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I want to make a very simple example for a PID controller (to learn and understand it). I thought of a controller for a water boiler. 1 liter of water in the boiler is in a 20°C room (fixed temperature), the water is at the beginning at 20°C and should get to 80°C (and stay at that temperature). The boiler has 2000 Watt, 100% efficiency.

Now the only thing that seems to be missing is how fast / how much energy gets away from the water. My first thought was to make it constant, but I want it to be a little more realistic. I guess the higher the difference between water temperature and room temerature, the more the water cools per second?

Could somebody please explain the relevant physics in this very simple scenario to me?

The main question is: How fast will 1 Liter of 65°C water get back to 20°C and how do I calculate it?

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  • $\begingroup$ By fast, if you mean the "time", then this might be relevant. $\endgroup$ Commented May 19, 2015 at 11:04
  • $\begingroup$ In most circumstances the cooling will obey Newton's law of cooling. $\endgroup$ Commented May 19, 2015 at 11:14
  • $\begingroup$ You don't mention the insulation of the boiler. $\endgroup$ Commented May 19, 2015 at 13:09
  • $\begingroup$ @MikeDunlavey This is meant to be a toy example. What are realistic values for the insulation? Does it make sense to say it is not insulated at all? $\endgroup$ Commented May 19, 2015 at 13:24
  • $\begingroup$ Not really. This is not a simple subject. Just google "boiler heat loss". $\endgroup$ Commented May 19, 2015 at 17:54

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You are way overthinking this. First, you can start with radiative cooling, but that's not the dominant process. At 65 C, in your workshop, convection cooling is the big dog. The total effect will depend on the shape and size of your container, how good a thermal insulator (or conductor, if you prefer) and even details like airflow. Without knowing these, it's impossible to even guess.

You are much better off just putting a pot full of water on the stove, get it to boiling, then pour a liter into your tank and measure how fast it cools. Make sure at the same time to measure the ambient temperature - if ambient is 20.5 you'll never get to 20.4.

You might object that this can take a long time, and you'd rather spend the time doing interesting stuff like calculation. In that case, your PID controller is going to be a very big disappointment to you, since water doesn't change temperature very quickly. I frankly doubt that using a 2kW heater in one liter of water is a great idea for a beginner. That is nominally 1 degree of heating per 2 seconds, which is pretty fast (as these things go), and if you don't include a mechanical stirrer you're likely to get convective currents in the water which will mess up your controller's response, particularly at degree level.

In order to characterize your controller, you'll need some sort of data acquisition system to allow you to accurately visualize the response of the system. You can get USB A/D converters quite cheaply these days.

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You can add Newton's law of cooling in to your model quite easily:

So we've said that Newton's Law of cooling tells you that the rate of cooling or temperature change is proportional to the temperature difference. Since your volume is fixed you can just look at this in terms of energy.

so your equation for heating power when the power is on is going to be 2000 - k(T-20), where k is some constant.

interestingly because the rate of cooling is proportional to the temperature difference, as you approach 20C the rate of cooling reduces. and when you are just a tiny bit away from 20 you have just a tiny rate of temperature change. so you never actually reach 20.

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  • $\begingroup$ Ok, lets say our measurement instrument can only register degrees. When would the water reach 20.4°C? What is a good value for k and what does k stand for? $\endgroup$ Commented May 19, 2015 at 11:29
  • $\begingroup$ So you need to determine k, an arbitrary constant, inversely proportionate to the insulation of your tank. by doing the maths. for a realistic cooling time, say 12 hours. you have the right model. I guess when you do the maths you will find that heating time and cooling time are functions of each other. $\endgroup$ Commented May 19, 2015 at 11:41

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