Proposed Solar Wireless Surveillance System

[RogerRetro]

Hope this is the correct section for this post...
I have done some research on building a single camera outside wireless solar powered setup for my house.
I would appreciate a sanity check on the following setup, as this is my first try, and I may very well not have it all figured out correctly. Any comments or advise appreciated.
Thanks,
Roger

Fremont, CA average sunlight hours = 5.5/day.
50W solar panel = 2.8A x 5.5 hrs. sun = 15Ah/day. 15A x 12V = 185Wh/day.
One 5V 4.5W PT Camera, running 24Hr/day, powered by a 12V/5V DC-DC Converter
Load – 4.5W x 24Hr = 108Wh/day@12V DC
Consumption – Ah/day = 108Wh/12V = 9Ah/day + Charging losses – 10% = 10Ah/day
Solar Ah required – 10Ah/5.5Hr @ min sun = 1.8A (min. for solar panel load in winter months)
50 Watt solar panel – optimum rated current = 2.8A
Battery – back-up power required ~4days, x 10Ah/day = 40Ah over 4 days = 96Hr
50Ah – Peukert discharge to 50% = 103Hr@0.375A = 4.3days
Panel orientation - 170° (from true north), vertical angle 52°, clear of shadows.

[sensij]

Is the solar panel powering the camera mounted south facing, at least 40 deg tilt, no shade and a clear view of the horizon? If not, your sun-hours estimate may be too high. If you haven't already, check out PVWatts (seems to be down at the moment, but is normally reliable), you can test different orientations. It will show average monthly insolation, and let you download the hourly breakdown as well. The model assumes "typical" weather, so actual weather will clearly vary, but it gives you an idea of what to expect. Sources of shade need to be considered separately.

Since your load is fixed, you may want to tilt the panel even more vertically, perhaps 50 deg or more, to improve the winter generation at the expense of summer, when the excess is less likely to be needed.

[RogerRetro]

Thanks for your quick reply...
I had found similar information to what you mention, and the panel will be oriented 170 degrees (from true north), at a vertical angle of 52 degrees, clear of shadows.
Thanks,
Roger

[DanKegel]

That's an interesting application. Can you give exact part numbers? I'd love to hear how it works out.
(And SunKing's right -- in winter, I've seen long runs of ~3 'sun hours' per day. Get that bigger panel he recommends.)

(Also, I'm skeptical that wireless is a good method for getting video out of security cameras.
Are you quite sure you can't run a cat5e cable to each camera? Using Power over Ethernet means not having to run a separate power cable...
and the video is a lot better with netcams than with old analog ones, compare
old analog cameras http://kegel.com/crimes/graffiti-cameras.html
with new network cameras, http://kegel.com/linux/trendnet-tv-ip311pi/
The old ones aren't usable for identifying faces or license plates at a distance, but the new ones are.)

[Sunking]

First you do not need to calculate battery Amp Hours as it is meaningless and just the final result of calculations.

Second where are you coming up with 5.5 Sun Hours? For a battery system demands you use the shortest Sun Hour Month

You panel wattage with 5.5 Sun Hours is going to result in a panel wattage to small to effectively charge the battery. Batteries require at least a C/12 charge current, and C/10 is perfect. So you calculated 108 Watt Hours a day which means you need a battery with a capacity of 5 x 108 AH = 525 Watt Hours. As 12 volts requires a 525 wh / 12 volts = 43 AH battery, lets call it 50 AH @ 12 volts. .

OK if you really had 5.5 Sun Hours to generate 108 wh per day with:


MPPT Charger is [108 wh x 1.5] / 5.5 SH = 30 watts.
PWM Charge Controller [108 wh x 2] / 5.5 = 40 watts

Problem is a 30 watt MPPT or 40 Watt PWM system only deliver 2.5 amps of charge current. 2.5 amps of charge current is only 50 AH / 2.5 Ah is only a C/25 charge rate which is way too low. You need 5 amps. To get 5 amps of charge current means you need either a 60 watt MPPT system or 80 watt PWM system. If it were me I would use a 100 watt panel with a 10 amp PWM charge controller and call it done. With a 100 watt PWMsystem you can even move up to a 60 AH battery.

