120V Reefer off Inverter?

[sailorchic34]

Originally Posted by kish

Assuming accuracy of 94 BTU (which it is suspect) and converting it to Amp-hours at 12 volt we have 94 BTU = 28 Watt-hour = 28 X (Volts x Amps)-hour = 28 x(12 x Amps)-hours
which is 336 Amp-hours at 12 volts

Something wrong in Sailorchick34 calculations

Oh dear, I have it wrong again. Everyone knows I don't know a amp-hr from an erg. Sorry. But how do you get 336 amp-hr's from 28 watt-hours. And you think my math is suspect.

Yes 94 BTU is equal to 28 watts. But you don't multiply watts by volts to get amps. You divide watts by volts to get amps.

So 28 watts divided by 12.5V (average system voltage) = 2.24 amp-hr extra per day. not 336 amp-hr.

Mind you, I was avoiding the whole enthalpy and moist air thingy and keeping things simple. I think my number is in the ball park anyway. Perhaps you can provide your own calc for a 3.5 CF fridge that is 3/4 full of food at 40 degrees F (4.4C)..

[Blue Crab]

This may help, and it's what I'd forgotten in the OP:

whats? / vim + vigor = ampules

After struggling with the maths and minimal cold air loss from my 12v+v toploader DammedFloss compressor = click clic chic ... carry the one ... /|\ ... ~5.5 ampules. Perhaps that's better than I thought considering it will freeze the ... er... paws off a polar bear and holds exactly one metric ton of canned adult beverages and manly-man meat. Rich, red, artistic-clogging meat. American meat.

Perhaps my primitive brain stem should be more careful watt it wishes for.

[svGenesis]

So i'm doing a complete refit on my boat. (the in-counter fridge compressor/plate is shot) I was going to refit with a norcold or isotherm fridge/freezer combo (7.0/7.8cf) at $1500 plus, and use the counter space as storage. Preferably I would like a 12/120v dorm fridge and a seperate chest freezer instead ($$$) (I'm putting 600 watts solar and already have 5k diesel genny) What are your thoughts on this? (carribean bound) Also the water heater is leaking, My thought was to install a small (1500watt) tankless as it will only be used for minutes a day showering (20-25 degree improvement) .

[newhaul]

Originally Posted by kish

But cold air has little mass so in reality not a lot of cold (actual BUTH) is loss. All the cold food stored in the fridge and freezer will easily recover the air temperature inside the fridge.
Incorrect as the specific heat capacity of air is low

sailorchic34 is similarly mistaken in describing energy usage as Amps/hour
Correct is Amp-hours
Her calculation of energy loss from opening door is suspect.

"At most, I'll loose 2 CF of air from the fridge. Air weights roughly 0.0855 pounds per CF so I would loose ~0.17 pounds of air, worst case. Assuming 90 degree air temperature (90-35 degree fridge=55 delta T) that would need (.17*55)= 9.45 btu's to recover. Allowing for 10 openings a day, you need to recover 94 btu's over a days time. That's roughly an extra 2.1 amp/hr's per day, worst case. It's just not that bad.
Allowing for 10 openings a day, you need to recover 94 btu's over a days time. That's roughly an extra 2.1 amp/hr's per day, worst case. It's just not that bad. "

"Not that bad" but very wrong in heat energy and electrical energy calculation,

Assuming accuracy of 94 BTU (which it is suspect) and converting it to Amp-hours at 12 volt we have 94 BTU = 28 Watt-hour = 28 X (Volts x Amps)-hour = 28 x(12 x Amps)-hours
which is 336 Amp-hours at 12 volts

Something wrong in Sailorchick34 calculations

The correct would be watts divided by volts equals amps so 28 wh divided by 12.5 nominal volts equals 2.24 ah
Where's stuM when we actually want his input.

[StuM]

Travelling.
SC's calc is the correct one - Watts / Volts.
And it's really nice to see that she has now grasped the hours part.

Now if we can all stop calling them Amp/hr and use the correct Ah or Amp hrs, I can enjoy the rest of my trip.

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[StuM]

SC is in the ball park with cost of opening a side door fridge Doing it in metric, I come up with:
SH of room temp air is about 1 Joule per gram per degree C
Density is about 1225 grams /m3
2 cubic feet is 0.057 m3 , so about 70 grams.
Assume 20C temp difference, that gives us 1400 Joules to re-cool the lost replacement air That is 0.38888 Watt hrs
So 10 openings would be about 39 Wh or 3.25 Ah at 12V

Near enough

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[kish]

Originally Posted by sailorchic34

True, But cold air has little mass so in reality not a lot of cold (actual BUTH) is loss. All the cold food stored in the fridge and freezer will easily recover the air temperature inside the fridge.

