# Running peltier thermoelectric cooler at low 'standby' voltage

#### wazman

Joined Oct 31, 2016
13
Please tell me more of this 'heat transfer coefficient' and how to calculate it. I imagine it involves things like the dimensions of the box, the materials used etc.?

Thinking of driving my Tec1-12706 at about 4 volts to act as the heat barrier but might be able to go lower I guess. Will have to experiment.

You shouldn’t have to reinvent the wheel. The data sheet for the TEC should give you the current versus voltage and the delta-T versus current information. For the current, a TEC resembles a resistor as long as you’re above about 1.5V.

The number I’d be looking for is the heat transfer coefficient of your box. Measuring that will allow you to relate the cooling load to your battery capacity.
lease

#### Yaakov

Joined Jan 27, 2019
1,611
Not really, and ice just adds dead weight, and takes up space, unless you end up drinking it of course
What it does is offer a way to store the power you have during the day and release it at night. It is actually done for commercial HVAC systems in very large buildings when extra electric capacity is greater. They freeze water at night and use it for AC in the day.

#### oz93666

Joined Sep 7, 2010
701
Not really, and ice just adds dead weight, and takes up space, unless you end up drinking it of course
Have you ever frozen any water with this peltier unit?? In my experience they are just able to keep a few cans cool.

#### wazman

Joined Oct 31, 2016
13
It is a moot point - I don't wish or intend to freeze water, just responding to another poster's suggestion.

Have you ever frozen any water with this peltier unit?? In my experience they are just able to keep a few cans cool.

#### Yaakov

Joined Jan 27, 2019
1,611
It is a moot point - I don't wish or intend to freeze water, just responding to another poster's suggestion.
Just to be clear, the idea is to use something with a high thermal storage capacity to act as a sort of “thermal flywheel” so when you lose the battery the cold is maintained not by the method of keeping the Peltier cool but by using the power of the battery stored in the cold object. I mentioned earlier I am very dubious of any efficiency gain by doing what you proposed (TANSTAAFL), but this method doesn’t just use the battery, it fully stores the solar cell output available during the day. That is, once the battery is fully charged, it continues to store the power in the form of a cold object for later use.

In any case, if you can work out the relevant math, do some calculation on whether you will actually gain anything with this idea. The insulation idea and thermal storage idea potentially offer more efficiency because they don’t rely on the battery. I am not at all clear that using the battery in different ways offers anything more than using the battery to exhaustion initially and letting heat takes its course. It may just be a net zero.

#### wazman

Joined Oct 31, 2016
13
Thanks, your thoughts are appreciated. Putting water inside the unit merely for the purpose of making ice is a non-starter for the reasons previously stated (weight and space), but certainly it will be possible and desirable to make the actual contents as cold as possible during runtime.

I am resigned to the contents warming up gradually while the unit is being run at a lower power setting to string the battery out for as long as possible. The expectation is that the next day, power will be restored, from solar or another source, and the unit cooled down again.

I began this conversation as an exploration about keeping the peltier running, at a low power level, to make sure the peltier assembly doesn't provide a direct route for ambient heat to get in. This would occur whether or not there was ice inside. But I don't completely discount the value of having ice or something else in play (a glycol water-jacket?) to store 'coolth'.

I'm learning interesting things - I am playing around with cascaded peltiers at varied voltages, and getting favourable results. Another thing is that trying to cool things in a peltier cooler in the same manner as a conventional compressor driven refrigerator, ie by cooling air and circulating it around the contents, is very inefficient. Much better to keep the contents in containers made of thermally conductive material, and have those containers in contact with the same thermally conductive material that forms the interior surface of the cooler, which is in turn in contact with the peltier assembly. Beer in cans rather than bottles, for starters!

Just to be clear, the idea is to use something with a high thermal storage capacity to act as a sort of “thermal flywheel” so when you lose the battery the cold is maintained not by the method of keeping the Peltier cool but by using the power of the battery stored in the cold object. I mentioned earlier I am very dubious of any efficiency gain by doing what you proposed (TANSTAAFL), but this method doesn’t just use the battery, it fully stores the solar cell output available during the day. That is, once the battery is fully charged, it continues to store the power in the form of a cold object for later use.

In any case, if you can work out the relevant math, do some calculation on whether you will actually gain anything with this idea. The insulation idea and thermal storage idea potentially offer more efficiency because they don’t rely on the battery. I am not at all clear that using the battery in different ways offers anything more than using the battery to exhaustion initially and letting heat takes its course. It may just be a net zero.

