[ppt]electrical machines - mvtafe -...
TRANSCRIPT
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Electrical Machines
LSEGG216A9080V
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TransformerTransformerLosses & EfficiencyLosses & Efficiency
Week 3Week 3
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ObjectiveObjectivess1. Describe the power losses which occur in a transformer
2. Describe the tests which allow the power losses of a transformer to be calculated
3. Calculate transformer losses and efficiency using test results
4. Define the all day efficiency of a transformer5. Calculate the all day efficiency of a transformer
6. Describe the relationship between transformer cooling and rating
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5. Calculate the all day efficiency of a transformer
6. Describe the relationship between transformer cooling and rating
7. Describe the methods of cooling
8. List the properties of transformer oil
9. Describe the tests conducted on transformer oil
Objectives
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Transformer RatingsTransformer RatingsTransformers are rated to supply a given output in
Volt Ampsor
VAat a specified frequency and terminal voltage.
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Transformer RatingsTransformer RatingsThey are NOT rated in Watts
The load power factor is unknown
IVS PFSPower
PFPowerS
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Transformer RatingsTransformer RatingsThey are NOT rated in Watts
The load power factor is unknown
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ProblemV1 = 6,351
VV2 = 230 V
S = 2 kVA
Power output at unity PF ?
P = 2 kVA x 1
P = 2 kW
Power = S x PF
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Problem
PFVSI
V1 = 6,351 V
V2 = 230 V
S = 2 kVA
Full load secondary current at 0.8 PF ?
I = 10.87 A0.8230
2000I
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Student Exercise 1
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PFSPower
V2 = 200 VV1 = 1270 V
S = 20 kVA
0.1000,20 P
P = 20 kW
(a) Power output at unity power factor
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PFSPower
V2 = 200 VV1 = 1270 V
S = 20 kVA
8.0000,20 P
P = 16 kW
(b) Power output at 0.8 power factor
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VxPFSI
V2 = 200 VV1 = 1270 V
S = 20 kVA
I = 100 A
(c) Full-load secondary current at unity power factor
200x1.020,000I
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VxPFSI
V2 = 200 VV1 = 1270 V
S = 20 kVA
I = 62.5 A
(d) Secondary current when transformer supplies 10 kW at 0.8 power factor
200x0.810,000I
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EfficiencyPowerInputPowerOutputη
LossesOutputInput
LossesPowerOutputPowerOutputη
Ratio between Input power and Output Power
PowerInputLossesPowerInputη
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Efficiency
100PowerInputPowerOutputη%
Efficiency is normally expressed as a percentage
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Transformer Efficiency
PowerIn
PowerOut
OvercomeIron
Losses
Overcome Copper Losses
Some Poweris used to:
η = 100%η = 95%η = 90%
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Student Exercise 2
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PFSPower
V1 = 230 V V2 = 32 V
S = 20 kVA
η = 90% PF = 0.85
(a) Power output of transformer
0.85100P
P = 85 W
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InOutη
V1 = 230 V V2 = 32 V
S = 20 kVA
η = 90% PF = 0.85
(b) Power input
P = 94.4 Wη
OutIn
0.9W 85In
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LossesOutIn
V1 = 230 V V2 = 32 V
S = 20 kVA
η = 90% PF = 0.85
(c) Losses
P = 9.4W
Losses8594.4
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Transformer Losses
Copper Losses (Cu)
•Varies with load current
•Produces HEAT •Created by resistance of windings•Short circuit test supplies copper losses
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Short Circuit TestCopper Losses
(Cu)
SecondaryShort Circuited
Limited
Supply Voltage ≈ 5-10 %
Wattmeter indicates Copper Losses (Cu)
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Short Circuit Test
1000.5lossCopper 2
Copper Losses (Cu)•Finds Cooper losses at full load
•Copper losses vary with the square of the load Full load Cu loss = 100
WTransformer loaded at 50%
PCu = 25 W1000.25lossCopper
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0102030405060708090
100110120130140150
0 10 20 30 40 50 60 70 80 90 100 110
Copper Losses (Cu)
% Load
Cu L
osse
s (W
)
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Transformer LossesIron Losses
(Fe)•Fixed•Always present•Related to transformers construction
Eddy CurrentsReduced by laminationsProduces HEAT
HysteresisReduced by using special steels in laminations
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Open Circuit TestFinds Iron Losses
(Fe)
Full Supply Voltage
SecondaryOpen Circuit
Wattmeter indicates Iron Losses (Fe)
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Transformer EfficiencyStudent Exercise 3
100PowerInputPowerOutputη%
LossesCuLoadFullload 2%Cu Losses
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100PowerInputPowerOutputη%
Sout = 30 kVA Fe = 220 WCu FL= 840 W
Calculate η%at Full Load
100Losses OutputPowerOutputη%
