Lab_1_Voltage_Sags_and_Harmonics
docx
keyboard_arrow_up
School
British Columbia Institute of Technology *
*We aren’t endorsed by this school
Course
3130
Subject
Electrical Engineering
Date
Apr 3, 2024
Type
docx
Pages
12
Uploaded by SuperHumanRiver3417
Lab 1 Voltage Sags and Harmonics Objectives
To Understand the fundamentals of voltage sags and harmonics, and their impacts on power systems
Part A – Voltage Sags Analysis Introductions Part A of the report explains the impact of voltage sags on the mills, and the reason their equipment has been tripped offline. In this file, I Have attached a Single line diagram as in Figure 1.
Single line diagram Figure 1Final Single Line Diagram
Using Load Flow Analysis Figure 2 Load Flow Analysis SLD
Power in W
Q in Var
BCH-WSN
28543
28166
T1
2285
-1267
T2
29984
24401
Generator
32000
17175
Figure 3 Results of Load Flow Analysis
As we can see in Figure 3 the results matched the required values of
a.
BC Hydro Williston Substation b.
Transformers T1, T2, and Generator
Short Circuit Analysis test in this test we be testing faults across a.
Remote Bus b.
BCC bus c.
CRD bus d.
Neighboring Plant bus e.
Load flow analysis with Neighboring-XFMR while it is energized Bus Voltage (% of nominal)
Bolted Fault or Disturbance Location
Bus A
Bus B
MCC-A
MCC-B
Remote Fault
96.7%
93.89%
96.66%
93.89%
BCC66 Bus
72.12%
57.85%
72.12%
57.85%
CRD66 Bus
66.76%
49.39%
66.76%
49.97%
Neighboring Bus
94.52%
90.76%
94.52%
90.76%
Transformer energization
67.27%
54.88%
67.27%
54.88%
Figure 4 Results of Short Circuit Test
Figure 5 Example of short circuit Test on Neighboring bus
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Figure 6 Fault at Remote bus
Figure 7Fault at BCC Bus
Figure 8 Fault at CRD Bus
Figure 9 Load Analysis with Load energized.
Discussion and analysis ANS: As the MCC Bus voltage drops, the Control voltage also drops. 2. Review the provided contactor technical specification (Eaton) and determine its drop-out voltage. When the contactor drops out, it means that the motor will stop, and the associated process will stop.
ANS: As in Page 3 table 2: Ratings the drop out voltage is 60% Rated Coil Voltage and the rated
voltage is 7.2kV. Drop-out voltage = 0.6 * 7.2kV = 4.32kV
3. The motors in MCC A and B drive critical steam processes and are controlled by these Eaton contactors. Based on the determined drop-out voltage, determine whether process interruption is expected
for faults at various locations. Highlight the results in the table in red if interruption to operation is likely.
ANS: Yellow Highlight = will drop out, Red Highlight = will not drop out Bus Voltage (% of nominal)
Bolted Fault or Disturbance Location
Bus A
Bus B
MCC-A
MCC-B
Remote Fault
96.7%
93.89%
96.66%
93.89%
BCC66 Bus
72.12%
57.85%
72.12%
57.85%
CRD66 Bus
66.76%
49.39%
66.76%
49.97%
Neighboring Bus
94.52%
90.76%
94.52%
90.76%
Transformer energization
67.27%
54.88%
67.27%
54.88%
4. Repeat #2 but for transformer energization in-rush at a neighboring facility. (Results from Step4)
ANS: Done in previous Question.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
5. The contactor-in-question can ride-through a voltage sag down to its drop-out voltage for 3 cycles. Assuming theBC Hydro breakers at Canreed and Williston carry a 3-cycle opening time, does your conclusion change in #3? Why or why not?
ANS: no it does not, as the minimum trip time is 1.5 cycles found in the datasheet. So since 3 cycles is more than 1.5, it will pick up and drop out any voltages above 60% 6.What is the reason behind differences in magnitude of voltage sags?
ANS: because it depends on fault location, system impedance and how long the fault is.
7.Is there any difference in voltage sags between Bus A and B for faults at the same location? Why?
