PROBABILITY OF ISLANDING IN UTILITY
NETWORK DUE TO GRID CONNECTED
PHOTOVOLTAIC POWER SYSTEMS
Bas Verhoeven
KEMA - the Netherlands
tel +31 26 356 35 81
fax +31 26 443 38 43
[email protected]
Bas Verhoeven
PVPS - Task 5
Workshop 24 - 25 January 2002
Arnhem - the Netherlands
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Contents
•
•
•
•
•
•
•
•
Introduction
Methodology
Residential Area
Test Setup and Measuring System
Determination of Ratio between PV and Load
Method to Determine Balanced Conditions
Balanced Conditions for Active Power
Balanced Conditions for Active and Reactive
Power
• Probability of Balanced Conditions
• Conclusions
Slide : 2
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Main Conclusion
Probability of encountering
an islanding is virtually zero
Let me prove you why
Slide : 3
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Introduction (1)
• Islanding is not allowed as the power quality
for loads in the island cannot be guaranteed
and where safety of maintenance crews is not
assured
• International discussions
–
–
–
Big Problem versus No Problem
High-end Protections versus Simple Protections
‘Personal feeling’ and ‘Intuition’ make the
discussions extremely difficult
• Information on probability and risk of islanding
is not available
Slide : 4
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Introduction (2)
• In 1999 the Netherlands started a research
project to determine the probability of
islanding in a low voltage power network
based on measurements of loads and the
power produced by a PV system
• Research is integrated in the Task V workplan
and other international research programs
• Work was funded by EnergieNed, Novem and
KEMA
Slide : 5
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Methodology (1)
• Residential area with high penetration level of
PV systems is not available
• Alternative method was developed to measure
the load of a distribution transformer and all
outgoing bays plus the power produced by a
PV-system
• Measuring system must ‘capture’ islanding
thus, data samples of load and PV power are
taken every second for one year to include
seasonal variations
Slide : 6
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Methodology (2)
• PV power is measured using a 100 Wp system
• Direct comparison with the load of the network
is therefore not possible
• PV power must be multiplied by a constant
factor before comparison can be made
• What constant factor should we use?
– Look at the available roof surface in the residential
area, or
– Determine from the measured data the ratio between
load and PV power and look for the ratio that occurs
most frequently
Slide : 7
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Methodology (3)
60000
At the crossing possible
islanding conditions.
50000
5-minutes average power
of bay 7
Power in Watt
40000
30000
20000
PV power multiplied by the
maximum correlation factor.
P = Ppv * 740
10000
Time [0 - 24 hrs]
Slide : 8
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16:00
15:00
14:00
13:00
12:00
11:00
10:00
9:00
8:00
7:00
6:00
5:00
4:00
3:00
2:00
1:00
0:00
0
Methodology (4)
• Various penetration levels of PV systems can
be evaluated when using different constant
factors for the multiplication of the PV power
• Comparison can be made for active power
– for indication only
•
Combination of active and reactive power
– The real situation!
Slide : 9
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Residential Area (1)
Residential area is located in ‘Rijkerswoerd’
a suburb south of Arnhem
Slide : 10
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Residential Area (2)
Area is representative as a modern housing
estate, approximately 7 years old
Slide : 11
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Residential Area (3)
Distribution transformer 10 kV / 230 V
Bay
1
Pub. lighting
2
3
4
5
6
7
7 houses
50 houses
78 houses
55 houses
56 houses
246 houses
Distribution cabinet of NUON
Slide : 12
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Test Setup and Measuring System (1)
Bay 1
SINGLE LINE DIAGRAM OF DISTRIBUTION CABINET
Bay 2
= Current probe on every phase
Bay 3
10 kV supply
Bay 4
U
Bay 7
Bay 5
PV system on roof of
the distribution cabinet
Bay 6
T12 transducers for all bays
and PV system
Wind
Transducer
T12 per
phase
Temperature
Slide : 13
Transducer
Transducer
T13 per
T12
phase
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Current
Phase angle
Current
Phase angle
Voltage
Digitizer and
computer data
logging
Test Setup and Measuring System (2)
Slide : 14
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Test Setup and Measuring System (3)
Slide : 15
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Test Setup and Measuring System (4)
• Measured and logged parameters
–
–
–
–
–
–
Red, Yellow and Blue phase voltage
