Whynter refrigerator just about killing my battery after 12 hours

luthj

Engineer In Residence
The actual energy content of the air in a fridge is tiny. Even if you swapped the air in the fridge out 30-50 times a day, its still less energy than it takes to cool down a bottle of water.

The specific heat of air is about 0.718kJ/kg C while water is 4.18kJ/kg C. A bottle of water is about 1kg, so 4kJ. With a 20C difference between room temp and fridge temp, a bottle of room temp water will need 80kj of energy removed to bring down to fridge temp. 80kJ=22 watt-hours.

To equal that amount of energy, you would need to cycle 5kg of air through the fridge. 5kg of air is about 4.5 cubic meters of air. For a 60 liter fridge that is completely empty, that would require replacing the air in the fridge 75 times.

Insulation is much more important than door type. High efficiency home units will have 4-6" of insulation, while many mobile units will only have 2-3"

If you wanted to offset the energy lost from air leaving the fridge, you could just put your condiment bottles away immediately after use. Or avoid putting warmed food directly into the fridge after eating.
 
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john61ct

Adventurer
Air has very little thermal **mass**.

That is what is important to reducing cycling, keeping the space full of cooled down **contents**.

The amount of energy needed to cool air is insignificant compared to that.

Finding ways to reduce the frequency of opening the unit, and the time spent with it open, will have **much** more impact than front opening vs chest style.
 

67cj5

Man On a Mission
The fridge ran 12 hours last night on a mostly charged new battery. Voltage this morning was 12.5 and the Jeep started right up. I'm annoyed I was chasing a battery problem but happy it was just a battery problem! My Whynter seems to draw around 4.5 amps with the compressor running at 12v.
If you can time it's on cycles and it's off cycles for a day and tell me what each one is in minutes and seconds I can tell you exactly how much power it is using if it helps.
 

67cj5

Man On a Mission
To see how the door style effects a freezers behaviour you only have to open them when they have been left closed for a day or two when set to -18*c/0*f and below because White Fog/Mist will pour out the door and fall as you open the door and with a chest type it will either rise slowly or just swirl around inside.
 

john61ct

Adventurer
The point is not what the air does, but that that doesn't much matter.

An AH meter, recording ambient & target temps vs AH used per 24hrs, allows anyone to verify these variables themself, it's not rocket science.
 

67cj5

Man On a Mission
The point is not what the air does, but that that doesn't much matter.

An AH meter, recording ambient & target temps vs AH used per 24hrs, allows anyone to verify these variables themself, it's not rocket science.

Yeah along with recording the duty cycles I also photograph the Watt meters.
 

OpenTrackRacer

Observer
If you can time it's on cycles and it's off cycles for a day and tell me what each one is in minutes and seconds I can tell you exactly how much power it is using if it helps.

Thanks for the offer but I have an Ah meter. I'm going to set it up on my bench supply and do a 24 hour test tomorrow night (going to run another overnight test for a longer period tonight).
 

luthj

Engineer In Residence
A difference that might matter with regards to the testing done by the MFGs of these fridges. The voltage of the power supply may vary. Some may use 12V, some may use 14V. Watt=energy With knowing the exact voltage the fridges are tested at, there could be a 10-20% variance in numbers. From that factor alone.

Personally I would prefer if the MFGs all reported their numbers in watt-hours.
 

OpenTrackRacer

Observer
A difference that might matter with regards to the testing done by the MFGs of these fridges. The voltage of the power supply may vary. Some may use 12V, some may use 14V. Watt=energy With knowing the exact voltage the fridges are tested at, there could be a 10-20% variance in numbers. From that factor alone.

Personally I would prefer if the MFGs all reported their numbers in watt-hours.

Agreed. That Whynter chart showed 12v and 4.5 amps is about right based on my testing at that voltage but the voltage is definitely going to vary in real world use.
 

67cj5

Man On a Mission
A difference that might matter with regards to the testing done by the MFGs of these fridges. The voltage of the power supply may vary. Some may use 12V, some may use 14V. Watt=energy With knowing the exact voltage the fridges are tested at, there could be a 10-20% variance in numbers. From that factor alone.

Personally I would prefer if the MFGs all reported their numbers in watt-hours.
most of them do give you the W/H per year, I know ARB does. As do Dometic/Waeco, I have tested mine and recorded readings.
 

DaveInDenver

Middle Income Semi-Redneck
The actual energy content of the air in a fridge is tiny. Even if you swapped the air in the fridge out 30-50 times a day, its still less energy than it takes to cool down a bottle of water.

The specific heat of air is about 0.718kJ/kg C while water is 4.18kJ/kg C. A bottle of water is about 1kg, so 4kJ. With a 20C difference between room temp and fridge temp, a bottle of room temp water will need 80kj of energy removed to bring down to fridge temp. 80kJ=22 watt-hours.

