Simple cooler

[MrAl]

I had a peltier cooled coolbox. It easily maintained a 25°C (45°F) differential and didn't take that much current (can't remember but it didn't worry the car battery).

Let's do some calculations,
A box of 1m x 0.5m x 0.5m and a wall thickness of 25mm of styrofoam would have 2.5m² of wall area. At an Rsi value of 1.25 it would require 2.5/1.25/1 = 2W per °K to cool it. So for a temperature differential of 20°C (36°F) would require 40W of cooling. Two reasonable sized units would easily suffice. An aircon unit would be total overkill.

Note, the R value is in m²·K/(W·in) hence 25mm = 1".

Edit, two of these should manage to move 60W at a 20°C differential and use 8A at 12V.

Mike.

Hi there Mike,

What kind of efficiency we talking here for the Peltier device?

BTW the link in your other post was very easy to spot this time. Bold text seems to work nicely like underlined text


MrRB:
Heating is much more efficient than cooling as im sure you know.

 

[zoner21]

You seem to have a basic lack of understanding of what a cooling unit does. It is impossible to create cool, it's the first law of thermodynamics... Compressor or thermoelectric modules are nothing more than heat pumps. Thermoelectric coolers have VERY limited temperature differentials, and it being solid state will mean nothing if you don't fully engineer the cold/hot side to properly dissipate the modules heat pumping capacity. This is ALL that AC units do, they have a hot side and a cold side exchange, I'm not sure why frosting was brought up as a problem if your cooling requirements are 55 degrees...

The logistics of this are NOT that hard to grasp. You're over analyzing, virtually any room AC unit wil do, you don't even have to redesign the temperature controller, you just have to move the sensing element to the most important spot in your cooler. Keep in mind air flow.

I don't lack any such understanding, buddy. I understand what a heat pump is. I get thermodynamics. I'm not a moron. I don't know where you're coming from if this is aimed at me...maybe I read it wrong.

I understand that we're just moving heat around...which is why I cite a potential use of this project as a heated box as well as a cooled box. Let me know where I've missed the physics on this, I'll rearrange my world view.

Also, no, any AC unit will not do. The requirements are that the enclosure be maintained at a set temperature, not just kept cool. An AC unit also wouldn't be easy to attach to a sealed box. And since I need temps below 60, as I said before, an AC unit will not work. I have never seen one that will go below 60. My understanding for this is (besides the fact that nobody wants a room that cold) the design doesn't allow for the constant running required to achieve these low temps. The system will ice over on the cool side.

If it were that simple to keep an enclosed space at a set temp they wouldn't sell dedicated units that are engineered specifically to do that one thing (yes I know I could just buy one of those units and tinker with it, but whatever).

The rest of you, thank you very much for your help so far.

 

[Sceadwian]

Sorry, didn't mean to come across so harsh zone, never do, but you've never looked up the efficiency ratings. TEMs are in the neighborhood of 5-10% efficient, where compressor based refrigeration systems are 40-60% efficient, TEMS do NOT scale well.

I would challenge the previous statement of a well insulated container making this practical, but the thought hasn't been put into the amount of latent heat that has to first be removed and how long it will take, and no mention has been made of the heat generated by the fermentation process itself.

 

[zoner21]

Those are actually good points. Fermentation generates a decent amount of heat (can raise temp 5F in a day for five gallons), but more important is the latent heat. Brew starts off at boiling, cools to ~80 by addition of cold water and just sitting out, then goes into the cooler after a day or two...there is no rush to get the temp down quickly, and it may actually be beneficial to drop the temp slowly, but five gallons of hot water has a lot of energy in it.

As I haven't looked at the efficiency issue yet, how big of a deal is this? I mean, is it going to show up on my power bill in a noticeable way? If it isn't going to use exorbitant amounts of power, I'm not too concerned that it isn't efficient. Also, will the hot side be throwing off a TON of heat for a couple of days straight to drop the temp inside initially? I really like the idea of a container with a set temperature (maybe cooler was the wrong word to use in the initial post) so I can brew different styles any time of the year. As it stands now, I can only brew iffy lager in the winter and ales in the summer. I prefer to drink them in their opposite season and they never seem to sit around long enough to make it.

