1. Dimensioning and Protection of PV Plants with examples of design
Session 2
Dr. Francesco De Lia - ENEA
C.R. ENEA Casaccia

2. 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

3. Calculation of Current Carrying Capacity of the cablesIZ

4. 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.

5. 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:

6. Dimensioning of electrical conduits
Not buried in the ground
Correction factor K1 for air temperature other than 30 °C
For more information:

7. Dimensioning of electrical conduits
Current carrying capacity: the factor K2
For more information:

8. IZ=IZ0 * K1 * K2 * K3 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

9. Dimensioning of electrical conduits
Installation buried in the ground
Table 1 - Current carrying capacity (IZ0) at Tground=20 °C
For more information:

10. Dimensioning of electrical conduits
Installation buried in the ground
Correction factors k1
For more information:

11. Dimensioning of electrical conduits
Installation buried in the ground
Correction factors k2
For more information:

12. PV Plant protection against overcurrents on DC side

13. Protection against overcurrents on DC side
Fault on string cable
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.
String
boxes

14. Protection against overcurrents on DC side
Fault on the cable that connects the StringBox to the inverter
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 Inverter switchboard.
String
boxes

15. Protection against overcurrents on DC side
Use of fuses
String
boxes
Fuse for cable string protection
SC in A is sensed by string/inverter fuses
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
Fuse for protection the cable that connects String box with the inverter.
Inverter
switchboard

16. Protection Devices against overcurrent
on DC side

17. 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
……..

18. 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).
Inf: is the current ensuring NO effective operation in the conventional time of the fuse 1h In
Inf If
Conventional Time
In ≤ 63 A
1.25 In
1.6 In 1h
63 A < In ≤ 160 A 2h
160 A < In ≤ 400 A 3h
400 A < In 4h
If=3.2 A
In=2A

19. Protection devices against overcurrent on DC side
The fuses
The fuses dissipate power

20. adjustable protective devices No adjustable protective devices
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
Inf If
Conventional time
adjustable protective devices
(CEI 17-5)
No adjustable protective devices
(CEI 23-3/1) -
In ≤ 63A
1.13 In
1.45 In 1h
In > 63A 2h
Ir ≤ 63A
1.05 Ir
1.3 Ir
Ir > 63A

21. 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

22. An example of circuit breaker derating
Thermal sizing of DC switchboards
Device derating
An example of circuit breaker derating

23. 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.

24. Protection devices against overcurrent on DC side
Why Diodes block are used
Shadowing phenomenon on one string
IT = absorbed current by the shadowed string in OPEN CIRCUIT conditions (I=0)
PARALLEL of two strings
NO SHADOWED string (1000 w/m2)
SHADOWED string (200 w/m2) = ~

25. Protection of electrical feeders

26. 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:
where:
IB : is the current for which the circuit is designed (for a string cable: IB =
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

27. 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 : is the rated voltage of the fuse
Voc : is the PV-array voltage in open circuit condition
1.2 : is a security factor

28. 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
Have rated current (Id) greater than
Be careful: block diode dissipate power (PD = VD*ID), consequently heat sink are required.

29. 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

30. Thank you for attention…..