ANS: yes because Bus A has a generator connected to it which causes the differences. 8.What other equipment in the mill may be sensitive to voltage sags? (Hints: recall what you’ve learnt from the very first lecture of Elex 3402)
ANS: VFDs and electronic and industrial equipment. 9.What conclusions can be made?
a. What are the most likely causes of interruption to the mill process?
b. Which bus/buses is/are more vulnerable and why?
c. Without modifying the mill’s or BC Hydro’s power system, is there anything that the Mill can do to avoid the interruptions as identified above?
ANS: After analysing the results I found that Remote Bus, Neighboring bus had an average bus voltage
of (90%-95%) while the BCC66 and CRD66 bus had an average bus voltage of (60%-70%) of Nominal. a.
The most likely causes of interruption to the mill process could include power outages, voltage sags or surges, equipment failures or malfunctions, and human error. These interruptions could be caused by factors such as severe weather, equipment maintenance or repair, load changes, or operator error.
b.
The BCC66 and CRD66 and the Transformer Energization Busses because they are more
prone to voltage sag.
c.
Scheduled maintenance plan for the whole system as well as monitoring systems that detect faults and trips.
Part B – Preliminary Harmonic Analysis
Introduction This part of the report will include an estimate to see if the utility requirements based on the IEEE 519 Standard are met based on the preliminary design.
Single line Diagram Figure 10 Processing Plant SLD
Load flow Current on Main Cable Motor FLA = HP
∗
746
√
(
3
)
∗
Vl
∗
n
∗
PF
=
193.26
Total Current = 210.52 * 2 = 386.52
Fault current and Current Ratio Calculation Zbase ¿
Vbase
2
s
=
0.36
Ω
Z = Zpu * Zbase = 0.36 * 0.0575 = 0.0207
Isc = V/Z = 600/ 0.0207 = 28.986 kA Ratio = Isc/ Total current = 28986/386.52 = 74.99 Voltage Distortion Limits Since the voltage is 600V which is less than 1kv we use individual harmonic = 5% Total Harmonic Distortion THD = 8% Current Distortion Limits Since the ratio is 74.99 which is around 75 we use Current Harmonic Spectrum Calculations Using a rated FLA of 386.52 and TDD of 12% so the fundamental = Total Current*
%fundamental current( 1 – TDD) Figure 11 Harmonic Current
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Cable & Transformer Impedances W = 2*pi*f , L= XL/2*pi*f = 0.239mH/Km , X= W*h*L Sample reactance calculation for 1
st
order harmonic = 377*1*0.239 / 1000 = 0.0901 H / Km
Z = sqrt(R^2 + X^2)
LT= 0.1503 mH
Harmonic Voltage
VH= IH * ( Zcable + Z transformer)
Items which don’t comply with Alabama Power’s requirements (IEEE 519)
The items that don’t comply will be highlighted in red
Questions and Analysis
14. If only one of the VFDs is online, does it change the voltage and current TDD and THD at the MCC? Please record the changes, if applicable, and describe whether it changes the
status of compliance of the MCC bus.
It does not change, since the TDD and THD are ratios of Voltage and current, so if it is only 1 the voltage and current values would be halved but the ratios would not change. 15. If the plant decides to add an equivalent amount (in terms of Amps) of motor loads, which doesn’t carry any harmonics, does it change your conclusion in #3? Why/Why not?
Yes it would cause a smaller percentage, because the ratio is divided by a larger number.
16. What is the effect on voltage and harmonic current distortion (%) by using a longer or shorter cable?
Shorter cable THD will Increase because there is less inductance, with a longer cable it have a decreased THD with more inductance. 17. What is the effect on voltage and harmonic current distortion (%) by using a larger or smaller transformer impedance (%Z)?
a larger transformer impedance can result in a low frequency harmonic and an increase in
high frequency harmonics. A smaller transformer impedance will have the opposite effect. 18. Why does IEEE 519 uses TDD for current distortion and not THD
Because TDD is more accurate for measuring Harmonic Distortion cause by PF< 1.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Related Documents
Related Questions
Q2\ Two generators G1 & G2 are rated 15 MVA, 11 kV and 10 MVA, 11 kV respectively. The
generators are connected to a transformer as shown in the following figure. Find the sub-
transient current in each generator when a three-phase fault occurs on the high voltage side of
the transformer.
