Not if the output is actually much higher than the input! 2011/11/19 Joshua Cude <joshua.c...@gmail.com>
> > > On Sat, Nov 19, 2011 at 6:10 PM, Daniel Rocha <danieldi...@gmail.com>wrote: > >> That means that part does not leave. > > > That could work for a while, but eventually the ecat would fill up. > Anyway, Rossi always uses the input flow rate to calculate the output > power, so he is assuming it is coming out. If it's not, then his > calculations are wrong. > >
