Solar: label watts => true average watts scaling factor?
Moderator: Peak Moderation
Solar: label watts => true average watts scaling factor?
If you have X watts of solar charging a lead acid battery, what fraction of X will actually be delivered by the battery?
This calculation must include day/night split, battery charging/discharge losses etc.
Now there MUST be a Rule Of Thumb for this factor.
Any ideas?
This calculation must include day/night split, battery charging/discharge losses etc.
Now there MUST be a Rule Of Thumb for this factor.
Any ideas?
Well... it depends on lots of things. In sub-Saharan Africa, an area I'm familiar with, the rule of thumb is that you need about 10x the 24-7 load.
If you had a 24-7 load of 1 kw (wanted 24 kwh from the system per day) then you would need 10 kwp of solar panel. That includes all the loses, the temperature effect, etc...
If you had a 24-7 load of 1 kw (wanted 24 kwh from the system per day) then you would need 10 kwp of solar panel. That includes all the loses, the temperature effect, etc...
I've got a 60w PV module flat on my van roof, so obviously at nowhere near the optimum angle, but there's no way it would run my laptop 24-7 throughout the summer, let alone winter. This summer was awful, and the battery was probably charged far more from the alternator when I was driving, than the PV.
The PV has produced 2.9kah since I installed it on 20th May 2008, and my laptop draws around 1.9 amps. I reckon that's about 63 days out of 564. I suppose with your lower power consumption and proper orientation, you'd do a fair bit better, but not that much.
The PV has produced 2.9kah since I installed it on 20th May 2008, and my laptop draws around 1.9 amps. I reckon that's about 63 days out of 564. I suppose with your lower power consumption and proper orientation, you'd do a fair bit better, but not that much.
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That gets complicated, and I'd be concerned about the wind lifting it when travelling. It's not practical to always park facing in a particular direction, so a simple hinge could mean it ends up facing north sometimes! I'd also have to carry some steps to get up to it!Neily at the peak wrote:JohnB
You could always mount it on a piece of board so it could be hinged upwards when stationary (is that with an e or a?) perhaps i should have said still!
My charge controller is twice the capacity of the PV, so I had thought of getting an extra module that could be free standing. The setup I've got does actually work though. The combination of driving and the time I stay in one place means I'm rarely short of power, when I don't need an electric hookup for heating.
It's too late now anyway. I'm supposed to be moving into one of those house things!
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You're right.Neily at the peak wrote:stationary (is that with an e or a?) perhaps i should have said still!
Remember: a stationer sells stationery.
I experience pleasure and pains, and pursue goals in service of them, so I cannot reasonably deny the right of other sentient agents to do the same - Steven Pinker
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A fair rule of thumb is that a PV module will put one amp into a 12 volt battery for every 20 watts of rated power
60 watt PV=3 amps into a 12 volt battery.
To determine the daily amp/hour production, multiply the amps produced by the daily hours of peak sun.
A rule of thumb for this is one hour a day in UK winter and four hours a day in summer.
Therefore a 60 watt PV module should put about 3 A/H a day into the battery in winter, and about 12 A/H a day in summer.
If the load is lighting, used only when it dark, then 10% should be deducted for battey losses.
If the load is partly supplied in daylight, such as a computer or fridge, then the losses are less, and 5% would be a fair estimate.
For a large or complex installation, a good textbook will give more detailed guidance, however for small and simple installations the above rules of thumb should suffice.
60 watt PV=3 amps into a 12 volt battery.
To determine the daily amp/hour production, multiply the amps produced by the daily hours of peak sun.
A rule of thumb for this is one hour a day in UK winter and four hours a day in summer.
Therefore a 60 watt PV module should put about 3 A/H a day into the battery in winter, and about 12 A/H a day in summer.
If the load is lighting, used only when it dark, then 10% should be deducted for battey losses.
If the load is partly supplied in daylight, such as a computer or fridge, then the losses are less, and 5% would be a fair estimate.
For a large or complex installation, a good textbook will give more detailed guidance, however for small and simple installations the above rules of thumb should suffice.
"Installers and owners of emergency diesels must assume that they will have to run for a week or more"
- emordnilap
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That's true. The stationery was stationary on the shelf, and the stationer was stationary behind the counter while selling the stationery to the customer, who took it to their home/office where the stationery remains stationary in the stationery cupboard until it's needed. Once used, the stationery is placed in a file that spends most of it's time stationary in the filing system .
Last edited by JohnB on 07 Dec 2009, 13:37, edited 1 time in total.
- emordnilap
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- emordnilap
- Posts: 14815
- Joined: 05 Sep 2007, 16:36
- Location: here
Vortex - just got my first Rutland 913 delivered Saturday. First quick measurement gave me around 4 amps at peak gusts (strong enough to rock my 14 ton boat about!). Will connect it up to the Morningstar regulator tomorrow to take some proper long term readings, but, (as always for me), am not over impressed with the damn thing - since Saturday it hasn't provided enough even for a single cuppa! I was all into buying 10 of the things - Rutland 913 seemed to come out tops value for money in a real life, 4 x day test of 4 different units, hence the choice.
Mitch - nb Soma