At most, I'll loose 2 CF of air from the fridge. Air weights roughly 0.0855 pounds per CF so I would loose ~0.17 pounds of air, worst case. Assuming 90 degree air temperature (90-35 degree fridge=55 delta T) that would need (.17*55)= 9.45 btu's to recover. Allowing for 10 openings a day, you need to recover 94 btu's over a days time. That's roughly an extra 2.1 amp/hr's per day, worst case. It's just not that bad.

So while a front loading fridge IS less efficient, It in reality is not that bad and has the advantage of not having a top load open more looking for that packet of food at the bottom. It's all trade offs.

Originally Posted by sailorchic34

Oh dear, I have it wrong again. Everyone knows I don't know a amp-hr from an erg. Sorry. But how do you get 336 amp-hr's from 28 watt-hours. And you think my math is suspect.

Yes 94 BTU is equal to 28 watts. But you don't multiply watts by volts to get amps. You divide watts by volts to get amps.

So 28 watts divided by 12.5V (average system voltage) = 2.24 amp-hr extra per day. not 336 amp-hr.

Mind you, I was avoiding the whole enthalpy and moist air thingy and keeping things simple. I think my number is in the ball park anyway. Perhaps you can provide your own calc for a 3.5 CF fridge that is 3/4 full of food at 40 degrees F (4.4C)..

There is a serious misinterpretation of Power and Energy here

To say "Yes 94 BTU is equal to 28 watts" is Incorrect

BTU is a unit of Energy and Watt is a unit of Power

so 94 BTU is 28 Watt-hours Not 28 Watts

28 Watt-hours is 28 *(Volts * Amps) * hours
and at 12 volts we have 28 *(12*Amps)*Hours = 336 AmpHours.

So opening the reefeer door 10 times per day will trash a battery if the calculation on the air changes is correct!!

[darylat8750]

Originally Posted by kish

There is a serious misinterpretation of Power and Energy here

To say "Yes 94 BTU is equal to 28 watts" is Incorrect

BTU is a unit of Energy and Watt is a unit of Power

so 94 BTU is 28 Watt-hours Not 28 Watts

28 Watt-hours is 28 *(Volts * Amps) * hours
and at 12 volts we have 28 *(12*Amps)*Hours = 336 AmpHours.

So opening the reefeer door 10 times per day will trash a battery if the calculation on the air changes is correct!!

When I add 2+2 sometimes I get 4, sometimes not. I try to avoid higher math like multiplying and dividing. I shan't get involved in this other than to say SailorChic has been living on the hook full time (with what? Less than 400w of solar power?) for several years and her system works. I use more energy a day than she does but have an apt. sized 10+ cubic foot off the shelf refrigerator powered by an inverter and a 5kw genset for 1 to 1.5 hours a day when on the hook and it has worked fine for me. I have a trawler and when moving the boat I need about 3 hours of engine run to avoid needing to run the genset. I have spent way less on extra fuel for the genset than a dual voltage reefer would have cost. Solar is on the wish list....

[sailorchic34]

Originally Posted by kish

There is a serious misinterpretation of Power and Energy here

To say "Yes 94 BTU is equal to 28 watts" is Incorrect

BTU is a unit of Energy and Watt is a unit of Power

so 94 BTU is 28 Watt-hours Not 28 Watts

28 Watt-hours is 28 *(Volts * Amps) * hours
and at 12 volts we have 28 *(12*Amps)*Hours = 336 AmpHours.

So opening the reefeer door 10 times per day will trash a battery if the calculation on the air changes is correct!!

Both BTU and watt are units of energy and easily converted between the two. Least wise with imperial units.

Note that the 28 watts is not watt-hours but simply watts required over ten door openings in a day. Yes we can call it 28 watt-hour. But that's not consumed per hour, but consumed in a day.

Also remember that a 12V system actually operates at levels a bit higher then 12V. Average voltage varies between 14.6V (when the sun is shining at noon ) to about 12.4V, least wise on my boat I used 12.5 volts as an average voltage.

That both StuM (who is a rather smart fellow) and I agree, gives great credence that we are correct. It's a simple calc and would vary a bit, based on cabin temperature, fridge temperature and amount of dense cold air in the fridge. When I start a voyage the fridge is packed and there is very little air space. After a week there is more and after 2 weeks lots O space.

Rather then just saying I'm wrong, perhaps you could post your own calculations for air exchange and cooling required.

[Dymaxion]

Originally Posted by sailorchic34

Oh dear, <snip>
.
.

