Project Overview

Thermoelectric coolers (TECs - as shown at left courtesy of Marlow Industries) and generators (TEGs) have been around more than 200 years, but they've been getting more press lately in the automotive industry, specifically BMW. They have been using TEGs for producing electricity from the temperature gradient from the exhaust pipe to the ambient air. In the reverse, by passing current through the two materials, typically n- and p-type semiconductors, you can pull heat from one side and reject it on the other, and use them in a cooling capacity.

I wanted to use this amazing effect for another hobby of mine, brewing beer. The main idea is to control the temperature of 5-6 gallons of fermenting wort without using a typical refrigerator system. This was as much for the challenge as it was for learning about these "cool" little devices. Overall for a system of this size or larger, a typical compression refrigeration system may be more efficient - but that wasn't the point of the project.

In this and subsequent blogs, I would like to trace my development of this system from start to finish. I will describe the system requirements, heat transfer analysis and resulting prototype circuit design and fabrication.

Stage 1 - A Little Background

During the brewing process, the stage of converting the sweetened wort, with the help of yeast, to alcohol is called fermentation. Fermentation typically does not require real close temperature control; for many styles of beers fermentation at room temperature is good enough. Where you get into trouble, however, is the fermenting lager style beers. This style of beer requires temperatures cooler than room temperature, roughly 50 deg F, and a refrigeration system to maintain this temperature for several weeks.

The refrigeration system consists of three main items - 1) the item to be cooled, wort, 2) the box/cabinet to contain and circulate the cooled air and 3) the device to do the cooling work. Items 1 and 2 are passively acting to counterbalance the work done by the cooling device (in this case a thermoelectric cooler). The energy balance needs to be calculated first of to size the cooling device.

Of course, all of my calculations can be sized for your application and could be used to cool or heat a number of various mediums.

Stage 2 - Heat Transfer Analysis

The refrigeration system has to be able to overcome three major sources of heat in it's worst case condition - when the warm fermentation vessel is placed in the refrigerator. The overall energy balance is as follows:

where Q_boxloss [W] is the heat lost through the insulation of the refrigerator box to the ambient atmosphere, Q_ferm is the heat generated by the yeast converting sugars to alcohol, Q_cool is the amount of energy required to bring the temperature down from the yeast-pitching temperature to our ideal fermentation temperature over a given time and finally, Q_pumped is the heat pumped out of the system by the thermoelectric device.

What you are most interested in is this worst case scenario. The cooling device needs to be able to provide enough heat pumping to deal with this worst case or you will never get the temperature of the your wort down to your ideal temperature. As the wort cools down to the final fermentation temperature, you only have the insulation losses and fermentation heat to contend with, which will be less than the worst case.

Let's go through how to calculate these values:

where A [m^2] is the total area of the outside of the refrigerator box , T_b [deg C] is the ideal inside temperature of the box , T_a [deg C] is the ambient room temperature (again don't forget to make this the worst case scenario, i.e. 85 deg F on a hot day), K [W/m-C] is the thermal conductivity of the insulation and x [m] is the insulation thickness. If you want to get real precise, you could also add in the wood/plastic/metal outershell and it's conductivity, but for simplicity here, I only am looking at the insulation for a rough estimate.

We must cool the wort down from it's initial temperature to a it's final temperature and this will take energy as shown here:

where m_w [kg] is the mass of the wort (density x volume), T_i [deg C] is the initial temperature of the wort, T_f [deg C] is the final idel fermentation temperature of the wort, c_p [J/kg-C] is the specific heat of the wort (i used a value slightly less than water here since it is comprised partially of sugars) and t [sec] is the amount of time that you would like to have the wort go from the initial to final temperature. If the volume you are cooling is large, say 5-6 gallons, 24 hours might no be unreasonable. For smaller volumes, say 12 fluid oz, 15 minutes might be reasonable; the time to cool will be dictated by your requirements.

For the beer experts, this next section might be useful, but if you are not cooling fermenting wort or your medium is not generating heat, you can skip this equation. I wanted to make sure that the fermentation process itself was accounted for. It turns out that the heat generated by yeast can be quite negligible compared to the insulation losses (unless you have an amazingly thermally tight refrigerator). I will add it here, however, for completeness. I would also like to give my due respects to Spencer Thomas for these following equations.

