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Discussion in 'Calculations' started by Sekocan, Aug 15, 2016.

1. SekocanMember

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Hi,

I am a mechanical engineer but mostly worked on mechanisms and mechanic things. Now I have a heat problem and I guess my heat knowledge is too rusty for this. So can you help me?

The problem is, I have a rectengular prism box with known thickness. Inside, there are some electronic devices. The heat generation of the devices is known. Ambient temperature is known. How can I calculate the temperature of the inside of the box.

Assumptions:
- The box is a flat kind. So the upwards convection inside the box can be neglected.
- The devices are not concentrated in some small space, they are placed homogeneously.
- The ambient is still, no wind.

If we need any numbers, let's say:
- The size of the box is: 1200x300x90mm (long flat box)
- Thickness is: 10mm
- Heat Generation is: 300 Watts
- Ambient Temperature: 25 C
- Inside and outside is air.

Thanks.

2.
3. Ganesh MoorthiMember

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Nov 2016
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Hi Sekocan,

The temperature inside your enclosure would be 55degc.

You can use the formula, heat required to raise the temperature of air:

Q = mcp(delta T)

mass = density * volume = 1.2754 * 0.0324

Cp = 1.005 Kj/KgK

Q = 300 W

Delta T = Treq - Ta

Solve using the above formula to get Treq as 55 degc.

Let me know, if you need any clarifications.

4. mike_atlasNew Member

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Wrong wrong wrong. First off, 300 W is the Rate of heat generated, not the total energy (which would be in Joules or BTU's). This is not the correct approach for this problem as the energy needed to heat the air has little to do with the eventual internal temperature.

This is a standard heat added vs. heat lost equilibrium problem. I.e. heat is added to the inside of the enclosure at a rate of 300 W. The enclosure's temperature will rise until 300 W is lost from the enclosure due to convection, conduction, radiation, etc. This will depend on lots of things, including the conductive resistance of the enclosure walls (i.e. if the enclosure was insulated, it will get MUCH hotter inside) the outside temperature, and the external dimensions of the enclosure (horizontal area, vertical area, etc.).

If the box is basically free standing almost all of the heat lost will be through convection unless the temperature reaches the point where things are glowing and radiation takes over.

There are plenty of formulas and calculators out there for things like this, but there are lots of assumptions to be made.

5. Ganesh MoorthiMember

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Hi Mike, Just to inform you, i have considered this as a complete adiabatic system. There is no heat and energy transfer between the system and the surroundings. If you see the problem statement, it talks about stand still ambient air (No wind)- So, only natural convection. So, much of heat transfer is not happening. Second, the thickness of the box is around 10mm, which is huge for conduction to transfer the heat to the outer surface.

So, my assumption of adiabatic system is correct to find out the Worst Case temperature inside the box. (Material information is not available) But still this is how you find out worst case temperature for such system.

6. s.weinbergWell-Known Member

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Like Mike said, this is a question of balancing heat generation vs heat loss. You'll need to work out the thermal resistance of your box. There are some basic formulae for working out thermal transfer through a plate, but this seems like the kind of thing that lends itself to cheap and easy prototyping.
Get a 300W heater, stick it in your box with a temperature sensor. Set the room temperature to 25C, and let 'er rip until you reach steady state.

7. Ganesh MoorthiMember

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Hi Weinberg, I agree that this is a heat balancing problem. But i'm sure that, to find out the Worst case temperature, adiabatic box assumption is a right approach. As i said, natural convection and Radiation are being the heat transfer mechanisms in this problem, the adiabatic approach will be near accurate. Let me know your feedback.
(Simply, assuming it as a wooden box)

8. mike_atlasNew Member

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Again. 300 W is NOT "Q". Q is a total amount of heat transferred. Watts is a Q per time. Nothing that the original poster said implies that this is an insulated box or that the electronics are only on for a limited time. The implication is that the electronics stay on, pumping 300 Watts into the box. The temperature inside the box will continue to rise until 300 Watts is lost to the outside and then will stay at equilibrium. The amount of heat needed to raise the air temperature within the box is not of great importance here.

What you suggest is more or less true if the box was insulated and you said something like "apply 300 Watts for 10 seconds and compute the temperature rise".

For a continuing supply of heat this is an equilibrium problem, as we have said. Look this up in your Heat Transfer textbook. It is a standard problem.

9. Ganesh MoorthiMember

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Hi Mike, True.. I mean the same Power = Heat/Time. By the time, it reaches steady state, the temperature inside the box will be 55degc, is what i meant.
If Power is continuous, it will be increasing continuously from 55degc.. based on the heat transfer with the outside world...

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