Balanced 3-Phase Power System Experiment for NEF2251 Course

This experiment is about studying a balanced three-phase Y-to-Y connected system. It includes a 3-phase Y connected voltage source connected to a 3-phase Y-connected load with various measurements taken to validate the relationship between line and phase voltages. The per-phase circuit, load impedance, and mathematical relationships between line and phase voltages are studied. The experiment is conducted at 50 Hz and all currents and voltages are RMS quantities.

Exp 1: Balanced 3-Phase Power System
NEF2251
FUNDAMENTALS OF ELECTRICAL AND
ELECTRONIC ENGINEERING
Laboratory Experiment 1
Balanced 3-Phase Power System
College of Engineering and Science
Name: Student Id

Exp 1: Balanced 3-Phase Power System
1.0 Introduction
The objective of this experiment is to study and get acquainted with/to a balanced
three-phase Y-to- Y connected system. In this experiment, a 3-phase Y connected
voltage source will be connected to a 3-phase Y-connected load with various
measurements taken to validate the relationship between line and phase voltages.
Students will so be expected to determine the load impedance from the
measurements taken during the experiment. The entire experiment will be conducted
at 50 Hz. All of the currents and voltages in this experiment are RMS quantities.
2.0 Background Summary and Preliminary
The power is generated, transmitted, and distributed in 3-phase (3-Φ) form. The 3-
Phase AC supply system has the following advantages:
• It is the most economical way of generating, transmitting, and
distributing electrical energy
• It can supply a wide variety of loads, those that require a three-phase
supply such as an induction motor, as well as others requiring only a
single phase supply
• A balanced three-phase system ensures a smooth flow of energy from source
to load since single-phase power produces a pulsating flow of energy
2.1 Balanced Three-Phase System
Figure 1 shows the circuit diagram of the three-phase circuit that will be connected up
for this experiment. A three-phase Y-connected RLC load bank is to be connected to a
Y-connected three- phase voltage source.
Figure 1 Balanced Y-to-Y connected three-phase experimental setup
Figure 2 shows the per-phase circuit (phase U) of the load bank. It is possible to alter
the phase impedances (ZU, ZV, and ZW) of the load bank individually by adjusting

Exp 1: Balanced 3-Phase Power System
the reactances XU, XV, and XW as well as the resistances RU, RV, and RW in each
phase. The reactances and resistances can be adjusted by using the three knobs
available on the RLC load bank. In this experiment, a balanced three-phase load is
required. Therefore, XU = XV = XW and RU = RV = RW. The settings for the
reactance and resistances will be provided to you by your lab supervisors during the
experiment.
It is also possible to set the load as an inductive or capacitive load by using the 3-way
switch available on the RLC load bank. If XL is chosen, the power factor will be a
positive value and a negative value if XC is chosen.
Figure 2 Per-phase circuit of the load bank
2.2 Preliminary
Please refer to your lecture and tutorial notes for more information in relation to 3-Phase
systems. Each student is expected to answer the following pre-lab questions in the space
provided by applying the theory gained in lectures and tutorials. The pre-lab exercises
must be completed before the start of the experiment. Students will not be given time
before the start of the experiment to do the preliminary work. Students who fail to do
their preliminary will not be allowed to carry out the experiment. Completion of the
preliminary is important, as this will help students to perform an analysis of the
experiment to bring them up to date with the theoretical background.

Exp 1: Balanced 3-Phase Power System
Pre-Lab Question1: Draw the circuit diagram of a 3-phase Y-connected voltage source
clearly labelling all phase and line voltages. Also, write down the mathematical formulas
that express the relationship between phase voltages, and phase and line voltages. Are
the line voltages greater in magnitude or phase voltages?
Solution:
Figure 3 Three phase star connected source
Line voltage: is the voltage measured between the any two-line segments, line voltage
are voltage between two live conductors.
Phase voltage: voltage measured between live and the neutral link of the three-phase
supply system.
V L=¿ √3 V ph ¿
For star connected circuit and for the relation between line and phase current given as
I line=¿ I phase ¿
During laboratory experiment we have recorded
Table 1 Observation line and phase voltage
Van 243 Volts
Vbn 242Volts
Vcn 239Volts
Vab 419Volts
And from mathematical relationship line voltage is 1.73 times the phase voltage so line
voltage is more than the phase voltage