The yeast will convert a portion of the sugars present in beer to alcohol and during this conversion will generate heat. If we assume that sugars being fermented release 140 kcal/kg or 586 J/kg (1 kcal = 4.184 J), we then must then figure out how much sugar is fermented and mulitply it by 586 J/kg and divide by the time they take to do this. The fermentation energy is then:

where t is the time of fermentation is seconds, which in our case would be maybe 1 week. The initial mass of the sugar, m_si [kg], can be calculated from the original gravity of the wort (OG) as follows:

where m_w [kg] is the mass of the wort and,

where OG is in points, i.e. 1.050 = 50 OG in points.

The final mass of the sugar, m_sf [kg] is a little more involved.

where FG is the final gravity of the beer in points. RE and AE deal with the attenuation of the beer, which is when the beer is no longer just water and sugar but is also alcohol as well. This composite has a different density and must be accounted for in the mass calculations when you are looking at the final gravity.

There you have it, all the heat transfer components to the left hand side of the energy balance equation. In my next blog, I will identify how to find Q_pumped and how to best to implement it. This will include specifying a thermoelectric cooler module and the heat sinks required to reach it's optimum performance.

The refrigeration system has to be able to overcome three major sources of heat in it's worst case condition - when the warm fermentation vessel is placed in the refrigerator. The overall energy balance is as follows:

where Q_boxloss [W] is the heat lost through the insulation of the refrigerator box to the ambient atmosphere, Q_ferm is the heat generated by the yeast converting sugars to alcohol, Q_cool is the amount of energy required to bring the temperature down from the yeast-pitching temperature to our ideal fermentation temperature over a given time and finally, Q_pumped is the heat pumped out of the system by the thermoelectric device.

What you are most interested in is this worst case scenario. The cooling device needs to be able to provide enough heat pumping to deal with this worst case or you will never get the temperature of the your wort down to your ideal temperature. As the wort cools down to the final fermentation temperature, you only have the insulation losses and fermentation heat to contend with, which will be less than the worst case.

Let's go through how to calculate these values:

where A [m^2] is the total area of the outside of the refrigerator box , T_b [deg C] is the ideal inside temperature of the box , T_a [deg C] is the ambient room temperature (again don't forget to make this the worst case scenario, i.e. 85 deg F on a hot day), K [W/m-C] is the thermal conductivity of the insulation and x [m] is the insulation thickness. If you want to get real precise, you could also add in the wood/plastic/metal outershell and it's conductivity, but for simplicity here, I only am looking at the insulation for a rough estimate.

We must cool the wort down from it's initial temperature to a it's final temperature and this will take energy as shown here:

where m_w [kg] is the mass of the wort (density x volume), T_i [deg C] is the initial temperature of the wort, T_f [deg C] is the final idel fermentation temperature of the wort, c_p [J/kg-C] is the specific heat of the wort (i used a value slightly less than water here since it is comprised partially of sugars) and t [sec] is the amount of time that you would like to have the wort go from the initial to final temperature. If the volume you are cooling is large, say 5-6 gallons, 24 hours might no be unreasonable. For smaller volumes, say 12 fluid oz, 15 minutes might be reasonable; the time to cool will be dictated by your requirements.

For the beer experts, this next section might be useful, but if you are not cooling fermenting wort or your medium is not generating heat, you can skip this equation. I wanted to make sure that the fermentation process itself was accounted for. It turns out that the heat generated by yeast can be quite negligible compared to the insulation losses (unless you have an amazingly thermally tight refrigerator). I will add it here, however, for completeness. I would also like to give my due respects to Spencer Thomas for these following equations.

The yeast will convert a portion of the sugars present in beer to alcohol and during this conversion will generate heat. If we assume that sugars being fermented release 140 kcal/kg or 586 J/kg (1 kcal = 4.184 J), we then must then figure out how much sugar is fermented and mulitply it by 586 J/kg and divide by the time they take to do this. The fermentation energy is then:

where t is the time of fermentation is seconds, which in our case would be maybe 1 week. The initial mass of the sugar, m_si [kg], can be calculated from the original gravity of the wort (OG) as follows:

where m_w [kg] is the mass of the wort and,

where OG is in points, i.e. 1.050 = 50 OG in points.

