Watts required to raise air temp

[Echoshill]

Help!

I am in a huge pinch for time and thus throw myself on the mercy of the group.

I have a box that I am shipping off to a -45C environment. It houses to small PC Boards and is enclosed in another larger equipment box. The larger box is not sealed from the environment yet there will be no direct wind blowing on the little control box. The box is 320 cuin (5243 cu cm) internal and is made of stainless steel. It is vented top and bottom.

Is there a direct way to figure how many watts I will need in heating to raise the ambient inside the box to -40C? Or better still, how many watts per degree C.

 

[desertfox]

Hi Ecoshill

Yes you can work out first the heat required to raise the temperature from -45 to -40C by using this formula:-

Q = m*Cp*(T1-T2) was m= mass of air in box

Cp= specific heat cap of
air

T1 and T2 are the final and
initial temps of air resp.

Now if you know your heat source power or how much time you want before the box gets to the reqiured temperature then the Q calculated above is related to power by:-

Q/(time) =power in watts

regards

desertfox

 

[MintJulep]

Desertfox has answered the question that you asked.

However, I don't think you asked the right question.

I suspect that you want to know the power that you need to MAINTAIN the box temperature at -40C.

 

[Echoshill]

First, thank God for you guys and the fast response.

Next, you are correct MJ. I need to maintain -40C.

How does that change things?

 

[desertfox]

hi Ecoshill

Well if its always in a -45 enviroment the heat calculated above is what you need initially to raise from -45 to -40, to maintain it you need to decide what tolerance on the -40 you can tolerate before a heat source kicks in to bring it back to -40, so if your heater cuts in at -42 you can rework the calculation to give you the heat required, the power then required is that heat energy calculated at -42 divided by the time required to bring it back to -40.
Basically you need some temperature control ie thermostat to turn heater on and off.

desertfox

 

[Echoshill]

The resistors kick on at 0 C.

So from the above calculations at -50C,

p of air =1.534,

C=1.005.

With a volume of .0052M^3, my air mass is then .0079Kg. Which gives a Q of .0039.

If I have 200W of heating, that means I should be able to change from -45 to 40 in .0017 hours?????

Something seems very wrong.

 

[IRstuff]

What you haven't stipulated is how much power your box dissipates and how long you have to warm up the box, and whether you can configure the box to automatically reset when the temperature is good enough.

Assuming that your box dissipates more than 3.8W, it'll self-heat the box and account for the heat loss to the larger box in about 20 seconds.

Joule heating:
(320ci)*(1.2kg/m^2)*(1005J/kg-K)*(5K)/(10s) = 3.16W

Convective loss
5*[(320ci)^1/3]*(5W/m-K)*(5K) = 3.77W

TTFN

FAQ731-376

 

[MrBTU]

You also need to consider the heat that the metal of the box and the circuit boards will absorb: mass x specific heat of these materials x delta t. You are heating up the box, the boards, and the air, not just the air.
There is also the heat loss from radiation and conduction from the outside of the box as you are heating it.

 

[Echoshill]

Guys,

I really wish I had all the information to solve the problem, but as is often the case we are being forced to guess and fast.

The system is designed to turn on the 200W worth of heaters at 0C (Ohmite series 270 100W at 10 Ohm). As for how much heat the box will dissipate, you got me. It is stainless steel and in the volume I stated. The PC boards are extremely small so my gut tells me that they are negligible considering the surface area of the box.

The components in the box are rated to -30C. So I guess the better question may have been will 200W of heaters be sufficient to keep that volume of air above -30C when the ambient may drop (over time) to -45C?

 

[desertfox]

Hi Echoshill

Just check your figure because I get 0.012 hours,
Its a very small mass of air your dealing with and temperature range.
IRstuff and MrBTU also make good points.

desertfox

 

[IRstuff]

Unless there are fans in the larger box, it should be no problem. With a 15°C delta, your box is losing only 11.3 W, so in steady state conditions, the 200W would be quite toasty. In fact, you'll have to turn down the power, otherwise, you'll probably overheat your boards.

TTFN

FAQ731-376

 

[MintJulep]

You don't have a fixed volume of air in your little box.

You state that it is vented top and bottom. As soon as you get a temperature differential you are going to start driving a convective flow of cold air in from the bottom.

 

[itsmoked]

Sadly this is a very tough issue. With very small differences in the assorted variables you have a large result difference.

You would do well to implement your 200W solution and then test it. Find someone with a deep freeze and pay them something to use it for a day.

I have worked in many deep freezes it's not a big problem.


MintJulep's warning is very true.



Keith Cress
kcress -

http://www.flaminsystems.com

 

[Echoshill]

Guys,

Thank you so much for all your help.

I revising the design to have 4 100W heaters. Once complete, I am going to send it off to Baltimore for a cold room test. Though this looks like a simple problem to management, I know enough to know it is not simple.

Good suggestions and thank you for the teaching.

 

[Compositepro]

You don't need a cold room test to measure what the heat rise is. It will be close to X degrees heat rise whatever the outside temperature is.