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re: inverter schematics
22 feb 1997
john francis wrote:
>i would appreciate it if anyone could post or mail a schematic(s) for an
>inverter in the 300 - 500 watt range running from either a 12 or 24 volt
>battery bank...
building one of these sounds like a bad idea, for the average person.
>i am not interested in the newest inverters on the market because they
>are not user serviceable.
perhaps you could replace a part with a soldering iron...
>we all know that thing's break when you need them the most.
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...
>a totally owner built system (or close to) would be the ideal way to go
>because you can have spare components to repair the system.
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
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