I've already presented the calculations in a number of previous posts, but what the heck, repetition is a mother of learning. Here goes.
Shaft power of a motor (any device producing torque, including turbines and windmills) is
P = 2 * pi * M * f
where M is the torque (in N*m) and f is the frequency of the spinning object (in Hz, or revolutions/s). According to this equation, a GM Vortec 6 liter engine, humming in a Hummer H2, running at peak-efficiency 4500 RPM and 502 N*m of torque outputs about 240 kW, which happens to exactly match the specs. So we're on track so far.
We have two unknowns that need to be found before we get to make any conclusions: the torque and the frequency.
The upper bound of the frequency will be dictated by the mechanical strength of our rig (because at some point the centripetal force will rip it to shreds). This is where we have to ballpark a little, because we don't have a CAD drawing of the mill at hand. You're saying that you want to have multiple smaller sail-like turbines, so let's say one will have a diameter of 20 meters (like the Ikaros) and have 8 triangular blades, looking something like this:
A little bit of trigonometry makes each blade an isosceles triangle with dimensions of 10, 10 and 8 meters and a height of 9 meters. Now let's assume the structure can withstand a pulling load of 100 kg (about 100× more than what the real sail can) - so we'll figure out when the centripetal force exceeds that. The center of gravity of such a triangle is about 6 meters from its tip - our mill's axis of rotation. Let's say it weighs only 1 kg - it's a real super material. The equation for centripetal force is:
F = m * r * omega²
where m is the mass, r is the radius and omega is the angular velocity in radians per second. Omega = 2 * pi * f, so we can modify the equation to
F = 4 * pi² * f² * m * r
and solve for frequency, since that's what we're after:
f = sqrt(F / (4 * pi² * m * r))
This equation says at which rotational frequency will the turbine disintegrate. Plugging in the values from the previous paragraph (F = 1000 N, m = 1 kg, r = 6 m), we end up with almost exactly 2 Hz. So the windmill will be able to barely withstand 2 revolutions per second.
The next unknown we need to find is the torque. According to this table, somewhere near Earth orbit the radiation pressure is little under 10 µPa (µN/m²). If we angle the blades at a 60° angle, exactly half of this pressure will contribute to our torque while still maintaining a sufficient surface area to capture sunlight. Each blade has a surface area of 37 m², but at a 60° angle the effective cross-section that's facing the sun in only 19 m². The torque is a product of force and a lever, in other words
M = F * d
plug in the pressures and the number of blades
M = A * p * d * 8
and for our numbers (A = 19 m², p = 10 µPa, d = 6 m) we get about 0.00912 N*m of torque. Going back to our shaft power:
P = 2 * pi * M * f
we get 114.61 mW. Not bad actually
looks like my LED statement was exaggerated a little. Sorry for that.
Which means you'd need 872 of these to power a 100W light bulb. Keep in mind, however, that all of this neglects any losses and assumes the generator runs at 100% efficiency. So it is in fact far from reality. Also the parameters were guessed in such a way to provide this idea with the gratest chance possible, e.g. there's no way you could have a rigid blade capable of withstanding 100 kg of force weigh only 1 kg. And who knows how long, if ever, would it take for this turbine to even start spinning and reach the two revolutions per second (120 RPM). And so on.
For comparison, a similarly sized solar panel with a run-of-the-mill (pun intended) efficiency of 20% would give you an energy output of 1366 * 0.20 = 80.9 kW, over 700,000x times as much, without a chance for a mechanical failure.
Another good comparison figure of merit is power per unit of weight - a 20% efficient photovoltaic cell gives you around 300 W/kg (assuming it's the same one ISS uses, at 1 kg/m²), while your 100% efficient machine (just the blades, I'm completely neglecting the gennies and other infrastructure) would give you about 14 mW/kg.
And in money terms - on an Atlas V Heavy ($4,000/kg) you'd pay about $285,714 per watt lifted into space, while a solar panel would cost just $13.35 per watt. That's over 20,000-fold difference.
