quote:

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> Ed Earl Ross <edearl@satx.rr.com> wrote:
>
>
>
>
> No.
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>
>
> I'm asking "What's the uptime fraction, over a year," which is less
> than 100%, or "How many minutes per year is the system unable to
> produce energy, on average?" This works best with some sort of alarm
> to tell people a panel is broken and a short repair time estimate,
> compared to a single panel MTBF, in a Markov chain calculation.
>
> "1 out of N" (eg 1/25) redundancy can be a lot cheaper than 1/2
> redundancy, with a lower $/min investment for a given decrease
> in UNavailability, altho it helps to attack the parts that fail
> more often, eg inverters.
>
>
>
>
> And the combination can be a lot more reliable than either, given
> a few batteries.
>
> Nick
>
>
> Article 1228 of alt.energy.homepower:
> From: nick@ufo.ee.vill.edu (Nick Pine)
> Subject: Re: Inverter Schematics
> Date: 22 Feb 1997 08:32:02 -0500
> Organization: Villanova University
> Keywords: redundancy reliability
>
> John Francis <johnfrancis@mindspring.com> wrote:
>
>
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>
> Building one of these sounds like a bad idea, for the average person.
>
>
>
>
> Perhaps you could replace a part with a soldering iron...
>
>
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>
> Seems true, if you mean when you are using them, but perhaps not, if you mean
> that things decide to break by reading your mind and deciding when to break,
> ie machine conspiracy theories...
>
>
>
>
> How about spare inverters? Jade Mountain has
>
> watts cost cost/watt efficiency warranty
>
> Porta Power 140 $75 $0.54 "over 90%" 90 days
> PowerStar 200 119 0.60 "over 90%" 1 year
> AC Genius 150 95 0.63 0.24A stby
> PROwatt 250 195 0.78 "over 90%"
> Trace 812 800 550 0.69 0.02A stby
> PowerStar 400 389 0.97 0.06A stby 2 years
>
> If an inverter fails once a year and takes a week to get fixed,
> this sort of (Markov) model might predict its availability:
>
> where L = is the failure rate, 1/52 week,
> L R = is the repair rate, 1/1 week,
> ---- --------> ---- P1 is the probability that the inverter works
> | P1 | | P0 | and P0 is the probability that it doesn't.
> ---- <-------- ----
> R
> Since it works or it doesn't, but not both P1 + P0 = 1, and P0 = L/R P1, so
> R/L P0 + P0 = 1, or P0 = 1/(1+R/L) = 1/(1+1/1/(1/52)) = 1/53, so we might
> expect the inverter to be out of service an average of 1/53 of each week,
> 165 hours per year, a bit less than one week per year.
>
> Add another inverter and this becomes
>
> 2L L P2 <--> both work
> ---- --------> ---- -------> ---- P1 <--> one works
> | P2 | | P1 | | P0 | P0 <--> none work
> ---- <-------- ---- <------- ----
> 2R R
>
> Again, P2 + P1 + P0 = 1, P1 = 2L/2R P2 and P0 = L/R P1, so
> 2R/2L P1 + P1 + P0 = 1, or R^2/L^2 P0 + R/L P0 + P0 = 1, or
> P0 = 1/(1+R/L + R^2/L^2) = 1/(1+ 52 + 52x52) = 0.00036, so we
> might expect both inverters to be out of service an average of
> 0.00036x8760 = 3 hours per year. (Reducing the repair time to
> 4 hours would decrease the unavailability to 1/2 hour per year.)
>
> Add another inverter and this becomes
>
> 3L 2L L P3 <--> 3 work
> ---- ------> ---- ------> ---- ------> ---- P1 <--> 2 work
> | P3 | | P2 | | P1 | | P0 | P1 <--> 1 works
> ---- <------ ---- <------- ---- <----- ---- P0 <--> 0 work
> 3R 2R R
>
> Again, P3 + P2 + P1 + P0 = 1, P2 = 3L/3R P3, P1 = 2L/2R P2 and P0 = L/R P1,
> so P0 = 1/(1+R/L+ R^2/L^2+R^3/L^3) = 1/143,365, so might expect all three
> inverters to be out of service an average of 8760/143,365 = 0.061 hours or
> 3.7 minutes per year.
>
> Nick
>