Exp 1: Balanced 3-Phase Power System
Pre-Lab Question2: Write down the relationship between the peak and RMS values of
a voltage. What are the peak and RMS values of the mains voltage in the Australia?
Solution:
The relationship between the peak and RMS values of a voltage is
V pk=¿ √ 2 V RMS ¿
The main voltage in Australia is 230 V at 50 Hz frequency therefore the RMS value of
mains voltage in Australia is 230 V
Thus, the peak value of mains voltage in Australia is
V pk=¿ √ 2 V RMS ¿
V pk =¿ √ 2(230)¿
V pk=¿325.27 V ¿
Pre-Lab Question 3: For the following Δ (delta) connected load, if the phase
impedance is 18 + 21j Ω/Φ, what would be the phase impedance of the equivalent
Y (wye) connected load. Draw the circuit diagram of the Y (wye) connected load.
Figure 4 Delta connected impedance
Solution:
RMS voltage is given as
V rms=¿V RMS ¿
Peak voltage is given as
V pk=¿ √ 2 V RMS ¿
The delta network can be converted into star

Exp 1: Balanced 3-Phase Power System
Figure 5 Star conversion
Given
z1=z2=z3=18+ j21
z A =zB=zC= z1 z2
z1+ z2 + z3
= Z
3
z A +zB + zC =18+ j 21
z A = Z
3 =(18+ j 21)
3
z A =6+ j 7
Similarly
zB =6+ j7
And
zc=6+ j7

Exp 1: Balanced 3-Phase Power System
Pre-Lab Question 4: What is the neutral current in a 3-phase 4 Wire Balanced
System? And why?
Solution:
The neutral current is zero in three phase Star configuration in balanced load while it gives some
value of current in unbalanced load. In a Star 4 wire system the neutral, in theory, carries only
the difference of current flow between the 3 phase conductors. So, when the current in phase A,
B and C are equal, then there is no current flow in the neutral. When current flow in phase A
only increases, then the neutral current will equal that increase. However, this is true only for
purely resistive (linear) loads. Non-linear loads cause harmonics, some of which will add up on
the neutral. Because when you add all the currents together at the neutral they add up to 0 if they
are balanced. If balanced current was 10A/ phase then add 10cos(0) + 10cos(-120)+10cos(120) =
0 for a 3 phase system. If it's not balanced then add them all together and there will be some
current at some angle left over
Pre-Lab Question 5: What is the per-unit system? How are the actual circuit quantities
(voltages, currents, impedances, etc) converted to their per-unit values? If the 3-Ф
apparent power (|S|) rating of a 3-phase Y-to-Y connected load is 5 kVA and line
voltage is 415 Volts, what would be the base impedance?
Solution:
The per unit system simplifies the analysis of complex power system by choosing a
common set of base parameters in terms of which all system quantities are defined
The definition of any quantity (voltage, current, impedance) in per unit system is given
as
Quantity ( per unit ) = quanity (normal unit )
Base value of quantity (normal units )
The per unit quanitites are defined as
Voltage: the ratio of actual voltage to the base voltage is called the per unit voltage
Vpu= V
Vbase
Where V-actual voltage
Vbase- base voltage or rated voltage
Current: the ratio of actual current to base current is called p.u. current

Exp 1: Balanced 3-Phase Power System
Ipu= I
Ibase
Similarly ration of actual impedance to the base impedance is called per unit
impedance
Zpu= Z
Zbase
Here for given problem
For the single-phase system
Zbase=V base
2
Sbase
And for three phase system
Zbase= V base
Ibase × √3 =V base
2
Sbase
And for current
I base= V base
Zbase × √ 3
Sbase= √ 3 x V base Ibase
Sbase=5 kVA 3 phase star connected system
V base=415 V
Base impedance
Zbase=V base
2
Sbase
Zbase= 4152
5000 =34.445 ohm