Forget Peukert in your application as your discharge rate is well below C/20

[RogerRetro]

Following are the components I'm starting with. The camera is a cheap camera that interfaces with Android only, and is very marginal in software and wireless performance, although it does have Ethernet capability. I am using this just to model the system/installation design. I prefer wireless if reasonable performance, otherwise wired. After validation of the solar part, I will upgrade the camera/software to something PC based.
Thanks for the link references.
Roger

Motorola FOCUS73-B Wi-Fi HD Outdoor Camera with Remote Pan, Tilt & Zoom

http://www.amazon.com/gp/product/B01..._detailpages00

Renogy 50 Watt 12 Volt Polycrystalline Solar Panel

http://www.renogy-store.com/50W-12V-...-p/rng-50p.htm

Renogy 20 Amp MPPT Solar Charge Controller

http://www.amazon.com/gp/product/B00..._detailpages00

55Ah 12V DC Deepcycle T6 Terminals SLA Solar Storage Battery

http://www.amazon.com/DEEPCYCLE-TERM...rage+batteries

SUPERNIGHT™ DC-DC 12V/ 24V to 5V 3A Converter Regulator 15W

http://www.ebay.com/itm/321461596997

[RogerRetro]

Below is my source for average of 5.5Hr/day:

http://solarelectricityhandbook.com/solar-irradiance.html

Solar Average Irradiance figures for Fremont CA, south facing
Measured in kWh/m2/day onto a solar panel set at a 52° angle:
(For best year-round performance)
5.613 = Average of below:

Jan	Feb	Mar	Apr	May	Jun
3.83	4.56	5.70	6.54	6.71	6.81
Jul	Aug	Sep	Oct	Nov	Dec
6.61	6.55	6.30	5.74	4.42	3.59

Is this not valid?
I can add another 50W panel easily...
Thanks,
Roger

[SunEagle]

Below is my source for average of 5.5Hr/day:

http://solarelectricityhandbook.com/solar-irradiance.html

Solar Average Irradiance figures for Fremont CA, south facing
Measured in kWh/m2/day onto a solar panel set at a 52° angle:
(For best year-round performance)
5.613 = Average of below:

Jan	Feb	Mar	Apr	May	Jun
3.83	4.56	5.70	6.54	6.71	6.81
Jul	Aug	Sep	Oct	Nov	Dec
6.61	6.55	6.30	5.74	4.42	3.59

Is this not valid?
I can add another 50W panel easily...
Thanks,
Roger

When you size your system you should use the lowest insolation hours (ie Dec @ 3.59) not the average of 5.5hr over the year.

If your system is expecting more sun hours then you get you will over discharge the battery system and could greatly shorten their life.

[Mike90250]

What happens in Dec and Jan when you have 3.5 hours of sun ? 4 days and the system dies, works for a sunny day and dies for 2 more, then the batteries are toasted, You have to size panels to work for your shortest months, and do you only have 2 days of clouds ? Last month, we had 11 days no sun, all cloud and rain. Lots of diesel burned to keep the batteries up when there is no sun.

[DanKegel]

When you size your system you should use the lowest insolation hours (ie Dec @ 3.59) not the average of 5.5hr over the year.
If your system is expecting more sun hours then you get you will over discharge the battery system and could greatly shorten their life.

Also, 3.59 is the average, but there will be years that are worse than average. IMHO planning for 2 or 3 sun hours / day would mean fewer days of observation lost to surprisingly bad winter weather.

[Sunking]

I can add another 50W panel easily.

Roger you must meet the minimum charge current of the battery and use worse case winter month Sun Hours If you use 5.5 Sun Hours two really bad things happen:

1: Every month with less than 5.5 Sun Hours you go dark and have to shut down or else you destroy the battery.
2. You fail to meet the battery minimum charge current requirement of your battery and you destroy your battery.

