Solar lighting system in our school

[pastmaster02]

It's a rough # for all the losses, PV efficency, battery losses (another 20%) charge controller losses......

Oh I see. How about the losses due to wire resistance? Is it already included or is it another part of the story? How would I know which type of wire, size and length is ideal for our circuit? Is there any formula to calculate that?

[Mike90250]

And about halving the 250 W-hr energy, what's that for? PV array inefficiency? Thanks.

It's a rough # for all the losses, PV efficency, battery losses (another 20%) charge controller losses......

[pastmaster02]

Incorrect answer. You only discharge 20% per day. [250 wh x 5] / 12 volts = 104 AH

Incorrect, you do not receive 7 Sun Hours in the winter months of December and January. I assume more like 3 to 4 hours. 250 wh / .5 / 3 h = 166 watts.

I will make adjustments to my calculations. Instead of 5 sun hours, I'm gonna use 4 hours of sun. So, 20% discharge would be the best option I'll have to prolong the battery life. I see. If that is the case, how long do you think my battery will last, considering I have one cycle of charge/discharge per day? And about halving the 250 W-hr energy, what's that for? PV array inefficiency? Thanks.

No.. the 4 hrs would be close to best case ..
Reason .. its never clear skys all day. you have to allow for some clouds. and clouds are solar energy killers.

Oh ok. So that's 4 hours of sun on my calculation. Thanks a lot!

[Sunking]

Battery: 250 W-hr x 4 = 1000 W-hr capacity. Is this equivalent to 1000 W-hr divided by 12 V = 83.3 Amp-hr? (does multiplying the energy spent by 4 mean we are using only 25% of the battery capacity? Is that good enough?)

Incorrect answer. You only discharge 20% per day. [250 wh x 5] / 12 volts = 104 AH

PV array: 250 W-hr x 2 = 500 W-hr capacity (assuming the the sun is giving full power for 7 hours, is the power our array gives equal to 500 W-hr/7 hrs = 71.43 W?

Incorrect, you do not receive 7 Sun Hours in the winter months of December and January. I assume more like 3 to 4 hours. 250 wh / .5 / 3 h = 166 watts.

[Sunking]

I'm in Luzon. 4 hours is a bit short. Is that the minimum sun hours? In my computation should I use that minimum? Or can I raise it to like 5 hours or something? Thanks!

You are getting yourself into trouble here. You cannot use the AVERAGE for battery systems. You have to use the shortest month of the year. If you fail to do that you will go dark and destroy your battery.

[john p]

No.. the 4 hrs would be close to best case ..
Reason .. its never clear skys all day. you have to allow for some clouds. and clouds are solar energy killers.

[pastmaster02]

If you are in Luzon 4 hrs Visayas 4.5 Mindano 5 hrs. im near Tacloban and that is Visayas about 4.5

I'm in Luzon. 4 hours is a bit short. Is that the minimum sun hours? In my computation should I use that minimum? Or can I raise it to like 5 hours or something? Thanks!

[john p]

If you are in Luzon 4 hrs Visayas 4.5 Mindano 5 hrs. im near Tacloban and that is Visayas about 4.5

[pastmaster02]

I dont know where you live but you certainly dont get 7hrs of good sunlight each day probably 4 hours. work onthat figure. if its winter it could be 2 to 3 hrs.

I live in the Philippines. From what I have searched through the net, our locality has an average of 5.8 sun hours a day. I think the array will produce power at its peak at around 10AM up until between 3-4 PM. So can I use 5 hours of sun? Thanks.

[john p]

I dont know where you live but you certainly dont get 7hrs of good sunlight each day probably 4 hours. work onthat figure. if its winter it could be 2 to 3 hrs.

[pastmaster02]

Well, since your professor is not grading me, I'm not going to give the answer away.

Oh, wait, I already did, in my first reply to you. You just have to do the math I laid out.

Plan on only a 20% discharge from full battery, or you will shorten the life considerably.

Thanks again for your reply.

Total power of lamps: 50 W
Number of powered hours: 5 hours (from 12 down to 5 hours)
Energy spent in 1 night: 50 W x 5 hours = 250 W-hr

Battery: 250 W-hr x 4 = 1000 W-hr capacity. Is this equivalent to 1000 W-hr divided by 12 V = 83.3 Amp-hr? (does multiplying the energy spent by 4 mean we are using only 25% of the battery capacity? Is that good enough?)

PV array: 250 W-hr x 2 = 500 W-hr capacity (assuming the the sun is giving full power for 7 hours, is the power our array gives equal to 500 W-hr/7 hrs = 71.43 W?

[Mike90250]

....

I know the idea of having a solar lighting system is really expensive but our professor required this project to us for our course curriculum. So we can't do much about it.
Thanks again.

Well, since your professor is not grading me, I'm not going to give the answer away.

Oh, wait, I already did, in my first reply to you. You just have to do the math I laid out.

Plan on only a 20% discharge from full battery, or you will shorten the life considerably.

[pastmaster02]

Thanks for your replies. We are about to use 10 pieces of 5-watt bulbs for a total of 50 W. My classmates decided to put a single LED lamp in every post in our catwalk.
Question: for a given rating of components 100 Wp solar array, 100 Amp-hr 12V battery, 30 A charge controller, and 50W lights, how long can we power the lights? Considering these components are somewhat fixed (in specs), what adjustments on powering time must we make? For example, we could just power the lights for 1 or 2 hours just like Sunking said. Is there any way to provide lighting for a minimum of 5 hours? Should we use a lower power lamps?

I know the idea of having a solar lighting system is really expensive but our professor required this project to us for our course curriculum. So we can't do much about it.
Thanks again.

[john p]

what is neing used?
50 watts of LEDs ?
50w LED lamps? if "yes" how many??

Mabe your calcs are all wrong on what he needs to provide the power.
50w LEDs would be overkill to just light a catwalk.
mabe just 6? 1w LEDs could do this adequately??
I use 12 of the to do the perimeter of a building and find that its plenty of light.

[Sunking]

100 watt panel will only power the light for about 1 to 2 hours. I agree you need a crash course in Math, real math, not green math.