Heating the air in a small box

Question

Background I am working on a project that involves heating air inside a small box, then measuring the temperature over time. The purpose of the box is to be a basic test chamber for PID experiments.

Initial Test I made a 10 cm * 10 cm * 10 cm box from 3 mm plywood. I then placed a small ~1 W heater inside with a temperature probe and observed the temperature rise... very slowly.

Proposed New Prototype I am going to construct a new prototype once I decide how to calculate the approximate time to warm the air inside the box to a set-point. Some specifications and assumptions are laid out below:

• The ~1 W heater consists of 4 parallel 100 ohm resistors and a 5 V power supply (there is also a 330 ohm resistor and an LED for visual indication). I don't want to change this heater design, as the intention is that this is easy to replicate and uses readily accessible 5 V power sources and electronic components.
• Currently no fan has been used but I have a small 20 mm, 5 V fan on order, so will integrate this into the design at some point.
• The box will be used at room temperature between 21–23 °C
• Ideally a temperature rise up to 30 °C (or more) would be possible over a 5-10 minute time interval
• The new proposed box size is 5 cm * 5 cm * 5 cm, 8 times smaller than the previous prototype. This is open to change depending on the previous 2 requirements of maximum temperature rise and time taken to achieve that change.

I am not after exact timings - I am aware that losses through the box will have an effect. But if this effect is minimal, then a simplified approximate solution is preferable as some design parameters may change slightly. I am looking for any guidance and calculations that will save me time instead of having to make different boxes and learn by trial and error.

Answer 1

A simple approach would be to just consider the heat added.

The specific heat equation gives the temperature change due to heating as

$$Q=mc\Delta T$$ where $Q$ is heat added, $m$ is mass. $c$ is the specific heat of the material in question (~1.005kJ/kgK for air at room temperature) and $\Delta T$ is the temperature change.

A 5x5x5 cm box will have an air mass of ~0.15 g (air density ~1.2 kg/m$^3$).

For a 1W (1 J/s) heater this gives a temperature change of $$\Delta T = \frac{Q}{mc}= \frac{1}{0.00015*1005}= 6.6 K/s$$

I will suggest this is much higher than you observed with the larger box (even accounting the increased volume) which suggests there is a factor not being considered.

Two factors leap to mind.

1) Heat losses from the box. Unless your box is well insulated it will lose heat via radiation and conduction. However for temperatures near room temp. I would expect these loses to be low.

2) Heating the box. Inevitably if you heat the air in the box this will transfer to the box itself. The box has much larger weight so will require more energy to heat (probably 100x or more). Intuitively I would think the transfer of heat from the air to the box would be slow compared to the heating of the air. But I am not an expert in this so may be wrong.

More likely I think the problem may be that most of the heat is going into the box initially if the heater is in contact with it, due to much much better thermal conductivity of wood than air. This is what I would look at to improve the rate of heating.