1) The Preliminary Work
Solution:
Measurement Results When X = XL (Inductive Load)
Quantity Value
XU = XV = XW 0.5pu=0.5pu=0.5pu
RU = RV = RW 0.5pu=0.5pu=0.5pu
Van 243 Volts
Vbn 242Volts
Vcn 239Volts
Vab 419Volts
ILine-a 6.7Amps
Ineutral 339Amps
PФ 780Watts
Cos θ 0.48
• Switch the load bank off by using on-off switch on the load bank.
• Change the position of the 3-way switch to point to the XC setting
• Re-energies the RLC load bank by turning on the on-off switch
• Redo the measurements and record the values of the circuit quantities in the table
provided below
Measurement Results When X = XC (Capacitive Load)
Quantity Value
XU = XV = XW 0.5pu=0.5pu=0.5pu
RU = RV = RW 0.5pu=0.5pu=0.5pu
Van 243 Volts
Vbn 242 Volts
Vcn 240 Volts
Vab 419 Volts
ILine-a 6Amps
Ineutral 1.9 Amps
PФ 775 Watts
Cos θ -0.5
2) Measurements recorded during the experiment including the nameplate
information in tabulated form
Solution:

Exp 1: Balanced 3-Phase Power System
Nameplate Information of the RLC Load Bank
Model
Apparent Power (|S|) 3 kVA
Voltage Rating 415 V
Current Rating 7 A
Frequency 50Hz
Power factor 0.1-0.98
3) The calculation of the XU = XV = XW and RU = RV = RW actual values, i.e. the
conversion from the per-unit values.
Solution:
% X= IX
V
0.5=7 xX
415
X =29.64 ohm
% R= IR
V
R=29.64 ohm
4) The calculation of the per-phase load impedance of the Y-connected load (Refer to
Class Example 2) for both the X = XL and X = XC cases. How does ZY for X =
XL differ from ZY for X = XC and why?
Solution:
Phase current: I1p = I1L, I2p = I2L, I3p = I3L
Line current: IL = I1L = I2L = I3L
Phase voltage:
Line voltage: VL = V12 = V23 = V31
For X = XL

Exp 1: Balanced 3-Phase Power System
The per phase impedance remains the same in start connected circuit no conversion is
needed so for the both cases it can be written as
Z=R+ jX
5) The computation of the per-phase load impedance of an equivalent Δ-connected
load
Solution:
The equivalent impedance is given as
z A =zB=zC= z1 z2
z1+ z2 + z3
= Z
3

Exp 1: Balanced 3-Phase Power System
As its balance load circuit all impedance are same so the per phase equivalent impedance is
give as
z A =zB=zC=9.88+ j 9.88 ohm
6) From the measurements, determine the mathematical relationship between the
magnitudes of the line voltage and line-to-neutral voltage for phase a. Does it
agree with the theoretical principles?
Solution :
V L=¿ √3 V ph ¿
According to readings
V an=243 V
V bn=242 V
V cn=¿ 239V ¿
So phase line voltage
V ab=¿ √3 x243 ¿
V ab=¿ 420.39¿
Theoretically it is 420.39 V and practically 419 V due to some measurement or
reading errors but its almost same as theoretical calculations.
7) Taking the phase-a voltage (Van) as the reference, calculate the magnitude and
phase angle of the neutral current. Taking Van as the reference means that the
phase angle of Van will be taken as 0◌. Does the magnitude of the calculated
current match with the measured value?

Exp 1: Balanced 3-Phase Power System
References:
[1] Louis, M.M., 2014. Elements of Electrical Engineering. PHI Learning Pvt. Ltd..
[2] Dorf, R.C., 2018. Pocket book of electrical engineering formulas. CRC Press.
[3] Bird, J., 2017. Electrical circuit theory and technology. Routledge.
[4] Mayergoyz, I.D. and Lawson, W., 1997. Basic electric circuit theory: a one-semester text. Gulf
Professional Publishing.