Notes from a session held at the Luminous Green Hands-On Workshop, 2nd of May 2007

1/ Calculate AC equipment loads in Watt hours per week

Example: Computer (500W)+ Projector (2000W) + Sensors AC-Loads: 500W x hrs/week (e.g.56) = 2 800W/wk + 13 200W/wk = 16 000Wh/wk

2/ Inverter Loss - Energy needs of the Invertor (12V) ⇒ multiply by 1,2

⇒ 16000W/wk x1.2 = 19 200 W/wk

3/ Calculate Amps/h load per week: Divide Wh/Wk corrected by inverter input voltage (usually 12V)

Ω Law: (V.A=W) –> 19200W/Wk / 12V = 1600Amph/Wk

4/ Calculate DC equipment loads in Watt hours per week /

5/ Divide DCWh/Wk by DC system voltage (12,24,48 or whatever it is) to get Amp hours per week load /

6/ Get total Amp hours per week load by adding 3 + 5

1600Ah/Wk + 0

7/ Divide by 7 to get amp hours per day

1600Amph/Wk /7days = 228,5Amph/day

8/ Days of storage needed

Multiply Amph/Day with the days of storage needed to get total system Amph storage needed. Besides the choice of machines, this determines the project cost. Hybrid systems are better than Solar only. Our example → 5 days capacity: ⇒ 228,5Amph/day x 5 = 1140 A/h

9/ Take into account a battery discharge limit factor

Divide Total system Amp hours needed with the discharge limit of batteries (0.5 for 50%, it can be from 0.2 to 0.8, depending on the batteries used) Deep cycle batteries like the one we use (car battery) do not like to go below 50% Factor 0,5 compensation (in this case). ⇒ Capacity 1140Ah/0,5= 2280A/h If you make an installation work that is not up and running all the time, batteries go dead.

10/ Multiply the TOTAL System Amph Corrected with the winter multiplier

Reserve storage capacity, depends on temperature and hours of sun

26,7° → 1,00 21,2° → 1,04 15,6° → 1,11 10,0° → 1,19 4,4° → 1,30 -6,7° → 1,59 Note: 1,11 = experience based “magic number” ⇒ 2280Amph/day x 1,11 = 2530Amph

11/ Get number of batteries needed to be connected in parallel

Divide the total system (winter corrected) amperage needed, with the amp hours rating of your batteries. 2530Amph / 75Amph ⇒ 34 batteries (of 12V)

12/ Get number of batteries wired in series needed

System Voltage (12,24,48…) / Battery Voltage= number of batteries wired in series =1

13/ Get total number of batteries needed

Multiply baterries parallel and batteries in series. =34

14/Correct the TOTALAmph/Day needed for battery loss, factor 1.2

228,5Amph/day x 1,2= 274,2 Amph/day

15/ Determine the Solar panel size

Calculate the total solar panel array amps needed for your system TOTAL Amph/Day Battery loss corrected / Average Sun Hours per day = 274,2 Amph/day / 15 (hours of sun/day) = 18,28Amph

16/ Get the total parallel number of modules needed

Divide your total solar array Amps with the Peak amps produced by each module. You calculate your Peak amps if you divide the module Wattage with the peak power point voltage (see the specs of your modules, PeakAmps/panel). Round off to the highest whole number.

Our Kyocera panel: 2,33Amp ⇒ 18,28Amph/2,33Amp= 8 panels

17/ Determine the number of modules in each series string needed to supply necessary DC battery Voltage

``` DC Battery Voltage 	 	Number of Modules
in each series string
48V 	   	-->    		4
24V    	   	-->    		2
12V	   	-->    		1```

The higher Voltage the more efficient (Ω Law) → you want to have lower Amperage. The higher the current, the higher heat and the more expensive the cables will be. Batteries put in parallel increase the Amperage; Heat will built up ⇒ Serial = better, although sometimes you need more amperage ⇒ combine sytems, put many serial banks in parallel f.i.

18/ Determine the total number of modules needed

Multiply the total number of modules if wired in parallel by the multiplier from the chart above according to battery voltage. 8 panels x 4 (12V) = 32

19/ Get the MINIMUM Amp for the solar charger

Multiply the peak amps by module by the number of modules take Peak Amps per panel: 2,33Amps → x 8modules = 18,64Amps Minimum

20/ Calculate your inverter power rating

Multiply your AC simultaneous loads and possible surges from electric motors and multiply by 1.2 for inverter loss to get total inverter power rating.

2500W continuous. Take into account that a stepper motor or kinetic solutions take a lot of energy for a short time. Therefore, add 1000W for the motor used in the installation in this example. ⇒ + 1000W = 3500W x 1,2= 4200

System set up hands on workshop

Components: Solar panel, Charger, Battery (simple car battery), Invertor ac-dc: (uses power, so for mini systems to avoid (or DC-DC Convertor))

Set up On the panel backside is the terminal, watch for blue (+) and brown (-) and connect to the charger Brown (+) Blue (-)

INSTALL RULE OF THUMB –> 90° > Sun In function/Inclination/Inclinometer > 51° Lattitude

Measure the power voltage

Connect to the battery

How it functions: light absorbed during the day, charging. At night particles go back, but charge has to be released.

solar

• luminous/solar_power_plant.txt
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