Solar Panel Installation

Your Questions About Solar Power Panels Efficiency

Laura asks…

can a 400-Watt 16.5-Volt Solar Panel Generator power my house?

I think that that solar panel will give 17,777.78 W at standard voltage (110V in the US), which comes out to be about double of what a standard household uses, taking into account efficiency (decrease to 70%) and cloud cover, night, etc (decrease to 20%). But I might be completely wrong. I would like to hear from those who know more about these things. :D

So is it equivalent to 17,777.78 W at standard voltage (110V in the US)? Or am I just lost? :P
Never mind—power doesn’t change, current does.

http://answers.yahoo.com/question/index?qid=20090218205052AA5PX3U
:D

Michael answers:

I dont understand your 17,777.78 figure.

Anyway, a 400 wat panel is unlikely to b enough to power your home unless your power requirement is vry low.
Those on solar power (off grid) generally start by reducing their power consumption as much as possible, by cooking on gas, using low enrgy light bulbs, and certainly not using electricity for any form of heating.

We live off grid in southern Spain with 860 watts of panels and it meets our requiremnts.
You can calculate your plower requirements (and find other info) at the link below

Thomas asks…

What is the physics definiton of “power”?Please answer the following questions:?

1. An 85kg ice skater at rest pushes off the wall of the rink, so that he is coasting at 5m/s. How much work has been done by the skater’s arms while pushing off the wall?
2. Every Kilogram of gasoline has about 45MJ of chemical potential energy stored in it. (1MJ = 10^6 J = 1, 000,000J)
a. If a car engine can convert 25% of the chemical energy into usable kinetic energy, how much energy could you get from a 25 L tank of gas?
b. What is the efficiency of this car engine?
c. Energy can not be destroyed, but a lot of it is not being converted in to kinetic energy. Where does the “lost” energy go, if it’s not used to move the car?
3. Modern solar power generators have an efficiency of about 15%. Every second, every square meter at the top of the earth’s atmosphere receives about 1300J of solar energy from the sun. If the international Space Station has 50m^2 of solar panels, how much useable energy do the solar panels collect every second?
What is the physics definiton of “power“?

Michael answers:

1. The skater has Kinetic Energy KE
=.5mv^2=.5*85*25=1062.5J This energy is the result of the skater pushing off the wall and so is equivalent to the work done.
2. You are told that exactly 1/4=25% of the energy in gas is converted to KE. 25% of a 25L tank is 6.25 a)
b. The efficiency is 25%
c. Heat, noise, plus exhaust fumes contain chemical energy.
3. GIven 1300J/s per square meter (Note: 1Joule per second is 1 Watt). Written
1300W/m^2. Now multiply by 50:
=65,000W=65,000J every second.

Mandy asks…

how do i pick out solar panels to match my battery bank?

if I have a battery bank of of 6 L-16 Trojan batteries, 6V 390AH, wired in series parallel to give me 12V at 1170AH (or 14,040WH) what kind of solar panels should I purchase assuming I have 8 hours of sunlight, a solar tracking system and I want to make sure I can fully recharge my batteries everyday if i use up all the power every night?

i understand there are efficiency issues to consider which i have not calculated in yet.

thanks.

Michael answers:

Hey Feemind, we have a home that is powered by the wind and sun, and we use L-16′s also. Our system has 20 batteries, a 1.4 kw solar array and H-40, 1000 watt wind turbine. I have a couple issues with your calculations. You do have about 14 kwh or storage, but if you are planning on using all of it each night, your electronics and inverter will not have access to all that power. The voltage curve of a lead acid battery drops quite low during the last 30% of discharge, even lower when you draw it all out in less than 20 hours. Batteries of this type are rated on a C-20, or 20 hour rate. Your system will shut down before you get through half of the battery capacity. In addition, even if you could discharge those batteries 80 or 90 % each night, they would be shot in less than 2 years. We only discharge ours 15 to 20 pecent each night, and after 11 years, my batteries are toast. We are replacing them even as I write this. That battery bank can reliable give you about 5 kwh each night, then it will need almost 6 kwh the next day to replenish it. Your array will have to produce the needed 6 kwh during the day, and carry all your loads at the same time, maybe another 8 or 10 kwh for your house. In the end you’ll need a large array and a quality charge controller.

Here’s the good news, since you selected 12 volts as your nominal system voltage, it doesn’t matter what panels you choose. Virtually all panels today are wired for 12 volt charging, and have an open circuit voltage of about 17 or 18 volts. You can buy 3 used Solarex VLX-53 panels this week, then find a great deal on BP-120′s next month and add them to the first three, and later wire in some Sharp-210′s. Solar panels all get along fine with each other as long as they are all set up for the same nominal charging voltage and wired in parallel to the battery or charge controller, and you can buy more of them over time until your reach nirvana.

There are some great resources to learn more about this sort of thing, I will list them below. Make sure you have DC rated overcurrent devices at all points exiting the battery, when you have over 1000 amp hours of battery storage connected to your homes electrical wiring, you have a tiger by the tail. A short circuit in that DC wiring without a DC rated fuse or circuit breaker can melt through monkey wrenches, catch 4/0 welding cable on fire and cause battery enclosures to explode. Also, try to install those batteries in some kind of plastic tub to catch any acid that gets loose for any reason. Make sure you water those L-16′s each month with distilled water too, more often in hot weather. You’ll be amazed how much water they go through in a month.

We started down the same road you’re looking down now some 12 years ago, the two best things I did back then was subscribe to Home Power Magazine and go to an energy fair. The fairs are all over the place, and most of them are listed in the back of Home Power each issue. You can hear talks on batteries, inverters, panels and wind turbines and buy discounted parts on the spot, that’s what we did. Spend some time getting more informed on this subject, it will be well worth it. Good luck Free, and take care, Rudydoo

Richard asks…

Maximum & normal power per square meter, can be obtained from the Sun?

Hi, can someone tell me what the maximum power (very strong sunlight during mid-day) and normal power (average power during day time over a day with sunshine) per square meter, which we can obtain from the Sun is? Assuming the efficiency of the collection method is 100%. I’m interested in this because I’d like to know the power limit that we can get from the Sun. I was told the efficiency of solar panels in the market nowadays is just about 15%. By improving the efficiency, we can certainly obtain more power from the Sun but the limit that I ask in the above will be the limit that we can get. Thanks!!!

Michael answers:

Actually this is incorrect. By using a heliostat, (mirrors tracking the sun) you can focus a great deal more than normal sunlight to some area. This has been used to increase the potential of solar collectors. So then the limit is simply an economic one. Do you want 9 mirrors focused on your solar collecter or 9000.

Your calculations may be useful and interesting but certainly not a limit.

Lisa asks…

what is the best way to measure the efficiency of a fuel cell?

a simple fuel cell that uses a solar panel to get the energy to electrolyze water and use the hydrogen to power a small fan.
what would be the best way?

Michael answers:

There are 4 sources of efficiency here: the solar cell, the electrolyzer, the fuel cell and the fan (also the propeller if you want to consider that, too). The easiest way to determine the total effeciency would be to measure each one individually and multiply the efficiencies. There are a lot fo equations for all of this, but basically you would need voltmeters for the solar cell and fuel cell, an ampmeter for the motor, a calorimeter or gas chromatograph for the fuel cell, and a wind speed detector if you want to measure the fan’s efficiency.

Powered by Yahoo! Answers

Leave a Reply

  

  

  

You can use these HTML tags

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>