0
\$\begingroup\$

I've got a polycarbonate enclosure and I'm going to place a heating element inside it to increase the air temperature in it from room to 50°C, plus a fan to maintain temperature. I need to find out how many watts the heating element needs to provide to heat up the air inside the enclosure so I can buy the right heating element.

The enclosure dimensions are: Length: 160 mm, Width: 120 mm and Height: 140 mm. The calculated wattage does not have to be very accurate as I will get something with slightly higher wattage to compensate if needed. I've found some formulas but they all vary and the answers are different. Ideally I want the air inside the enclosure to reach 50°C in less than a minute but I'd like to know how I can calculate the wattage so can I vary the time and find the most suitable one. Thanks.

\$\endgroup\$
7
  • 2
    \$\begingroup\$ The most difficult part is knowing the rate at which you lose heat to the surrounding air. But good thing you don't need to know since your temperature rise low, your volume is small, and time fairly long. I'd just get an element I know is vastly overpowered like a 500W or 1000W element and use feedback to turn the element off when the temperature is reached. \$\endgroup\$
    – DKNguyen
    Dec 10, 2020 at 23:43
  • 1
    \$\begingroup\$ Put a 60W lamp and a thermometer in the enclosure and see if it does what you want. You may need a dimmer... \$\endgroup\$
    – user16324
    Dec 10, 2020 at 23:47
  • 1
    \$\begingroup\$ Get an incandescent light bulb of some wattage, plug it in, put your hand near it and use the mechanical intuition and gut feel you've gained over your life experiences and imagine one, or several of those in your tiny box. I initially misread your box as being 16"x12"x14". So 500W is wayyyyy too much and would flash heat your box (which is fine if you want to go super fast since you turn it off anyways when temperature is reached). \$\endgroup\$
    – DKNguyen
    Dec 10, 2020 at 23:49
  • 2
    \$\begingroup\$ "Air has a specific heat capacity of slightly more than 1kJ/kgk at room temperature So it takes a 1kW heater 1 second to heat 1kg (roughly 1 m^3) of air 1 deg C" from physics.stackexchange.com/questions/45349/…. \$\endgroup\$
    – Kyle B
    Dec 10, 2020 at 23:52
  • \$\begingroup\$ You left out two of the most important enclosure specs: thickness and material. \$\endgroup\$
    – hobbs
    Dec 10, 2020 at 23:58

2 Answers 2

2
\$\begingroup\$

An extreme simplification of the case is to use the next analogy to find a solution:

enter image description here

Current source I1 pumps certain heating power into the box. Its current in amps is the heating power in watts. The thermal capacity of the box +what's inside is the capacitance of C1. The temperature of the box air as kelvins is the voltage of NODE1.

There's leakage between the ambience and the box interior. The thermal resistance Rt, where ohms can be directly seen as kelvins per watt, determines together with the capacitance the time constant of the temperature change. The thermal resistance and the ambient temperature (=VA) determine how high a certain heating power (=I1) can finally lift the box interior temperature (=voltage of NODE1)

You do not know the capacitance nor the resistance but you can measure them. Put a few watt known heating power and follow how high the temperature finally rises. It and the ambient temperature give how much is the resistance.

When you know the resistance you can calculate the capacitance by letting the system cool. The temperature decays towards the ambient temperature with time constant R1 * C1.

After you know the resistance and the capacitance you simply calculate how much power (=I1) you need to the wanted temperature rise in the wanted time and how much power you need to maintain the wanted temperature. A circuit simulator can be used to search the initial heating power if you hate calculations with exponential and time constant.

You still need a thermostat to be sure no overheating occurs and to take into the account automatically the variations of the ambient temperature.

BTW. The capacitance in farads directly presents the the thermal capacity= watts per (kelvins per second) or as well =joules per kelvin.

\$\endgroup\$
1
\$\begingroup\$

With that volume, you'd almost be alright going with some nichrome wire. It would be able to supply enough heat for your purposes. I've done a few projects like this, and I personally believe that nicrome wire is a perfect solution.

If you don't want to melt your whole design, then heating element it is.

Based on the amount of heat the nichrome wire would generate, I'd suggest 100W of heating power (or more). It doesn't really matter how large the wattage is, because you can just throttle it with lower voltage.

Here's a fan with a built-in radiator and heating element that would probably suit your design: https://www.aliexpress.com/item/32963635545.html

I understand you want to calculate the wattage, but I personally wouldn't even bother.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.