Should the temperature of a heated metal cone decrease in a plastic enclosure?

Question

Summary

A steel cone is heated by a nichrome ribbon to 45–50 °C. I placed a Type K thermocouple over the surface of the metal to measure the heat transferred over the metal surface. The thermocouple reads about 45 to 50 °C, which is expected. The issue is when I put a plastic cap (enclosure) over the metal cone, the temperature drastically decreases.

Procedure

There is a metal cone made of mild steel.

A nichrome ribbon wire of 0.09 mm thickness is glued as a vertical strip from the tip to bottom on the metal surface of cone.

On the opposite side of the metal surface a beaded thermocouple wire of 40 awg (0.08 mm) is glued from the tip to bottom.

Both of them are placed on the metal using a superglue.

A plastic cap that has a similar cone shape is put to the metal. The air gap between the metal and the plastic structure is 0.1 mm. In this gap the Nichrome wire and a Type K thermocouple are placed.

The structure is:

plastic cap ->Thermocouple(or Nichrome wire)->glue(Superglue or Loctite)->Metal surface.

Observations

1. Without the plastic cap, the thermocouple measures about 45 °C after 4 to 5 minutes of heating.

2. When the plastic cap is added, the temperature increases up to 40 °C and struggles to reach 45 °C. The approximate time it takes to reach 45 °C is 10 to 12 minutes.

Questions

1. How is it possible to ensure that the thermocouple is measuring the temperature of the metal and not of the plastic?

2. We do not want the temperature measurement to change when the plastic cap is inserted. Is there any quick way of heating the plastic side and also getting the right temperature measurement?

I am open to any suggestions to improve this topology.

Answer 1

For plastic thermoforming, or heating behavior, this kind of behavior would be expected. Quite simply, when you form the plastic you typically heat up the plastic rather than the metal.

Say you were working with LDPE, which is the only plastic I can think of that even changes shape at those temperatures. Each material has a heat capacity - 452 J/kg K for steel, and the LDPE has 1800 J/kg K - 3400 J/kg K. For this answer, I'm using 2600 J/kg K. The heat capacity is how much energy it would take to increase 1 kg of the material by 1 K (or 1 degree Celsius), assuming the system is completely insulated. As you can see, the LDPE is larger.

Using a fluid mechanics analogy, your wire is a hose, supplying energy to whatever you need it to warm up. It will supply your system with so many Joules of heat per second - that's the Watts of electricity used. Each of your materials is like a bucket that needs to be filled up. But for some materials, it takes a lot more energy for them to register a single degree change of temperature - think of it as a large diameter bucket v. a narrow diameter bucket - to fill the larger diameter bucket to the same height requires a lot more water. Same concept.

For the steel (say a 1 kg cone) you need 9.04 kJ of thermal energy to raise it from 25 °C to 45 °C. With a 1 kW wire, that's 9.04 seconds - no time at all.

Adding the mass of say 2 kg plastic onto the 1 kg cone, it now takes:

$$(2\:\mathrm{kg} * 2600\:\mathrm{\frac{J}{kg\cdot K}} + 1\:\mathrm{kg}* 452\:\mathrm{\frac{J}{kg\cdot K}})* (45\:\mathrm{°C} - 25\:\mathrm{°C}) = 113.04 \:\mathrm{kJ}$$

That is 12 times the energy to heat it to the same temperature. As a result, it takes 12 times as long - plus, while your original cone heated so fast you didn't have to worry about losses, the plastic now will take longer when heat transfer effects are taken into account.

In short, always heat the plastic, and form it around the metal - not the other way around.