It is correct because otherwise, it would leave the pipe without bubbling! :)
2011/11/19 Peter Heckert <[email protected]> > Am 20.11.2011 01:38, schrieb Daniel Rocha: > > Not if the output is actually much higher than the input! > > He doesnt measure the output mass flow. > He always assumes this equals the input mass flow and it is all vaporized. > From this assumption he calculates the output energy. > If this assumption is wrong, then the calculation is wrong. > Peter > > > 2011/11/19 Joshua Cude <[email protected]> > >> >> >> On Sat, Nov 19, 2011 at 6:10 PM, Daniel Rocha <[email protected]>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. >> >> > >
