Assessment of Plug-in Electric Vehicles Charging on Distribution Networks
Master Thesis Defense - Tsz Kin (Marco) Au
Committee Chair:
- Dr. M. Ortega-Vazquez
Committee Co-Chair:
- Dr. M. El-Sharkawi
Committee Member:
- Dr. D. Kirschen
Assessment of Plug-in Electric Vehicles Charging on Distribution - - PowerPoint PPT Presentation
Assessment of Plug-in Electric Vehicles Charging on Distribution Networks Master Thesis Defense - Tsz Kin (Marco) Au Committee Chair: Dr. M. Ortega-Vazquez Committee Co-Chair: Dr. M. El-Sharkawi Committee Member: Dr. D. Kirschen
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– Charging time – Battery state of charge (SOC) – Charging method – Customer load variation
6/5/2012 Electrical Engineering Department - University of Washington Master Thesis Defense - Tsz Kin (Marco) Au 7 Read data and initialize parameters Generate random scenarios Run deterministic system
– Battery capacity – Energy consumption per unit distance
– Residential area – Commercial area
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Type 1: Charge at home only Type 2: Charge at home and work
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6/5/2012 Electrical Engineering Department - University of Washington Master Thesis Defense - Tsz Kin (Marco) Au 13 Commute distance (miles) Percentage (%) 0 – 4.0 19.19 4.1 – 8.0 22.95 8.1 – 12.0 16.67 12.1 – 16.0 13.77 16.1 – 20.6 9.37 20.1 – 24.0 6.07 24.1 – 28.0 4.59 28.1 – 32.0 2.69 32.1 + 4.70
5 10 15 20 25 Percentage (%) Commute Distance (Mile)
Commute Distance Distribution
y = 353.04x5 - 725.13x4 + 526.87x3 - 140.15x2 + 22.691x - 0.0038 R² = 0.9997
5 10 15 20 25 30 35 0.2 0.4 0.6 0.8 1 Commute Distance (Mile) Probability
Quantile Function of Commute Distance
# 𝑝𝑔 𝑄𝐹𝑊 𝑗
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2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 2 4 6 8 Hour Power (kW) 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 2 4 6 8 Hour Power (kW)
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𝑔 𝑄
𝑐𝑣𝑡,𝑢𝑗 =
1 𝜏𝑐𝑣𝑡,𝑢𝑗 2𝜌 𝑓
−1 2∙(𝑄𝑐𝑣𝑡,𝑢𝑗−𝐵𝑤𝑄𝑐𝑣𝑡,𝑢𝑗) 𝜏𝑐𝑣𝑡,𝑢𝑗
2
𝐵𝑤𝑄𝑐𝑣𝑡,𝑢𝑗 = 𝑄𝑢𝑧𝑞𝑓,𝑢𝑗
𝑜𝑝𝑠𝑛 × 𝐵𝑤𝑄𝑐𝑣𝑡
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1 2 3 S2 S3 3000’ 4000’ 7200V
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1 2 3 S2 S3 3000’ 4000’ 7200V
Forward sweep:
1) Assume voltage at node 3 is 7200V 2) Compute 𝐽3 𝐽3 = 𝑡3 𝑊
3 ∗
= 143.0∠ − 29.0 𝐵
𝑊
3 = 7200 𝑊
𝐽23 3) Compute 𝐽23 𝐽23 = 𝐽3 = 143.0∠ − 29.0 𝐵 𝐽3 4) Compute 𝑊
2
𝑊
2 = 𝑊 3 + 𝑎23 ∙ 𝐽23 = 7260.1∠0.23 𝑊
5) Compute 𝐽2 𝐽2 = 𝑡2 𝑊
2 ∗
= 231.0∠ − 26.3 𝐵 6) Compute 𝐽12 𝐽12 = 𝐽23 + 𝐽2 = 373.9∠ − 27.3 𝐵 7) Compute 𝑊
1
𝑊
1 = 𝑊 2 + 𝑎12 ∙ 𝐽12 = 7376.2∠0.97 𝑊
8) Compute mismatch between 𝑊
1and 𝑊 𝑡
𝑁𝑗𝑡𝑛𝑏𝑢𝑑𝑖 = 𝑊
𝑡 − 𝑊 1
= 176.2 𝑊 𝐽12 𝐽2
𝑊
2 = 7260∠0.23 𝑊
𝑊
1 = 7376.2∠0.97 𝑊
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1 2 3 S2 S3 3000’ 4000’ 7200V
Backward sweep:
1) Assume voltage at node 1 is 7200V, and use the line currents computed from forward sweep 2) Compute 𝑊
2
𝑊
2 = 𝑊 1 − 𝑎12 ∙ 𝐽12 = 7085.4∠ − 0.68 𝑊
𝑊
3 = 7026.0∠ − 1.02 𝑊
𝐽23 𝐽12
𝑊
2 = 7085.4∠ − 0.68 𝑊
𝑊
1 = 7200 𝑊
3) Compute 𝑊
3
𝑊
3 = 𝑊 2 − 𝑎23 ∙ 𝐽23 = 7026.0∠ − 1.02 𝑊
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1 2 3 S2 S3 3000’ 4000’ 7200V
Perform forward sweep again:
1) Use the voltage at node 3 from the backward sweep 2) Compute 𝐽3 𝐽3 = 𝑡3 𝑊
3 ∗
= 146.5∠ − 30.1 𝐵
𝑊
3 = 7026.0∠ − 1.02 𝑊
𝐽23 3) Compute 𝐽23 𝐽23 = 𝐽3 = 146.5∠ − 30.1 𝐵 𝐽3 4) Compute 𝑊
2
𝑊
2 = 𝑊 3 + 𝑎23 ∙ 𝐽23 = 7087.6∠ − 1.02 𝑊
5) Compute 𝐽2 𝐽2 = 𝑡2 𝑊
2 ∗
= 236.6∠ − 27.2 𝐵 6) Compute 𝐽12 𝐽12 = 𝐽23 + 𝐽2 = 383.0∠ − 28.3 𝐵 7) Compute 𝑊
1
𝑊
1 = 𝑊 2 + 𝑎12 ∙ 𝐽12 = 7206.5∠0.0 𝑊
8) Compute mismatch between 𝑊
1and 𝑊 𝑡
𝑁𝑗𝑡𝑛𝑏𝑢𝑑𝑖 = 𝑊
𝑡 − 𝑊 1
= 6.535 𝑊 𝐽12 𝐽2
𝑊
2 = 7087.6∠ − 1.02 𝑊
𝑊
1 = 7206.5∠0.0 𝑊
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= residential area = 85 households = commercial area = 1 small shopping plaza = 81.6 + 40.8j (kW+kVar) = 100 + 75j (kW+kVar)
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Type 1
Type 2a (charge at residential area)
Type 2b (charge at commercial area)
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0% Penetration 30% Penetration 50% Penetration 100% Penetration
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