OPV design breakthrough saving its cost and land area
OPV design breakthrough saving its cost and
land surface
10.02.2018
Author: David Judbarovski, principle
inventor, retired engineer, Israel
The OPV is supposed to be flexible organic
solar panel made by roll-on-roll for its maximal cheapening. Let the OPV
backside be a steel foil support. The said support is fixed to a steel wire
fixed by its upper ends to a pair of special invented towers being a key
element of the my design. Its lower ends are fixed to the ground. The said OPV
is under a small angle to vertical and faced toward the midday sun, and for
regions of middle- or high latitude the said design is saving a lot of land
surface area, but harvesting practically maximal solar power.
The cheaper OPV, the less is energy
efficiency, the shorter its livetime. For example, if being 5%, it can serve
about two-three years, but can be 100-1000 times cheaper than Si-PV being in
its turn about 50 cents USA per Watt, or about USD 500.0/m2.
So such OPV is about USD 1.25/m2 and harvests
55.0 kWh/m2 for a season.
Really, if being 1300 kWh/m2 of annual
insolation * 5% = 65 kWh, or 55 kWh for 9 month of a season, or 110 kWh for 2
years, and it is 125/110 = 1.1 cents/kWh
Such small efficiency ordinary demands a huge
area for the OPVs’ deployment and it is a main obstacle to their introducing in
solar powering.
We can’t use ordinary aerostats to hang the
said PV, because the aerostats cost would be additional 150% to a cost of the electricity
harvested. For example, if OPV being 15 m height and 10 m width = 150 m2, and for
designed wind of 6 m/s, its load is 9 kg/m2, or 1350 kg totally. So two aerostats
lifting force must be (1350/2) * 1.25 * 7 = 5900 kg each = 1.25 * 0.5 * D^3 * (29-
16)/29, so D = 27 m diameter of each aerostat, so two ones are 2 * 0.5 * 27^3 =
20,000 m3, or even if USD 120/1000 m3, it is USD 240.0, while 110 kWh/m2 * 150 m2 = 16500 kWh, or 24000/16500 = adds 1.45
cent to OPV kWh. QED.
The said tower invented by me, is a hollow
cylinder with thin walls of plastic film, e.g. polyamide one, filled by natural
gas (NG) under pressure P, while consist of some vertical sections. Let the
tower diameter be 2 m
Let the tower under wind (20 m/s) pressure
is tilted up to 10 grades to vertical. It is loaded by the wind more than 35
m/s by 4100 kg = 150 m2 * 55 kg/m2 /2, and its back side approximately 0.34 m
shorter than its front side to the wind, so the tower resists to the wind load
of 4100 kg ~= P (NG) * (15 m * 0.2) m, so P = 0.12 kg/m2 = 0.12 kg/cm2. If the
walls’ specific strength being 1000 kg/cm2, so the walls thickness is 2000 mm *
(0.12/1000) = 0.24 mm = 240 micron can resist against the hurricane!!!
Moreover, it uses 3 m2 * 15 m = 45 m3 natural
gas vs. 20,000 m3 if aerostats, and uses 3 m2 * 10 m = 30 m2 land area for
16500 kWh electricity production, so adds to its cost negligible.
For Ukraine electricity production of 150
milliard kWh/year, the total area for such solar power units would be 3 m2 * 150
* 10^9 /(0.5 * 16500) = 54 km2 totally being
extremely negligible!!!
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