Your Questions About Solar Panel Efficiency

James asks…

Lithium Ion battery and Solar Panel Questions…NEED HELP ASAP!!?

1) An environmental monitoring station is battery powered using a Lithium-ion battery pack which stores 4.3 kJ of useable electrical energy before its output drops below 2.7 V. The monitoring station requires an input voltage of 2.7 V or greater, and draws a current of 12 mA.

(a) How many seconds (and how many hours) will the battery pack last, assuming no internal resistance?
Before rushing to calculate, you check the battery specifications and realise that the output is initially about 4.3 V but it drops steadily as current is drawn. Now: draw a sketch of output emf versus time, and decide what voltage you will use in your calculation. Then do the calculation.

2) A solar panel is used to recharge the battery pack. Data for this solar panel and its location are:
· 11% of the Sun’s energy reaching the panel is converted to electricity
· in this location the mean solar energy per day onto 1 square meter (allowing for the season and typical weather patterns) is 10 MJ.

(b) What total electrical energy is produced per day if the solar panel has effective photovoltaic area of 0.15 m H 0.10 m? (Is this enough to keep the battery pack recharged?)
Note: Efficiency is defined as energy output/ total energy or power output/ total power
where ‘total energy’ here would mean the total solar energy incident on the panel.

ssadmin answers:

1) Take “drops steadily” to mean a linear decline in voltage. Also assume that the 12 mA is constantly drawn, no matter what the voltage. Then you can draw a graph of voltage vs. Time, starting at time t=0 with v=4.3 volts, and going down in a straight line to v=2.7 volts at time t=T. The power being drawn at time 0 is 4.3 * 0.012 = 0.0516 watt, and the power at time T is 2.7 * 0.012 = whatever. The area under the curve is the energy dissipated, and if t is in seconds, then the energy will be in Joules. You can use calculus to find the area under the curve as a function of t, or just plain geometry. One you have that, set it equal to 4.3 kJ = 4300 J, and solve for t in seconds.

The assumption for linear decline of the voltage is unrealistic in real life, by the way. The lithium battery will more likely maintain its high voltage until the end, then fall off a cliff, resulting in shorter life than calculated above.

2) I don't understand exactly – is 0.15 x 0.1m the dimension of the solar panel? If so, multiply the area by 10 MJ per day, and by 0.11 (efficiency), then compare the result with 4.3 kJ.

Betty asks…

I have a 60w solar panel and a 12v large deep cycle battery, how do I get it to charge faster?

From total discharge (11.65v) the battery charges to 12.03v in one day, 12.24v the second day and 12.26v the third day (all sunny days). I have the solar panel (I built) hooked directly to the battery (there are blocking diodes in place). The panel tops out at 17.5 volts or so (using 36 1.8w cells). I know the panel delivers charge, take the 12.26v reading as a baseline, if I plug in the solar panel it jumps to 12.34 volts. So is there anything else I can do to increase the efficiency of the charging at this point?

ssadmin answers:

Hey Savage, Jeff is right on the angle, but I imagine if you've built the panel yourself, you're tuned into that. I'm guessing that your panel does not have the amperage to push that deep cycle battery up very fast. Lead acid batteries go through a charging voltage curve, the volts curve up quickly at first, and around 12.6 level off until you are above 90% state of charge, then you see much higher voltage as the battery begins actively gassing at the top end of its charge cycle. Starting at 11.65 volts, if you battery is 50 or 60 amp hour, and your panel is putting out 1 amp, you will need about 12 sunny days to get to full charge. Have you checked your short circuit current on the panel? Take an ammeter, unhook the panel from the battery, set it in the sun and put the ammeter across the output wires. If your completed panel is 2 square feet, you might see 1.5 amps. You can get your panel wattage by multiplying the open circuit voltage by the short circuit amperage. 18 volts, and 1 amp would give you a 18 watt panel. That panel will take a couple weeks of sun to bring that battery up to full again, and then you will see voltages above 13.5 while it is charging. There isn't any magic elixer that will improve the efficiency of your solar cells. Silicone cells are around 10 – 13 % efficient, that's all the chemistry will convert. Some people have tried using mirrors to focus energy on them, but they just overheat and get damaged that way. You can't change the physics.

