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Q & A: Solar Panels for Jet booster

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Most recent answer: 10/22/2007
Q:
How many solar panels are required to generate enough electricity to power a jet booster? And how long is the life-span of a solar panel?
- Vivien
A:
Well, we aren't going to work this out explicitly for you, particularly since there are a few variables which you may have to measure or estimate somehow to get a reliable answer. But here is a strategy for solving this problem.

A jet engine works by burning fuel in a combustion chamber and propels itself and whatever it is attached to by expelling hot exhaust gases out the back end. There really aren't such things as electrically-powered jet engines. Nonetheless, you can build something close to what you want with motorized fans.

The amount of power available in sunlight is about 1000 Watts per square meter on a sunny day with the sun right overhead. Naturally it will be less on a cloudy day and you will need more solar panels to run your booster on such days. Also, sunlight will be strongest at noon. You can tilt the panels to collect as much light as possible, but there will still be the effect of the sunlight passing through more air on its way in.

Solar panels can typically change about 10% of the incident sunlight into electrical energy. The other 90% gets reflected or heats up the solar panel. Some solar panels are better than others at converting these kinds of energy.

To determine how much energy you need, you will need to solve a problem of balancing forces. Your fans will have to run and give some energy and momentum to bits of air as they are blown downwards, in order to provide a force upwards to counteract gravity. The force of gravity is mass*g, where the mass is the total mass of the jet booster, solar panels, plus payload (whatever's being lifted by the jet booster), and g is the gravitational constant, 9.81 meters/second^2.

The force from the air is the change in momentum of the air in a unit of time. So, say, let V be the amount of air the fans blow per second, rho is the density of the air, and s is the speed at which the air goes out of the fans in the downwards direction. In one second, rho*V*1second*s is the change in momentum of the air, if the air was initially stationary before being grabbed by the fans. Dividing by one second to get the force, rho*V*s = mg gives the relationship between V and s which you need to satisfy to stay aloft. So you need to pick an s and a V for your fans which solves this.

The power to do this is the change in energy of the air per unit time. The energy change of a bit of air moved in one second is 0.5*rho*(V*1 second)*s^2. The power is energy per second, and is 0.5*rho*V*s^2. Using s and V from the above constraint, you can compute the power required. The motors will be mostly efficient (90% of the electrical energy will be converted into mechanical energy, and the rest is dissipated as heat), and I am not sure about the fans -- there may be losses due to air friction and turbulence in the fans. But at least this is a start. There is also an issue of optimization here. Because many choices of V and s will make the jet booster float, you may want to pick the one that corresponds to the least amount of power needed. Choosing a bigger V and smaller s means making bigger fans with fan blades turning more slowly. Bigger s and smaller V can be had with small, high-speed fans. A large fan will weigh more. If you can make very large fans which are lightweight, this may be the best way to optimize the energy consumption.

As for the lifetimes of photovoltaic cells, they usually last for decades unless they are damaged. Putting them on a jet booster may increase their likelihood of being damaged, however!

Tom

(published on 10/22/2007)

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