#### Yaakov

Joined Jan 27, 2019
1,611
Conduction will always be more efficient than convection.

One more thought, not necessarily useful, but...

Assuming the cooling assembly is on the bottom. Imagine a lightweight "cover" the cooler can fit into, made of foam, like a standard cooler. Imagine the cooler stands on four short, retractable legs that fit the inside corners of the cover, and when extended provide enough room for airflow from the cooling fan.

At the start of the day, when the sun is available, you manually extend the legs with are latched using a solenoid, which, when operated, allows the legs to retract. It's then place into the cover. When the battery is on its last, the peltier is turned off and the fan suns until the hot side is at ambient, then, with its last breath (poetic exaggeration here) it operates the solenoid and gravity drops the cooler into the insulating cover. A dashpot could keep it from coming down too hard.

The same thing could be done on the side, you'd just replace gravity with a spring.

Its lightweight, and doesn't occupy and space in the cold compartment.

Just a thought.

#### wayneh

Joined Sep 9, 2010
16,128
Please tell me more of this 'heat transfer coefficient' and how to calculate it. I imagine it involves things like the dimensions of the box, the materials used etc.?
For your purposes it can simplified. The typical formula is:

dH/dt = k•A•∆T
Rate of heat flow equals the heat transfer coefficient constant k, times the surface area, times the temperature differential.
The units of k are, for example, BTU per minute per square foot per °F. The value obviously changes if you choose a different set of units.
H is heat, eg. BTU or Joules.
The temperature inside your box depends on the heat content and the heat capacity Cp of the box; ∆T =Cp•∆H
It's typical to define a zero energy state, such as H=0 at 0°C.​

You don't really care about k unless you're building boxes of different sizes, you can just measure the combined k•A.
You can estimate Cp by observing the temperature change of your box after a known cooling time with your TEC. (Or you could insert a resistive heater to measure the temperature change as heat is supplied.) Plot temperature versus time and there will hopefully be a nearly-linear region. In that region the change of ∆T/∆t (the slope) is a measure of heat capacity Cp.

To get a real measure of Cp you'll have to guesstimate how much heat the TEC moves. But you could make up you own units of energy if you want, such as TEC-minutes.

I know this is probably a little befuddling but keep the goal in mind: You want to model how the temperature of the box changes when the TEC is running and when it is not. Combined with the predictable behavior of the TEC, you will be able to approximate how much average power is needed to hold a given ∆T.

#### MrChips

Joined Oct 2, 2009
19,768
Cold cannot escape.
You need to stop the heat from getting in.
Solution: pack some ice on the hot side,
or cool it down with another peltier.
Now that I come to think of it, this is not such a crazy idea.
When you attempt to run the TEC at full rated power, the heat transfer backwards via conduction will overwhelm the peltier effect.
Cascading two or more TECs helps to reduce the temperature differential between each TEC stage.

#### wazman

Joined Oct 31, 2016
13
Yes I am already cascading, to achieve a greater temperature drop below ambient. I am running both plates at significantly below rated voltage. From what I have read you should only go to about 70% of rated voltage anyway. As long as you tweak the voltages carefully and can shed the heat from the outermost hot side it is viable to cool one peltier with another.

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#### wazman

Joined Oct 31, 2016
13
This is exceptionally helpful. Thank you

For your purposes it can simplified. The typical formula is:

dH/dt = k•A•∆T
Rate of heat flow equals the heat transfer coefficient constant k, times the surface area, times the temperature differential.
The units of k are, for example, BTU per minute per square foot per °F. The value obviously changes if you choose a different set of units.
H is heat, eg. BTU or Joules.
The temperature inside your box depends on the heat content and the heat capacity Cp of the box; ∆T =Cp•∆H
It's typical to define a zero energy state, such as H=0 at 0°C.​

You don't really care about k unless you're building boxes of different sizes, you can just measure the combined k•A.
You can estimate Cp by observing the temperature change of your box after a known cooling time with your TEC. (Or you could insert a resistive heater to measure the temperature change as heat is supplied.) Plot temperature versus time and there will hopefully be a nearly-linear region. In that region the change of ∆T/∆t (the slope) is a measure of heat capacity Cp.

To get a real measure of Cp you'll have to guesstimate how much heat the TEC moves. But you could make up you own units of energy if you want, such as TEC-minutes.

I know this is probably a little befuddling but keep the goal in mind: You want to model how the temperature of the box changes when the TEC is running and when it is not. Combined with the predictable behavior of the TEC, you will be able to approximate how much average power is needed to hold a given ∆T.