Here is what we have used:
TL Energy Factor
13 1
14 2.5
15 5
16 10
17 25
18 50
19 100
20 250
21 500
22 1000
23 2500
24 5000
25+ continue progression
The energy factor is the amount of kW of power per square meter of the PV panels. Using the numbers provided by snrdg051306, the ISS panels produce about 14kW per square meter placing the TL between 16 and 17, which is about right for today.
Anyone got a better way?
rmfr
Wrote the above from memory, when I looked it up and I forgot one thing about PV panels. We also had an equation that compensated for how close you were to a star and its luminosity factor in terms of Sol.
P = s × F × L; where P = power generated, s = size of panel in square meters, F = factor as above, and L = luminosity factor in terms of Sol.
Even if the star is equal to Sol (our star), if you were in orbit about Jupiter, then L would equal 0.037 since Jupiter receives only about 51 W/m^2 compared to Earth's at about 1367 W/m^2. Thus, if you had TL20 PV panels and in orbit about Jupiter, they would produce 9.25 kW power per square meter (1 × 250 × 0.037).
If you were at the equivalent distance of Earth from an F2-V star, then those same PV panels would produce 3250 kW per square meter since the F2-V star would have a luminosity factor about 13× Sol (1 × 250 × 13).
Here is a table for each planet in our system showing the value for "L". Remember, the "L" factor is using the
mean solar irradiance.
Object | | "L" factor |
Mercury | | 7.577907827 |
Venus | | 1.91038771 |
Earth | | 1 |
Mars | | 0.441477688 |
Jupiter | | 0.037198244 |
Saturn | | 0.01100951 |
Uranus | | 0.00271763 |
Neptune | | 0.001100951 |
Pluto | | 0.000775421 |
Now you can kind of see why the Voyager probes needed those nuclear reacters for power out there at Saturn, Uranus, and Neptune, and beyond.
To find the flux (energy received) received from a star: F = L ÷ (4 × PI × d^2); where F = flux received, L = luminosity in terms of Sol, PI is the constant, d = distance from emitter in AUs.
rmfr