Enclosure heat losses

[prmmel]

I have (7) enclosures going to a site in Kentucky along the Ohio river. The enclosures are 36"D x 72"W x 72"H and are fabricated aluminum skinned enclosures with an open bottom. We will have water pipes inside which are critical so we need to place a heater in each enclosure. I am trying to accurately calculate what I can expect for required heat.

The bottom of the enclosures which are bolted to a concrete slab will have and open bottom (30" x 60") which may radiant some ground temperature into the box. The enclosures are insulated with 1" thick sheet insulation boards.

Can anyone suggest a calculation or a website that offers calcs to figure this out.

I know that good design will consist of a large safety factor and cycling the heater. I had an engineer tell me today that it would take 500W continuously at 0F outdoors to maintain 32F inside the enclosure. I think this is a bit extreme. I was estimating about 150-200W.

Thanks,

 

[IRstuff]

In still air? What's the thermal conductivity of the insulation?

TTFN

FAQ731-376

 

[prmmel]

Looks like .033 W/mk

 

[Compositepro]

Just like when you want a warm house, it is just as important to not have any openings that allow for drafts as it is to have insulation.

 

[prmmel]

Yes, I understand. These enclosures are very tight. Fully gasketed as well with no penetrations. Just trying to figure the basic thermal requirements.

Thanks,

 

[KiwiMace]

I reckon for the box:

Q=UAdT
Q=0.44W/m2K x 11.7m2 x 17.8K
Q~90W

Your insulation value seemed high so I threw in one of my own. Probably a couple hundred w the base and losses.

 

[IRstuff]

the minimum heat required to maintain 42°F is 234 W, assuming that your insulation is 1 inch of Styrofoam tc at 0.033W/m-K thermal conductivity.

total area = 11.7m^2, assuming/ignoring open bottom
thk = 1 in
assume still air htc = 2.5 W/m^2-K

solve:

dT1 + dT2 = 42°F
htc*area*dt1 = tc*area*dT2/thk

dT1 = 14.4°F dT2 = 27.6°F
Q = htc*area*dT1 = 234 W

I specifically chose a net dT of 42°F, but that's still marginal if the objective is to prevent condensation or icing, since your original target of 32°F is just at freezing. There are inherent errors in thermistors/thermocouples, unknown moisture content, ignored thermal radiation losses, etc.

Nonetheless, we can see that anything below 234W is on the hairy edge of being insufficient.

A semi worst-case estimate of radiation to the night sky puts it at about 246 W, so 500 W is pretty much spot on:

radiation area.rad = top +0.2*sides = 2.68 m^2

sigma*area.rad*[(275K)^4-(253K)^4] = 246 W

246W + 234W = 480 W, with dT1 = 1.77°F and dT2 = 40.2°F

TTFN

FAQ731-376