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re: misleading arguments in wind vs. fossil fuel debate
21 jan 2003
richard bell  wrote:

>>hatunen  wrote:

>>>once conventional backup has to be built, then the cost of the solar
>>>or wind facility becomes an extra unnecessary cost.

>>conventional plants also fail unpredictably, with predictable failure rates.
>>they also require backup. it's a matter of numbers...
 
>some numbers:
>
>suppose that we are trying to build a 1000 megawatt facility.  the
>conventional approach would be five 200 mw gas combined cycle units.
>for redundancy, a sixth unit is installed and the plant is insured
>against the unlikely, simultaneous failure of two, or more, units.

this is a good start, except for the word "unlikely." if "system failure" is
"being unable to deliver at least 1000 mw," we need at least 5 units working
for success. you imply below about 1 failure per year (l = 1/8765h) per unit.
with 1 week repairs (r = 1/168h), we have something like this:

                                                        p6 <--> 6 work
     6l      5l      4l      3l     2l      l           p5 <--> 5 work
 --  --> --  --> --  --> --  --> --  --> --  --> --     p4 <--> 4 work
|p6|    |p5|    |p4|    |p3|    |p2|    |p1|    |p0|    p3 <--> 3 work
 -- <--  -- <--  -- <--  -- <--  -- <--  -- <--  --     p2 <--> 2 work
     r       2r      3r      4r      5r      6r         p1 <--> 1 works
                                                        p0 <--> 0 work

p6+p5+p4+p3+p2+p1+p0 = 1, p5 = 6l/r p6, p4 = 5l/2r p5 and so on, and
we are unable to deliver 1000 mw for (1-p6-p5)8765 hours per year... 

>the windfarm consists of 200 5mw turbines.

large numbers are good here. we might use a binomial distribution.

>if the wind varies from 40 to 100 per cent of expected...

that sounds high. suppose the expected output is 25%, with no output
half the time and 50% during the other half...

>there only needs to be a hydro installation with six 120mw turbines
>to take up the slack of the wind dying down...

how do you figure that, for the same availability? and why not wind
backup, with a larger number of widely-dispersed wind turbines?

>assuming that the wind turbines have an expected life of fifty years, and
>a mean time between failures of four years, one is down for repairs every
>week, and four times a year, the unit must be replaced.  for the same
>reliability, the combined cycle plant seldom experiences more than one
>failure a year, and two failures at the same time are nearly impossible,

let's use numbers, rather than words like "seldom" or "nearly impossible."

>the important thing is the backup for the thermal plant is a small fraction
>of its total capacity, and the backup for the windfarm is a large fraction
>of its capacity...

we also need to remember that wind fuel is non-polluting and free.

a system with lots of small widely-dispersed windmills can supply 100%
of a grid's peak capacity with the same availability and just a tad more
expected capacity, ie just a few more windmills percentage-wise.

we can quantify "lots" and "a few more" if we know the unavailability spec
for the grid (eg an average of 5 minutes outage per year) and wind power
probability distributions (poisson, rayleigh, etc.) the confidence level
grows with the number of windmills... a network of n windmills each having
mean output m and standard deviation s has a total mean output nm with
an s/sqrt(n) standard deviation. 

we might supply a grid with a large number of independent power sources
and no energy storage, assuming the sources really are independent. in the
example below, we replace a 1 gw nuke (booo) with an unavailability of
5 minutes per year with lots of windmills sprinkled all over a large area,
eg 1076 4 mw units with an average output of 1 mw (2 mw 50% of the time,
otherwise 0, like a coin flip), or 10,236 400 kw units, or 100,740 at 40 kw,
or 1,002,334 at 4 kw, or 10,007,380 400 watt windmills.

the overcapacity needed for the same grid unavailability decreases as
the number of sources increases, because of the magic of large numbers.

10 q=5/(8765*60)'grid unavailability (5 minutes per year)
20 p=.2316419'estimate error function using the technique of
30 t=1/(1+px)'m. abramowitz and i. stegun as described in the
40 a1=.127414796#'handbook of mathematical functions, 
50 a2=-.142248368#'national bureau of standards,
60 a3=.710706871#'applied mathematics series no. 55,
70 a4=-.726576013#'washington, d.c.; u.s. government printing office,
80 a5=.530702714#'1964, page 932.
90 t=1/(1+px)
100 f=((((a5*t+a4)*t+a3)*t+a2)*t+a1)*t
110 xl=x
120 x=sqr(2*log(f/q))'erfc(x) = q
130 if abs((x-xl)/x)>.01 goto 90'iterate to 1%
140 print "grid unavailability:",,q
150 print "number of standard deviations from mean:",x
160 p=1e+09
170 print"desired total power (w):",,p
180 for lpw=2 to 6'log of average source power
190 pw=10^lpw
200 print "peak source power (w):",,4*pw
210 print "average source power (w):",,pw
220 m=(x/2+sqr((x/2)^2+4*p/pw))/2
230 n=m^2
240 print "number of sources:",,n
250 next lpw

grid unavailability:                     9.507511e-06
number of standard deviations from mean: 4.662678

desired total power (w):                 1e+09

peak source power (w):                   400
average source power (w):                100
number of sources:                       10007380

peak source power (w):                   4000
average source power (w):                1000
number of sources:                       1002334

peak source power (w):                   40000
average source power (w):                10000
number of sources:                       100740

peak source power (w):                   400000
average source power (w):                100000
number of sources:                       10236

peak source power (w):                   4000000
average source power (w):                1000000
number of sources:                       1076

nick




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