Re: OT: GO SOLAR ! Was: go nuclear. Was: Detecting ETI via CO2

OK Matt,

I found at least one discrepancy between our numbers :

First : Based on the 130 acres per 10MW from the 1977 paper
you found an area of 138 miles squared to generate 1TW :

> So we need to generate 948 GW peak. Here is 10 MW so we need 94,800 sites
of this size. At 130
> acres per 10MW
[...]
> 138 miles square
>
> Ohio is roughly 200 miles square or 40,000 sq. miles.
>

Let me for the moment not contest this (half of Ohio) number.
But then you start adjusting :

> As the peak power was determined for the summer where the latitude is only
15 degrees
> north of the sun we subtract cos(15) from cost(15+2*23.5) or .496. This
requires us to double the
> size of the solar field.

Peak power in the winter is same as peak power in the summer.
Summer/winter is already included in the land-use estimate of the plant, so
no need to compensate for that again ! This is why :

Peak power (1000W/m^2) is accomplished if all mirror are full and faced in
the sun.
It does not matter if it is morning, midday, evening, high or low lattitude,
summer or winter.

The only thing that reduces power is if they are in each others shadow.
And THAT depends on the factor of land area over mirror area (land-use
factor).
For example, it makes no sense to pack mirrors closer to each other than a
land-use factor of cos(alpha) where alpha is the highest point of the sun at
mid-day in the summer.
Likewize, for an 8 hour 'work' day, it makes no sense to spread the mirrors
further out than the cos(beta) with beta the angle of the sun 4 hours before
or after highest point in the winter.

Given a certain land area, computing the size (packing factor), amount and
location of mirrors, depends on many variables like cost of land, cost of
mirrors (how much shadowing is allowed), lattitude of the plant, local
circumstances, power requirements during a day (during summer-day, or winter
day) etc etc.
Also note that the 'front' row of mirrors is always illuminated, from dawn
to dusk.

For trough systems, the land-use factor is about 2, for a practical working
days during winter and summer, and average 8-hour work days during the
average days of spring and fall.

So there is your land-use factor : land required is double the collecting
area.

>
> That gives us 38,000 sq miles. If a miraculous insulation capability is
achieved of only 50% lossy

Storing heat from the summer for the winter is simply not a good idea.
Storing it for 16 to 24 hours (at average output power) is more than enough,
and often that is not even needed.

And if you mean heat loss for short-term (<24 hour) storage, that is very
easily in the range of 99% efficient. I'll find the solar I / II
experimental plant web-links for you that show this.

> we 78,000 sq miles, 2 square Ohios. Roughly 280 sq. miles.
>
> And still no transmission losses considered.
>
> Do you disagree with my assumptions or my arithmetic?
>
> If I remember correctly I assumed only 50% transmission losses and came up
with four square Ohios.

So, this is how you got to 8 times the original estimate :
- You doubled size for winter (which was already designed-in)
- You doubled size for heat storage losses (which is simply not needed)
- You doubled size for 50 % transmission losses (which is irrelevant for
many years to come)

With the 4-Ohio system, you will create 8 TW power even if your area
estimate of 130 acres/10MW peak is accurate. And I don't think it is. I
think peak-power will actually be even higher (like 20 TW or so), and
average (24/7,365days/year) power will be a bit lower (7 TW or so). Let me
address that in the next post.

Also, I'll look a bit more into transmission losses.

I like this stuff...

Cheers !

Rob


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