Newbie project

System · March 2010

Eventually, we are going to move to an off-grid location, so I figured I'd try a little project to get started. Before I go further, I know I could do this less expensively by just trenching from the house's grid system - I just want to attempt this. Anyhow...

We have just purchased a wireless pet barrier for the dog. Basically, it shoots a radio frequency to the collar. When the collar is out of range, it starts to beep. If the dog continues past the beeps, he gets a correction shock. The system radiates from a central location from a few feet in diameter, to about 1/2 acre. My thought is to build a doghouse in about the center of the yard, giving him the maximum room to roam. I am thinking of designing the size of the dog house around the solar panel(s) if need be. I'd pitch the roof to the proper declination.

Here are my calculations, so feel free to correct anything you feel is wrong - thanks!

I measured the maximum draw on this unit, which is .17amps at 120VAC. Since the dog will not be outside 24 hours/day, I found a wireless remote control with a
6 amp/12V relay. My calculations so far figuring some inverter loss are based on .2amps @120Volts.

120V X .2a = 24 watts

24 watts X 12 hours(maximum) = 288 watts/day so I'll use 300watts

Location = 4.29 hours per day (avg) sunlight

Assuming the 300 watts/day and 4 hours of sun, I'm thinking a 100 watt panel would suffice. I already have a 12VDC to 120VAC converter (200 watt - sine wave) that I can use.

Here's where I start getting lost (I think).

Battery size: .2amps X 12V = 2.4AH. Now that I'm using a 12 volt system, I must multiply the 2.4AH by 10, which equals 24AH. I would assume that a 50% load would bring me to 48AH. If I wanted to store a couple days worth of charge, I'm looking at a 110AH battery.

A couple of other points:

1) The system would not run 12 hours/day 90% of the time
2) I always have the option of charging the battery to 90+% via a battery charger

Questions:
1) Would this panel be big enough to juice up the battery?
2) What type/size of charge controller would be recommended

Any other comments would be appreciated. Thanks in advance!

Would 1 - 100watt panel charge the battery effectively?

BB. · March 2010

Re: Newbie project

Using the PV Watts program, assuming 1,000 watts of panels (sorry, smallest program will accept), 0.52 for system derating (yes, you really get 1/2 the rated panel power out the end of a 120 VAC inverter for an off-grid system). Assume Memphis, Tenn:

"Station Identification"
"City:","Memphis"
"State:","Tennessee"
"Lat (deg N):", 35.05
"Long (deg W):", 89.98
"Elev (m): ", 87
"PV System Specifications"
"DC Rating:"," 1.0 kW"
"DC to AC Derate Factor:"," 0.520"
"AC Rating:"," 0.5 kW"
"Array Type: Fixed Tilt"
"Array Tilt:"," 35.1"
"Array Azimuth:","180.0"

"Energy Specifications"
"Cost of Electricity:"," 6.9 cents/kWh"

"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
1, 3.93, 62, 4.28
2, 4.54, 64, 4.42
3, 5.34, 81, 5.59
4, 5.78, 81, 5.59
5, 5.80, 81, 5.59
6, 6.01, 80, 5.52
7, 6.10, 83, 5.73
8, 6.29, 87, 6.00
9, 5.21, 71, 4.90
10, 5.53, 81, 5.59
11, 4.10, 59, 4.07
12, 3.48, 54, 3.73
"Year", 5.18, 884, 61.00

So, for a 100 watt panel, you are really looking at 1/10 of the above, or 5.4 to 8.7 kWhrs per month or:

5.4 kWhrs per month / 30 days = 0.18 kWH = 180 Watt*Hours per day winter
8.7 kWH per month / 30 days = 0.29 kWH = 290 WH per day summer peak

Also, you appeared to have measured 120 VAC Current into the wall transformer... Please note that P=V*I=Watts only for AC loads that look like a resistor.

Many wall transformers look very inductive and tend to take current, but not power (Power Factor)... Basically the Cosine of the phase angle between voltage and current if you remember your physics or electrical classes from years past.

It is very possible that the wall transformer is using 1/2 the true power to operate. You can get a Kill-a-Watt meter to measure the true power (also good for measuring 120 VAC / 15 amps maximum loads throughout your home).

Your other option is to go with a DC conversion of the input power--it may use a lot less energy and you do not have the (very significant) inverter losses (inverters may have 6-8 watts of power just because they are on--will kill you at this low power levels). If you are going to be measuring a lot of smaller DC loads for your off-grid planning, one of these DC Amp*Hour/Watt*Hour meters may be very nice to have too.

Anyway, I have to go now--Questions?

-Bill

System2 · March 2010

Re: Newbie project

Thanks Bill!

Quick question.

The output of the power converter on the wireless fence system is 19VDC. If I measured the output side (19VDC), would I get the true power rating? If so, I'm not sure where I would get a 12VDC (or even a 24VDC) to 19VDC converter. Any ideas anyone?

BB. · March 2010

Re: Newbie project

The 19 VDC from a wall transformer is always a guess what the true voltage is when connected to the load (with filter capacitors, possibly rectifier).

And for volt meters, some measure the AC voltage and assume it is pure sine wave AC--and when you measure DC (like a battery), it will still read voltage on AC setting, etc... You almost need an oscilliscope on the the load's input and first stage power conditioning circuit to figure out what is going on (12 volts, 19 volts, 24 volts, etc.).

There are lots of various converters out there--You can get some of them from our host (NAWS 12/24 bidirectional converter). Without knowing the details of the load--it is very difficult to pick an appropriate device that is efficient (for the loads) and cost effective. You can also find a bunch of DC converters on www.digikey.com -- But you need to be somewhat electronically minded to pick one to try.

It is very possible that your "base station" will work just fine on 12 volts DC battery bank (really ~12-15 volts). But--I have no idea and would hate to see you smoke it experimenting.

You might try calling the mfg. and see if they have a battery option/suggestions on how to make a "portable" unit.

-Bill

mike95490 · March 2010

Re: Newbie project

2 ideas about your 19V "wall wart"

1) look carefully at the back, molded in, should be text that says
12V 3W or DC 15V .2A

2) hook it up, and run power as normal. Get 2 sewing pins, about an inch apart, poke the pin into the low voltage wire from the wall wart to the wireless. Stick the other pin in the other wire, and measure voltage. Remove the pins, and the nearly invisible holes will be OK till the cord falls into a tank of water. But you know what the voltage is feeding the wireless shock collar.

System2 · April 2010

Re: Newbie project

Thanks for the info everyone!

Newbie project

Comments

Categories