As I'm looking at parts to determine if this is as stupid as it seems right now, does anybody have any idea how good computer CPU coolers would be for heatsinks w/fans for this application? I have several Zalman coolers with fans on them that I could clamp on both sides of the Peltier with thermal paste.

 

[Sceadwian]

You need to determine the wattage rating of 5 degrees of temperature for 5 gallons of water, and translate the wattage, OUTSIDE of the wattage required for static stability once the desired temperature is reached.

Look up the Joule rating of 5 degrees temperature differential in 5 gallons of water, figure out the watt hours you have to deal with. The science will get more complicated from here.

 

[zoner21]

BTUs seemed simpler than joules to start with...idunno why. So 5 degree shift for 5 gallons of water is 207.5 BTUs = 218 925 joules. 218 925 joules over 24 hours gives ~2.5 watts. N'est-ce pas? If I did that wrong, let me know. It's been one hell of a long time since I was in physics class. So...anyway that's 61 watt-hours. If my cooler is 5% efficient (worst case scenario?), that gives 1220 watt-hours. That's 37 kWh a month, and that's what I got for containing fermentation heat. So now I need to figure out how much heat will get through the insulation and I'll have a decent idea of worst-case cost for power.

I'm not all that confident about my math, its been a long time since I've played with power units. Let me know if I done goofed.

 

[Pommie]

The efficiency is very easy to work out. Quote "to move 60W at a 20°C differential and use 8A at 12V". That is 96W to move 60W.

I hadn't thought that the process itself may produce heat but as I allowed 50% spare capacity (60W when 40W required), I figure it'll work ok.

So now I need to figure out how much heat will get through the insulation and I'll have a decent idea of worst-case cost for power.

That's the calculation I did above.

Look up the Joule rating of 5 degrees temperature differential in 5 gallons of water, figure out the watt hours you have to deal with. The science will get more complicated from here.

Not compilcated at all if you use joules. Five gallons is ~20Ltrs. To cool 20 Ltrs by 1 degree requires 4*20000*1=80,000 Joules. If we remove 60W per hour, that is 60*3600 = 216,000J per hour. So, we can cool our liquid by about 2°C per hour.

Mike.

 

[zoner21]

I'd have to adjust that calculation you did above for the actual size of the container, but that makes it very simple. If it takes 96W to move 60W...isn't that around 60% efficient?

 

[Pommie]

The efficiency depends on the temperature differential. Assuming 20°C difference then it'll need 40W and consume 12*5=60W of electricity (from graph in data sheet). At 15c per kWh it will cost 0.15*60*24/1000 = 22c per day to run. (15c just plucked from the air, insert actual cost for a better estimate)

To calculate watts required you do, Temperature difference (°C) * Surface Area (m²) / Wall Thickness (inches) / R (1.25).

Note, you can half the watts required by doubling the wall thickness.

Mike.

 

[Mr RB]

...
Note, you can half the watts required by doubling the wall thickness.
...

Or wrap the keg in a few layers of bubblewrap, then put it in styrofoam. Bubblewrap is an awesome insulator in still air.

...
Heating is much more efficient than cooling as im sure you know.

Agreed! But I believe both Pommie and myself were talking actual cooling watts, not power consumed by the peltier. I think Pommie's right on the money with 40W actual, ie a 60W or 80W peltier, or a couple of 50W units. And don't forget the big heatsinks and PC fans on the outside.

 

[zoner21]

Okay, so assuming I wanted to go ahead and do this, how would I go about wiring my 2 TEMs to a temperature controller so that they can both heat and cool?

 

[MrAl]

Or wrap the keg in a few layers of bubblewrap, then put it in styrofoam. Bubblewrap is an awesome insulator in still air.



Agreed! But I believe both Pommie and myself were talking actual cooling watts, not power consumed by the peltier. I think Pommie's right on the money with 40W actual, ie a 60W or 80W peltier, or a couple of 50W units. And don't forget the big heatsinks and PC fans on the outside.

