B How to estimate the heating power of warm air entering a room?

[guyburns]

TL;DR Summary

The variables required to calculate the heating effect of warm air blown into a room

In winter I heat my workshop with a radiant heater and an oil heater. Easy to know how much heat is entering the room.

But in spring and summer I don't use those; I blow hot air from the ceiling cavity into the room, and I was wondering how I would estimate the heating power and the formula to use. I assume it would depend on air flow rate and temperature difference between roof cavity and workshop, but do other variables that come into it?

And does it really depend on flow rate? When I open a window (about 1.5m above floor level) so the air flow is at full speed, would that have a different heating effect from closing the window – so the air can only escape by the small opening under the door? It seems to me the latter might be more effective because the warm air is then rubbing up against the cold concrete floor.

I use a car radiator as a fan, blowing air vertically down from the ceiling, about 3m high.

 

[hutchphd]

There are folks here who actually know this stuff...until they answer I will make suggestions. It seems to me your best bet is to circulate your room air through the attic space. Blow one fan from the attic to your primary work spot. Use a duct (maybe with fan?) to circulate air from floor level (where it is probably cooler) back up into attic. Balancing these two will minimize cold air infiltration from outside. Some turbulance in the attic space will increase heat transfer from the roof inside surface.

 

[jrmichler]

If you know the:

Flow rate of the hot air
and
Temperature of the hot air
and
Temperature of the room

Then the heat flow due to the hot air can be calculated. In English units, air flow is in cubic feet per minute (CFM), temperature is in degrees Fahrenheit, and heat is in British Thermal Units (BTU). The equation is: ${BTU}/{hour} = (T_{hot} - T_{room}) * {cubic feet}/{hour} * 0.018 {BTU}/({cubic foot}-{deg F})$

I once built such a system. It had a fan to blow hot air down from the attic, and I built a controller to run the fan whenever the attic was hotter than the shop below. It did not work very well, then the controller died, and the fan was taken out. My shop now has two feet of insulation in the ceiling. The insulation does more good than the fan ever did or could.

 

[guyburns]

Thanks for the replies.

Right now…

• the temperature in the ceiling is 40.7C
• 30.0C in the workshop (a bit warm, I admit, but I'm testing)
• estimated air flow with window open (based on data from this website), say conservatively, 200 cubic metres/hr (but I reckon it could be double that)…

BTU/hr = 19.3 x 7000 x 0.018 = 2400

That's about 700Watts.

So, in round figures, if I've got a 10ºC difference in temperature, I can expect at least 1KW of heating. Sound right?

 

[jrmichler]

Yes, that's right. HOWEVER, keep in mind that warm air wants to leak out when the shop is warm and the ceiling air is cold.

 

[hutchphd]

And that the temperature of the attic air is unlikely to remain at 40C when you are sucking out at 200$m^3$/hour. How big is the attic space (how long for a complete air exchange)?. Where does the replacement air come from and how fast is it reheated?
You have no real control over the total influx of external heat into the attic space. If you circulate the air faster, the air you suck out will be less warm. Make some turbulence in the attic and limit the net influx of cold outside air into the system. The rate of exchange is secondary to the design performance.

 

[Lnewqban]

Consider that you must have the same volume of make up cold air flowing into the roof cavity, which may reduce the estimate available temperature of 40.7 C after some time.
Also, the warmer air pushed down by the fan will tend to occupy the higher levels of the workshop, flowing into the open window, while remaining heavier cold air will tend to occupy lower levels.

 

[guyburns]

Thanks for all the feedback.

It's not an attic I'm sucking the air from, but the roof cavity, the space between the corrugated iron on the roof and the plaster board ceiling, probably about 100m^3. Air intake is whatever is sucked in under the eaves.

I haven't gone to much bother to implement this. Just put the radiator fan (about 10 years ago) in the manhole that happens to be in the workshop.

It has never been apparent to me that I am removing hot air faster than it can be reheated. Sometimes I just blow the air through the open door to warm the garage (open to ambient via a 3x6m entrance), and the incoming air will typically stay above 35ºC all afternoon.

That brings up another question: is it possible to estimate the energy being transferred by the sun to the air in the roof cavity at midday this time of the year?

1. Location: 41º south latitude (Tasmania). Sun rising to about 70º at midday.
2. Roof lying north-south
3. Corrugated iron, 10m wide, 15m long (N-S)
4. Sloped to the middle, with a rise of about 1m in 5m
5. Ambient air temperature say 20ºC

The maximum temperature I've recorded in the roof cavity, was 47ºC, when I was installing antenna wiring. That's when I decided I should put the heat to good use. But it was a dangerous place to be in high summer (see Marcus Wilson story).

 

[hutchphd]

The very rough ballpark for direct summer sunlight is 1kW per m^2. If you are off direct angle just use the projected area towards the sun. And as you said $$1kW=3400BTU/hr$$ Its hard to say how much gets transferred to the air...but it sounds like there is plenty of heat available. Suck it out of the ridge (how? ) and be warm. Seems like your calculations make sense
Of course its always easier to turn the flow rate down rather than up after installation! And typically the exhaust port (cold air exit) is put at lowest level.