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Old 08-27-2015, 04:21 PM   #1
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Solar panels on a short bus?

I recently bought a short bus and want to put solar panels on the roof. I plan on having a mini fridge and interior lights run off of it 24/7. I would also like to be able to use a toaster oven, blow dryer, and phone/laptop charger connected to it every now and then as well. Ive been looking at some solar panels online, found some decently priced ones. But id like to know a general idea of how many watts im going to need, how many batteries i should use for my battery bank. And i also would like to know what type of solar panel i should get. Im afraid they will break due to weather or if i hit a big bump in the road, will they be sturdy enough to stay intact? Any advice?
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Old 08-27-2015, 05:53 PM   #2
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This is a good website to learn about solar : https://handybobsolar.wordpress.com

For more info, Google "RV solar" or "RV boondocking".
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Old 08-28-2015, 10:36 AM   #3
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OK, so here's a crash course "back-of-the-envelope" solar estimation. "Back-of-the-envelope" means in engineer-speak a very rough approximation to quickly estimate whether something is even close to feasible. It uses arbitrarily chosen reference points, and even some outright guesses. But that's the point: it's fast and easy. If the numbers land close to being feasible, then a proper design can be done in which components are chosen more carefully and the numbers are more reliable.

First, the fridge. It's the biggest consumer. I couldn't find any power consumption numbers for mini fridges on Home Depot's web site; the only data point I have is my own fridge sitting in the shop waiting for its day to be installed in my bus. It has been plugged into a Kill-A-Watt meter for a "long time" which now reads 99.5 kWh and 1342 hours -- its average is 1.8 kWh per day. Maybe a mini fridge will be 70% of that; 1.8*0.7=1.26 kWh.

Lights. Having no other data, I'll guess the bus needs interior light from 6-10 pm and 6-8 am. Maybe three of the 8-watt LED lamps will do. 24 watts times 6 hours is 0.144 kWh.

Phone. Negligible.

Computer. Again, very little data.. suppose 60 watts for 3 hours a day. 0.18 kWh.

Total daily load: 1.584 kWh. It's all 120 volt load, so it runs through an inverter. Suppose 85% efficiency, then 1.86 kWh are needed at the input (ie, from the battery).

Battery sizing. Make it big enough to run through two days, in case of cloudy weather and no solar production. Battery needs 3.7 kWh output. Suppose it's a 12v bank, then 3.7 kWh/12 v=310 Ah.

Battery type. Cheap lead acid? Don't discharge them fully. Normally the rule of thumb is to size it to double the anticipated load. In this case I already doubled the load to account for a cloudy day, so let's let the battery go to 33% state of charge on these infrequent days (you're not in Seattle, right?). Multiply up by 1.5; bank should be around 450 Ah. Check the instant power output. Trojan seems to recommend not more than 0.2C output, 450*0.2=90 amps. Times 12v is 1080 watts. The regular stuff, is well below this, but the occasional toaster oven or hair dryer, could be pushing it a little.. Charge efficiency. Lead acid battery charge efficiency varies inversely with state of charge (http://windandsunpower.com/Download/Lead Acid Battery Efficiency.pdf). Suppose we go with 40% efficiency; then the 3.7 kWh output from this bank is going to require 3.7/0.4=9.25 kWh input. But it's only that bad if we need to catch up a full day's consumption with no solar input; let's allow two days for the catch-up so we need 1.584 kWh*1.5/0.4=6 kWh charging input.

Solar panels. Suppose we use 315 watt monocrystalline panels and we have 5 hours of good sun each day. The 6 kWh/315/5=4 panels. These panels are about 78x40 inches, so they'll cover 87 square feet and cost about 350*4=US$1400. Plus racks, wiring, charge controller, batteries, inverter......

