Electric Heater using Light Bulbs as Heat Source-French Patent

[F6FLT]

Hi SkyWatcher123

Nice setup but inconclusive measurement.
May I suggest you immerse the light bulbs in a given quantity of distilled water in a plastic basin. Even if there is some current leakage in the water, it will not account for much due to the high resistivity of distilled water, but be careful about safety.
By measuring the temperature increase over time while stirring water to equalize the temperature, then you can calculate the minimum thermal energy that will be produced, thanks to the known heat capacity of water, and compare it to the electrical energy that will be consumed. A 5 minutes test should be enough.

[lancaIV]

Hi SkyWatcher123

Nice setup but inconclusive measurement.
May I suggest you immerse the light bulbs in a given quantity of distilled water in a plastic basin. Even if there is some current leakage in the water, it will not account for much due to the high resistivity of distilled water, but be careful about safety.
By measuring the temperature increase over time while stirring water to equalize the temperature, then you can calculate the minimum thermal energy that will be produced, thanks to the known heat capacity of water, and compare it to the electrical energy that will be consumed. A 5 minutes test should be enough.

J'accuse: by such a kind of measurement the experimentor will loose fairly 50% from the thermodynamic infraredemission effect which is based by radiation reflection.( more : Prof. Dr. Dipl- Ing. Claus Meier and Prof. Dr Dipl.-.Ing. Klaus Matschiner ).
In the U.S. and Canada there were several experiments whith comparison from light bulbs and heatpads which have shown 2/3 energy savings ( more with controler) for the heat pads.(lower surface temperature and greater radiation area )
It is also the kind of radiation( near or far infrared) which let us consumer feel well or freezing.The negative for the light bulbs, now envelopped will be that the 750 hours function validity will get shorten.The bulb and filament itself will heat up cause the transparency for the bulb external radiation is neutralized by the aluminium foil.
Sincerely                OCWL

[znel]

Measure inlet and outlet temperatures - Outlet - Inlet = Temperature difference.
If you have an Anemometer your can measure air speed at the outlet.  ( Amazon anemometer $13.00 )
Airspeed in mph x 5280 / 60 = feet per min.
Measure outlet dimensions and calculate sq ft.    Square ft outlet x feet per min = Total CFM.
CFM x 1.08 x Temp diff = BTU   3.41 BTU is the equivalent to 1 watt.   

[lancaIV]

Measure inlet and outlet temperatures - Outlet - Inlet = Temperature difference.
If you have an Anemometer your can measure air speed at the outlet.  ( Amazon anemometer $13.00 )
Airspeed in mph x 5280 / 60 = feet per min.
Measure outlet dimensions and calculate sq ft.    Square ft outlet x feet per min = Total CFM.
CFM x 1.08 x Temp diff = BTU   3.41 BTU is the equivalent to 1 watt.

BTU or British Thermal Unit
https://worldwide.espacenet.com/publicationDetails/biblio?FT=D&date=19900213&DB=EPODOC&locale=en_EP&CC=US&NR=4900898A&KC=A&ND=4#
11000 BTU ~ 3225 Watts

[lancaIV]

I agree with that. Any electrical heater has an efficiency of 100%, including a bulb that is prevented from spreading light due to a housing.
The issue of electric heaters for a home is more about how the heat will be distributed (for example large surfaces at lower temperature give a better feeling than very hot point sources).

Related ".. has an efficiency of 100% ...."
out-/input or only output

http://www.canlaser.com/en/Hhm.aspx.   How did the Uni engineers measured the performance ?

https://translate.google.com/translate?hl=de&sl=uk&tl=en&u=http%3A%2F%2Faverage.com.ua%2F
http://average.com.ua/

https://patents.google.com/patent/US6741805B2/en    ".....Energy savings of up to about 75% ...."