1
\$\begingroup\$

I have an equipment which draws 550W at a constant level, no peaks. I want to power this unit with solar cells only during the day when there is light using a simple car battery. Because I don't want to invest in a expensive battery I put it down to run only during the day. How can I calculate the exact cell power I need and the controller. Battery is 63Ah/12V, would it be enough ?

edit: Sorry, forgot to mention: 1. device run on 220V and i'll use power converter from 12 to 220. 2. I'll run the equipment as long as there is sunlight, summer hours are average is 10hrs a day winter average is 7hrs Thanks,

\$\endgroup\$
5
  • \$\begingroup\$ What kind of power does your device need? Can it run from 12Volts, or does it need AC mains power? \$\endgroup\$
    – JRE
    Commented Jul 24, 2015 at 15:17
  • \$\begingroup\$ How long are you running the equipment for? 12 hrs? 20 hrs? During the day is vague, we have no idea what region of the world you are in, how much sunlight per day? Need more details to help answer your question. \$\endgroup\$
    – Adam
    Commented Jul 24, 2015 at 15:20
  • \$\begingroup\$ Sorry, forgot to mention: 1. device run on 220V and i'll use power convertor from 12 to 220. 2. I'll run the equipment as long as there is sunlight, summer hours are average is 10hrs a day winter average is 7hrs \$\endgroup\$
    – Hristo Damianov
    Commented Jul 24, 2015 at 15:34
  • \$\begingroup\$ If you only run the load when the sun is shining, why have a battery?. Or are you planning to run the load during the entire daylight period and want the battery to supplement solar power early and late in the day? And do you want to run the load during daylight is the sky is overcast? And if so, how many days of overcast in a row are you willing to accept? \$\endgroup\$
    – WhatRoughBeast
    Commented Jul 24, 2015 at 16:49
  • \$\begingroup\$ Correct question, but let's image i have a a "sun outage" because of a cloud for 2min... if I don't use battery the equipment will shutdown. That's my idea for the battery. \$\endgroup\$
    – Hristo Damianov
    Commented Jul 24, 2015 at 16:51

1 Answer 1

Reset to default
1
\$\begingroup\$

The battery will not be enough.

Assuming all of your power will go from the battery to the load with 100% efficiency (no losses in the conversion from 12V to 220V), 550W requires the following current output from your battery:

550W / 12V = 45.83A

Your battery is 63AH, so the device can be powered from the battery alone for:

63AH / 45.83A = 1.4 Hours

After which the battery would be completely drained and would have to be fully recharged from the solar panels.

Solar power at the Earth's surface is (roughly) 1KW/square meter and solar panels are (pessimistically) 10% efficient, so you get 100W/square meter. Again, assuming perfection, that gives you:

100W / 12V = 8.33A

That means, to power your load ONLY you'd need:

45.83A / 8.33A/Square Meter = 5.5 Square Meters of solar panels.

That's a square area 2.3 meters long on each side (7.5 feet if you swing that way).

Although, realistically you'd need more area because it's likely you'll need to charge the battery and power the load at the same time at some point.

There's not enough information in your question to suggest what system parameters might work for you, but my guess is it's a large, expensive system.

\$\endgroup\$
2
  • \$\begingroup\$ 550W / 12V = 45.83A so using a 12->220 converted don't change this math ? because i don't have 550W at 12V but at 220V which is 2.5A ? \$\endgroup\$
    – Hristo Damianov
    Commented Jul 24, 2015 at 17:41
  • \$\begingroup\$ All of the power has to come out of your battery or solar panels, so to save time I calculated everything at 12V. Power will be transformed at varying efficiency as it is converted, but you'll never gain any power - only lose it. Thus, for rough order of magnitude calculations it's acceptable to calculate everything at 12V \$\endgroup\$
    – AngryEE
    Commented Jul 27, 2015 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.