Hello, I'm the R&D Manager for a company that make Electric heaters for hazardous areas, be they process heaters or air warmers.


As many have said before, it is all about knowing the insulation value of the construction materials, or U-Value.  Once you have these, it's very east to calculate the duty required to heat a room. The accuracy shall of course depend on the level of detail you want to achieve. We heat a lot of varied hazardous areas from aircraft hangers, munition stores, flour mills, spray shops and anything offshore of onshore for the oil and gas industry. 


It's not so much what is required to heat the room, but what it required to combat the losses of that room. I may be over simplifying to start with for you, but an example I would offer is as follows:


A small freestanding storage room 5m x 5m x 3m with ambient temperature 10°C, target 20°C.


To heat the air inside the room

Volume works out at 75m³ - Volume can be corrected to account for objects withing the room, but for ease, this room is empty.

Air @ 10°C has Specific Heat of 1.005Kj/kg.K and density of 1.227kg/m³. With a target Delta of 10K, we can work out (Q=mcΔt) that you require:

257W to heat this volume of air from 10°C to 20°C in an hour.


So now to work out your losses, it depends how much detail you want to go into.

For simplicity, we have a flat roof, 25m², flat metal with 2" fibre insulation. We lookup a U-value for this and ascertain losses of 0.48W/m².K for instance.

25m² x 0.48W/m².K x 10K (Δt) = 120Watts


To rush through the others:

25m² floor, poured concrete @ 0.4W/m².K = 100W losses

12m² of double glazing @ 0.7W/m².K = 84W

8m² of uninsulated steel door @ 1.2W/m².K = 96W

40m² remaining wall, corrugated steel with 1" insulation @ 0.6W/m².K = 240W


Total Losses for the room are 640Watts. Obviously, I'm a lot more industrial biased, but the principles are the same in homes and offices.


In short, if you put 640W into the room, then eventually you will get you 10°C delta that you are trying to achieve.

If you then add the 257W required to hear the volume of air, then with 897W will heat that room by 10°C within an hour.


So where do you stop... Now consider that someone is going to open a door a few times every hour and let some of the heat out, and no one has or really wants an airtight room, so we would consider that over the course of an hour, the air in the room is replaced with cold, 10°C air 3 times. So no we are having to heat 225m³ of air every hour, and not the original 75m³. Making the 257Watts now 771W.


The total for our room is now 1.4kW, we would usually look to add a margin for incorrect, or rather 'real life' U-values, rather than assumed, approximated and estimated based on lab testing. So add 10% and round a little. We need 1.5-1.6kW to heat this room.



Now, if you need to actually calculate what your radiator is kicking out, that is a different matter. As someone has already mentioned the Stefan Boltzmann Constant, we will start with radiant losses, the larger part of what a heater kicks out.
q = ε σ (T_h⁴ - T_c⁴) A_c      (Temperature in °K)


For this calculation, you need to know the emissivity of the object, a number between 0 (does not emit any energy) and 1 (perfect black body, 100% losses) Radiator paint is designed to be pretty good, and will have a reasonably high emissivity, 0.8-0.9 (0.85 for example).  People have also discussed the temperature variance of a radiator, if you use the average value, you will get the most accurate answer.

So when you first turn on the heating (10°C ambient), and once the heater is up to temperature - average 55°C  in this example, that equation would look something like this:

0.85 x 5.6703e10^-8 x (328.15^4 - 283.15^4) x 2(metres sq surface area)  =  498Watts from Radiation


Now there is also a good convection from Radiators, and this is where it is much easier to assume than calculate. Calculations here will start to look at the flow cause by heating air over what is effectively a flat plate, natural convection, warm air rising over the surface of the radiator starts to escalate the hotter it gets, so as air from the very bottom of the radiator heater heats and travels upwards, it can leave the top at speeds of 1-2m/s. this creates flow which then helps to 'cool' the radiator, or rather transfer the heat away from it.


To calculate the actual losses, use equation:
q = h_c A dT

However, to work out h you need to go into all sorts of calculations to work out all manner of things, Rayleigh number, Grasshof number, and even then you shall likely lose yourself in the numbers that are required to calculate these. So, typical values for h, free convection in air. I would use 3-5W/m².K (4 in example)

So here we would look at 4 x 2 x 45 = 360Watt from convection.


