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)
