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

Rob Dekker wrote: 
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 :

That depends upon what is meant by power rating. It is traditional to rate power plants by their max output. What you say is true for fossil as it can be run at peak year around. It is not clear that is meant to apply to these measures. If rated like fossil then it would be max output with max "fuel" input. You don't have the max input in winter.

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.

This is clearly not the case. We do not feel the same warmth from the sun in summer and winter. Homes are not equally sun warmed in summer and winter. A fixed area gets less solar heating the further it is rotated from the sun. The cross-section intercepting sunlight is what matters. A piece of paper flat on to a light gets total light than when on edge to the light. As the paper rotates away the intercepted light becomes less as the cross-section actually facing the light diminishes. The function is a sine or cosine depending on which angle you consider zero.

	If this were the case there would be no temperature difference between summer and winter.

If you want to go to tilting mirrors then the field gets larger to prevent shadowing as they have to tilt through 47 degrees with extremes of the latitude degrees assumed to be 15 and 62 degrees in winter. That will take more space to prevent shadowing.

Rule one of engineering, moving things stop moving whenever they damn well feel like it. So in addition to being more expensive to construct they are more expensive to maintain.

	If the simplicity of fixed orientation is chosen then there is annual variation in insolation.

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.

Here is where there has to be room for maintenance vehicles between mirrors. A minimum separation is reasonably wider than the width of the vehicle. They don't have to turn around but there has to be a reasonable excess width the get a replacement whatever off the truck and into position. As some point one has to consider replacing mirrors which would mean a crane. I did not find any pictures which had objects to give an idea of scale compared to a truck. The bigger the mirrors the bigger the crane.

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 not stored then we have to have a larger field to produce it in real time meaning a larger field. Yes, season does matter.

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)

I do not think it should be read as being designed in because that is not the way fossil plants are rated.

- You doubled size for heat storage losses (which is simply not needed)

Of you need to increase the size to collect more sunlight in winter. I expect the result to be about the same. The total energy over a year has to be a constant whether it is collected in the summer and stored or genterated in winter.

 - You doubled size for 50 % transmission losses (which is irrelevant for
many years to come)

It is not irrelevent now when plants are only a few miles from cities today. Everything has to be considered long distance transmission with a centralized generation.

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.

Try this. Transformers are needed to increase and decrease the voltage for transmission. If transformers are 90% efficient that throws away 19% of the power just to get the higher voltage for less loss. If it only gets 19% less loss it is a wasted exercise.

I like this stuff... 

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