A/Y
15 MVA, 11 kV, X"gl=j0.10 pu
G2
15 MVA, 11/66 kV,
Xr-j0.06 pu
10 MVA, 11 kV, X"g2-j0.10 pu
arrow_forward
Please solve the problem i want urjent
arrow_forward
A magnetic circuit has a m.m.f. of 400
AT and a reluctance of 2 x 105 AT/Wb.
The magnetic flux in the magnetic circuit
is_
mWb.
6
X
arrow_forward
a core type transforem is constructed with 0.014-in. thick laminations
having a uniform width of 2.75 in. If the maximum flux and flux density are,
respectively, 6.2 X 10' maxwells and 8.2 X 10 maxwells per square inch and the
air spaces between laminations occupy 8 per cent of the stacked core, calculate
the number of laminations in the transformer.
I
arrow_forward
10. In a transformer,
the leakage flux of each wind-
ing is proportional to the current in that winding
because
(a) Ohm's law applies to magnetic circuits
(b) leakage paths do not saturate
(c) the two windings are electrically isolated
(d) mutual flux is confined to the core.
11. In a two-winding transformer, the e.m.f.per turn
in secondary winding is always.........the induced
e.m.f. power turn in primary.
(a) equal to K times
(b) equal to 1/K times
(c) equal to
(d) greater than.
12. In relation to a transformer, the ratio 20: 1 indi-
cates that
(a) there are 20 turns on primary one turn on
secondary
(b) secondary voltage is 1/20th of primary
voltage
(c) primary current is 20 times greater than the
secondary current.
(d) for every 20 turns on primary, there is one
turn on secondary.
arrow_forward
A transformer has primary and secondary turns of 1250 and 125 respectively. It has core cross-sectionof 36 cm2 and its rms flux density is to be limited to 1.4 T (to prevent core saturation). Whatmaximum 50 Hz voltage can be applied on the primary side and the corresponding open-circuitsecondary voltage?The core has a mean length of 150 cm and its relative permeability can be assumed to be 8000. Whatwould be the rms exciting current when the transformer’s primary winding is excited at a voltage ascalculated above? Also calculate the magnetizing susceptance as seen from primary and secondarysides.If the transformer were to be excited at 60 Hz, 1. What should be the maximum primary voltage for thecore flux density limit not to be exceeded? 2. What would be the magnetizing susceptance as seen oneach side in this case?
arrow_forward
All subparts
arrow_forward
Q.4/ choose true or false:
1. The reluctance of any core is proportional directly with the length of it and
inversely with both the permeability and the cross-section area of it.
2. The number of parallel paths in wave windings is equal to twice of the number
of poles.
3. In dc machines with same values of Z, N, Ø, the emf for wave winding type is
more than emf of lab winding type.
4. If we insert a current in such a conductor subjected to a constant magnetic
field, a movement will achieved giving the principle of electric generator.
arrow_forward
What is phase angle of voltage induced in primary and secondary windings of transformers due to mutial flux?
arrow_forward
(a) different number of turns
(b) same size of copper wire
(c) a common magnetic circuit
(d) separate magnetic circuits.
10. In a transformer, the leakage flux of each wind-
ing is proportional to the current in that winding
because
(a) Ohm's law applies to magnetic circuits
(b) leakage paths do not saturate
(c) the two windings are electrically isolated
(d) mutual flux is confined to the core.
11. In a two-winding transformer, the e.m.f.per turn
in secondary winding is always.........the induced
e.m.f. power turn in primary.
(a) equal to K times
(b) equal to 1/K times
(c) equal to
(d) greater than.
12. In relation to a transformer, the ratio 20: 1 indi-
cates that
(a) there are 20 turns on primary one turn on
secondary
(b) secondary voltage is 1/20th of primary
voltage
(c) primary current is 20 times greater than the
secondary current.
(d) for every 20 turns on primary, there is one
turn on secondary.
13. In performing the short circuit test of a trans-
former
(a) high voltage side is…
arrow_forward
I need an experienced tutor to solve this, avoid using Ai
arrow_forward
Solve all with type solution. plz don't copy
arrow_forward
SEE MORE QUESTIONS
Recommended textbooks for you

Power System Analysis and Design (MindTap Course ...
