LVD = Low Voltage Disconnect.
LVD is a circuit to disconnect a load before it over discharges the battery.
Proposed Solar Wireless Surveillance System
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Logan005,
I don't understand... Would you please explain LVD (I'm assuming low voltage differential), with respect to supplying 5V to the load (from the controller's load output of 12V)? And how to implement a separate LVD?
Thanks,
RogerLeave a comment:
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In an Off-Grid system, everything revolves around the batteries and location. Get the batteries and location wrong, and everything else falls apart. Your application is rare with a consistent load around the clock which makes it very simple. Only thing you have to look out for due to your location being better than average, is making sure you provide the batteries with enough charge current. In your case more panel wattage than required to meet the load demands. You had to meet the batteries minimum demand rather than the load maximum demands. That is a good problem to have.
Take a look at the poor SOB's north of you like Portland Oregon and Gloomy Doomy Seattle with winter sun hours lower the 2 Sun Hours. The panel wattage has to be very high to where roughly Panel Wattage will be equal to daily Watt Hours. Example for 1 Kwh requires a 1000 watt panel which produces 80 amps of charge current at 12 volt battery. Well for 1 Kwh of daily power @ 12 volts is a 400 AH battery. Not only do FLA batteries have a minimum charge requirement of C/12, they also have a maximum current they can accept. For FLA batteries is roughly C/8 and on a 400 AH battery is 50 amps. Oops you get 80 amps at 1000 watts which is way too much. So WTF do you do?
You have two options. Make the battery larger, or use AGM. AGM can handle up to roughly C/4 charge current. However AGM batteries cost twice as much as FLA for a given size and last about half as long making them some 400% higher in cost over FLA. So from an economic POV you would use a larger FLA battery and up the size to 80 amps of charge current x 8 Hours = 640 AH rather than 400 AH you originally calculated to handle the higher wattage panels. So where you are at a 1Kwh/day system cost you roughly $2000. For the SOB's up north are looking at $3500. Want to use 400 AH AGM instead of 640 AH FLA the cost goes up to $4000, and you will be replacing AGM batteries ever 2 or 3 years instead of every 5 to 6 years if you used FLA.Last edited by Sunking; 03-08-2016, 01:16 PM.Leave a comment:
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Also do not count on the LVD in the SuperNight Dc to DC converter as it is as low as 10.5 volts. I would use a seperate LVD and set it no lower than 12voltsLeave a comment:
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OK, Sunking, I get it now...
Battery charge/discharge characteristics have not included these details in online discussions I had found on solar system design (until I got here!).
Thanks,
RogerLast edited by RogerRetro; 03-08-2016, 12:43 PM.Leave a comment:
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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
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.
Leave a comment:
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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,
RogerLeave a comment:
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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.Last edited by Sunking; 03-08-2016, 01:03 AM.Leave a comment:
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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,
RogerLeave a comment:
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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.Last edited by Sunking; 03-08-2016, 12:50 AM.Leave a comment:
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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.Leave a comment:
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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.Leave a comment:
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Below is my source for average of 5.5Hr/day:
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:Is this not valid?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
I can add another 50W panel easily...
Thanks,
Roger
If your system is expecting more sun hours then you get you will over discharge the battery system and could greatly shorten their life.Leave a comment:
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Below is my source for average of 5.5Hr/day:
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:Is this not valid?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
I can add another 50W panel easily...
Thanks,
RogerLeave a comment:
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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 Leave a comment:
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