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This is kind of a follow-up question about determining the heat dissipated from equipment based on power consumption.

I have a situation where I have to install a router and UPS system in an enclosed particle board cabinet, in an RV style vehicle. It seems to me that we should also install ventilation fans.

I've been asked to justify that opinion based on how long they should be able to run the system before the router and UPS reach their max temp of 104°F.

I thought a good back-of-the envelope method would be to use specific heat capacity of air, dimensions of the cabinet, and the max. wattage of the equipment inside the cabinet (175W) to calculate how long it would take the air in the cabinet to climb from 80°F to 104°F, but the answer I'm getting is an absurd 5.5 seconds.

So, I'm thinking since my method ignored the equipment itself as a heat sink, I need to somehow factor that in. Thinking back to chemistry and physics a couple of decades ago, I don't remember ever having to find the rate of temperature change in a situation that involves multiple materials. It seems that the thermal conductivity the equipment would become relevant. I have found straightforward methods for calculating fan requirements to remove heat in this situation, but my assignment now is to determine how long the system can be operated before requiring ventilation.

Is there a standard method I can use based on the information I have available in equipment spec sheets, or based on dimensions that I could use to estimate how long it would take to raise the temperature past the max. operation temperature in the enclosure?

PS. I was going to show my calculations where I got 5.5 s from, but that doesn't seem relevant at this point.

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    \$\begingroup\$ To calculate the heat loss through the wood, we need to know the thickness and dimensions of the box. And understand that this is a first order calculation (estimation), not including convection. Accurate results will require a simulation like Hearth said. \$\endgroup\$
    – Mattman944
    Sep 4, 2022 at 23:41
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    \$\begingroup\$ You should be able to find the U value of the box material, in W/m²K, where K is the temperature difference across the box material. Given the U value, the surface area of the box, and the heat being transferred (150W), you should be able to find the steady state temperature difference, or temperature rise between the outside and the inside of the box. Depending on the size of the box (surface area) your equipment may not ever reach 104 deg F. \$\endgroup\$
    – SteveSh
    Sep 4, 2022 at 23:52
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    \$\begingroup\$ I did a back of the envelope calculation using U=W/m²K. If your enclosure is made of 1/2" plywood (U~2), and the surface area is 2 m^2, then with 150 W dissipation in the box, the temperature rise from the outside to the inside, steady state, will be about 37.5 deg C. So your problem is a transient one, and not steady state. Feel free to check my math. \$\endgroup\$
    – SteveSh
    Sep 5, 2022 at 1:11
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    \$\begingroup\$ @PeterJennings I don't want want the equipment to reach its max temperature. But I need to justify to my management and customer why they need to pay the extra cost for the ventilation. Since it's on a vehicle that is used on limited deployments, they are going to question if they can use it for a few hours without overheating. I need to come back with a reasonable answer. \$\endgroup\$
    – Frank
    Sep 5, 2022 at 4:16
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    \$\begingroup\$ Can you just measure it? How hard would it be for you to mock up the box, and put a resistor inside the box and a small fan (just to stir the air inside the box... not a ventilation fan) and cause the resistor to dissipate 175 Watts while you measure the internal temperature? It will give you a rough idea of how long it is going to take. I am on your side, by the way. I think you should plan for the fan. \$\endgroup\$
    – user57037
    Sep 5, 2022 at 6:01

2 Answers 2

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My gut feeling was that it would get too hot. But, not as hot as the first estimate by @SteveSh. My first order calculation is almost the same as Steve's revised estimate (slightly different assumptions).

But, this is a first order calculation. It assumes that the outside of the cabinet is the same as the room temperature. If there are dead spaces around the cabinet, this may not be true.

It also assumes that there is enough convection inside the cabinet so that the temperature is fairly even. There could be hot-spots near the equipment.

Based on my assumptions, it is marginal in my opinion. I would leave the door off the cabinet. Strap the equipment down, the door won't hold it anyway unless you have a strong latch. Or, add vent holes like BoxFlux suggested. Maybe a louvered door.

I will be glad to substitute actual numbers for my estimates.

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"Show formulas" enabled: enter image description here

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Heat transfer formula in thermodynamics,

Q = CxMxdT 
Q energy in joules 
C specific heat of air
M mass in kg
dt change in temperature Celsius

This formula calculates heat transfered from one system to another. There is a specific formula for radiation at this link : https://byjus.com/heat-transfer-formula/

But the above formula gives a good approximate to the time it takes to reach max temperature.

If you want to solve you will start by finding the weight of air enclosed an use it as mass, then get the specific heat of air, and finally plug in your temperature difference in C, then you will get Q which is in joules.

Now to calculate time until the temperature reaches max temperature, find out total energy lost in your circuit as heat you will get a value in watt/hour, you will then divide Q/energy loss that will give you approximate time it would take until the temperature reaches max temperature, for example if Q was 15kj which is the same as 15kw and your circuit losses 1kw of power each hour it will take 15 hours to reach max temperature.

Note that this formula calculates when the whole system temperature reaches your max meaning that the air will be at this temperature as well, while probably the circuit will reach this temperature some time before the whole system gets to the same temperature , in addition if there were any cooling that gets inside from the walls of the enclosure it will take more time for it to reach the same temperature and you can use the same method to calculate how to design the ventilation as in the above example of 1kw of energy lost you will have to cool it by transferring 1kw of heat through ventilation or any other method of cooling.

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  • \$\begingroup\$ I believe that this is what the OP did to come up with 5 seconds (because the air has almost no heat capacity). If you include the equipment and assume that the equipment has a mass of 10 kg and an average specific heat capacity of 0.5 J/(g-K), then the time is 371 seconds. \$\endgroup\$
    – Mattman944
    Sep 5, 2022 at 12:20
  • \$\begingroup\$ @Mattman944 I believe as in the above calculations 66.6kj with 175w/h dissipated it would take ~380 hrs not seconds, I'm not sure though, no you are right 175w/h is 630kj \$\endgroup\$
    – AhmedH2O
    Sep 5, 2022 at 16:26

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