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Calculating temperature increase in a building as a result of solar radiation 
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#1
Dec1812, 11:53 AM

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I'm working on a thermal balance of a building. In particular, I'm trying to calculate the effect energy gains (mostly due to solar radiation) have on the inside temperature of the building. So I calculated the energy gains per month, for instance, 20 kW for the month of February (it's an extremely poorly insulated building). The only way I know of relating heat, temperature, and mass, is through the equation:
Q=m.cp.dT where:  Q is the heat in kJ  m is the mass of air in kg  cp is the specific heat of air in kJ/kg.K  dT is the temperature difference in K (Note: I'm using the dot to represent thousands and the comma to represent decimals) So, in one hour, we have 20 kWh, which is 72.000 kJ. For a volume of 637 m3 (the volume of the building) and a density of air of 1,2 kg/m3, we end up having 764 kg of air. If we consider the specific heat of air to be 1,0 kJ/kg.K, then dT is 94 K. I must admit the number shocked me, so much so that I think I'm doing something wrong. Is there something I'm missing here? Theoretically, if I had a perfectly insulated volume of 637 m3 of air, and I heated it during 1 hour with 20 kW, then it's correct that said volume would experiment a temperature increase of 94°C? Finally, in the case of a poorly insulated building (it has three slidingglass walls and a corrugated steel pitched roof), how much could I expect the inside temperature to rise in summer? In other words, how can I accurately calculate temperature increase in a building as a result of solar radiation gains? Thanks. 


#2
Dec1812, 04:59 PM

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You can neglect the air, unless you care about shortterm effects from ventilation. The heat capacity of solid materials will dominate the total heat capacity. There is an easy way to see this: The mass of the building itself is much more than 760kg.



#3
Dec1812, 05:24 PM

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If you had 20kW worth of central heating and you ran it for an hour, an average house would be 'very cosy' and that would be a real building with many heat loss mechanisms (which are going to lose heat proportional to the temperature difference). After an hour the walls and contents would have started to warm up and the heat capacity of those and all the other contents.
Years ago, I decided to get central heating installed in my home and I estimated the size of radiators needed for each room for a given temperature rise etc, etc. The answer I got was laughably small  until I realised that you need to take air exchange into account. I then got some sensible answers. The guy who did the installation then went and ordered different ones and the house was always very comfortable!! So much for theory. (I think he bought the cheapest sizes so I'm not complaining.) 


#4
Dec1912, 04:40 AM

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Calculating temperature increase in a building as a result of solar radiation



#5
Dec1912, 04:41 AM

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#6
Dec1912, 04:50 AM

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Going to the situation of high summer in a hot climate, you can expect to have a similar amount of heat to get rid of in order to stay comfortable. 


#7
Dec1912, 04:58 AM

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#8
Dec1912, 07:02 AM

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P: 11,617

Heat capacity of buildings is dominated by the heat capacity of the building itself and not its air. The heat capacity of air can be relevant for heating/ventilation. You can quickly cool (or heat) the air by several degrees if you open a window. But as soon as you close the window again, temperature will quickly return to its previous value (with some very small drop) as the walls heat the air again. 


#9
Dec1912, 07:05 AM

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#10
Dec1912, 02:03 PM

P: 3,107

An "energy gain" is measured in Joules or possibly kWH if you insist. It's not measured in kW, that would be a power gain. If you are getting into this for real I recommend joining The Green Building Forum and posting requests for info on passive houses over there. I think there might be a small one off joining fee but can't remember how much it is. http://www.greenbuildingforum.co.uk 


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