Tuesday, November 16, 2010

Thought experiment, continued

This is a continuation of the thought experiment started yesterday.  Let's reduce
this down to just one car and the amount of solar panels you would need to produce
the electricity for that one car.  From my study of batteries, you can get about
3 miles per kilowatt hour with a battery.  This calculation is from my memory and
may not be entirely accurate nor up to date.  But this isn't exact, a round figure
will suffice for now.

The Chevy Volt has a 40 mile range for its battery pack.  Using the 3 mile per kwh
number, you would need about 13 kwh of power to drive with electricity.  Again, this
may not be exact, but it is using a figure from the real world in order to get a
rough estimate.

From this we can calculate how much solar panelling will be necessary in order to
produce the 13 kwh.  Using the web page from yesterday, we have 80 watts of continuous
power from a 1000 sq inch panel.  At 95% availability in geosynchronous orbit, the
panel will produce  (.95*80 watts * 24 hours)/ 1000 watt hours per kwh = 1.824 kwh.
Why 95% availability?  A solar panel in geosynchronous orbit at an altitude of
22,000 miles (approx) will be in the sun almost 95% of the time.  From this number,
we can now calculate the number of panels needed in order to produce the electricity
for 1 auto.  13 kwh/1.824 kwh per panel gives a little over 7 panels.

What would be the cost of these 7 panels?  This is about 49 square feet of panels.
From this webpage, we can calculate it (again a bit rough calc) it figures out
to be about 56 dollars per square foot.  So, 7*56 = 392 or about 400 dollars.

How long will it last?  Maybe it could last 30 years.  Perhaps that would be way
too optimistic.  Let's say 10 years.  Then the cost per year would be 400/10
which would be about 40 dollars per year.

Sounds pretty cheap, still.  What is the cost per kwh if you use electricity from
the wall plug?  At 10 cents per kwh, you and 13 kwh per day you get the following
.10*13*365= which is about 475 bucks per year.  The above solar panel configuration
would only have to be 1/10th as efficient in order to yield the same cost.

Is it worthwhile to launch from the moon?  You have to have the infrastructure in
place on the moon in order to make the solar panels and to put them into orbit.
That would take a considerable investment.  If you can get the costs down of
manufacturing the solar panels to anything near what it takes on earth, then what
you have left is the cost of launch.  How much would that be?  About 5% of what
an earth launch would cost.  Again, rough figures.  That would be about 10,000 *
5 percent or 500 dollars per kg.  You'll still have to do better than this because
7 solar panels would definitely weigh more than 1 kg.  The 10,000 number is based
upon shuttles which are expensive even for earth launches.

The launch problem still exists even on the moon, or at least in this example.
Can you do better than the 5% number?  Maybe.  It takes energy to launch from
anywhere.  Getting stuff off the moon would require an efficient way to use
energy for launches.  If you figure that out, you figure out this problem.

What about mass launchers?  It would only use energy and little to no fuel.  This
could be a start.  I pick this up again in a future post.

5 comments:

Greg said...

Looks like I goofed. It will take 7 panels of 7 square feet each at 56 dollars per square foot. That gives results of $2744 cost, instead of the just a little less than $400. Left out that last 7 panels. If the panels lasted 30 years as in the original guesstimate, it would be slightly less than using the 10 cent per kwh from the wallplug.
In other words, it would be competitive if the thing will last and if you could get it launched and installed for the difference in price. Probably not likely.

Greg said...

Ok, goofed again. In the wallplug example at 10 cent per kwh, you get $475 per year. For 30 years, you spend 30 times 475 which equals $14250.
That compares with the $2744. This is a factor of 14250/2744 which equals 5.19. The payback would be in a little over 5 years- not counting launch and installation charges. Hopefully, I got it right this time.

Greg said...

The Tesla batteries hold 53 kWh and have a range of 244 miles. This give about 4.6 miles per kWh. I pulled a 3 mile per kWh from memory. What the Volt will do, I don't know just yet. But this number about the Tesla was as recent as 2008. This would change some of the calculations. The key here is that you could have another choice. Batteries vs. Fuel Cells. Which would be cheaper? Fuel cells could get their hydrogen from electrolysis of methanol. The methanol could be fossil fuel based or synthesized.

Greg said...

It costs $9513.93 in order to power a Tesla using electricity at a cost of .10 kWh for 30 years.

What about from a solar panel in geosynchronous or bit over a 30 year lifespan? The $2744 calculated above. But the 2744 doesn't include other costs to manufacture and launch from the moon. What if the manufacturing could be done for less and the difference could pay for lunar launch? Then you are back to 2744. Take this principle to lower priced vehicles and you can see that this could be mass marketed.

Greg said...

Another post here to clear up another oversight I made. Instead of 7 panels needed for the Tesla, you would only need 5. That's because the Tesla gets 4.6 miles per kWh instead of 3, which is a number I pulled out of memory. Each panel generates 1.8 kWh per day and the Tesla needs 40/4.6 kWh; this gives just under 9 kWh needed. This divided by 1.8 per panel per day gives just under 5 panels. The cost of 5 panels at 56 dollars per square foot and 35 square feet of panels yields 1960 dollars. If the panels lasted 30 years, the cost would be 65 dollars per year. This compares to 329 dollars per year at 10 cents per kWh from the wallplug. That's 9500 vs 1960 over 30 year period. Or 329-65 or
264 per year. Enough to buy all the panels in 5 years.