To equal that amount of energy, you would need to cycle 5kg of air through the fridge. 5kg of air is about 4.5 cubic meters of air. For a 60 liter fridge that is completely empty, that would require replacing the air in the fridge 75 times.
Your calculations don't include the compressor coefficient of performance and motor losses, the heat removed to regain equilibrium has to occur through the compressor and motor doing some non-zero amount of work. It may be a small difference, I don't know a typical real efficiency of a fridge. Are these losses trivial?

Assuming the mechanical and electrical losses are significant, is there hysteresis in typical controllers so they know to let the stored mass attempt to cool the exchanged air rather than cycling the compressor immediately upon sensing the change in air temperature?

A chest internal temp should be more stable without those exchanges and so they probably only cycle when it actually needs to cool something. Perhaps abrupt, abbreviated cycles accumulate into real differences?
 

luthj

Engineer In Residence
Your calculations don't include the compressor coefficient of performance and motor losses, the heat removed to regain equilibrium has to occur through the compressor and motor doing some non-zero amount of work. It may be a small difference, I don't know a typical real efficiency of a fridge. Are these losses trivial?

Assuming the mechanical and electrical losses are significant, is there hysteresis in typical controllers so they know to let the stored mass attempt to cool the exchanged air rather than cycling the compressor immediately upon sensing the change in air temperature?

A chest internal temp should be more stable without those exchanges and so they probably only cycle when it actually needs to cool something. Perhaps abrupt, abbreviated cycles accumulate into real differences?

You are making this more complicated than needed. The fridge has a temperature sensor. It takes several minutes for it to adjust to changes in the fridges internal temp. If you open the door the air temp in the fridge will change for a minute or so, but the contents of the fridge will bring the temp back down quickly. This is generally not enough to cause an immediate cycle. The energy you just gained (cold air lost) will be removed from the fridge at the next cycle.

Most chillers (AC systems, fridges etc) have a coefficient of performance between 2-4. That means that for every watt hour of electrical energy they consume, they remove 2-4 watt hours of power from the fridge (as heat). So that bottle of water which needs 22 watt hours of cooling will need 5.5-11 watt hours of electricity to recover. That is about 0.5-0.8AH at 12V.
 

67cj5

Man On a Mission
You are making this more complicated than needed. The fridge has a temperature sensor. It takes several minutes for it to adjust to changes in the fridges internal temp. If you open the door the air temp in the fridge will change for a minute or so, but the contents of the fridge will bring the temp back down quickly. This is generally not enough to cause an immediate cycle. The energy you just gained (cold air lost) will be removed from the fridge at the next cycle.

Most chillers (AC systems, fridges etc) have a coefficient of performance between 2-4. That means that for every watt hour of electrical energy they consume, they remove 2-4 watt hours of power from the fridge (as heat). So that bottle of water which needs 22 watt hours of cooling will need 5.5-11 watt hours of electricity to recover. That is about 0.5-0.8AH at 12V.

You can't work it out like that regarding a bottle of water because under real world usage you would not run a fridge for just one bottle of water, When in reality you are trying to cool all the items, So if you had 31 items in the fridge and you added a bottle of water then you would have to divide the wattage by the number of Items but that is only useful if you are trying to work out the cost of cooling each item, but in real world we just want it all to cool down no matter if there is 30 items in there or 55 items.

How the fridge is packed does have a big impact on how much power it uses, As does the setting and the Ambient temp along with the type of packaging that the food's are packed in, One thing about all this testing is that I can normally tell how it is packed once it has cycled once and by the second run time you can workout how it is going to run.
 

luthj

Engineer In Residence
You can't work it out like that regarding a bottle of water because under real world usage you would not run a fridge for just one bottle of water, When in reality you are trying to cool all the items, So if you had 31 items in the fridge and you added a bottle of water then you would have to divide the wattage by the number of Items but that is only useful if you are trying to work out the cost of cooling each item, but in real world we just want it all to cool down no matter if there is 30 items in there or 55 items.

How the fridge is packed does have a big impact on how much power it uses, As does the setting and the Ambient temp along with the type of packaging that the food's are packed in, One thing about all this testing is that I can normally tell how it is packed once it has cycled once and by the second run time you can workout how it is going to run.

No. If the fridge is stabilized (steady state) adding a bottle of water etc will increase the cooling load by the amount I calculated. Obviously you have the existing cooling load + 22 watt-hours. This energy usage (about 8 watt hours of electricity) will be spread out over an hour or two. It is a specific and easily calculated value. I work with this type of stuff for a living, this is an engineers bread and butter really.

The whole purpose of this exercise in physics is to describe/define the size of various factors on a fridges cooling load. As described above air loss is a minor impact on cooling load for fridges that are not opened and closed constantly.

A fridge is system, in physics energy in minus stored energy must equal energy out.
 

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