PV System Design and Installation

LO 8 – PV Electrical Design

Electrical Design (8% of test questions)

Task/Skill

8.1. Determine series/parallel PV array arrangement based on module and inverter specifications

8.2. Select BOS components appropriate for specific system requirements

8.3. Determine voltage drop between major components

Step 1

Determine inverter “constraints”

a) Maximum DC Input Voltage

b) MPPT Range

PV Array String Sizing

Step 2

a) For maximum number of modules in a string use the “Coldest Record Temperature”

b) For minimum number of modules in a string need Average Monthly High Temperature plus 30 degree F

Avg High

Tempurature

Record Low Tempurature

Determine OperatingTemperature Conditions

Temperature conversion formulas

Tc = (5/9)*(Tf-32)

Tf = (9/5)*Tc+32

Tc = temperature in degrees Celsius

Tf = temperature in degrees Fahrenheit

Step 3

Apply NEC Voltage Temperature Correction Factor to Open Circuit Voltage (Voc) of PV Module

PV module string size must not exceed maximum DC input voltage (usually 500V or 600V).

In Albany, Record Low Temp = -22 F, therefore must use a temperature correction factor of 1.25

Vmax string = Voc x Temp x Modules in String Correction Factor

Reference Class Book by Dunlop

Determine Maximum String Size

Step 4

Determine Minimum String Size

Apply Manufacturer’s Temp Correction Factor to Maximum Power Point Voltage (Vmp)

Make sure that Vmp (temp corrected) stays within MPPT tracking range of inverter.

For example, temperature correction factor for SunPower Modules is -0.1368 V

Vmin = Vmp + (High Temp – 25 C) x Modules in String

Where,

High Temp = Average High Temperature + 17 CReference Class Book by Dunlop

Determine Minimum String Size

String 1

String 2

String 3

Project: ____________________________________ 24 Modules x 210 W = 5.04 kW PV System (grid-tied)

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

Junction Box

Junction Box

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

Junction Box

Positive and negative conductors on roof will be USE-2. Equipment grounding conductor will be #6 bare copper.

Source Circuit (NEC 690.7, 690.8 and 690.9)

Imp = ____________________A Isc = ______________A

Max. Current = Isc x 1.25 = ______________ A

Sizing of Conductors and Overcurrent Devices

Max. Current x 1.25 = _______________ A

Vmp = _________ x ___________modules in string = ________________VDC

Voc = __________ x __________modules in string = ______________VDC

Max. PV system voltage = Voc x 1.25 = ________ VDC

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

SunPower 210WImp=5.25 A

Vmp=40 VDCIsc= 5.75 A

Voc= 47.7 VDC

+-

5.25 5.75

7.19

8.98

40 8 320

47.7 8 382

478

Step 5

Power = Series String Voltage (Vmp) x Parallel String Current (Imp)

Where,

Parallel String Current = Number of Strings in Parallel

Determine Number of Strings necessary to achieve PV array capacity requirements (kW)

Choose inverter

In general, PV array DC input (watts) can be up to 1.2 x Inverter AC output (watts) rating in Upstate New York

Max Number < I max DC input of inverterof strings I max string current

For example, using SunPower SPRm 5000 inverter, using 230W module (Imp = 5.61 amps)

Max Number < 21 amps = 3.74 stringsof string 5.61 amps

PV Array SizeMaximum (3 strings x 8 modules) = 24 x 230W = 5.52 kW

On-line inverter configuration tools

Wire Sizes

Look at wire layout board in HVCC lab

Voltage Drop (volts) = Current (amps) X Resistance (ohms) V = I x R

% Voltage Drop = 0.2 x D x I x R

V

D= One-way distance in feet

I= Current (use Imp for solar circuits)

R= Resistance of conductor (Ohms/1,000 feet). From NEC Chapter 9, Table 8

V= Voltage (use Vmp x the # of modules)

Voltage Drop Calculations

Typically design for less than 1 % voltage drop

NEC Requirements

Maximum from Service to the Load of 5%

Maximum from final overcurrent protection device to the load of 3%