PV Specifications

[siddhartha]

Hello everyone.

I am very new to Solar Tech and was hoping someone could explain a few things about solar PV module and battery to me.

My query is... I have the following specifications with me...

"PV Module 12 Wp under STC, measured at 16.4V as load. Module V(oc) minimum of 21V.

Sealed maintenence free battery with 12V-7AH @C/20 with Max DoD=75%"

Avg daily insolation =5.5kWh/sq m."

Can anyone explain to me, as how much power I can draw from the PV.. and .. I intend to use the following specs to make LED luminaire which can be used for an avg of 12 hours, so what can be my system wattage.

I will be very grateful if anyone could demystify this for me.

Kind regards,
Sid

[Naptown]

about a 3 watt bulb.

[Mike90250]

Chose your location, and caculate your harvest - see this thread

http://www.solarpaneltalk.com/showthread.php?4147-Sunlight-power-day-length-expected-harvest-PV-Watts

Then read the posts here:

http://www.solarpaneltalk.com/showthread.php?5334-Solar-Off-Grid-Battery-Design

&

http://www.solarpaneltalk.com/showthread.php?2144-How-Many-Batteries-amp-Why-Tutorial

[Mike90250]

"PV Module 12 Wp under STC, measured at 16.4V as load. Module V(oc) minimum of 21V.

12watt panel - not very large

Sealed maintenence free battery with 12V-7AH @C/20 with Max DoD=75%

12v 7ah battery - not very large Avoid regular discharge below 70% of capacity.

Avg daily insolation =5.5kWh/sq m."

That's the only good news, but you need to look up the shortest days, and plan your system around that.

[siddhartha]

Thanks

about a 3 watt bulb.

Thanks Naptown.

But could you also explain how you arrived at that??

[siddhartha]

Thanks MIke

12watt panel - not very large



12v 7ah battery - not very large Avoid regular discharge below 70% of capacity.


That's the only good news, but you need to look up the shortest days, and plan your system around that.

Thanks for the Help mike. Will surely go thru the links u posted and learn more abt it.

Kind regards,
Sid

[Naptown]

Thanks Naptown.

But could you also explain how you arrived at that??

Well you have a 12 watt panel
You will only get half of this into the battery
so you will get about 6 watts for say the 5.5 hours a day ( I assume you live in the tropics in the desert so insolation is steady year round.
So you can put 33 watt hours into the battery per day
You want to operate something for 12 hours a day
Therefore 33/12 = 3
This gives you 0 days of autonomy
If you want 1 day double the panel wattage
2 days triple it.
You have a 12 volt 7 AH battery
This at total discharge would be 84 watt hours
One night of use would take 36 Watt hours given 3 watts on for 12 hours.
That is close to a 50% DOD
To stay at 30% DOD you should reduce the bulb size to 2 watts. or even better to 1.5W

[RCinFLA]

The insolation is the equivalent number of hours of full 1000 watts/square meter of illumination, usually for a south (or north for southern hemisphere) pointed direction at tilt equivalent to latitude. It is NOT the hours between sunrise and sunset. It changes for seasons.

Earth's axis is tilted at 23.45 degrees that account for seasons. In northern hemisphere the sun is lower to horizon during winter with less hours of sunlight the farther north latitude you go. For a fixed south mount (again, ref. to N. hemisphere) you can gain some equivalent hours of insolation with a simple tilt adjustment that you change 2 to 4 times a year based on season. Less tilt in summer, more tilt in winter.

Some data sources have insolation numbers for various tilt angles and some average fudging for local weather, like historical rainy season. The data won't account for any particular site obstructions like nearby mountains or tree shading.

Here is a site with U.S. data.

http://rredc.nrel.gov/solar/pubs/redbook/

Even if you are not in the U.S. the number will be close for an equivalent latitude anywhere in the world, with exception for local weather, like a monsoon season where it rains every day.

[siddhartha]

Please check my calculations.

Hello All...

Can anyone go thru my calculation and tell me if I am wrong/right

I have a 12 V, 7AH @C/20 battery, with maximum Depth of Discharge=75%.

I have to use this battery to make a system run for atleast 14 hours.

Hence, max current I can draw= 7AH/14H = 0.5A.? Correct??

Max Wh from battery= 12*7= 84 Wh.

With Max DoD of 75%, Max Wh that I can draw= 84 * 0.75= 63Wh.

To maintain autonomy of 14 hours (over 3 days time), the maximum rating of the system can be= 63Wh/14h= 4.5 W.

Is the calculation correct???

Thanks in advance.

[Sunking]

You are going about it wrong, and your logic is flawed.

a 12 volt 7 AH battery does have a theoretical capacity of 12 volts x 7 AH = 84 watt hours.

First a lead acid battery should never be discharge below 50% DOD. 2nd is you design for no more than 20% DOD per day. So for a 12 volt 7 AH battery with 3 real days of autonomy with no more than 50% DOD the most you can draw per day on a 12 volt 7 ah battery is 84 wh / 6 days = 14 watt hours per day, or 7 ah / 6 = 1.16 AH per day.

[siddhartha]

@Sunking

You are going about it wrong, and your logic is flawed.

a 12 volt 7 AH battery does have a theoretical capacity of 12 volts x 7 AH = 84 watt hours.

First a lead acid battery should never be discharge below 50% DOD. 2nd is you design for no more than 20% DOD per day. So for a 12 volt 7 AH battery with 3 real days of autonomy with no more than 50% DOD the most you can draw per day on a 12 volt 7 ah battery is 84 wh / 6 days = 14 watt hours per day, or 7 ah / 6 = 1.16 AH per day.

Would u approve of the following calculation??

Battery is 12V-7Ah, Max DoD= 75%. (but as u say shud keep it around 50%)..

I need to find out the system wattage which I can feed from the battery.

Assuming the system wattage = X.
Daily use= 5 hrs.
Hence, daily Wh= 5X Wh.

for 3 days autonomy, 50% DoD, I must design for 6 days right??

Hence, (5X * 6)/12= 7 ( as per a post from Sunkin on battery amp hrs),

Hence X= 2.8 W, correct???

Or have I gone wrong again?

Also, I think is this the wrong place to ask abt Batteries?? If so, moderators please correct me. Will shift this to Battery related forum on the same site.

Kind regards,
Sid

[Sunking]

for 3 days autonomy, 50% DoD, I must design for 6 days right??

Yes sir correct.

Hence, (5X * 6)/12= 7 ( as per a post from Sunkin on battery amp hrs),

Hence X= 2.8 W, correct???

The math is correct for solving for X