line current I and the phase angle.
8451
0-500V
1
2
3
4
E
2
0-500V
E
1 686
8329
4 1 4 1 4
P
1
Q
1
P
2
Q
2
686
5
6
2
3
5
6
2
3
5
6 A
8821 686
8446 8446
0-415V
3-phase
E
3
0-250V
8311
8331
Fig.3 Connection diagram for Steps 6 – 8
9.
Using the results of step 8, draw the phasor diagram of per phase values of E
1
and E
2
to scale and draw E
3
. From the diagram compute E
3
and compare it with the measured value. Also compute the real power, reactive power and apparent power consumed by the line. From the apparent power compute the line current and compare it with the measured value.
Table 1: Results of procedure step 2
R


4800
E
1
V
P
1
W
Q
1 var
E
2
V
P
2
W
Q
2
Var
2400
1600
1200
960
800
686
5

Table 2 Results of procedure step 3
R


4800
2400
1600
1200
960
800
686
X l


4800
2400
1600
1200
960
800
686
E
1
V
P
1
W
Q
1 var
E
2
V
Table 3 Results of procedure step 4
R


4800
2400
1600
1200
960
X c


4800
2400
1600
1200
960
800
686
E
1
V
P
1
W
Q
1 var
E
2
V
800
686
Table 4 Results of procedure step 6
R


4800
X c

E
1
V
2400
1600
1200
960
800
686
Results of procedure step 8
E
1
= E
2
=
P
2
= P
1
=
Q
1
=
Phase angle =
Q
2
=
P
1
W
Q
1 var
E
3
=
E
2
V
Line current, I =
P
2
W
P
2
W
P
2
W
Q
2 var
Q
2 var
Q
2 var
Angle degree
6

Sample calculation
This sample calculation is to help you to answer Exercise 3.
Let
E
1
= 350 V
P
1
= 600 W
E
P
2
2
= 350 V
= 510 W
E
3
= 165 V
Q
1
= 170 var
Phase angle = 48
Q
2
= -280 var o
and Line current, I = 0.95
E
1 per phase = 350/

3 = 202 V
E
2 per phase = 350/

3 = 202 V
E
3
= 165 V
P
1 per phase = 600/3 = 200 W
P
2 per phase = 510/3 = 170 W
Q
1 per phase = 170/3 = 56.7 var
Q
2 per phase = -280/3 = 93.3 var
The phasor diagram of voltages to scale is shown in Fig.4.
E
1
-48 o E
3
=165
E
2
Fig. 4 Phasor diagram
From the figure E
3
= 165 V which is the same as the measured value. [The voltage E
3
may also be calculated using the formula, E
3
= 2*E
1
*sin(24 o )]
Real power consumed = 200 –170 = 30 W
Reactive power consumed = 56.7 –(-93.3) = 150 var
Apparent power in the line = 150
2 
30
2 
153 VA
Current through the line = 153/165 = 0.93 A
The difference between the calculated value and the measured value is 0.02 A.
1.
Draw the graphs asked in the procedure steps 5 and 7 and discuss your results. From the graphs plotted calculate the voltage regulations for load powers of 60W, 70W and 80W respectively under different loading conditions (resistive, resistive-inductive and resistive-capacitive) and compare the results.
2.
Do the calculation asked in procedure step 9 and discuss your results.
7

3.
A three-phase transmission line has reactance of 100

per phase. The sending-end voltage is 100 kV and the receiving-end voltage is also regulated to be 100 kV by placing a bank of static capacitors in parallel with the receiving-end load of 50 MW. Calculate
(a) the reactive power supplied by the capacitor bank
(b) the reactive power supplied by the sending-end side
(c) the voltage drop in the line per phase
(d) the phase angle between the sending-end and receiving-end voltages and
(e) the apparent power supplied by the sending-end side.
4.
If the 50 MW load in Exercise 3 is suddenly disconnected calculate the receiving-end voltage which would appear across the capacitor bank. What precaution, if any, must be taken?
5.
If a transmission line were purely resistive, would it be possible to raise the receiving-end voltage by static capacitors?
6.
State briefly what you have learned from this experiment.
8

