Dummy load construction and heat dissipation

Question

I'm building my own dummy load, consisting of resistors submerged in liquid paraffin (or something similar, not 100% decided).

I had some questions around heat dissipation and how I can calculate and avoid the thing over-heating.

1. Is there a particular fluid that is better than other in both performance and cost?
2. What sort of volume of fluid should I use assuming I would be using up to 100W, such that it won't start heating up quickly? Is there even a way to accurately calculate this?
3. Are there any non-obvious safety concerns I need to think about when constructing and using this? I don't want to miss anything simple but obscure.
4. In a sealed system like this, is there any risk of explosion due to pressure from the liquid heating?

Answer 1

Is there a particular fluid that is better than other in both performance and cost?

Water is hard to beat in both respects. The trouble is it has this tendency to become a conductor as it dissolves salts, so it must be insulated, or somehow kept very pure. Sometimes the trouble of this is too much, so some sort of oil is easier.

What sort of volume of fluid should I use assuming I would be using up to 100W, such that it won't start heating up quickly? Is there even a way to accurately calculate this?

Heat capacity is perhaps the relevant physical property here. The specific heat capacity of liquid water is about $4.8J/(gK)$, and since 1mL of water weighs about 1g, we can also say the volumetric heat capacity of liquid water is about $4.8J/(mLK)$. We know that $W=1J/s$, and from these two things, given a power, and some volume of water, we can calculate at what rate the temperature will rise, if the temperature is uniform throughout, and no heat energy is being lost elsewhere. Let's use 100W and 1L as an example:

$$ \require{cancel} \frac{100\cancel{J}}{s} \cdot \frac{\cancel{mL}K}{4.8\cancel{J}} \cdot \frac{1}{1000\cancel{mL}} = 0.21\frac{K}{s}\\ $$

Since we are talking about a rate of change, you might as well consider $K=^\circ C$.

Of course, this involves two assumptions we know to not be true, the first being that the temperature of the water (or whatever coolant) is uniform everywhere. If the coolant is actively stirred, then it might be close enough to true. Otherwise, you are dependent on conductive currents and conduction to distribute the heat energy throughout the coolant, some parts are hotter and some are cooler, but the average temperature increases at this rate.

But this is somewhat moot, given the second assumption: that no heat energy is otherwise being lost. Probably you won't wrap the dummy load in blankets then operate it until it overheats, but rather construct it with a surface designed to radiate heat into the ambient environment well enough that it can be operated indefinitely without overheating. The job of the coolant is really just to provide a good thermal coupling between the heat sink and the heat sources, and to average out any transient thermal loads.

The relevant physical property for the heatsink is absolute thermal resistance. This quantity has units $K/W$ and tells you, for a given constant power, what the temperature increase above ambient will be. It will be specified by the manufacturer's datasheet. You also need to add to this the thermal resistance of everything between the heat sources and the heat sink. Unfortunately calculating the absolute thermal resistance of your tank of coolant is hard, because it has a complex geometry, and the things likely to be used in the tank (water, oil) are actually not especially good thermal conductors: they move the heat around mostly by convection.

So, the general approach is this: figure out, for your design power and allowable temperature rise, how big the heatsink would need to be assuming the best case, then make it bigger to account for other factors. TLAR is probably the most economical method, and if you need something more precise, then pump a known power into a prototype, and if it gets too hot, make it bigger.

Are there any non-obvious safety concerns I need to think about when constructing and using this? I don't want to miss anything simple but obscure.

Besides the obvious ones of having a potentially large quantity of possibly flammable, probably very hot fluid, potentially heated beyond its boiling point, connected to an electrical energy source that might be capable of generating sparks or heating (especially under fault conditions) materials above said fluid's flash point? No, I can't really think of any. I don't think it's the non-obvious safety concerns that are going to kill you.

In a sealed system like this, is there any risk of explosion due to pressure from the liquid heating?

Very yes, see last question. I wouldn't seal it if I could help it, and if I did, I'd be sure it wasn't sealed very well, so an overpressure fault would just dribble rather than explode. Also couldn't hurt to make sure the pressure release mechanism indeed dribbles and not shoots a hot jet of oil.

Answer 2

100W isn't very much power, compared to the load fluid dummy loads were meant to handle. 1kW dummy loads might use a 100W rated tubular resister inside the fluid. You could, in fact, use a 100W 50Ω resister in air with a heatsink and skip the fluid altogether. You may want to enclose it in a Faraday cage of some sort (another reason loads were often made in paint cans) such as a mesh screen so you still have good airflow through the resister. The kicker here is that you need a non-inductive resistor, and most tubular 100W resisters are coils. Still, you can get thick film 100W resisters and use a regular cheap heatsink, or a chunk of random metal you probably already have lying around to get the same load rating. Use it as a hotplate to warm your beverage this winter.

There is little reason to go to the trouble of calculating the fluid needed for a 100W intermittent load. If you still want to use a fluid dummy load, then just use a liter or more, watch the temperature if you need to transmit for more than a minute at a time, and move on with your testing.

Keep in mind that most fluid dummy loads are only designed for intermittent use. They do not actually move the heat into the air at the same rate they receive it, they simply absorb it. They are meant for short periods of testing, with the transmitter running, at most, 1 minute out of every ten minutes.

They will overheat and explode if used beyond their design.

So the information you really need to provide is your duty cycle, and how long a testing period would you be using it for.

If you're going to run 10-20 hundred watt tests, each lasting a minute or less with a break between each test of 10 minutes or more, then nearly any simple liter fluid dummy load will be able to handle the 10 to 20 watt hours of total energy you'll be putting into them.

If you need to do continuous testing, and need to dissipate 100W for hours at a time, then you'll have to look at a different solution.

But, as mentioned above, 100W simply isn't that much. Getting a 100W non-inductive power resistor, attaching it to a heatsink that can dissipate 100W continuously isn't going to be very expensive. Then you can test as much as you like and never worry about your duty cycle, or the dummy load temperature and risk.

When you want to test kW level transmitters at a low duty cycle, you can't beat the price of a fluid filled DIY dummy load. For any other situation (lower power, or continuous use) fluid filled dummy loads aren't as attractive as other solutions.