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I want to build a small ice cannon that uses pressurized air (either through an air compressor or a CO2 cartridge) to shoot small ice pellets at my coworkers. I plan to use a thermoelectric cooler to freeze the water but I don't understand how to spec it.

For example, what are the key parameters of this product that will let me know if the cooler will achieve what I'm hoping it to?

Thermoelectric Cooler - 40x40mm

COM-10080 In Fritzing Library

Description: Thermoelectric coolers (TEC or Peltier) create a temperature differential on each side. One side gets hot and the other side gets cool. Therefore, they can be used to either warm something up or cool something down, depending on which side you use. You can also take advantage of a temperature differential to generate electricity. The thermal tape listed below works very well to attach heat sinks to the hot side.

This Peltier works very well as long as you remove the heat from the hot side. After turning on the device, the hot side will heat quickly, the cold side will cool quickly. If you do not remove the heat from the hot side (with a heat sink or other device), the Peltier will quickly reach stasis and do nothing. We recommend using an old computer CPU heatsink or other block of metal to pull heat from the hot side. We were able to use a computer power supply and CPU heatsink to make the cold side so uncomfortable we could not hold our finger to it.

Features:

  • 40 x 40 x 3.6mm
  • lmax - 7A
  • Umax - 15.4V
  • Qcmax - 62.2W
  • Tmax - 69C
  • 1.7 Ohm resistance
  • 127 thermocouples
  • Max Operating Temp: 180°C
  • Min Operating Temp: -50°C

If I want to freeze 10 mL of water from room temperature within a certain time period, how would I spec out the cooler? What are the base equations that govern how long it takes to freeze the water?

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    $\begingroup$ I'm voting to close this question as off-topic because the ultimate aim of the OP in asking the question is to conduct an unsafe act. shooting pellets at people could result in someone having a permanent eye injury & it would contravene the health & safety policy of reputable employers. $\endgroup$ – Fred Apr 8 '15 at 6:59
  • $\begingroup$ Interesting. Ultimately the goal is actually a more sustainable desk nerf gun replacement and I was being facetious about shooting anyone in the face. But if the way I posted goes against community guidelines, than by all means close it. $\endgroup$ – Ben Hoff Apr 8 '15 at 11:16
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    $\begingroup$ While I don't advise making this gun and shooting it at people, I don't think the nature of the question makes it unacceptable. The device could be used in safe ways too, and the information could be useful to people building something else. Most practical engineering projects have significant safety risks, and I don't know where we'd draw the line between what's acceptable and what's not. $\endgroup$ – Ethan48 Apr 8 '15 at 13:15
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There seems to be two questions here, how to evaluate thermoelectrics and how much cooling is required to freeze water. I'll lead off with some instruction on heat transfer and thermoelectrics and then discuss your design question.

Heat Transfer, Really Quickly

When heat conducts through a solid, the heat flow is proportional to the temperature difference such that $q = K \Delta T$, where $q$ is the heat rate (usually in Watts, energy moved per unit time); $K$ is the conductance, a property of the solid; and $\Delta T$ is the temperature difference

  • Heat sink manufacturers will specify an effective conductance of their heat sinks, usually in °C/W
  • The "R-value" of insulation is the inverse of conductance in really weird units

Thermoelectrics

Thermoelectric datasheets contain values that are easy to measure (for the manufacturers) but not easy to put into system models (for you). The first step is to get the Seebeck coefficient, $\alpha$, and the thermal conductance, $K$, from the data provided. These properties are important because they feed into the standard model of thermoelectrics1:

$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$ $$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$ $$ V = \alpha (T_h - T_c) + IR $$

Where $q$ is heat rate (more on that in the design section), $I$ is current, $R$ is electrical resistance, $T_h$ and $T_c$ are the cold and hot side temperatures, respectively and $V$ is the applied voltage.

$K$ can be found easily by dividing Qcmax by Tmax (which properly should be called $\Delta T_{max}$). Tmax is the maximum temperature difference the module can sustain when the hot side is held at a reference temperature. At this point, all of the electrical power input is being used to pump the heat which was conducting through the device, i.e. $q_{c,max} = K \Delta T_{max}$

The seebeck coefficient can be found by using Qcmax. At max heat pumping, the temperature difference has to be zero (from the first equation), so $\alpha I_{max} T_c = q_{c,max} + \frac{1}{2} I_{max}^2 R$$

Designing a System

Thermoelectric System

Your system, I imagine, will look a bit like this. Your thermoelectric will need to be able to pump out both the heat that comes in from the outside (how much will depend on how well you insulate your compartment) and whatever heat is released by the water during the freezing process.