To compare the flow of real and reactive power when sender and receiver voltages are different, but in phase.
To perform the flow of real and reactive power when sender and receiver voltages are equal, but out of phase.
To analyze the flow of real and reactive power when sender and receiver voltages are different and out of phase.
Transmission lines are designed and built to deliver electric power. Power flows from the generator (sender end) to the load (receiver end). But, in complex interconnected systems, the sender and receiver ends may become reversed depending upon the system load conditions which, of course, vary throughout the day. Power in such a line may flow in either direction. The character of the load also changes from hour to hour, both as to kVA loading and as to power factor. How, then, can we attempt to understand and solve the flow of electric power under such variable loading conditions, further complicated by the possible reversal of source and load at the two ends of the line?
We can obtain meaningful answers by turning to the voltage at each end of the tine. In Fig.1 a transmission line having a reactance of X

(per phase) has sender and receiver voltages of E
1
and E
2
V respectively. ( A transmission line is both resistive and reactive, but we shall assume that the reactance is so much larger that the resistance may be neglected ) If we allow these voltages to have any magnitude or phase relationship, we can represent any loading condition we please. In other words, by letting E
1
and E
2
take any values and any relative phase angle, we can cover all possible loading conditions which may occur
Sender and receiver voltages are different and out of phase.
Referring to Fig. 1, both E
1
and E
2
are phasors with different magnitude and out of phase.
X
SENDER
E
1 I
E
2
RECEIVER
Fig.1: Transmission line
The voltage drop along the line is E
1
- E
2
; consequently, for a line having a reactance of X Ω, the current I is given by I =
E
1
 jX
E
2 when E
1
– E
2
is the phasor difference between the sending- and receiving-end voltages.
If we know the value of E
1
and E
2,
and the phase angle between them, it is a simple matter to find the current I, knowing the reactance X of the line. From this knowledge we can calculate the real and reactive power, which is delivered by the source and received by the load.
Suppose, for example, that the properties of a transmission line are as follows:
9

Line reactance per phase, X = 100

Sender voltage (E
1
) = 20 kV
Receiver voltage (E
2
) = 30 kV
Receiver voltage lags behind sender voltage by 26.5°.
These line conditions are represented schematically in Fig. 2. From the phasor diagram in Fig. 3, we find that the voltage drop (E
1
– E
150 A and it lags behind (E
1
2
) in the line has a value of 15 kV. The current I has a value of 15 kV/100

=
– E
2
) by 90°. From the geometry of the figure, we find that the current leads E by 27°. The active and reactive power of the sender and the receiver can now be found.
1
S
X=100

26.5
E
1
=20kV
E
1
=20k
V
E
2
=30k
V
E
2
=30kV
I = 150 A
27°
90°
53.5°
E
1
= 20 kV
26.5°
Fig. 3: Phasor diagram
E
1
– E
E
2
2
= 15 kV
= 30 kV
Note: When determining the sine and cosine of the angle between voltage and current, the current is always chosen as the reference phasor. Consequently, because E
1 lags behind I by 27°, the angle is negative.
The real power delivered by the sender is, 150 A x 20 kV x cos (-27°) = +2670 kW.
The real power received by the receiver is, 150 A x 30 kV x cos (-53.5

) = +2670kW.
The reactive power delivered by the sender is, 150 A x 20 kV x sin (-27°) = -1360 kvar.
The reactive power received by the receiver is, 150 A x 30 kV x sin (-53,5°)= -3610 kvar.
(Note that equations for real power and reactive power given in the lab sheet for
Experiment-1 can also be used for the above calculation.)
Based on the results calculated above, if wattmeters and varmeters were placed at the sender and receiver ends they would give readings as shown in Fig. 4. This means that active power is flowing from the sender to the receiver, and owing to the absence of line resistance, none is lost in transit.
150A
+2670 -1360 +2670 -3610 kW kvar kW kvar
S
Real Power Reactive Power Real Power Reactive Power
Fig. 4: Direction of real and reactive power flow
R
However, reactive power is flowing from receiver to sender and, during transit, 3160 - 1360 = 2250 kvar are consumed in the transmission line. This reactive power can be checked against
Line kvar = I 2 X = 150 2 x 100 = 2250 kvar.
10

It will be noted that this is not the first time that we have found real power and reactive power flowing simultaneously in opposite directions.
Sender and receiver voltages are different, but in phase.
When the voltages at the sender and receiver ends are in phase, but unequal, reactive power will flow. The direction of flow is always from the higher voltage to the lower voltage.
Consider a transmission line in which the voltages at the sender and receiver ends are 30 kV and 20 kV respectively and the line reactance is 100

