Nils,

You have to establish the energy balance on the inside air (gain, losses,..). After that you can deduce the power required to maintain your desired temperature (15°C).

Good luck

Hassan

The energy needed to cool down is simply calculated by multiplying the mass of water by its calorific value (4180 J/kg·K or 1.161 Wh/kg·K for water and 720 J/kg·K or 0.2 Wh/kg·K for glass ) and the diffefrence betweeen the initial and final temperature. The cooling power is simply the ratio of this energy by the cooling time you want.

For example, if I forget the glass vials and I assume that half of the volume is occupied by water (i.e 250 grams), cooling down from 20°C to 15 °C needs 0.250 kg*1.161 Wh/kg·K*5 K= 1,45 Wh. The cooling power is then about 3 watt to cool down in 30 minutes, for example.

Keeping this cooled depends on the insulation of the box and the outside-inside temperature difference. Assuming that the box is cubic, its sizes are 8 cm long, and the envelope area of the cube is 6*8*8 = 384 cm² or 0.0384 m².

Neglecting the slightly increased losses at the corners, the cooling power for maintaining a given temperature difference across this envelope is the product of the heat transfer coefficient U , the envelope area and the temperature difference. A safe value of the U-value is simply the thermal conductivity of the insulating material (in W/m·K) divided by its thickness in meter. For example, the U-value of 2 cm polystyrene foam is better (lower) than 0.04 W/m·K * 0.02 m = 2 W/m²K and the cooling power for maintenance will be 0.0384*2*5 = 0.384 Watt for keeping the samples at 15 ° in a room at 20 °C.

To be safe and sure that the increased losses at the corners are taken into accout, the external dimension of the insulated box can be taken. With 2 cm thick this gives an envelope area of 12*12*6=864 cm² or 0.0864 m² and the maximum maintenance cooling power will be 0.864 W, that is about half the power needed to cool down in one hour. So if the cooling power is actually 1,5 W, the cooling time will be 1.45 Wh/(1.5-0.864)W= 2,28 h. After this time, the cooling power should be controlled by a thermostat to keep the samples at 15°C.

Note that the calculated powers in these examples are for cooling the box only. If the box is cooled by a cold water circuit, the power to keep this circuit cool should be added.

A way for cooling such a small box is to make it in a conductive material (e.g. cooper) and to cool it with Peltier elements glued on its walls. The warm side of these elements shall be cooled, either by a water circuit, or with air, using cooling radiators such as those used in power electronics. In this case, no thermal insulation shall be put on the box walls.

Hi Claude-Alain Roulet I was looking through your answer and am a little confused by your units. You are calculating the volume of the cube, but leave your units as m2 rather than m3. What is the reasoning behind this? Also a note for others looking at this, to make units work for U value, make sure you divide thermal conductivity by wall thickness, not multipy it. The answer is shown with division but written out as multiplication.

Hi Abby

I did not compute the volume of the cube, but the area of its envelope, that is 6 squares of 8 (internal) or 12 (external) cm size. This is because the heat losses are roughly proportional to this area, the U-value of the envelope and the inside-outside temperature difference.

Thanks for your contributions.

You have been to explain to us that 1.45 Whr (for example) can be converted to 3W by dividing it 30 minutes (the cooling period). If the cooling period is 45 minutes, then the power becomes 1.93W.

I will like to know the approach to take to make the power to be the same, irrespective of the cooling period.

I will be expecting your response.

You are right: Power is energy divided by time or reciproqually, power provided during a certain time is energy. Therefore, if an amount E of energy is to be delivered (or retired in case of cooling) during a limited time t, the required power P(W)=E(Wh)/t(h).

If the power P is given (e.g. the power of the Peltier cells) then the necessary time is the energy to be delivered (or taken out) divided by the power: t(h)=E(Wh)/P(W). The higher the power, the shorter the time required to cool down the sample.

To make the power to be the same, irrespective of the cooling period requirees that the required energy (e.g. the amount of cooled material) be adapted accordingly.