Re: OT: GO SOLAR ! Was: go nuclear. Was: Detecting ETI via CO2
- From: "Rob Dekker" <rob@xxxxxxxxxxx>
- Date: Sun, 31 Jul 2005 07:59:37 GMT
Hi Matt,
Below some more comments on the winter/summer differences and other stuff.
But first something important : A solar power plant is not like a fossil
fuel plant in that the "plant factor" (the ratio of average power supplied
versus peak capacity) of a solar plant is much lower than in fossil fuel
plants. In fact, the whole reasoning of looking at peak power usage (which
we did so far) is sort of useless, since peak power is not that important :
A solar plant is a bit like a hydro-electric plant : It can run at peak
power for a while, but not forever. The lake (energy storage) for the
hydro-plant will be empty at some point.
A solar plant is also quite unique in that it has almost constant input
power, which has to go somewhere. In fact, a solar plant resembles a fossil
fuel plant with a burner-control valve stuck at some number during the day,
and goes off at night. So we never want to put so much solar power in place
that it will exceed the minimum usage of the US during any given day. We
would waste energy if we did (cannot store heat for an entire nation for
more than 24 hours). The peak usage will have to be filled in by other
energy sources, or (if peaks last shorter than 24 hours) can be provided out
of heat storage.
I propose that for now, we insead look at the average usage. That way, we
will get a fair number for the total energy that can be produced by solar
energy, and the total area of land which would be needed for that.
Average electricity usage in the US was 3.88 trillion kWh per year (to use
the Dep.of.Energy numbers). That is 453 GW 24/7. So how do we generate 453
GW per year with solar energy ? Lets calculate the average energy output
provided by 1m^2 of mirror first :
- 1000 W/m^2 input at clear day.
- Solar to electrical efficiency : 0.2 (20%, as shown by exp. solar
plants).
- Average work day 8 hours : 8/24 = 0.33 (7 days in winter, 11 hours in
summer)
- Average 30% cloudy days : 0.7
- Shadowing effects average : 0.9 (for 50/50 spacing)
The average work day (conservatively set at average 8 hours) and shadowing
effects (trough design), are obtained from this info :
http://www.powerfromthesun.net/chapter4/Chapter4Word.htm
Check out figure 4.8/4.9 for the summer and winter production for 1m^2 of
mirror (with a 2-axis tracking device).
Note that the peak power is almost the same (a bit lower in winter due to
the longer path the sunlight travels through the atmosphere).
Check out figure 4.13 for the shadowing effects (0.9) of a trough-design
with 50/50 spacing.
So, 1m^2 of mirror on average produces average of 42W of electricity 24/7.
Shadowing and winter/summer differences included in this number.
To produce 453 GW, this requires 10 billion m^2 of mirrors.
How much land is needed ? Depends on spacing. 50/50 spacing seems
reasonable, and is already used in the 450MW plant of Southern Edison (by
the way, this gives ample space for mirror-cleaning robots). So we would
need a total of about 20*10^9 m^2 of land for these solar plants. That is
about 100 miles squared.
If you consider another 10 percent area for the power train, cooling towers,
heat storage and transformers and stuff, we get to 110 miles squared. Still
less than half of Ohio...
Now, once again, this is the maximum we need, since anything more than this
will surely go wasted. We simply don't use enough electricity for more solar
plant size than this.
In reality, we will not even need to get close to this, since other energy
sources (hydro, nuclear, wind etc) will also take up a portion of the
electrical needs. But for the sake of the argument here, a total of (less
than) half of Ohio size is all we need.
And, we can currently produce solar energy for 10ct/kWh, and this is going
down with the improved cost-effectiveness of heliostats. And the cost of
heliostats is a matter of engineering and volume production only, so cost
WILL go down.
Now are you somewhat convinced now that solar-thermal energy is not to be
ruled out ?
Or should I start mentioning the downsides of nuclear ? (Even though I am
not against nuclear, there are some serious problems with it).
Some more notes below
Rob
"Matt Giwer" <jull43@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:2YdGe.53024$mC.39578@xxxxxxxxxxxxxxxxxxxxxxxxxx
[...]
> > 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.
>
We are both right here. Peak power production of a solar plant is not much
different in summer or winter (happens at mid-day), but average power is
certainly lower in the winter than in the summer.
> > 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.
>
Right. If the surface remains fixed (like the Earth surface), the cos()
function applies for energy 'observed'.
That's why winters are colder than summers, and higher lattitude is colder
than lower.
But solar mirrors are not fixed. They are rotated to absorb as much solar
energy as they can at all times. As long as they have their face into the
sun, they will reflect about 1000W/m^2, summer or 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.
>
Now you are getting into it. It's a tradeoff between area and shadowing
effects.
Peak power is the same, but what makes winter production of a solar plant
lower than summer is caused by shadowing effects (which are worse if the sun
is low). Also the design of the plant (power-tower or trough) and the
spacing and position of the mirrors causes differences in summer and winter.
Probably overall, the cos function will probably be about correct for power
plants, as long as we are talking about total energy absorbed per day, or
therefore : the average power input (joule/day) is lower in winter than in
summer.
[...]
> > 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.
Well. Solar plants are different. Like hydro plants are different.
You can't and should not just discard hydro because it cannot sustain peak
power indefinitely. Similarly, you cannot discard solar because its peak can
be set at anything, but what really counts is its average power. If you
design them based on their peak, you will have way too much (like a factor
4 or so) energy in the summer, during mid-day.
>
> > - 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.
Similarly as above, if you design solar power plants to provide peak power
in winter, then you will have way too much power in the summer.
>
> > - 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 did not have the time to look at transmission losses, but every power
plant needs transformers, and every power plant needs to transmit energy to
their consumers, so I really dont see how solar plants are different there.
I WILL check on extra losses on high-voltage lines over longer distances,
because increased distance between plant and consumer is they only thing
that could differentiate solar plants from any other plant. But I will not
look into transformer losses, because these are the same for any other power
producing method.
.
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