(Refer to Fig. 5).
X = 100

I
E
S
1
3 E
1
=30 kV E
2
=20kV
E
2
=20kV E
1
=30kV
The voltage drop in the line is 10 kV and the current is 10 kV/100Ω = 100 A as shown in Fig. 6.
E
2
= 20kV E
1
= 30kV
E
1
– E
2
= 10kV
I = 100A
Fig. 6 Phasor diagram showing current and voltages
The real power delivered by the sender end is, 100 A x 30 kV x cos (+ 90°) = 0 W.
The real power received by the receiver is, 100 A x 20 kV x cos (+90°) = 0 W.
The reactive power delivered by the sender end is,
100 A x 30 kV x sin ( + 90°) = + 3000kvar.
The reactive power received by the receiver is
100A x 20 kV x sin ( + 90") = +2000kvar.
If wattmeters and varmeters were placed at each end, the readings would be as shown in Fig. 7.
100A
0 kW
+3000 kvar
0 kW +2000 kvar
S
Real Power Reactive Power Real Power Reactive Power
R
Fig. 7 Direction of real and reactive power flow
Reactive power flows from the sender to the receiver, and 1000 kvar are absorbed in the transmission line during transit. As can be seen, reactive power flows from the high-voltage to the low-voltage side.
Sender and receiver voltages are the same, but out of phase.
Real power can only flow over a line if the sender and receiver voltages are out of phase. The direction of power flow is from the leading to the lagging voltage end. Again, it should be noted that this rule applies only to transmission lines, which are mainly reactive. The phase shift between the sender and receiver voltages can be likened to an electrical "twist", similar to the mechanical twist which occurs when a long steel shaft delivers mechanical power to a load. Indeed, the greater the electrical "twist" the larger will be the real power flow. However, it is found that it attains a maximum when the phase angle between the
11

sender and receiver ends is 90°. If the phase angle is increased beyond this (by increased loading) it will be found that less real power is delivered.
Consider a transmission line in which the voltages at each end are equal to 30 kV and the receiver voltage lags behind the sender by 30°. The line reactance is 100 
, and the circuit is shown in Fig. 8.
S
E
1
X = 100
= 30kV

I
E
2
=30kV
30

R
Fig. 8 Line with phase angle difference between E
1
and E
2
E
2
=30kV
The voltage drop in the line (E
1
- E
2
) is found to be 15.5 kV. So the current I =15500/100 = 155 A and it lags (E
1
- E
2
) by 90 o , as shown in Fig. 9.
E
1
= 30kV
E
1
– E
2
= 15.5kV
15 o
I
15 o
E
2
= 30kV
Fig. 9 Phasor diagram showing voltages and current
Taking the current as the reference phasor, we can find the real and reactive power associated with the sender and the receiver ends as shown in Fig.10.
155A
S
+4500 kW
+1200 kvar
+4500 kW
-1200 kvar
R
Sender End
Real power delivered = 30 kV x 155 A x cos (+15°) = +4500 kW,
Reactive power delivered = 30 kV x 155 A x sin (+ 15°) = +1200 kvar.
Receiver End
Real power received = 30 kV x 155 A x cos (-15°) = + 4500 kW.
Reactive power received = 30 kV x 155 A x sin (- 15

) = - 1200 kvar.
The sender delivers both active and reactive power to the line and the receiver absorbs active power from it.
However, the receiver delivers reactive power to the line, so that the total reactive power received by the line is 2400 kvar.
This example shows that a phase shift between sender and receiver voltages causes both real and reactive power to flow. However, for angles smaller than 45

