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.
The power is generated, transmitted, and distributed in 3-phase (3-Φ) form. The 3-Phase AC supply system has the following advantages:
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 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 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.
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.
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?
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.
For star connected circuit and for the relation between line and phase current given as
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
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
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
Pre-Lab Question 3: For the following Δ (delta) connected load, if the phase impedance is 18 + 21j ?/Φ, what
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-? 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
The
Voltage: the ratio of actual voltage to the base voltage is called the per unit voltage
Vbase- base voltage or rated voltage
Current: the ratio of actual current to base current is called p.u. current
Similarly ration of actual impedance to the base impedance is called per unit impedance
Here for given problem
For the single-phase system
And for three phase system
And for current
Base impedance
|
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 |
|
P? |
780Watts |
|
Cos θ |
0.48 |
|
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 |
|
P? |
775 Watts |
|
Cos θ |
-0.5 |
1) The Preliminary Work
2) Measurements recorded during the experiment including the nameplate information in tabulated form
|
Nameplate Information of the RLC Load Bank |
|
Model |
|
Apparent Power (|S|) |
|
Voltage Rating |
|
Current Rating |
|
Frequency |
|
Power factor |
3) The calculation of the XU = XV = XW and RU = RV = RW actual values, i.e. the conversion from the per-unit values.
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?
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
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
5) The computation of the per-phase load impedance of an equivalent Δ-connected load
The equivalent impedance is given as
As its balance load circuit all impedance are same so the per phase equivalent impedance is give as
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?
According to readings
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.
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