Hope this clears up a thing or two
ChangeLog:
corrected the "For comparison" paragraph
Shaft power of a motor (any device producing torque, including turbines and windmills) is
P = 2 * pi * M * f
where M is the torque (in N*m) and f is the frequency of the spinning object (in Hz, or revolutions/s). According to this equation, a GM Vortec 6 liter engine, humming in a Hummer H2, running at peak-efficiency 4500 RPM and 502 N*m of torque outputs about 240 kW, which happens to exactly match the specs. So we're on track so far.
We have two unknowns that need to be found before we get to make any conclusions: the torque and the frequency.
The upper bound of the frequency will be dictated by the mechanical strength of our rig (because at some point the centripetal force will rip it to shreds). This is where we have to ballpark a little, because we don't have a CAD drawing of the mill at hand. You're saying that you want to have multiple smaller sail-like turbines, so let's say one will have a diameter of 20 meters (like the Ikaros) and have 8 triangular blades, looking something like this:
A little bit of trigonometry makes each blade an isosceles triangle with dimensions of 10, 10 and 8 meters and a height of 9 meters. Now let's assume the structure can withstand a pulling load of 100 kg (about 100× more than what the real sail can) - so we'll figure out when the centripetal force exceeds that. The center of gravity of such a triangle is about 6 meters from its tip - our mill's axis of rotation. Let's say it weighs only 1 kg - it's a real super material. The equation for centripetal force is:
F = m * r * omega²
where m is the mass, r is the radius and omega is the angular velocity in radians per second. Omega = 2 * pi * f, so we can modify the equation to
F = 4 * pi² * f² * m * r
and solve for frequency, since that's what we're after:
f = sqrt(F / (4 * pi² * m * r))
This equation says at which rotational frequency will the turbine disintegrate. Plugging in the values from the previous paragraph (F = 1000 N, m = 1 kg, r = 6 m), we end up with almost exactly 2 Hz. So the windmill will be able to barely withstand 2 revolutions per second.
The next unknown we need to find is the torque. According to this table, somewhere near Earth orbit the radiation pressure is little under 10 µPa (µN/m²). If we angle the blades at a 60° angle, exactly half of this pressure will contribute to our torque while still maintaining a sufficient surface area to capture sunlight. Each blade has a surface area of 37 m², but at a 60° angle the effective cross-section that's facing the sun in only 19 m². The torque is a product of force and a lever, in other words
M = F * d
plug in the pressures and the number of blades
M = A * p * d * 8
and for our numbers (A = 19 m², p = 10 µPa, d = 6 m) we get about 0.00912 N*m of torque. Going back to our shaft power:
P = 2 * pi * M * f
we get 114.61 mW. Not bad actually
looks like my LED statement was exaggerated a little. Sorry for that.Which means you'd need 872 of these to power a 100W light bulb. Keep in mind, however, that all of this neglects any losses and assumes the generator runs at 100% efficiency. So it is in fact far from reality. Also the parameters were guessed in such a way to provide this idea with the gratest chance possible, e.g. there's no way you could have a rigid blade capable of withstanding 100 kg of force weigh only 1 kg. And who knows how long, if ever, would it take for this turbine to even start spinning and reach the two revolutions per second (120 RPM). And so on.
For comparison, a similarly sized solar panel with a run-of-the-mill (pun intended) efficiency of 20% would give you an energy output of 1366 * 0.20 = 80.9 kW, over 700,000x times as much, without a chance for a mechanical failure.
Another good comparison figure of merit is power per unit of weight - a 20% efficient photovoltaic cell gives you around 300 W/kg (assuming it's the same one ISS uses, at 1 kg/m²), while your 100% efficient machine (just the blades, I'm completely neglecting the gennies and other infrastructure) would give you about 14 mW/kg.
And in money terms - on an Atlas V Heavy ($4,000/kg) you'd pay about $285,714 per watt lifted into space, while a solar panel would cost just $13.35 per watt. That's over 20,000-fold difference.
Hope this clears up a thing or two
ChangeLog:
corrected the "For comparison" paragraph
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