Smidge204 said:
TimeHorse said:
680 W m-2 average over the sunlit half of the Earth, with high-latitudes getting much less and equatorial getting much more. Anyone claiming more than that, especially in Texas or farther from the equator is definitely all smoke and mirrors.
Not sure where you got that figure. It's not like we have to guess or approximate about the insolation values here... Middle of Texas? ~6kWh per day per square meter.
680 W m^-2 averaged over the area of the sunlit Earth: http://en.wikipedia.org/wiki/Earth%27s_energy_budget#Incoming_energy
680 W m^-2 corresponds to 16.32 kWh day^-1 m^-2 using your units, and as I said Texas would not be getting that at such a high latitude, which agrees with your calculation of 6 kWh day^-1 m^-2. Thus, your calculations should be a little more than 1/3 of mine.
Smidge204 said:
TimeHorse said:
So what is the Theoretical Max?
6kWh/day/sq.m * 4046 sq.m/acre = 24,276kWh/day/acre, * 365 days/year = 8.86 GWh/year/acre, or 31.9 TJ.
31.9 TJ -- I believe you mean TJ year^-1 acre^-1.
Smidge204 said:
I honestly can't tell if your calculations are off because they are impossible for me to follow. :?
I agree with your calculations Smidge; I've explained my Solar radiation calculation, which does not include atmospheric effects and other losses of power for the solar radiation so I am happy to go with 250 W m**-2 in the calculation. Also, I have no problem with using the Energy Density of Ethanol over Diesel because we're looking for an order of magnitude here, not an exact calculation.
Then, I took things from a different direction; instead of calculating energy per acre, I calculate energy per bacterium since I want to see what the biological output would have to be to achieve theoretical max. How much material must a cyanobacterium generate per second to achieve theoretical max.
So, using your 6 kWh day**-1 m**-2, or 250 W m**-2, multiply by the top surface area of a cyanobacterium, about 3.14E-8 m**2 to get the total power received by the bacterium during the day: 7.85E-6 W or 7.85 microwatts. My prior calculation came to 21.4 microwatts, a little less than 3 times yours, which so far agrees given our different starting values for the solar radiation. Of course, 7.85E-6 W is the same as 7.85E-6 J sec**-1, which is 127 thousand seconds for each Joule in your case or about a little under 1 1/2 days for each cyanobacterium to produce a Joule of energy -- actually, about 3 days because half of the
day is night.
Finally, to go from J/s to L/s, we need to divide by the Energy Density of our destination material. In this case, we've chosen Ethanol at 33 MJ/L or 33E6 J/L or 3.03E-8 L/J. Multiplying 3.03E-8 L/J * 7.85E-6 W gives us 2.38E-13 L/s -- which I think is where my calculation went off. Maybe I multiplied by accident?
Anyway, so we get 2.38E-13 L/s or 2.38E-16 m**3 / s or 1.03E-8 L per 12-hour day and 3.75 microliters per year (of 12-hour days).
Since bacteria can be relatively densely packed, the real question I have is can a biological entity produce that much material?
In the end, I think you're right Smidge and sorry for the typo in the final calculation. Seems we basically agree now, it's just a matter for me of breaking it down to 2 levels:
- How much material can a genetically-engineered cyanobacterium create in a day
- How densely can you pack cyanobacteria
- How many molecules of Ethanol does 2.38E-13 L actually represent? How much carbon dioxide? How much water?