As you noted, it takes 88 J/g of water to freeze water. If you multiply 88J/g of water by 10g (at 1g/mL) and divide by the number of seconds you want it to take, you will have your heat rate requirement $q$ to freeze the water. As Olin said, add in some amount of heat that will conduct through the walls and you have an estimate of $q_c$, the amount of heat the thermoelectric needs to move.

Finally, the heat sink that you buy will have some conductance rating $C$ in °C/W, we know that the hot side of the thermoelectric $T_h$ will be equal to $C q_h + T_{ambient}$.

Now we have enough information to solve the problem. Effectively, what we need to ask is whether there is a current $I$ that will satisfy the three equations below for the known $q_c$, $T_c$, $\alpha$, $R$, and $K$.

$$ q_{c} = \alpha I T_c - \frac{1}{2} I^2 R - K (T_h - T_c) $$ $$ q_{h} = \alpha I T_h + \frac{1}{2} I^2 R - K (T_h - T_c) $$ $$ T_h = Cq_h + T_{ambient} $$

Although you might be able to solve those equations by hand, I would just plug them into Excel and use the solver functionality to see if it is possible. If it isn't possible, try adjusting your $q_c$ requirement, e.g. by freezing the ice slower or adding a second thermoelectric module (which would cut $q_c$ per module in half.

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You ultimately need to know the totaly electrical engergy required to go into the cooler per ice pellet. However, even without calculating this, it seems fairly obvious that too much energy will be required for reasonable battery operation.

To calculate the energy, follow these steps:

  1. Assume a starting temperature for the water, like 20°C. Using the specific heat of liquid water, calculate the energy required to cool it down to 0°C.

  2. Now look a the heat released by phase change from liquid to solid for your mass of water, and add that to the total.

  3. Look up or guess the efficiency of your Peltier cooler at, say, 25°C in and 0°C out. Divide the total energy required by this value to get the electrical input energy. Note that Peltier coolers are very inefficient devices, most likely well under 10% in this application.

This gives you the theoretical minimum required electrical energy input. There will be various losses in the system and other inefficiencies. For example, the cooler won't cool just the ice and nothing around it. And, no insulation is perfect so there will be some heat coming from ambient that needs to be removed, in addition to removing the heat from the cooling and solidifying water. All around as a rough SWAG I'd probably figure at least 4x more energy required overall than just the minimum from the basic physics of cooling the water thru a Peltier device. Ultimately only real measurements will get you good numbers because there are too many variables that are too hard to quantify.

Now look at how big the Peltier needs to be, what you need to do to remove heat from its hot side, the size of the battery you need to support this, and realize why this is a totally unrealistic idea. It is conceivably possible with a device that sits on a desk and is plugged into the wall. But, shooting hard pellets, like ice, around a office sounds like a fundamentally dumb and irresponsible idea in the first place.

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  • $\begingroup$ Who said anything about battery operation? I was never looking at this to be battery powered. A quick google search tells me that the energy to freeze water from room temp is 88J per gram water. This doesn't help me in terms of the specifications of the thermoelectric cooler (what are the key parameters.. Qcmax? Tmax?) or give me the base equations for how long this would take. I never took heat transfer, is there a simple model that involving the time domain that would answer if this is even feasible with thermoelectric coolers? $\endgroup$ – Ben Hoff Apr 8 '15 at 23:15
  • $\begingroup$ @Ben: At one point you likened this thing to a nerf gun, which is a portable hand-held device. The energy you need is a start. To get a power rating you have to decide how long it should take to freeze the pellet. Obviously the cooler has to be able to produce a temperature below the freezing point of water. $\endgroup$ – Olin Lathrop Apr 9 '15 at 13:44
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Calculate latent heat Q= ML , using amount of water you want to freeze . Divide by time you want to want the cannon to be made. You Get cooling capacity . Add losses to this.

Make trials of TEC from calculator below. https://viveksilwal.wordpress.com/2015/04/15/28/

Once this is done, Install a software Ansys Icepak, you can learn it in a week, there you can simulate TEC easily. Analyse your system there.

Although thermoelectric is very less efficient , but I think a good heat sink and tec combination can make you freezed balls.

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