Download PWMSP_delia2

Dimensioning and Protection
of PV Plants with examples
of design
Session 2
Dr. Francesco De Lia - ENEA
C.R. ENEA Casaccia
Outlook
• Module Datasheets
• Electrical mismatch in PV array
• Electrical conduits sizing on DC and AC sides
• PV Plant protection against over currents on DC side
• Thermal sizing of electrical switchboards
• Lightning protection of PV plants
• Description of PV Plant design
Ing. Francesco De Lia - ENEA
Calculation of Current Carrying
Capacity of the cables
IZ
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Current carrying capacity (Iz)
Current carrying capacity IZ is defined as the maximum current which can be carried
continuously by a conductor under specified conditions without its steady-state
temperature exceeding a specified value (Tmax).
the maximum temperatures allowed are:
Tmax=70 °C for cables with PVC insulation
Tmax=90 °C for cables with EPR insulation
the Current carrying capacity IZ is obtained by using the formula (installation not
buried in the Ground):
IZ=IZ0 * K1 * K2
where:
IZ0 is the current carrying capacity of the single conductor at 30 °C reference ambient
temperature
k1 is the correction factor if the ambient temperature is other than 30 °C
k2 is the correction factor for cables installed bunched or in layers or for cables installed in a layer on
several supports.
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Not buried in the ground
Table 1 - Current carrying capacity (IZ0) at Tamb=30 °C
•The table1 provides
Izo for installation
method indicated in
the picture.
•There are a lot of
other tables that
provide Izo,
depending on the
specific installation
method
For more information:
Ing. Francesco De Lia - ENEA
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
Dimensioning of electrical conduits
Not buried in the ground
Correction factor K1 for air temperature other than 30 °C
For more information:
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Current carrying capacity: the factor K2
For more information:
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Installation buried in the ground
The current carrying capacity Iz of a cable buried in the
ground, is calculated by using this formula:
IZ=IZ0 * K1 * K2 * K3
where:
IZ0
is the current carrying capacity of the single conductor for
installation in the ground at 20°C reference temperature
K1 is the correction factor if the temperature of the ground is other
than 20°C;
K2 is the correction factor for adjacent cables
K3 = is the correction factor if the soil thermal resistivity is different
from the reference value 2.5 K·m/W
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Installation buried in the ground
Table 1 - Current carrying capacity (IZ0) at Tground=20 °C
Ing. Francesco De Lia - ENEA
For more information:
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
Dimensioning of electrical conduits
Installation buried in the ground
Correction factors k1
For more information:
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
Ing. Francesco De Lia - ENEA
Dimensioning of electrical conduits
Installation buried in the ground
Correction factors k2
Ing. Francesco De Lia - ENEA
For more information:
http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf
PV Plant protection against
overcurrents on DC side
Ing. Francesco De Lia - ENEA
Protection against overcurrents on DC side
Fault on string cable
String
boxes
•Hypothesis: Short Circuit in A
• from
•B to A flows an electric
current equal: (m-1)*Isc
•C to B flows an electric
current equal: (m-n)*Isc
•These currents could damage
the cables
•ALL STRINGS supply their short
circuit current to the A point.
•Single inverter: the parallel of
the strings can be indifferently
made in String Boxes or in the
Inverter switchboard.
Ing. Francesco De Lia - ENEA
Protection against overcurrents on DC side
Fault on the cable that connects the StringBox to the inverter
String
boxes
•Hypothesis: Short Circuit in A
• from
•B to A flows an electric
current equal: n*Isc
•C to A flows an electric
current equal: (m-n)*Isc
•The current from C to A could
damage the cable.
•The current from B to A could
NOT damage the cable.
•ALL STRINGS supply their short
circuit current to the A point.
•Single inverter: the parallel of
the strings can be indifferently
made in String Boxes or in the
Ing. Francesco
De Lia - ENEA
Inverter
switchboard.
Protection against overcurrents on DC side
Use of fuses
String
boxes
Fuse for cable string protection
Fuse for protection the cable
that connects String box with
the inverter.
•SC in A is sensed
string/inverter fuses
by
•SC in B are NOT sensed by
string/inverter fuses. However,
this event has low-probability to
occur because it would occur
inside the switchboard
Inverter
switchboard
Ing. Francesco De Lia - ENEA
Protection Devices against overcurrent
on DC side
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
On the DC side of PV-Plants we can found several
protecion devices:
fuses
circuit breakers
block diodes
……..