There is a great book at the library by Richard Perez called, “The Complete Battery Book.” I read just the chapters about lead acid batteries and skipped over the litiums, nicads and so forth. Richard also happens to be the editor of Home Power Magazine, a great periodical on solar, wind, hydro and other home made energy sources. Using his information and going to a couple energy fairs years ago got us where we are now. Our home is completely powered by the wind and sun, has been for almost 10 years. It's been interesting and fun, and we made plenty of mistakes, but we made it here anyway. There are also some great non profit groups with websites you can learn from, I will list some names below. Good luck Savage, and take care, Rudydoo

Joseph asks…

12V DC brushless fan – I want to run it directly off a PV solar panel – but what size panel?

I have a 12V DC brushless fan, rated at 0.16 Amp, 1.92 Watt (extracted from a dead desktop PC). I want to run it directly from a solar panel (not via a rechargeable battery), so that it will only operate during daylight hours and when there is a reasonable amount of light from the sun (i.e. not at times of heavy cloud).

I would like to know what size or type of PV solar panel (that generates – presumably – about 2 Watts of power in full sunlight and peak efficiency) I would require to run the fan and, perhaps just as importantly, how to calculate the size required.

The panel would be located on a 45 degree, south facing roof at 51 degrees North (UK).

Finally, is it feasible to run a fan directly from a solar panel or would more sophisticated electronic hardware be additionally required?

N.B. I want the fan to boost the output of a thermo-syphon solar air heater:

ssadmin answers:

Due to inefficiencies, you would most likely prefer to slightly oversize the solar panel. That provides more operating headroom during hours when the sun is not peaking, like in the winter and outside 10 am to 2 pm.

Based on the 2 watt rating, somewhere around 5 to 10 watts for the panel rating is reasonable. If the motor is okay with slight over voltage (like 15 V) it should be okay to directly connect it.

However, during evening hours when there is, say, only 0.25 watt available, it would be wise to interrupt the circuit so the motor won't be damaged by low current which is insufficient to turn, though that should not be a problem due to the smarts of most BLDC motors.

The easiest way to protect the motor would be to use a voltage regulator, a simple electronic component (like this ) available at electronic component outlets. Choose one rated for at least three times the highest expected current. Most regulators require a filter capacitor, and a choke wouldn't be a bad idea either since stepper motors tend to generate power supply noise.

Mark asks…

I am doing a Solar Panel DIY?

I am doing a big project for school, and I have decided to do it on solar panels. I am new to the subject but have done some research on it; I would like to buy a single panel, a charge controller, an inverter and a battery, to charge a simple lamp. My question is if the items in the links will work together and if the panel will produce electricity?




I haven't yet found a battery. Also, if you know any way to get cheaper items but for the same efficiency, please tell me. Thanks.
I've changed the inverter to this one:

ssadmin answers:

Sounds like fun.

A 1-watt panel doesn't need a charge controller, though. In fact, it may take more than 1 watt to power the charge controller. With a low-wattage panel, you can just connect the panel directly to the battery. A 7 amp-hour gel cell (the kind used in burglar alarms) would be good for this purpose. The panel is too small to charge a car battery.

The gel cell should power that inverter, but only for a few minutes. If you want to be more impressive, look for a low power, 12-volt device. An iPod plus a 12-volt car adapter would be an example.

Steven asks…

thermodynamic question that includes heating water with solar panels?

A homeowner is considering putting a horizontal solar panel on her roof to heat water for domestic use. An average increase of water temperature each day from 60oF to 120oF is wanted for 150 gallons. How large a panel would be needed on a clear winter day at 40o latitude if the overall efficiency is 50%?

ssadmin answers:

You need to figure out the BTUs needed. A BTU is how much it takes to heat 1 pound of water one degree F. They are trying to heat 150 gallons 60 degrees. Each gallon weighs 8.34 lbs. 150 x 8.34 = 645lbs. X 60 degrees = 38,700BTUs.

Now, I'm not sure if this question is for real world or a math/science question, because the next step may be different for each. You can go to the SRCC, an independent testing lab, and find the standard tested rating of a solar collector, to see how many BTUs each one generates a day. Http:// Here's a typical 4′ x 8′ collector rating. 40 degree latitude would be a cool climate, so you look at row D, water heating in a cool climate. Probably take the middle rating, mildly sunny day, and it is 10,000BTU per panel per day (it's 19,000 on a sunny day). 38,700 BTU / 10,000 BTU = 3.87, they'd need about 3 to 4 panels. If you used the sunny day number, you'd only need 2, but 3 is more realistic, as it is not always a sunny day.

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