Electrical Engineering
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:Cengage Learning
Related Questions
- Q2\ Two generators G1 & G2 are rated 15 MVA, 11 kV and 10 MVA, 11 kV respectively. The generators are connected to a transformer as shown in the following figure. Find the sub- transient current in each generator when a three-phase fault occurs on the high voltage side of the transformer. A/Y 15 MVA, 11 kV, X"gl=j0.10 pu G2 15 MVA, 11/66 kV, Xr-j0.06 pu 10 MVA, 11 kV, X"g2-j0.10 puarrow_forwardPlease solve the problem i want urjentarrow_forwardA magnetic circuit has a m.m.f. of 400 AT and a reluctance of 2 x 105 AT/Wb. The magnetic flux in the magnetic circuit is_ mWb. 6 Xarrow_forward
- a core type transforem is constructed with 0.014-in. thick laminations having a uniform width of 2.75 in. If the maximum flux and flux density are, respectively, 6.2 X 10' maxwells and 8.2 X 10 maxwells per square inch and the air spaces between laminations occupy 8 per cent of the stacked core, calculate the number of laminations in the transformer. Iarrow_forward10. In a transformer, the leakage flux of each wind- ing is proportional to the current in that winding because (a) Ohm's law applies to magnetic circuits (b) leakage paths do not saturate (c) the two windings are electrically isolated (d) mutual flux is confined to the core. 11. In a two-winding transformer, the e.m.f.per turn in secondary winding is always.........the induced e.m.f. power turn in primary. (a) equal to K times (b) equal to 1/K times (c) equal to (d) greater than. 12. In relation to a transformer, the ratio 20: 1 indi- cates that (a) there are 20 turns on primary one turn on secondary (b) secondary voltage is 1/20th of primary voltage (c) primary current is 20 times greater than the secondary current. (d) for every 20 turns on primary, there is one turn on secondary.arrow_forwardA transformer has primary and secondary turns of 1250 and 125 respectively. It has core cross-sectionof 36 cm2 and its rms flux density is to be limited to 1.4 T (to prevent core saturation). Whatmaximum 50 Hz voltage can be applied on the primary side and the corresponding open-circuitsecondary voltage?The core has a mean length of 150 cm and its relative permeability can be assumed to be 8000. Whatwould be the rms exciting current when the transformer’s primary winding is excited at a voltage ascalculated above? Also calculate the magnetizing susceptance as seen from primary and secondarysides.If the transformer were to be excited at 60 Hz, 1. What should be the maximum primary voltage for thecore flux density limit not to be exceeded? 2. What would be the magnetizing susceptance as seen oneach side in this case?arrow_forward
- All subpartsarrow_forwardQ.4/ choose true or false: 1. The reluctance of any core is proportional directly with the length of it and inversely with both the permeability and the cross-section area of it. 2. The number of parallel paths in wave windings is equal to twice of the number of poles. 3. In dc machines with same values of Z, N, Ø, the emf for wave winding type is more than emf of lab winding type. 4. If we insert a current in such a conductor subjected to a constant magnetic field, a movement will achieved giving the principle of electric generator.arrow_forwardWhat is phase angle of voltage induced in primary and secondary windings of transformers due to mutial flux?arrow_forward
- (a) different number of turns (b) same size of copper wire (c) a common magnetic circuit (d) separate magnetic circuits. 10. In a transformer, the leakage flux of each wind- ing is proportional to the current in that winding because (a) Ohm's law applies to magnetic circuits (b) leakage paths do not saturate (c) the two windings are electrically isolated (d) mutual flux is confined to the core. 11. In a two-winding transformer, the e.m.f.per turn in secondary winding is always.........the induced e.m.f. power turn in primary. (a) equal to K times (b) equal to 1/K times (c) equal to (d) greater than. 12. In relation to a transformer, the ratio 20: 1 indi- cates that (a) there are 20 turns on primary one turn on secondary (b) secondary voltage is 1/20th of primary voltage (c) primary current is 20 times greater than the secondary current. (d) for every 20 turns on primary, there is one turn on secondary. 13. In performing the short circuit test of a trans- former (a) high voltage side is…arrow_forwardI need an experienced tutor to solve this, avoid using Aiarrow_forwardSolve all with type solution. plz don't copyarrow_forward
arrow_back_ios
arrow_forward_ios
Recommended textbooks for you
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning

Power System Analysis and Design (MindTap Course ...
Electrical Engineering
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:Cengage Learning