The final mass of the sugar, m_sf [kg] is a little more involved.

where FG is the final gravity of the beer in points. RE and AE deal with the attenuation of the beer, which is when the beer is no longer just water and sugar but is also alcohol as well. This composite has a different density and must be accounted for in the mass calculations when you are looking at the final gravity.

There you have it, all the heat transfer components to the left hand side of the energy balance equation. In my next blog, I will identify how to find Q_pumped and how to best to implement it. This will include specifying a thermoelectric cooler module and the heat sinks required to reach it's optimum performance.

I will follow for Joe. He will like this nerdy stuff.

ReplyDeletei am not very electronic-but i am trying this anyway-using a peltier module i want to heat something ,not cool it.ihave some 40x40 peltiers,some heatsinks,a dcsupply a few types of fans,i'm trying for70/75 degrees c.i think i can only go to 15 v max and 8 amp max,that's my knowledge,what else would i need?an how to doit,any and all help appreciated-thanks

ReplyDeleteHi anon -

ReplyDeleteFirst off, if you are only heating, it may be best to go with a resistive heater. This can be made from a plain old resistor of the appropriate wattage rating. But if you want to use a peltier device, that's okay too. A peltier does give you the option of precision temp control because if it overshoots on heating, it can then cool your system to bring it back in range.

Second, do you want to control the temp at all, or do you just want to be in that 70/75 deg C range? The reason I ask is that you could use a basic potentiometer in series with the peltier device and power supply. Adjust the current via the pot can control temp manually.

A little more about what you are trying to accomplish would be helpful to give you the best answer. I would be glad to help you further.

hi-villacherman,thanks for the reply.my power supply will be dc-cig lighter in a car.i'm not really after big temperature fluctuations,if i could get a constant 70/75 that would be good.i live in the bush and after a city trip,long day shopping,it would be nice to take home some take-away food,kfc,maccas,chinese etc,but my coleman only goes to 60 degrees so after my 2 hour trip most food is off the boil and not worth eating,i really don;t feel like pulling it apart and tinkering with it ,so i made a box

ReplyDeleteiput 2 peltiers inside the cavitywith an ally plate onthe outside of the box(cool side)and a heat sink attached to the peltier(using thermo paste)inside,a fan to dissipate the heat and let it flow inside the box,but i have no idea how to get it hotter ,ie:use a resistor maybe?and how to keep amps down so not to fry things,and buy as little as possible,any clearer??

ReplyDeletep.s. i'm in the australian outback

ReplyDeleteOutback -

ReplyDeleteYeah, that seems clearer, thanks for the info. To get the interior of your box warmer, you can do a few things.

1) Put a fan on the cold side heat sink, this will allow the cold side of the peltier so it becomes warmer which in turn can produce a warmer hot side (delta temp across the peltier device remains about the same, just shifts in the warmer direction).

2) Insulate the heck out of the box.

3) The other route is that you could use cartridge heaters or just plain resistors. If you do use plain resistors, the wattage rating has to be high enough that you don't fry the resistor. Example: use 10 resistors in series, each resistor would be 0.2 Ohms and rated to at least 8 W and this should give you 72 Watts of heating. Here are my calcs in .ods and .pdf. I highlighted my recommendation. You may have to increase the power if you are heating a big box, then you may be limited by the current output of the outlet. I'm not familiar with auto outlet limits.

http://adam.heintzelman.googlepages.com/ResistorPower.ods

http://adam.heintzelman.googlepages.com/ResistorPower.pdf

As a disclaimer, these are recommendations only. I am not liable for any fires, electrical shocks, burnt food, pirate attacks or snake bites.

Also, as an adjustment, I would also put a potentiometer in series with the resistors. This way you can do final temperature adjustment.

ReplyDeleteha ha -very good.the box is about 1.5-2 sq ft.thanks for the info and help,i've got a potentiometer thingy--will try all--and will keep in touch and let you know if dinner was served or trashed

ReplyDelete