Am 07.10.2011 16:59, schrieb Jed Rothwell:
peter.heck...@arcor.de wrote:
If the heat exchanger has only 60% efficieny, then the energy loss is
5kW * 0.4 = 2kW.
Where does the enrgy go? Energy cannot vanish magically, it must go
into the ambient.
I think even if the heat exchanger at this size (as visible in the
video) has no insulation, it cannot lose 2kW.
It is well isolated and the loss must be much lower.
I believe the heat exchanger plus the reactor itself can radiate 2 kW.
They look crude to me. Such things are inefficient. See photo of the
two of them (in one box):
http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29
I cannot calculate this, I can only estimate it by comparison with known
devices:
I live in rooms directly under the roof.
I have a gas boiler 10 kW. This heats water, that is pumped through
copper pipes and these are connected to 5 radiators.
Because I live under the roof, the pipes are partially on the outside of
the walls. They are still under the roof but exposed to the cold winter
air. They are isolated by glass wool and alu foils, just as the e-cat.
The isolated copper pipes are in a length of about 5m exposed to the
winter air.
The water temperature is max. 80 centigrade. Now imagine it is outside
under the roof -10 centigrade. Then I must loose several kilowatts of
heating energy in winter! Possibly I should check this.
Peter
