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1 Off-grid PV technologies Irradia2on Series and parallel config Cable calc. Charge controller DC Disconnect Volt drop Inverter Voltage Calcula2on s Ba=eries P r Moun2ng Structures

PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

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Page 1: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

1

Off-gridPVtechnologies Irradia2on

Seriesandparallelconfig

Cablecalc.

Chargecontroller

DCDisconnect

Voltdrop

InverterVoltageCalcula2ons

Ba=eriesPr

Moun2ngStructures

Page 2: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Systemsizingwhenstorageisincluded

Page 3: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

3

Step1-Calculateconsump<on

Step1.b)DailyConsump<oninWhStep1.a)PowerRequirementinW

Page 4: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

• LeadAcidSystemLossesvarybetween±22to30%• Li-ionSystemLossesvarybetween±18to22%•  hMp://www.baMerysizingcalculator.com

4

Step2-BaMerySizing

Inverters

5

5

2

15

Inverter

Chargecontroller

Cabling

System(BaMeries+connec<ons)

Page 5: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required

Step 2.b) Divide Daily storage by DC voltage = 4003Wh / 48V = 83Ah of Average Daily Ah needed

Step 2.c) Adjust to depth of discharge = 83Ah / 0,5 = 166Ah Sub Total of storage required

5

BaMerySizing–LeadAcid

Thisvaluerepresents27%Losses

The“48V”valueisfoundinthedatasheetStep1.ashowedusweneed414WofPower.Atleastan800Wto1000Winverterwouldbe

required.Forthiscalc.itisassumedthecustomerwouldwanttoincreasehissystemover<me.FromthedatasheetPageB1,wewillseethe5000Winverterusesa48VbaMery/baMerybank.

Use0,5for50%D.o.D,0,4for40%D.o.Dandsoon.....

Page 6: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Step 2.d) Make provision for rainy days =166Ah x 2 days = 332Ah Total Storage required

The battery bank needs to be 48V 332Ah according to our calculations.

Step 2.e) Select battery to match both 48V and 332Ah as close as possible.

6

BaMerySizing–LeadAcid

Thisvalueisjustaguessandwouldvarybasedonthenumberofrainydaysforthatarea

12VoltBaMery200Ah4baMeriesinseriesto

reach48V200Ah2stringsin

paralleltoreach

400A

h

Page 7: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

BaMerySizing–LeadAcid

•  Step2.f)ChecktoensurethatthechosenbaMerycanhandletherateofdischarge(c-ra<ng)

=400Ah@C20=400AhataCapacityof20hrs=400Ah/20hrs=20A

ThechosenbaMerycanhandlearateof20Aforaperiodof20hoursandthenitwillbecompletelydischarged.

WedonotwanttodischargethebaMerycompletelyandideallyshouldsizeaccordingtolessthanhalfofthenumberofhourswheretheD.o.Disintendedtobe50%

Power =VoltagexCurrent =48Vx20A =960W 960WiswhatthebaMerybankcanhandleinPOWERdischargerate

Page 8: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

8

Dischargerate

FromStep2f.–ThebaMerybankcanhandle960W.Atmaxwewillonlydischargeat414W

Page 9: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

9

Conductorcurrentcarryingcalcula<on

5000WInverterI = __PV

TV

400Ah48VoltBaMeryBank

Page 10: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Stepscon<nuedStep 3.a) Calculate Power required to charge batteries

Power = Volts x Amps - Where: Volts = The battery’s absorption voltage Amps = Maximum Current the battery bank can handle

This information will be found in your inverter datasheet or charge controller datasheet or charge controller rating – Depending on what is being used to charge the batteries.

The rate of charge from the charge controller and AC charger has to be balanced with rate at which the battery can accept charge. See the next slide. . . . .

Page 11: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

BaMerybankchargerate

•  BaMerybankchargerate

•  =20%ofC20•  =20%oftheCapacityofthebaMeryat20hrs•  =20%of400Ah•  =80A

400Ah48VoltBaMeryBank

From Page B5 we can see the battery can handle a maximum charge rate of 20% at C20.

Thisisourmaximumchargerateaswellasthecurrentvaluerequiredforourcalcula<on

Page 12: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Choosemethodofcharge

•  ACCharger•  Chargecontroller•  Othermeansofcharging

Page 13: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

BaMerybankchargeVoltage

•  Power(WaMs) =Volts x I(Amps)•  =57,6Vx80Amps

Thevoltagevaluealsocomesfromthedatasheet.PageB5

•  Power(WaMs) =Volts x I(Amps)•  =57,6V x 80Amps•  =4608WaMs

This 4608 Watt value is the absolute maximum power required for the battery bank.

Page 14: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Module/PanelSelec<on

•  Step 3.b) Select modules to match –  Power Requirement, –  Roof space –  Ease of handling –  And stock availability

From page B2 the 60 Cell 260W module is selected

Page 15: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

•  Objec<ve:Calculatethemodule’ssummerandwintervoltages

•  SummerVoltageisthelowestpossiblevoltageatthehighesttemperature

•  WinterVoltageisthehighestpossiblevoltageatthelowesttemperature

Tempco-efficientcalcula<ons

Winter Summer

ModuleVoltagesareHigh

Temperaturesarelow

ModuleVoltagesareLow

Temperaturesarehigh

Page 16: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

ModuleTempcoefficientcalc

•  Weneedtoknowwhattheminimumandmaximumvoltagesforthemoduleisgoingtobe.ForthisstepweusefactorsfromtheSANS10142-1-2standard.

•  Summervoltagefactorof0,75•  Wintervoltagefactorof1,15

•  SummerMinVoltage

•  30,9Vx0,75•  23VmpMin

•  WinterMaxVoltage

•  30,9Vx1,15•  =36VmpMax

Page 17: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

•  -0,45%/℃260W60cell

•  Effoftemp=30,9Vmpx0,45%/℃

•  =0,139V

17

Temperatureco-efficientcalcs

23Vmp

80℃25℃-15℃STC

Mul<plyco-efficientfactor

30,9Vmp

36Vmp

MaximumVoltage(VDC-max)=VmpMax_STC(SolarPV)x1.15=VmpMin_STC*0.75

Mul<plyco-efficientfactor

Page 18: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

•  MicrocarevsVictronrecommendedchart

•  Pleasecheckothermanufacturerspecifica<ons

18

Step4-SelectChargeController

Page 19: PV technologies Series and parallel config · Step 2.a) Multiply daily consumption with losses (as a factor) = 3152Wh x 1,27 = 4003Wh Daily Storage required Step 2.b) Divide Daily

Stepscon<nued

Step 4.a) Module layout and design is done according to charge controller power, voltage and current limitations. (Group Exercise)