Can anyone tell me how many kW needed

[mointainwalker]

.......to raise the temperature of 300 m3 of air from -6 C to + 7C ?

I don't know how significant altitude is, but this air is at 1400 m and I don't know current air pressure.

I can work out heat-loss myself, so please ignore that and as for " over what time-period " , let's just say "reasonable or most economic "

Equally if anyone can give me the formula, I'm happy to do it myself. Have looked at various conversion web-sites but not found info.

Thanks all.

 

[ban-all-sheds]

http://en.wikipedia.org/wiki/Specific_heat_capacity

 

[mointainwalker]

Well since I posted I have viewed that and other sites and arrived at a figure which seems far from reasonable - approx 5.2 kW .

Even after going through the calcs several times and trying it with calories and joules ( just to see if I was making a silly error with some conversion factor ) I come back to a similar result.

 

[RedHerring]

My calculation: (s'cuse typing but can't find superscripts)
Air weighs about 1.2Kg per cubic meter (Sea level)

Temp change is 13degrees C (-6 to 7)
Volumetric Specific Heat of air is .001297 (J/cubic cm.K)(sea level)
thus 1297 (J/cubicM.K)

Air pressure does make a difference but I think not sufficiently for this example.

Formula is:
Heat Energy(Joules) = Mass Kg(1.2 X 300) X Specific Heat(1297) X Temp Change (13degrees) =
6069960 Joules
Kilowatt is power, Joules is energy, Kilowatt Hour is energy
Joules to KilowattHours, To obtain kilowatt-hours when joules are known, multiply by 2.778 x 10(to the power of -7)
= 16,862,348.88 X 10 to the power of -7 =
1.686234888, say 1.7 KwH
Or approx 1.5 Kw for 1 hour
or 3Kw for half an hour
or 6 Kw for 15 minutes.

But then the heat is being transferred to all the solid objects in the room, asuming they were at -6 degrees before you started.

 

[ban-all-sheds]

Well since I posted I have viewed that and other sites and arrived at a figure which seems far from reasonable - approx 5.2 kW .

What time period did you choose?

Even after going through the calcs several times and trying it with calories and joules ( just to see if I was making a silly error with some conversion factor ) I come back to a similar result.

Must be right then.

If you think that what you would need to achieve the heating you have specified is unreasonable then I suggest you forget the whole thing, because what you would actually need in practice would be far more than 1.7kWh. There will be inefficiencies in the system, there will be other objects to heat up, as RedHerring has observed, and the building fabric itself, and there will be losses from the building.

To put this into a real-world scenario, if we assume a ceiling height of 2.4m a 2-storey house with a floorplan of 7m x 9m has a volume of 302.4m³.

Do you really think you could effectively heat that with only a few kW?

 

[RedHerring]

1.7KwH was only for the air and only for going from -6 to 7 degrees! Or any 13 degree change.
A long way from heating a 2 storey house.

 

[r.bartlett]

2.2.1 Gas Loads
Using the BSRIA Rules of Thumb 2003, the following heat loads have been applied for the
various building types:
a) Offices 70 W/m2
b) Residential 60 W/m2
c) Retail 110 W/m2
d) Café/Bars 110 W/m2
e) Community Space 90 W/m2
Peak gas load has been calculated by summing


heating R.O.t is 60w/m2.

9 x 7 = 63 x 2 = 126 x 0.6 = 7.5kw.

So in all honesty that is not far off judging by the R.O.T ?

 

[ban-all-sheds]

1.7KwH was only for the air

Unless the OP can find a way to stop the air from touching the building fabric and any objects inside then he cannot only heat the air.

and only for going from -6 to 7 degrees! Or any 13 degree change.

Like, for example, a 7m x 9m house unheated for a while which has got down to 5° and you'd like to raise it to a tolerable 18°? Could that be done in just over an hour with a 1.5kW heater?

A long way from heating a 2 storey house.

What makes heating the OP's 300m³ of air a long way from heating the 300m³ of air in a 2-storey house?

what you would actually need in practice would be far more than 1.7kWh.

 

[ban-all-sheds]

heating R.O.t is 60w/m2.

9 x 7 = 63 x 2 = 126 x 0.6 = 7.5kw.

So in all honesty that is not far off judging by the R.O.T ?

No, but the OP thinks that 5.2kW is far from reasonable.

And a CH calculator picked at random on the web has just given me 28,875 BTU/8463W for a room 10m x 12.5m x 2.4m with the best-case selections related to heat loss (i.e. insulation, no windows etc.)

So everything is in the same ball-park.

Dunno what the OP was expecting.

 

[mointainwalker]

Thank you all for the time and trouble - particularly Red Herring with his calculation - you have taken with this question.

By the way RH, if you're interested the air-pressure difference at 1400 m means 15% reduction in kg/m3 vs sea-level.

I see now that the major flaw in my appreciation of the matter is that when looking at the physics I was forgetting the real world where this air is not just a theoretical mass being heated in a vacuum (sic) but something that is filling a large building with many hundreds/thousands of kilos of wood, stone, pb all at ambient - 6C that is sucking the heat out of it BEFORE there being the question of losses to the atmosphere.

BAS I got the impression you thought I was asking a question relevant to a permanent heating solution - not the case.

My unheated w-i-p barn is currently at -6C and likely to stay below 0 C until early April. If I want to carry on working, I need to get the self-levelling screed for the floors pumped in , but the contractor says I have to have min 7 C internal before they can/will do it.

My query was to enable me to get a rough idea of what size heaters I would have to hire to allow me to proceed with confidence but I really can't see how I can have that now.

I know what my heat losses ( through the structure ) will be , but I can't see how I can even guesstimate the time/energy needed to heat up the internal structure by 13 C.

Anyway thanks again for your time and interest

Maybe I should try and install a wood-burner and stoke it up to see what I can reach or maybe go skiing somewhat more often until it's a lot warmer

 

[ban-all-sheds]

And how airtight and well insulated is this WIP barn?

And how high is the ceiling? If you need 7° at ground level and it is not airtight and you get convection currents....

I think rather than try and calculate what will happen in the real world (which would be very hard to do), try an empirical approach.

Just hire a couple of large propane/butane industrial space heaters for the weekend, and see what happens.

And if you can temporarily fix sheets of ply/chip/OSB to the beams to reduce the volume which needs to be heated, that might be worth trying.

 

[mointainwalker]

Well BAS

Before I started thinking about the mathematical approach, I considered the empirical way, but was somewhat taken aback by the hire-cost of these space-heaters ( 12 - 15 kW ?) which came out around pnds 70/day plus fuel as well as 120 miles of driving to pick up and return.

Hence my desire to try and get some idea of energy required before blindly going down that route which I woud have to take twice - once for the try-out and once for the screed-install.


Thanks again for your interest.