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
The fuses
for a fuse, we define following parameters:
In: is the rated current
If: is the current ensuring effective operation in
the conventional time of the fuse (1h/2h/3h/4h).
1h
Inf: is the current ensuring NO effective
operation in the conventional time of the fuse
In
Inf
If
Conventional
Time
In ≤ 63 A
1.25 In
1.6 In
1h
63 A < In ≤ 160 A
1.25 In
1.6 In
2h
160 A < In ≤ 400 A
1.25 In
1.6 In
3h
400 A < In
1.25 In
1.6 In
4h
If=3.2 A
In=2A
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
The fuses
The fuses dissipate power
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
Circuit breakers
Also for a circuit breakers, we define following parameters:
In: is the rated current
If: is the current ensuring effective operation of the circuit breaker
in the conventional time (1h/2h)
Inf: is the current ensuring NO effective operation of the fuse in
the conventional time
Circuit breakers
adjustable
No
protective
adjustable
devices
protective
(CEI 17-5)
devices
(CEI 23-3/1)
In ≤ 63A
In > 63A
Ir ≤ 63A
Ir > 63A
-
Inf
If
Conventional
time
1.13 In
1.13 In
1.05 Ir
1.05 Ir
1.45 In
1.45 In
1.3 Ir
1.3 Ir
1h
2h
1h
2h
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
Circuit breakers
Also the circuit breakers dissipate power
the manufacturers provide PD,pole
Dissipate Power per pole (W)
1.5…6.4
1.8 – 7.2
Ing. Francesco De Lia - ENEA
Thermal sizing of DC switchboards
Device derating
An example of circuit breaker derating
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
Block Diodes
Block diodes are traditionally used in Stand Alone PV Plants
In Stand Alone PV-Plants, during the sunset, without block
diodes, the current would flow into the PV-module instead of
into the loads because of the presence of batteries
Diodes block are installed in series of strings
Sometimes, in grid-connected PV Plants, diodes block are
installed because, in presence of shadowing phenomena's on
PV array, one or more PV string could absorb current from
the others PV string.
Ing. Francesco De Lia - ENEA
Protection devices against overcurrent
on DC side
Why Diodes block are used
Shadowing phenomenon on one string
I
PARALLEL
of two
Parallelo delle
strings
stringhe
NO SHADOWED
Stringa
1000 W/m2 string
(1000 w/m2)
=
SHADOWED
string
2
Stringa
200 W/m
(200 w/m2)
~
IU
U
IT = absorbed current
by the shadowed string
in OPEN CIRCUIT
conditions (I=0)
S
V
IT
T Francesco De Lia - ENEA
Ing.
Protection of electrical feeders
Ing. Francesco De Lia - ENEA
Protection of electrical feeders
If in an DC cables can flow current greater than their current carrying
capacity (IZ ), the cable must be protected by an proper protective
device.
In order to protect the cable, must be satisfied the following formulas:
IB  IN  IZ
; I f  1.45*I Z
where:
IB : is the current for which the circuit is designed (for a string
cable: IB = 1.25*Isc @STC)
IN : is the rated current of the protective device; for adjustable
protective devices, IN is the set current
If :is the current ensuring effective operation in the conventional
time of the protective device
IZ : is the current carrying capacity
Ing. Francesco De Lia - ENEA
Protection of electrical feeders
How-to choice the protective devices on DC side
Choice of fuses
if we choose Fuses for protection of electrical feeders, they are “gG type”
(that are fuses used for cable protection). Moreover:
The fuses must be compliant with DC installations
must be satisfied the formula:
Vn > 1.2* Voc
Where:
Vn
Voc
1.2
: is the rated voltage of the fuse
: is the PV-array voltage in open circuit condition
: is a security factor
Ing. Francesco De Lia - ENEA
Protection of electrical feeders
How-to choice the protective devices on DC side
Choice of Block Diodes
• If shadowing phenomenon on the PV array can occur, must be
evaluate the possibility of installing block diodes. If so, the diodes
must:
• Preferably be low drop voltage
• Have max reverse voltage greater than 2*Voc,@STC
• Have rated current (Id) greater than 1.25*Isc@STC.
• Be careful: block diode dissipate power (PD = VD*ID), consequently
heat sink are required.
Ing. Francesco De Lia - ENEA
Protection of electrical feeders
How-to choice the protective devices on DC side
• Be careful:
in order to reduce loss power, choose the low-voltage version
Ing. Francesco De Lia - ENEA
Thank you for attention…..
Ing. Francesco De Lia - ENEA