the real power considerably exceeds the reactive power.
12

EQUIPMENT REQUIRED
DESCRIPTION MODEL
Resistive Load 8311
Inductive Load 8321
Three-Phase Transmission Line 8329
Capacitive Load 8331
Three-Phase Regulating Autotransformer
AC Voltmeter
8349
8426
Three-Phase Wattmeter/Varmeter
Phase Meter
6446
8451
Power Supply 6821
Connection Leads 9128
PROCEDURE
WARNING.
High voltages are present in this Laboratory Experiment! Do not make any connections with the power on!
In order to convey a sense of realism to the terms "sender" and "receiver" two consoles will be used in the following experiments. A transmission line will connect the two consoles (Station A and B) and the active and reactive power flow between them will be studied. The experiment will be conducted in three parts.
Part-I: Sender and receiver voltages unequal, but in phase.
Part-II; Sender and receiver voltages equal, but out of phase.
Part-III: Sender and receiver voltages unequal, and out of phase
Note that there will be minor changes only in the connections between the parts of the experiments. Don’t remove all the connections, simply do the changes only. Verify your connections for each part with your lab supervisor
PART-I: Sender and Receiver voltages unequal, but in phase
1.
Connect a three-phase transmission line between terminals 4, 5, 6 (variable AC output) of two consoles, one of which is designated as Station A and the other, Station B. Connect the two three-phase wattmeters/varmeters
(6446) at each end as well as a phase meter (8451) as shown schematically (single line diagram) in Fig. 11. Note that watt/var meters and phase meter need 24V AC supply provided in the power supply unit. Verify your connections with the lab supervisor before switching on the power supply.
STATION A
0 – 500 V
E
1
8451

E
2
0 – 500 V
STATION B
0-415 V
3-phase
4
5
6
O
O
O
8821
P
1
Q
8446
1
8329
P
2
Q
8446
2
O
O
O
8821
4
0-415 V
5
3-phase
6
Fig. 11 Sender and Receiver voltages unequal, but in phase
2.
With the transmission line switch S open, adjust the line-to-line voltages so that E
1 and E
2
are both equal to 300
V and observe that the phase angle difference between terminals 4-5 of station A and terminals 4-5 of station B is zero. Is phase angle zero?

Yes

No
13

3.
Without making any changes, measure the phase angle between terminals 4-5 of station A and terminals 5-4 of station B.
Phase angle is __________
4.
Without making any changes, measure the phase angle between terminals 4-5 of station A and terminals 5-6 of station B.

Phase angle is lagging

Phase angle is leading
5.
Measure the phase angle between terminals 4-5 of station A and terminals 6-4 of station B.

Phase angle is lagging

Phase angle is leading
6.
By measuring all phase angles between line and neutral of station A and B prove that the phasor diagrams for both stations are as given in Fig. 1 2.
The purpose of this preliminary phase angle check is to familiarize with the phase angles between the voltages at the two stations.
4A 4B
Rotation Rotation
120°
N
120°
120° 120°
N
120°
120°
6A
5A
6B
5B
Fig. 1 2.
The phase angles between the voltages at the two stations.
7.
Close the Three-Phase Transmission Line switch, S with E
1
= E
2
= 300V, and the transmission line impedance =
200

. Observe the three-phase wattmeter/varmeter readings. There should be no significant power exchange.
P
1
= _________ W
P
2
= _________ W
Q
1
= _________ var
Q
2
= _________ var
8.
Raise station A voltage to 350 V and observe power flow.
P
1
= _________ W
P
2
= _________ W
Q
1
= _________ var
Q
2
= _________ var
Which of the two stations would be considered to be the sender?
_________________________________________________________________
9.
Reduce station A voltage to 300 V and raise station B voltage to 350 V. Observe power flow.
P
1
= _________ W
P
2
= _________ W
Q
1
= _________ var
Q
2
= _________ var
Which station would be considered to be the sender?
________________________________________________________________
14

0-415 V
3-phase
10.
Vary the voltages of both station A and station B and check the truth of the statement that reactive power always flows from the higher voltage to the lower voltage.
PART-II: Sender and Receiver voltages equal, but out of phase
Use the Three-Phase Regulating Autotransformer to shift the phase of station A by 15°. The phase shift (lag or lead) is obtained by changing the connections of a three-phase transformer by means of a tap switch.
When the position of the tap-switch in the regulating transformer is altered, the secondary voltage will a) be in phase with the primary, b) lag the primary by 15° or, c) lead the primary by 15".
11.
Connect the above phase-shifting autotransformer to the variable AC terminals 4,5,6 of station A as shown schematically in Fig. 13. Open the switch (S) in the transmission line or d isconnect the transmission line .
Adjust the voltage at stations A and B to 350 V. With the Phase Meter determine the phase angle of the secondary voltage 4, 5, 6 of the phase shifting transformer with respect to the variable AC terminals 4, 5, 6 of the Power Supply of Station B. Record your readings for the three positions of the phase-shift tap switch in
Table 1.
4
5
6
O
O
O
Phase shifting
Autotransformer
P
1
0 – 500 V
Q
1
E
1
8451