Mind you, I was avoiding the whole enthalpy and moist air thingy and keeping things simple. I think my number is in the ball park anyway. Perhaps you can provide your own calc for a 3.5 CF fridge that is 3/4 full of food at 40 degrees F (4.4C)..

uhmm, is that 40 degrees a wet bulb or dry bulb temp?

My issue around domestic fridges is that the coils that reject the heat are just under the side skins of the unit, foamed in place. So the fridge rejects heat out sideways. (PS: never use a 2" hole saw to cut an opening through the side for your beer keg tap.... don't ask ) If the fridge is slid in between galley cabinets it simply can't dump out the heat properly, and you can't just add a fan either. Not to plug Isotherm, but the coils are at top/rear and have a fan, so heat rejection is easy, you just need somewhere for the air to blow out. Oh... domestic fridges don't generally like to be heeled over at 30 degrees for sustained periods. Not that I do either.

[sailorchic34]

Originally Posted by Dymaxion

uhmm, is that 40 degrees a wet bulb or dry bulb temp?

My issue around domestic fridges is that the coils that reject the heat are just under the side skins of the unit, foamed in place. So the fridge rejects heat out sideways. (PS: never use a 2" hole saw to cut an opening through the side for your beer keg tap.... don't ask ) If the fridge is slid in between galley cabinets it simply can't dump out the heat properly, and you can't just add a fan either. Not to plug Isotherm, but the coils are at top/rear and have a fan, so heat rejection is easy, you just need somewhere for the air to blow out. Oh... domestic fridges don't generally like to be heeled over at 30 degrees for sustained periods. Not that I do either.

Dry bulb of course....

You are correct that the condensers are mounted to the exterior skin and you really can't mount them inside a cabinet as they need some air space on each side. As to the heeling, I've sailed at 20+ degrees for hours on end and have had no issues with the 120V fridge. Really the basic design and construction features of the 12V danfoss compressor is almost the same as the typical 120V compressor.

I've found that heeling thingy to be a non-issue.. Mounting the fridge does take care. I have it bolted through the rear compressor frame and with angle brackets at the top back. I also have a SS clasp to hold the door closed when healing. Finding a 120V with an external condenser would be nice, but I don't think they are made. Least wise I've not found one yet.

[StuM]

Originally Posted by sailorchic34

Both BTU and watt are units of energy and easily converted between the two. Least wise with imperial units.

Note that the 28 watts is not watt-hours but simply watts required over ten door openings in a day. Yes we can call it 28 watt-hour. But that's not consumed per hour, but consumed in a day.

Oh dear, and you were doing so well

BTUs are energy. Watts are power. The SI energy unit is indeed the Watt hour.

Although the rest of your calculations are OK , on this point you are again mistaken.


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[StuM]

Originally Posted by Dymaxion

uhmm, is that 40 degrees a wet bulb or dry bulb temp?

My issue around domestic fridges is that the coils that reject the heat are just under the side skins of the unit, foamed in place. So the fridge rejects heat out sideways. (PS: never use a 2" hole saw to cut an opening through the side for your beer keg tap.... don't ask ) If the fridge is slid in between galley cabinets it simply can't dump out the heat properly, and you can't just add a fan

Those "coils" do very little cooling - they are,as you say, surrounded by foam - which is a very poor conductor. What do you think that big radiator is for on the back of a domestic fridge?

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[newhaul]

Originally Posted by StuM

Those "coils" do very little cooling - they are,as you say, surrounded by foam - which is a very poor conductor. What do you think that big radiator is for on the back of a domestic fridge?

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Stu actually most of then domestic fridges stateside have the coils either inside the back skin or more often a smaller unit in the bottom with the compressor

[sailorchic34]

Originally Posted by StuM

Oh dear, and you were doing so well

BTUs are energy. Watts are power. The SI energy unit is indeed the Watt hour.

I know, we were doing so well too. Actually both BTU's and watts are power. Power is measured in watts, kilowatts, horsepower, Btu/h, tons of cooling, and foot-pounds/minute.

Maybe is a metric/imperial thingy...


Edit: OK BTU is energy and btuh (common U.S. engineering term for btu-hr) is power. Yet I was calculating BTU's and not btuh. That is Btu's required per door opening. Then converted to watts for 10 door openings. So is it a watt or a watt-hr in that case? If it is watt-hr, where did the time unit come into play in the calculations?

To be honest I use BTU and BTUH as pretty much everything I do, building energy systems wise is hour to hour anyway. In this case we're talking heat gain from air exchange and energy (or is it power) required to overcome that heat gain.