Red, Yellow and Blue phase current per bay
Red, Yellow and Blue phase angle per bay
PV current and phase angle
Speed of wind
Temperature
• 750 MB of data per month = 9 GB per year
• Measuring period from May - 1999 to April 2000
Slide : 16
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Determination of Ratio PV and Load (1)
• We must determine a reasonable factor for the
multiplication of the PV power before the
comparison between load and PV power can
be made
• Calculate ratio for every second in whole year
Pload
Ratio =
Ppv
(for Ppv > 0)
• Plot ratio in frequency distribution chart for the
whole year
Slide : 17
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Determination of Ratio PV and Load (2)
5000
4500
4000
Bay 7 / 246 houses
3500
Seconds
3000
2500
2000
1500
1000
500
0
0
500
1000
1500
2000
2500
Ratio
Slide : 18
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3000
3500
4000
Determination of Ratio PV and Load (3)
12000
11000
Bay 7
10000
9000
Peak value reduces and moves to higer ration
when going from summer to winter
8000
Seconds
7000
6000
5000
4000
3000
2000
1000
0
0
250
500
750
1000
1250
1500
1750
2000
2250
2500
2750
Ratio
May
Slide : 19
Jun
Jul
Aug
Sep
Oct
Nov
Dec
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Jan
Feb
Mar
Apr
3000
Determination of Ratio PV and Load (4)
25000
22500
20000
Bay 6 / 140 houses
17500
Seconds
15000
12500
10000
7500
5000
2500
0
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
Ratio
Slide : 20
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1300
1400
1500
1600
1700
1800
1900
2000
Determination of Ratio PV and Load (5)
Peak value of ratio for the different bays
Slide : 21
Bay
Peak value
at ratio
Houses
to bay
Bay 2
Bay 3
Bay 4
Bay 5
Bay 6
Bay 7
20
110
350
150
140
740
7
50
78
55
56
246
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Determination of Ratio PV and Load (6)
90
Houses per Bay on phase level
80
y = 0.1047x + 0.4861
R2 = 0.9577
70
Linear trend
line fits best!
60
50
Equivalent with
913 Wpeak on
every house
40
30
20
10
0
0
100
200
300
400
500
Peak Ratio in Category per Bay
Slide : 22
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600
700
800
Determination of Ratio PV and Load (7)
5000
4500
Ratio is zero and islanding
cannot occur
4000
3500
This is the minimum load of
the network during daytime
Seconds
3000
2500
2000
1500
1000
500
0
0
500
1000
1500
2000
2500
Ratio
Slide : 23
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3000
3500
4000
Determination of Ratio PV and Load (8)
90
Houses per Bay on phase level
80
y = 0.3806x0.9009
R2 = 0.9505
70
Power trend
line fits best!
60
50
Equivalent with
413 Wpeak on
every house
40
30
20
10
0
0
50
100
150
200
250
First Ratio above zero
Slide : 24
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300
350
400
450
Method to determine balanced conditions (1)
Actual PV power times
the fixed multiplier
Power in Watt
Error bar 5% on
load power
Actual load of
the network
Balanced
condition for
3 seconds
Balanced
condition for
2 seconds
No balanced condition,
just crossing
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Seconds
Slide : 25
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17
18
19
20
21
22
23
24
25
Method to determine balanced conditions (2)
For the comparison we have two variables:
• Multiplier of the PV power to vary the
penetration level of the PV systems
– Base is at the peak of the ratio times an offset to
simulate lower and higher penetration levels
• Error margin for accepting balanced conditions
– This error margin is used to simulate the mismatch of
active power and reactive power that may be present
in an island for which the islanding remains stable
Slide : 26
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Method to determine balanced conditions (3)
5000
1x
4500
0,9x
4000
Selected Offset Multipliers
3500
Seconds
3000
2x
2500
2000
1500
3x
0,75x
1000
500
0
0
500
1000
1500
2000
2500
Ratio
Slide : 27
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3000
3500
4000
Method to determine balanced conditions (4)
Bay
.75x
.9x
1x
2x
3x
2
3
4
5
6
7
15
147
263
113
105
555
18
99
315
135
126
666
20
110
350
150
140
740
40
220
700
300
280
1480
60
330
1050
450
420
2220
• Analyses are made for every bay separately
Slide : 28
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Method to determine balanced conditions (5)
Mult. =
0.75x
Mult. =
0.9x
Mult. =
1x
Mult. =
2x
Mult. =
3x
Slide : 29
P = ±2%
Q = ±2%
v
P = ±5% P = ±10% P = ±15%
Q = ±2% Q = ±5% Q = ±10%
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
v
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Balanced conditions for active power (1)
10000
Bay 2 = 2-2-3 houses
1000
P margin = 5%
Number
100
Red
Offset = 1x
Yellow
Blue
10
1
0.1
0
5
10
15
20
25
30
35
40
45
50
55
Time
Slide : 30