Booyah and others -- how'd I do? Any corrections?
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Old 08-28-2015, 10:48 AM   #4
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Quote:
Originally Posted by family wagon View Post
OK, so here's a crash course "back-of-the-envelope" solar estimation. "Back-of-the-envelope" means in engineer-speak a very rough approximation to quickly estimate whether something is even close to feasible. It uses arbitrarily chosen reference points, and even some outright guesses. But that's the point: it's fast and easy. If the numbers land close to being feasible, then a proper design can be done in which components are chosen more carefully and the numbers are more reliable.

First, the fridge. It's the biggest consumer. I couldn't find any power consumption numbers for mini fridges on Home Depot's web site; the only data point I have is my own fridge sitting in the shop waiting for its day to be installed in my bus. It has been plugged into a Kill-A-Watt meter for a "long time" which now reads 99.5 kWh and 1342 hours -- its average is 1.8 kWh per day. Maybe a mini fridge will be 70% of that; 1.8*0.7=1.26 kWh.

Lights. Having no other data, I'll guess the bus needs interior light from 6-10 pm and 6-8 am. Maybe three of the 8-watt LED lamps will do. 24 watts times 6 hours is 0.144 kWh.

Phone. Negligible.

Computer. Again, very little data.. suppose 60 watts for 3 hours a day. 0.18 kWh.

Total daily load: 1.584 kWh. It's all 120 volt load, so it runs through an inverter. Suppose 85% efficiency, then 1.86 kWh are needed at the input (ie, from the battery).

Battery sizing. Make it big enough to run through two days, in case of cloudy weather and no solar production. Battery needs 3.7 kWh output. Suppose it's a 12v bank, then 3.7 kWh/12 v=310 Ah.

Battery type. Cheap lead acid? Don't discharge them fully. Normally the rule of thumb is to size it to double the anticipated load. In this case I already doubled the load to account for a cloudy day, so let's let the battery go to 33% state of charge on these infrequent days (you're not in Seattle, right?). Multiply up by 1.5; bank should be around 450 Ah. Check the instant power output. Trojan seems to recommend not more than 0.2C output, 450*0.2=90 amps. Times 12v is 1080 watts. The regular stuff, is well below this, but the occasional toaster oven or hair dryer, could be pushing it a little.. Charge efficiency. Lead acid battery charge efficiency varies inversely with state of charge (http://windandsunpower.com/Download/Lead Acid Battery Efficiency.pdf). Suppose we go with 40% efficiency; then the 3.7 kWh output from this bank is going to require 3.7/0.4=9.25 kWh input. But it's only that bad if we need to catch up a full day's consumption with no solar input; let's allow two days for the catch-up so we need 1.584 kWh*1.5/0.4=6 kWh charging input.

Solar panels. Suppose we use 315 watt monocrystalline panels and we have 5 hours of good sun each day. The 6 kWh/315/5=4 panels. These panels are about 78x40 inches, so they'll cover 87 square feet and cost about 350*4=US$1400. Plus racks, wiring, charge controller, batteries, inverter......

Booyah and others -- how'd I do? Any corrections?

I was going to say the same thing...but I was too lazy to type. LOL. Good job.

P.S. I accidentally hit the edit not quote button.
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Old 08-28-2015, 12:24 PM   #5
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You are WAY ahead of me on this stuff. Kudos!
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Old 08-29-2015, 07:17 PM   #6
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Yay

Hey thanks for all the replies... Alot of information but very useful. I will need to do some further research on all of this but thanks to all of you who replied 😄
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Old 08-30-2015, 08:18 AM   #7
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battery

some very rough estimates,

just for lights and charging a few items:
100 to 200ah of battery with 100 to 200 watts of solar

just for the lights and fridge:
400 to 500ah of battery with 400 to 500 watts of battery

your cooker probably uses 1000 to 1500 watts, which is about 10-15 amps AC, and 10 amps AC equals 100 amps DC and since you dont want to pull more than 10% of your battery bank at any one time you would need at least a 1000ah battery bank, a little less than 10% is usually better.

most people will have 0.5 to 2 watts of solar per AH of battery, since your cooker wont run a lot you might be able to get away with ~700 watts of solar but still ~1000 watts of solar is probably better.
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