Monday, Dec 27, 2004 at 19:40
Nudenut,
No one is, has, or does deny the actual inverter uses energy to work..
READ as; NO ONE, no person and no company, now or in the past!
however the amount of power the actual inverter uses is miniscule!
The fact is;
an inverter does use electrical energy to power a 240v compressor, HOWEVER that compressor in actual FACT pushes MORE refrigerant around the fridge, because the 240v compressor capacity is far larger, therefore cooling the inside product, food, far quicker than a smaller 12v compressor, and so in any reasonable time period eg; 24hours the inverter driven fridge is working for far LESS TIME (not screaming - just letting you see the truth) repeat, less time, and does use less
battery power over a longer period of time, than a smaller 12v compressor that in the higher ambient temps of Northern Aussie can be working for 18 hours daily if not full time, in the back of a vehicle.
#Sealed Compressor Efficiency
Refrigeration compressor types have a very differing ability to convert the INPUT energy consumed into energy OUTPUT.
This in fact is called the co-efficient rating.
The co-efficient factor of a typical piston compressor operating as a freezer would be 0.92, yet as a refrigerator this figure would increase to 1.55.
Cabinet Temp -- (Piston Compressor) --- {Rotary Compressor}
-15ºc Freezer, -- (0.92) --- {1.09}
+ 2ºc Refrigerator, -- (1.55) --- {1.82}
#Calculating Power Consumption
Having selecting the correct co-efficient factor, it is now possible to calculate the amount of
battery power to be consumed.
(Amp hours at 12 volts, per day) Amp hours consumed per day is the total heat load divided by the relevant co-efficient factor, then divided by 12 (Power supply volts)
Example:
A 50 litre cabinet in a 40ºc environment (616 THL), used as a refrigerator with a rotary compressor system (1.82 CEF) powered from a 12 volt battery, would calculate as;
616 divided by 1.82, divided by 12, equals 28.2 Amps per day
or an average consumption of 1.18 Amps per hour (on a daily basis).
The hours of running time per day depend on the capacity and efficiency of the refrigeration unit applied.
A small system with low rated current draw will ALSO have a low refrigeration OUTPUT, therefore the co-efficient factor and adequate OUTPUT should be the major considerations for BEST ECONOMY and assurance of required performance.
NudeNut,
I do hope you can actually understand the above information and as you so vehemently believe you are right; I ask you to "prove beyond any reasonable doubt" the above information is incorrect..
IF you can, then yes I will shut up as you have requested of me.. if not an apology will be accepted
(however not expected as it takes a man to admit when he is wrong)
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