Hi MrRB,

Ok that's fine in a way, but it's also misleading. If we had a Peltier that was 10 percent efficient and we were talking "40W actual cooling" that would mean we would have to input 400 watts, which is nothing even remotely close to 40 watts. Im sure you can see where i am going with this and that's why i brought up the efficiency. Im not up on the most recent Peltiers though, so if someone wants to slip a data sheet into this thread now would be a good time to do it

 

[Pommie]

I posted a data sheet and all calculations earlier. The peltier I linked to would be 60% efficient. I don't understand why people think this is a bad idea!!

Mike.

 

[MrAl]

Hi Mike,

Well, you posted that link as regular text again so it was easy to miss. That's very non standard for web usage, where the typical link is underlined. If you're going to go through the trouble to type it at all why not make it underlined? That way it is very apparent that it is a link.

According to the data sheet the efficiency can be between 15 and 61 percent, depending on the required temperature differential.
What final temperature is he looking for here, and what is the max ambient in his environment?

 

[zoner21]

The lowest temp I'd be going for would be (let's say) 50 with an ambient of 75. That would be unusual but within the realm of brewing possibility.

 

[Sceadwian]

I see thermal efficiency based on area, but nothing to mention the amount of power to cause the physical mass to reach the equilibrium point in the first place, and again the brewing heat itself is completely overlooked. If the wattage's given here so far are used to create a real world device, it will fail.

Pommies calculations require the entire mass of the cooling area to be AT temperature before it's placed in the cooling vessel and assumes it generates no heat, both requirements are not available in the real world, unless you use a fridge in the first place to set up the thermocouple unit to maintain it.

 

[zoner21]

In what way will it fail? Will it be unable to maintain temp? Or do you mean an actual hardware failure? Anybody else agree? Also, anybody want to chime on on how to wire such a doomed device?

 

[Reloadron]

When it comes to Peltier-Electric for creating a Mini Fridge this link may be worth a read. Additionally here is another example. The latter link showing some power requirements. Worth noting is how small both units actually are and then weigh in the power consumption.

Personally I would just buy a small fridge and if necessary modify the thing. While Peltier-Electric is interesting most devices like those linked to never seemed to take off as a consumer product. Anyway, just my take on the things.

Ron

 

[ronv]

Will it Work?

It can certainly work, but it is not a toss together project either. You have posed a lot of good questions that have no answers yet.

Do you have a peltier in mind?
What are the part numbers of your heat sinks. The trick is to keep the hot side cool. You will also need a small fan on the cold side to circulate the air.
There needs to be some space between the hot heat sink and the cold heat sink with insulation in between or the 2 sides will fight. This is usually accomplished with an aluminum spacer and foam insulation.
What is the cold temperature range - hot temperature range.
I'm thinking just a simple thermistor and comparator circuit for the temperature control. A large FET for the switch.

 

[Pommie]

I see thermal efficiency based on area, but nothing to mention the amount of power to cause the physical mass to reach the equilibrium point in the first place, and again the brewing heat itself is completely overlooked. If the wattage's given here so far are used to create a real world device, it will fail.

Pommies calculations require the entire mass of the cooling area to be AT temperature before it's placed in the cooling vessel and assumes it generates no heat, both requirements are not available in the real world, unless you use a fridge in the first place to set up the thermocouple unit to maintain it.

I addressed both these points in post 27!!! And please, in future, provide some logical argument rather than "it will fail". There is nothing wrong with any of my calculations. I took the given scenario and added 50% (20°C differential rather than 25°F) and worked out the power required to maintain the temperature. I then added another 50% for good measure. When the bucket is placed in the cooler the temperature differential will be zero and so it will cool at 2°C (3.5°F) per hour. Unless there is a heat wave, the desired temperature will be reached in a few hours.

Show me where these calculations are wrong!! Or even optimistic!!

Edit, on the question of brewing heat, the whole point of cooling the beer is to slow the metabolism of the yeast and so I think the amount of heat produced is negligible and can be ignored.

MrAl, will try and remember to bold links. Maybe EM can look at changing the default colour.

Mike.