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60
65
70
75
80
85
90
Balanced conditions for active power (2)
10000
Bay 7 = 82-83-81 houses
1000
P margin = 5%
100
Number
Offset = 1x
Red
Yellow
Blue
10
1
0.1
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
Time
Slide : 31
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80
85
90
95
100 105 110 115 120
Balanced conditions for active power (3)
10000
Bay 2
1000
Offset = 1x
100
Number
M = 2%
M = 5%
M = 10%
M = 15%
10
1
0.1
0
10
20
30
40
50
60
70
80
90
100
110
120
Time
Slide : 32
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130
140
150
160
170
180
Balanced conditions for active power (4)
10000
Bay 7
1000
Offset = 1x
100
Number
M = 2%
M = 5%
M = 10%
M = 15%
10
1
0.1
0
10
20
30
40
50
60
70
80
90
100 110 120 130 140 150 160 170 180 190 200 210 220 230 240
Time
Slide : 33
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Balanced conditions for active power (5)
100000
Bay 2
10000
Margin = 5%
1000
Number
Off=0.75x
Off=0.9x
Off=1x
100
Off=2x
Off=3x
10
1
0.1
0
5
10
15
20
25
30
35
40
45
50
55
60
Time
Slide : 34
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65
70
75
80
85
90
Balanced conditions for active power (6)
10000
Bay 7
1000
Margin = 5%
100
Off=0.75x
Number
Off=0.9x
Off=1x
Off=2x
Off=3x
10
1
0.1
0
10
20
30
40
50
60
70
80
90
100
110
120
Time
Slide : 35
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130
140
150
160
170
180
Balanced conditions for P and Q (1)
• In power networks balanced conditions for
active and reactive power has to be present
• The reactive power consumption of the
network is fairly constant
• The PV inverter must produce the reactive
power as required by the load
• Balanced conditions for both active and
reactive power occur very seldom
Slide : 36
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Balanced conditions for P and Q (2)
1000
Bay 2
Offset = 1x
100
Number
M = 2/2%
M = 5/2%
10
M = 10/5%
M = 15/10%
1
0.1
0
5
10
15
20
25
30
35
40
Time
Slide : 37
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45
50
55
60
Balanced conditions for P and Q (3)
1000
Bay 7
Offset = 1x
100
Number
M = 2/2%
M = 5/2%
10
M = 10/5%
M = 15/10%
1
0.1
0
5
10
15
20
Time
Slide : 38
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25
30
Balanced conditions for P and Q (4)
1000
Bay 2
Margin = 5 / 2%
100
Number
Off=0.75x
Off=0.9x
Off=1x
10
Off=2x
Off=3x
1
0.1
0
1
2
3
4
5
6
7
8
9
10
Time
Slide : 39
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11
12
13
14
15
Balanced conditions for P and Q (5)
1000
Bay 7
Margin = 5 / 2%
100
Number
Off=0.75x
Off=0.9x
Off=1x
10
Off=2x
Off=3x
1
0.1
0
1
2
3
4
5
6
7
8
9
10
Time
Slide : 40
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11
12
13
14
15
Probability of balanced conditions (1)
• Probability of islanding is defined as:
∑ balanced condition x time
P=
Relevant seconds in year
• When assuming the relevant number of
seconds as from 8.00 to 18.00 hour, we obtain
for the year 13,14E6 seconds
Slide : 41
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Probability of balanced conditions (2)
10000.0
Offset = 1
Probability [1E-6]
1000.0
100.0
10.0
1.0
0.1
Bay 2
Bay 3
Bay 4
M = 2/2%
Slide : 42
M = 5/2%
Bay 5
M = 10/5%
M = 15/10%
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Bay 6
Bay 7
Probability of balanced conditions (3)
1000.00
Margin = 5 / 2 %
Probability [1e-6]
100.00
10.00
1.00
0.10
Bay 2
Bay 3
Bay 4
Off=0.75x
Slide : 43
Off=0.9x
Bay 5
Off=1x
Off=2x
Off=3x
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Bay 6
Bay 7
Conclusions (1)
• Islanding does never occur up to a certain level
of PV power
– For Dutch situation this is 413 Wp per house
– For loads with air-conditioners this will be
significantly higher
• The maximum PV power that may be installed
in a power network for which islanding does
not occur is estimated at three times the
minimum night load in that power network
Slide : 44
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Conclusions (2)
• The allowable mismatch between load and PV
power significantly determines the number of
balanced conditions
• Once above a threshold, the number of
balanced conditions is not significantly
influenced by PV penetration level
• Inverters should be fixed to operate at unity
power factor
• A low voltage power network should not or
only partially be compensated for its reactive
power consumption
Slide : 45
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Conclusions (3)
The probability of balanced conditions in a
certain part of a network is in the range
of 1E-6 to 1E-3 per year
x
The probability of that part of the network
being disconnected with balanced conditions
present and the protection is not working well
is very small
x
Probability of encountering an islanding
is virtually zero
Slide : 46
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