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My objective is to heat a liquid inside a bottle. I have a air gap of 2 cm between the bottle and the heating element (resistive layers). The heater and the bottle is surrounded by a bottle insulator, so the volume of air to heat is fixed.

I did the calculation to know how many power I need in order to heat my liquid from 18°C to 37.5°C in 5 minutes with these formulas, taking into consideration weigh of the bottle, delta temperature and specific heat of glass material.

Q = cp * m * dt

  • Q is heat required (kJ)
  • Cp = specific heat (kJ/kg K, kJ/kg°C)
  • Weight of bottle and liquid inside (kg) = 750 g

Then I deduced power needed: P = q / t

  • P = power (kJ/s, kW)
  • T = time (s)

P = 9.75W

My questions now are:

  • How can I calculate the thermal dissipation between my heater physical element and my bottle of liquid as I have a air gap of 2 cm between both elements
  • How can I deduce the real power I need with my heater element to be able to heat my liquid ?
  • What is for you the best solution to heat my liquid in term of heater technology ?

I looked on google and I was able to determine my R factor for the whole assembly = 0.5 (due to air which is playing as insulator). But it's really hard to understand how to do this calculation and the steps of the calculation...

Can you help me ?

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    \$\begingroup\$ This will be quite a complicated thermodynamic problem... I'll be interested to see if you get an answer. But you're probably better off on physics.se or engineering.se; it's well off topic here. \$\endgroup\$ – user_1818839 Nov 11 '20 at 16:51
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    \$\begingroup\$ I’m voting to close this question because it is not a problem of electrical engineering. In theory if your insulation is good, the air gap inside the insulated volume may be tolerable but it will mean running your heater at an elevated temperature; you may want to consider improving thermal transfer by placing the heater in contact, using a band heater, etc. \$\endgroup\$ – Chris Stratton Nov 11 '20 at 18:59
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Something that will simplify your job is if you can ensure that the thermal connection between your heater and your water is much stronger than that between the heater and the outside environment (through your outer insulating layer). If you can achieve that, then to a decent approximation, all the heat generated by the heater goes into heating everything in your assembly, rather than being dissipated to the outside world, and simple specific heat calculations will be all you need. This would eliminate all the complications of trying to deal with heat transfer coefficients, thermal pathways, and all kinds of thermodynamics.

Of course, eventually the heat leaks out, but if your objective is to calculate the power required to heat your water in five minutes, then this will be a valid approximation if you achieve that relationship between the two thermal paths. Now to find your way to this situation, you still may need to do some work with heat transfer coefficients and thermal conductivity, but you won't need to be all that accurate. You just need to make one number much bigger than another.

As for your heater technology, you will want something that spreads the heat out to as large an area as you can, as you have to worry about getting that heat off the heating element. You could burn it (and other things) up if you can't remove heat from the heating element fast enough. Heat can be spread by conducting it away from the heating element with a high-thermal-conductivity material, or by spreading out the heat generating element to a large area. I don't have specific recommendations there, but maybe someone else could help.

I should add that I am slightly sweeping one issue under the rug, which is the issue of whether the heat can get to your water fast enough without heating your heater up to unreasonable temperatures. This will depend on how spread out you can make your heater, and to really do this part of the design right, you should try to do the heat transfer calculations. But if you can get into this regime I am suggesting, the requirements on your accuracy are a lot lower because you just need to do enough calculation to know that you are in a regime where the heat can get to the water, and you don't need the heat transfer calculations to calculate how much power you need.

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