- #1
Kiwaho Kilowatthour
- 12
- 1
It is said that the Madrid Spain solar chimney project occupies 110 acres land, but only output humble 50 kw: https://en.wikipedia.org/wiki/Solar_updraft_tower
The fact: the average yearly solar power density of above powerplant = 0.11 W/m^2
Interestingly, there are many countries still planning next solar chimney projects, such as the billions dollar project of solar chimney with land coverage of circa 100 square kilometers in the desert of China who is the biggest PV solar panel production country in the world.
Why the harvestable solar power density is so tiny, and still sought-after?
To answer this question, we need a reference frame.
In comparison with planting cash crops, what is the minimal yet profitable power density PD of a land size dependent power generator?
Assuming energy price just as same as Hydro company $0.10/kwh, and given 1 acre = 4047 m^2, harvestable $653/acre/year soybean from USA agriculture statistic data.
Solve from this equation: 0.1 x (4047 x PD x 24 x 365 / 1000) = 653
We get: PD = 0.18 W/m^2
It means that every new invented method of solar energy harvest must have at least 0.18 W/m^2, despite the transient powerful sunshine 1360 W/m^2.
The power density 0.11 W/m^2 of the Spanish project is very close to this threshold, and obviously there is still potential for improvement.
Perhaps that is the possible answer for the question.
What is your opinion?
The fact: the average yearly solar power density of above powerplant = 0.11 W/m^2
Interestingly, there are many countries still planning next solar chimney projects, such as the billions dollar project of solar chimney with land coverage of circa 100 square kilometers in the desert of China who is the biggest PV solar panel production country in the world.
Why the harvestable solar power density is so tiny, and still sought-after?
To answer this question, we need a reference frame.
In comparison with planting cash crops, what is the minimal yet profitable power density PD of a land size dependent power generator?
Assuming energy price just as same as Hydro company $0.10/kwh, and given 1 acre = 4047 m^2, harvestable $653/acre/year soybean from USA agriculture statistic data.
Solve from this equation: 0.1 x (4047 x PD x 24 x 365 / 1000) = 653
We get: PD = 0.18 W/m^2
It means that every new invented method of solar energy harvest must have at least 0.18 W/m^2, despite the transient powerful sunshine 1360 W/m^2.
The power density 0.11 W/m^2 of the Spanish project is very close to this threshold, and obviously there is still potential for improvement.
Perhaps that is the possible answer for the question.
What is your opinion?
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