
re: rolling blackouts hit houston again, today...
25 jul 1999
gigg a watt wrote:
>i always preach, that reliability can be applied to almost everything...
>even a wife.... what good does it do, if she is not around when
>you need her...
that's "availability," which can be dramatically enhanced by redundancy :)
>the same with electricity... you are depending on others, to provide
>you with this important commodity... and it these providers become
>unreliable... then alternative sources of electricity has to be found...
grid electricity is becoming more unreliable under deregulation. our local
utility peco now deliberately undersizes pole transformers (fused at 10x the
power rating), figuring that they have little to lose, economically, on the
average, when a transformer catches on fire in the statistically unlikely
event that a halfdozen neighbors all turn on their hair dryers, electric
dryers, ranges, air conditioners and toasters at the same time.
>...the byproduct of home, pv produced electrical energy... is
>energy independence....
one might argue that the world needs more interdependence, vs teenagers,
rogue armies and rugged individualists.
>you can now live in places that you might like better, without
>considering the local power monopoly's, electrical lines...
but pv electricity is often less reliable than the grid, by itself.
otoh, having pv and the grid (or 2 complete pv systems, or a generator
and pv, candles, drums full of water, etc.) makes for a much more available
system than either alone, if you can tell that one subsystem is broken
and send it out to get fixed before the others fail.
if an inverter fails once a year ("mtbf=8760 hours") and takes a week
to get fixed, this sort of (markov) model might predict its availability:
where l = is the failure rate, 1/(52 weeks),
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 the time,
ie 165 hours, or 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
decreases the average unavailability to 1/2 hour per year.)
add another inverter and this becomes
3l 2l l p3 <> 3 work
 >  >  >  p2 <> 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

