Mechanical losses in an electrical system? You mean thermal losses. There are higher losses is a belt driven system due to heat. Heat generated in the belt by flexing around the pulley. Also heat from the compression of the air. Since you are generating 6 psi instead of .5 psi, there are also much higher gains. You are only focusing on the losses and are missing the big picture. Losses only take away from the gains. If the gains are very small the losses will be small. If the gains are big the losses could be larger. The net gain is what actually makes a difference. So it doesn't matter how efficient the electric fan is because the net gains are so small. To put it another way, you would probably only be satisfied with minimum losses to the system. You would need a very long extension chord to power the fan. This would represent a 0 loss because no alternator power would be used to power the fan. But I have news for you, even with 0 system loss you are still only generating .5 psi for a 2 HP gain. Still way off from a more traditional supercharger that can produce 6 psi or more for say 50 hp gain.
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