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The device being cooled isn't submerged in water, it has hoses running in and out of it carrying water through an internal heat exchanger. The device itself is just surrounded by empty air inside of the enclosure. I have been assuming that all heat generated by the device will be removed by the...

Just to clarify, the 85 °C is the temperature of the air inside of the enclosure that I have calculated, not the outside surface of the enclosure. I calculated the external surface temperature of the enclosure in the original post, and it was about 58 °C. Seems reasonable given that the...

Assuming 1/3rd of the enclosure surface area is exposed to 1000 W/m2 and using the expression T∞1 = (α⋅qsun⋅As - 10)⋅Rtot + Ts, I get an internal air temperature of approximately 85 °C. As here is the surface area of the cylinder, Rtot is the sum of the conductive and convective thermal...

Hmm yes I suppose that it would it. That is an excellent suggestion and something that could be easily implemented, thank you for pointing it out. I will re-check my calculations and post the internal air temperature that I calculated.

Apologies for not replying sooner. The device has an inlet and outlet for coolant and some type of internal heat exchanger, thermoelectric I think. At the moment, I'm just assuming that all heat generated by the device is removed by the cooling loop. When I calculated the heat of the...

I want to estimate what the air temperature inside of the enclosure will reach given an initial temperature of 25 °C, when exposed to 40 °C ambient air and direct sunlight. If it gets too high then the design might need to change, or be scrapped altogether. If the air temperature changes...

Hello all, I’m looking for some guidance on a calculation that I’m trying to do. I posted a similar recent thread titled 'Heating of air in enclosure - calculation and experiment discrepancy'. This question is in relation to the same device as the one in that other thread, but the enclosure is...

I bought one of these today with the same functions as the one you described. I've only done some quick tests (I haven't ran the apparatus at full-load, which will eventually operate at, but close to) and I can see that it will dissipate close to what the datasheet listed. Thanks for making the...

I'll see if I can buy a power meter and measure the input as you both suggest. I will update when I have measured it. Thank you both for your help, I really appreciate it.

I should have been more clear in that the 'electronics' is actually a small off the shelf apparatus that generates, according to the datasheet, ~ 60 W of heat. I wasn't too sure how accurate that was and wanted to measure it, but as you say my setup isn't really appropriate. Unfortunately, I...

Hello everybody, I would like feedback please on a large discrepancy between a theoretical calculation and an experimental measurement I have taken. I ran a 470 Ohm 5 Watt ceramic (at 50 V and 0.1 A, both confirmed using a multimeter) resistor inside of an 8 litre insulated cooler box and...

Yes very close to my answer as well. Thank you all for your contributions and advices. Miran

Yes you are right, thank you. I will try to estimate somehow. Yes I suppose it is no longer a wall and more like rectangular block with tapered ends. Why not accurate? There might be opportunity soon to test internal temperature with thermometer. I am trying to estimate heat entering box...

Yes thank you very much. Think I was just tired and thinking about what I was doing at the time too much and confuse myself. Thank you everybody for your answers. Miran

Qdot is rate heat transfer (Watts). cp is specific heat capacity of fluid (kJ/kg C), water in this case. ρ is fluid density (kg/m^3). ΔT is temperture difference (°C). Vdot is volume flow rate (m^3/s). When arrange like this, Qdot = Vdot * cp * ρ * ΔT, I see that more flow is more heat. If...

Hello, I try to use equation to calculate how much flow rate needs to be to removed heat from apparatus, but I confuse myself. I thought more flow means more heat taken away? I take this equation: Qdot = Vdot * cp * ρ * ΔT. I rearrange to find Vdot: Vdot = Qdot / cp * ρ * ΔT If Qdot kept...

Hello, Much apologies for the delay. No, box is sealed and not ventilated. Here is sketch of box. Box is 0.5 m in length, width and height. Apparatus will be attached to front of box but on little posts so not making directing contact. I will assume all heat generated by apparatus will...

Much apologies, I did not know. Should I ask moderator to move thread? Yes now that you say it, I think it is very obvious. Thank you for making it clear. As you say, my box is steel and polystyrene so I will calculate heat transfer coefficient. I think next step is for me to calculate...

Hello, New user here, excited to draw on knowledge of others. I have a small box to contain a science apparatus. The science apparatus needs to be kept cool and has inlet and outlets for fluid tubes. Sorry for this question if it is silly: do I need to cool the inside of the box as well? If I...