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

Calculation of Current Carrying Capacity of the cables IZ

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.

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: 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

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

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

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

Dimensioning of electrical conduits Installation buried in the ground Correction factors k2 For more information: http://www05.abb.com/global/scot/scot209.nsf/veritydisplay/ae2141fea4bfa9d748257a700024a579/$file/1SDC010002D0206.pdf

PV Plant protection against overcurrents on DC side

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

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

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

Protection Devices against overcurrent on DC side

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 ……..

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

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

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

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

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

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.

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) = ~

Protection of electrical feeders

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 = 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

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

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.

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

Thank you for attention…..