The expanded model
A previous article introduced the basic model of a Peltier as used in this website. This article covers the expanded model to include external components to a system such as hot side heatsinks or heat exchanges, the cold side object and thermal insulation.
The purpose of this article is just to introduce the external circuit and typical components without much analysis, since the effects can be different depending on the application.
The hot side (Zh)
In general on the hot side it’s only necessary to consider the thermal resistance to ambient, which is referred to as “Zh” in this website. The units for thermal resistance is “K/W” (Kelvin per Watt), and basically is a measure of the temperature drop as a function of heat flow. For example, a heatsink and fan rated at 0.5K/W means that with 40W flowing through it there will be 20K temperature difference. If the room ambient is 23°C, this means the “hot side” of the heatsink will be 43°C.
For heat pump applications, typical values for Zh range from 0.05K/W for a 3 fan water-to-air heat exchange to around 0.5K/W for a large 12cm x 12cm heatsink with a cooling fan as is often used in medium sized Peltier “fridges” (pictured below). All the options have a fan; using heatsinks without a fan generally won’t work except for special applications. The thermal resistance for heat exchanges shown were all measured in open air. The Peltier “fridge” heatsink was actually installed in an enclosure with fairly small ventilation slots, and sitting next the switch mode power supply, so that the Peltier hot side would have an even less tenuous relationship with the outside ambient air than the number suggests.
The hot side can also have heat capacity such as aluminium heatsinks or water in the water cooling systems, but realistically these can be ignored as the time constants involved are fairly short, typically less than 10min. If required, these can be represented as a capacitor. For reference, a 500g aluminium heatsink has a heat capacity of about 450J/K. In a thermal circuit with 1K/W between the heat source and the aluminium, the resulting time constant would be 450s (7.5min). Although it can be ignored for long term applications, it’s important to note that in the short term it makes the system appear much better than it is, as will be discussed in a later article.
The cold side
The cold side usually has a few more components and varies a lot depending on the application.
In this website, “Zc” is used for stuff that connects direct to the cold side of the Peltier. For an application like air conditioning or Peltier capacity testing, this element might be all that is required, similar to the hot side.
For cooling objects like the inside of a cooler box or a can of beer, a key factor is the insulation between object and the outside ambient, referred to as “Zi” in this website. Typical values would be 1K/W to 5K/W, which for electrical engineers might seem low for something referred to as “insulation” given than an insulator would normally be measured in MΩ or GΩ while a good conductor is usually 1Ω or less, separated by several orders of magnitude and easy to maintain. In a thermal context, 0.01K/W would be an excellent conductor while 10K/W would be a pretty good insulator, but the reality both figures can be hard to obtain.
Finally, when cooling time is important, then the heat capacity of the target object together with the thermal resistance will set this time. Heat capacity is measured in J/K, and thermal resistance is in K/W which is in effect Ks/J. Multiplying the two leaves just time in seconds, and is the time constant of the circuit. For example a 350mL can of beer has a heat capacity of about 1500J/K. In a cooler box you might expect around 2K/W between the Peltier cold side and the can, which means time constant of 3000s (50min). This explains why beer can take a few hours to cool, and it’s important to note that this has little to do with the efficiency of the system.
The combined circuit, worked example
Let’s take a small foam cooler box bought at the hardware store, with a TEC 12708 run at 3A, with classic Al heatsink+fans on both hot and cold sides. With some experience the circuit can be modelled as follows (the following gallery shows the circuit with the actual parts, then just the circuit components, and finally the component values):
The values are based on:
Xc and Xh of 0.9K/W is based on measurements of typical 40x40mm Peltiers, and includes the thermal resistance of a good quality contact between the Peltier and metal of the heatsink
The core temperature difference of 60K is based on a K factor of 20 K/A which is typical for TEC 12708 Peltiers (8A), and a current of 3A
The waste heat is 13.5W based on I²R for 3A and 1.5Ω
The heatsink+fans are 0.5K/W based on measurements, and the 1500 J/K for the can is based on 4.2J/K for water, and a volume of 350mL
The insulation of 1K/W is estimated based on cheap foam 25mm thick and box dimensions of 30x40x40cm. The calculation is not complicated but there’s already too many numbers for for now, but the critical point is the surface area (0.66m²) and the thickness (25mm) and the thermal conductivity of the foam (0.035W/m/K).
If we are only concerned with the steady state (final) temperatures, it should be possible to see that we can follow the same technique as the basic Peltier model. On the cold side Zi, Zc and Xc are all in series and just make one big resistor, and similarly on the hot side Xh and Zh combine. Using this approach:
Step 1: Assume the cold side is disconnected from ambient, and the waste heat flows only through the hot side. This results in an available temperature difference of:
Ta = 60 - 13.5 x (0.9 + 0.5) = 41.1K
Step 2: Reconnect the cold side to ambient, and measure the cooling power based on Ta / Zt, where Zt is the total resistance of the circuit:
Pc = 41.1/(1.0+0.5+0.9+0.9+0.5) = 10.8W
Finally we can now estimate the temperature at the various points in the circuit, same as we would V= IR in a normal electrical circuit. Of key interest is the temperature inside the chamber relative to ambient which is found by simply multiplying the insulation resistance Zi by the cooling power (which is negative in polarity):
Tc = -10.8 x 1.0 = -10.8K
This means, if we allow enough time, the chamber will stabilize at about 11K below ambient, or about 16°C for a room of 25°C.
Using a circuit simulator
Instead of stepped approach above, it’s also possible to just put the components in circuit simulator. Currently, I’m using CircuitLab, which is easy to use and gets results quickly:
In the above circuit, the “temperatures” (voltages) are relative to ground so the chamber at -10.82V means roughly 11K below ambient, as was found in the above analysis. The grounds at each end can be replaced by voltage sources allowing the ambient to be simulated as well.
Is 16°C in a 25°C room good? Maybe for a wine cooler, especially for just 13.5W in (plus fans). But not so good for beer drinkers, luke warm at best. But we’ll leave the analysis and improvements for another article. For now, it’s good to concentrate on understanding the method. Also for now we’ll skip the beer can’s heat capacity (1500J/K) since affects how long it takes to cool, which is a different subject and will be covered later.