PV System Design and Installation LO 8 – PV Electrical Design.
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PV System Design and Installation LO 8 – PV Electrical Design
Transcript of PV System Design and Installation LO 8 – PV Electrical Design.
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
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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 inverter
of 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 strings
of string 5.61 amps
PV Array Size
Maximum
(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%