On the Flip Side if you live in some place like Tuscon AZ where December Sun Hours are above 3, you do not calculate panel wattage by Sun Hours. You calculate it to give the battery optimum charge current which is C/10. The result will be your panel wattage is more than required to recharge the battery every day which is a good thing. This applies to you.

Determine your watt hours you need in a day. Say too watt hours. Determine battery = (Daily Watt Hours x 5) / Battery Voltage or 100 ah x 5 / 12 volts = 50 AH.

Next determine panel wattage required to provide C/10 charge current or 5 amps in this example. For MPPT Panel Wattage = Battery Voltage x Charge Current or 12 volts x 5 amps = 60 watts. For PWM choose a 36-cell panel with an Imp of 5 amps which is going to be 90 watts (18 volts x 5 amps).

So if you have a 20 amp MPPT controller, you can use up to a 250 watt panel on a 12 volt battery. Get a 100 watt panel and a 60 AH battery and call it done.

[RogerRetro]

OK, starting out in complete ignorance on solar stuff, I didn't find battery charge rate information as a part of the process, or how to apply solar insulation chart data. Thanks for that...
A question... If Peukert discharge is to be forgotten, how to determine limiting discharge to 50%?
Thanks,
Roger

[Sunking]

A question... If Peukert discharge is to be forgotten, how to determine limiting discharge to 50%?

Peukert has nothing to do with it. 50% SOC = 12.06 open circuit voltage.

Here is great example of Peukert at work. Take a Rolls 24 HT 80 a 12 volt 80 AH battery. The battery is specified at the 20 hour discharge rate or a load of 4 amps for 20 hours. Take a look at the Chart. That same battery is 106 AH if discharged at 1.06-amps, or a 29 AH if discharged at 29 amps. That is Peukert.

You will not be discharging faster than 20 hours, thus you can ignore Peukert. You load is pretty constant right. If you use 100 watt hours a day, means you are only taking .34 amps at 12 volts.

[RogerRetro]

Interesting...
"You will not be discharging faster than 20 hours, thus you can ignore Peukert. You load is pretty constant right. If you use 100 watt hours a day, means you are only taking .34 amps at 12 volts."

The Peukert chart I used gave me essentially the same info (I think)... "50Ah battery - Peukert discharge to 50% @ 108Wh = 103Hr@0.375A"
I took this to indicate that it would take 4.3 days to reach the 50% point without any solar output (no sun).
is this just a coincidence? I'm confused.
Thanks,
Roger

[Sunking]

Interesting...
"You will not be discharging faster than 20 hours, thus you can ignore Peukert. You load is pretty constant right. If you use 100 watt hours a day, means you are only taking .34 amps at 12 volts."

The Peukert chart I used gave me essentially the same info (I think)... "50Ah battery - Peukert discharge to 50% @ 108Wh = 103Hr@0.375A"
I took this to indicate that it would take 4.3 days to reach the 50% point without any solar output (no sun).
is this just a coincidence? I'm confused.
Thanks,
Roger

No there is a correlation. Without seeing the discharge curves for the battery in question, I can only guess.

Bu there is the deal, consumer batteries are specified at C/20, higher quality commercial and industrial batteries are rated from C/4 to C/8. Take the Rolls 24HT80 above. As a consumer battery is rated 80 AH, as a commercial battery is rated 60 AH. From a design point of view you only worry about Peukert is when you discharge faster than the specified rating.

Battery size is determined by the number of days antinomy with 5 being the minimum. You are running roughly 100 watt hours in a day, so you are looking at a 500 wh battery. 50% is 250 wh or roughly 3 days. So what AH is 500 watt hours? You don't need or want to know that until you have determined battery voltage. So at 12 volts 500 wh / 12 volts = 41.66 AH. You are not going to find a 41.66 AH battery. So you bump up to the next available size which is likely 50 AH at the C/20 rate. In your case you are only pulling .4 amps or there about which is around a C/125. So the battery is going to last longer than you calculated. However that battery has a minimum charge current requirement and your panels must be able to deliver that current.