E
2
0 – 500 V
P
2
Q
2
O
O
O
8821
8349
8329
8446
Fig. 13 Phase shifting of voltages at Station A
8446
8821
CAUTION!! KEEP THE SWITCH OF THE TRANSMISSION LINE OPEN FOR STEPS 11 AND 12
Table 1
4
5
0-415 V
3-phase
6
Tap switch position in Phase angle E
1
in V E
2
in V degree
0
(Lag/Lead)
+ 15
-15
Note: The buck-boost tap switch must be kept at zero and the correct phase sequence must be applied to the primary of the transformer.
12.
Check that the phase-shift is the same for all the three phases, and that all vol tages are balanced.
13.
Connect a three-phase, 400-

transmission line between secondary terminals 4, 5, 6 of the threephase phase shifting autotransformer and the power supply terminals of station B by closing the switch (see Fig.14). Change the tap switch position and record your results in Table 2.
15

0-415 V
3 phase
0 – 500 V 8451 0 – 500 V
4
5
6
O
O
O
REGULATING
AUTOTRANSFORMER
8821
Tap Switch
Position in degree
8349
P
1
Q
1
E
1 
E
2
P
2
Q
2
8329
8446 8446
Fig. 14 Sender and Receiver voltages equal, but out of phase
Table 2
E
1
V
P
1
W
Q
1 var
E
2
V
P
2
W
Q
2 var
Phase angle in degree
8821
O
O
O
4
5
0-415 V
3 phase
6
0
+ 15
-15
Does this experiment proves the statement that real power flows from the leading voltage towards the lagging voltage side of a transmission line?

Yes

No
(Caution!! First DIRECTLY SWITCH OFF the main power supplies of both sending end and receiving end and then set the supply autotransformers to zero voltage. Make changes in connection for PART-III)
PART-III: Sender and Receiver voltages unequal, and out of phase
In the following steps we shall connect passive loads (resistive, inductive, and capacitive) at the receiver end of the line. The object of the experiment is to show that a phase shift between sender and receiver voltages occurs only when real power is being delivered to the load.
14.
Using only one console, set up the experiment as shown in Fig. 15, setting E
1
= 380 V and using a star-connected Resistive Load of 1200

per phase and a 200-

Transmission Line. Take readings and record your results in Table 3
0 – 500 V
8451
0 – 500 V
E
1
 E
2
0-415 V
4
5
O
3-phase
O
P
1
Q
1
P
2
Q
2
LOAD
6
O
8329
8446 8446 8311
8821
Fig. 15 Transmission line with different loadings 8321
8331
15. Repeat procedure step 14 using an Inductive Load of 1200

/phase. Take
readings and record your results in Table 3.
16

16. Repeat procedure step 14 using a Capacitive Load of 1200

/phase. Take
readings and record your results in Table 3.
Step Load E
1
V
Table 3
P
1
W
Q
1 var
E
2
V
P
2
W
Q
2 var
Phase shift degree
14
15
RESISTIVE
INDUCTIVE
16 CAPACITIVE
Exercise
1.
A three-phase transmission line has a reactance of 100

and at different times throughout the day it is found that the sender and receiver voltages have magnitude and phase angles as given in Table 4. In each case calculate the real and reactive power of the sender and receiver and indicate the direction of the power flow. The voltages refer to line-to-line voltages.
Table 4
E
S kV
E
R kV
Phase angle P
MW
SENDER
Q
Mvar
P
MW
RECEIVER
Q
Mvar
100
120
100
120
120
100
100
120
100
100
60

E
S
Leads E
R
60

E
S
Leads E
R
60

E
S
Leads E
R
30

E s
Lags E
R
0°
2.
In Question 1 assume that E
S
= E
R
= 100 kV at all times but that the phase angle between them changes in steps of 30° according to the Table 5. Calculate the value of the real power in each case as wall as its direction of flow, knowing that E
R lags E s
in each case.
Table 5
 degree
SENDER, P
MW
RECEIVER, P
MW
0
30
60
90
120
150
180
Plot a graph of real power vs phase angle.
Is there a limit to the maximum power which such a line can deliver under the static voltage conditions?

Yes

No
3. State briefly what you have learnt from this experiment.
17