100k 0.22k 84k 30k 30η%
.0
η% = 96.6%
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100Losses OutputPowerOutputη%
Sout = 30 kVA Fe = 220 WCu FL= 840 W
Calculate η%at 75%Load
1000.220.75 22.522.5η% 2
84.0
η% = 97%
1000.220.4725 22.522.5η%
5.223075.0 outS 5.47284075.0 2 75%Cu
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100Losses OutputPowerOutputη%
Sout = 30 kVA Fe = 220 WCu FL= 840 W
Calculate η%at 50%Load
1000.220.5 1515η% 2
84.0
η% = 97.21%
1000.220.21 1515η%
15305.0 outS
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100Losses OutputPowerOutputη%
Sout = 30 kVA Fe = 220 WCu FL= 840 W
Calculate η%at 25%Load
1000.220.25 7.57.5η% 2
84.0
η% = 96.49%
1000.220.0525 7.57.5η%
5.73025.0 outS
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100% η = 96.6%75% η = 97%50% η = 97.21%25% η = 96.49%
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0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0 10 20 30 40 50 60 70 80 90 100 11096.00
97.00
% Load
Loss
es (
W)
η%
Cu Losses
Fe Losses
η%
Fe = Cu =Max η
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Maximum Efficiency
Fe = Cu =Max η CuLoadFe 2
2LoadCuFe
LoadCuFe
Load840220
Fe = 220
Cu = 840
Load %= 51.18%
1000.220.5118 300.5118300.5118η% 2
84.0
1000.2215.35η%
22.035.15
η%= 97.21%
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0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0 10 20 30 40 50 60 70 80 90 100 11096.00
97.00
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All Day Efficiency
• Most Transformers are connected permanently
• The time that the transformer has to be calculated when determining efficiency
• Able to determine the best transformer for the application
by its efficiency
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All Day EfficiencyTransformer A
Sout = 300 kVA Fe = 1.25 kVA Cu = 3.75 kVAHours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh
1.00 6.00 5 100 500.0 33.33 0.42 2.08 6.25 8.33 508.336.00 7.00 1 200 200.0 66.67 1.67 1.67 1.25 2.92 202.927.00 8.00 1 300 300.0 100.00 3.75 3.75 1.25 5.00 305.008.00 9.00 1 360 360.0 120.00 5.40 5.40 1.25 6.65 366.659.00 12.00 3 300 900.0 100.00 3.75 11.25 3.75 15.00 915.0012.00 14.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0314.00 18.00 4 300 1200.0 100.00 3.75 15.00 5.00 20.00 1220.0018.00 20.00 2 360 720.0 120.00 5.40 10.80 2.50 13.30 733.3020.00 22.00 2 280 560.0 93.33 3.27 6.53 2.50 9.03 569.0322.00 1.00 3 200 600.0 66.67 1.67 5.00 3.75 8.75 608.75
5900.0 5998.02
98.37
Time Period
P out kWh = P in kWh =
% Eff =
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All Day EfficiencyTransformer B
Sout = 300 kVA Fe = 2.5 kVA Cu = 2.5 kVAHours Load kW kWh % Load Cu Loss Cu kWh Fe kWh Losses kWh Input kWh
1.00 6.00 5 100 500.0 33.33 0.28 1.39 12.50 13.89 513.896.00 7.00 1 200 200.0 66.67 1.11 1.11 2.50 3.61 203.617.00 8.00 1 300 300.0 100.00 2.50 2.50 2.50 5.00 305.008.00 9.00 1 360 360.0 120.00 3.60 3.60 2.50 6.10 366.109.00 12.00 3 300 900.0 100.00 2.50 7.50 7.50 15.00 915.0012.00 14.00 2 280 560.0 93.33 2.18 4.36 5.00 9.36 569.3614.00 18.00 4 300 1200.0 100.00 2.50 10.00 10.00 20.00 1220.0018.00 20.00 2 360 720.0 120.00 3.60 7.20 5.00 12.20 732.2020.00 22.00 2 280 560.0 93.33 2.18 4.36 5.00 9.36 569.3622.00 1.00 3 200 600.0 66.67 1.11 3.33 7.50 10.83 610.83
5900.0 6005.34
98.25% Eff =
Time Period
P out kWh = P in kWh =
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Transformer Cooling• Transformer ratings can be increased if their windings are cooled
by some external means
• The most common cooling mediums are in direct with transformer windings;
and/orAir Oil• The most common methods of circulation are
Forced
and/or Natural
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Transformer Classification• Transformers are allocated symbols which indicate the type of
cooling used• Can consist of up to 4 letters indicating the cooling system
1st Letter 2nd Letter 3rd Letter 4th Letter
The cooling medium in contact with the windings
The cooling medium in contact with the external cooling
system
Kind of Medium
Circulation type
Kind of Medium
Circulation type
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Transformer ClassificationType AN
Dry Transformer withNatural Air Flow
Air Natural
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Transformer ClassificationType AF
Dry Transformer withForced Air Flow
Air Forced
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Transformer ClassificationType ONAN
Oil Tank Cooling Natural Oil Flow - Natural Air Flow
Oil Natural Air Natural
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Transformer ClassificationType ONAF
Oil Tank Cooling Natural Oil Flow - Forced Air Flow
Oil Natural Air Forced
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Transformer ClassificationType OFAF
Oil Tank Cooling Forced Oil Flow – Forced Air Flow
Oil Forced Air Forced
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Transformer Oil
• Low Viscosity• High Flash point• Chemically inert• Good insulator
Acts as Coolant & Insulator
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Transformer Oil Tests• Dielectric Strength• Acidity • Power factor • Interfacial tension • Dissolved Gas
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THE END