[email protected] 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.
Correct. It radiates into the surroundings, from the reactor and the
heat exchanger. Lewan reported the reactor surface was "between 60°C and
85°C." I presume he means at different times. I do not know how he
measured that. It has a lot of surface area so it is radiating a lot of
heat. Someone better physics and I can estimate how much.
With a really good calorimeter having a high recovery rate, nearly all
the heat ends up captured by the calorimeter. With the flow calorimeter
it ends up heating the water. With a Seebeck calorimeter it may radiate
out into the room, or if there is a water bath on the outside shell of
the chamber to ensure a stable background, it will be captured by the
water bath.
This reactor is too large for most Seebeck calorimeters.
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
- Jed