So for a radiator with 2m² surface area, we get about 860 Watts. Simplest thing here.... double the size of your radiator. at 4m², we are looking at 1.7W which is just about the total 1.5-1.6kW required.


I apologise for the length and detail in this if not necessary, I often get accused of over complicating things.


Good Luck, Regards James

Thanks Michael, but I was just answering Arran's initial question with quite a comprehensive example.


Didn't even get onto calculating the U-values, but thought that would have gone a few steps too far.


Regards

James

After running through the calculations a question arises as to how much to overspecify the power output of a radiator.


For example, the total surface and ventilation losses of a ground floor room is 1240W and 195W are required to raise the air temperature from 5°C to 21°C in one hour. Therefore a radiator with a minimum power of 1435W is required. If a non-thermostatic radiator valve is used then the power output of the radiator should not be significantly higher in order to avoid overheating the room. If a thermostatic radiator valve is used then it provides an opportunity to install a radiator with a considerably higher power output to provide extra heat in the event of a big freeze up when the valve is on a high setting but on an averagely cold day the valve is turned down to shut off the radiator when the room has been heated to its desired temperature.


Would a 1800W radiator be a realistically sensible choice or overkill?

I'm no HVAC expert but I did do some work on building heat loss as part of an MSc University Project.

The CIBSE guide was very useful and if its good enough for a Master project I'm sure it will be sufficient for your needs.  The method is the same as what James Brown described, essentially the Steady State - Simple model.  There were of course Dynamic models but these are far too complex for manual calculation and require software.


As others have said why look for something overly complex?  In my field of expertise (Electrical Engineering) we often use relatively simple methods to calculate various aspects, of course more complex methods exist but outside of academia are of little use within an industrial context.  I'm sure the situaiton here is the same, why not use tried and tested simple(ish) methods that can be easily done manually and give reliable results without the need for software?  Indeed you said earlier you were not interested in software, yet what you were asking for is calling for it.

Also I'm surprised nobody else has said that some of your comments come across belittling and condescending towards trained and qualified persons - eg plumbers or HVAC Engineers.  The rules of thumb methods may work very well for individuals on site who may not have access to more detailed information at the time or have the notion to carry out calculations (even if more simple than the 'advanced' mehtods you mentioned at the start of the post).  Coming back to my own field, rules of thumb (or empirical) do have their place in science and work for on site persons, being backed up by more detailed calculaitons if required.

It is quite insulting therefore for someone who may have many years experience on site or of a practical nature and be well qualified to then read this post by someone that it appears has less knowledge than they do, does not really seem to know what they are asking for but is yet dismissing their tried and tested methods as irrelevant.

If you have no interest in software you may as well use the average figures for heat loss per square metre based on the standard components used in the building at the time of construction.


If it's a leaky 1970's building shoot for 120w/sq m for example. Add up the square meterage, multiply that by the required power density and you have your radiator output. Generally 20% is added as a design margin.


If you are fitting a combi-boiler and want to use low flow and return temperatures like 60/40 you'll need to size your radiators up by about 50% for the same total heat output.


Always fit TRVs to every radiator except one so that the boiler can always bleed heat off through an open circuit when required. This generally goes in the coldest room in the house or the bathroom.


An oversized radiator allows the room to heat up quicker before the TRV throttles down the flow to maintain whatever temperature you set it at.


You are sizing the system for the coldest weather conditions, at anything less the system will be running at far less than maximum output so even if you get it wrong you'll only have to worry about it for a few days each year when it's -5 outside.


The total heat loss for a house is anything between 2.5kW for a modern, well-insulated small house and 20kW for a leaky old large house. Any boiler you choose will be able to output that level of heat with no issues.


You can make it horrendously complicated if you like but there is really no point, it's actually very simple. You would have to try really, really hard to do something bad because all the radiators will have TRVs.


If you have a look at Table 18 of BSRIA "BG 9" Rules of thumb 5th edition you'll see the sort of thing i mean.


Hope that helps.

In June 2019 Quinn Radiators of Newport went into administration. Information I have is that the designs and tooling for the Barlo designer ranges have been bought by Myson and are now manufactured in north east England. The sale did not include the panel radiators including the Barlo Delta Compact.


The Barlo Delta Compact is believed to be the most efficient of all (mass market?) radiators sold in Britain with the highest heat output in relation to its physical size. They are still available from central heating suppliers whilst their stocks last but it appears that a potentially unique product has now been discontinued.


I have been verbally informed by a central heating installer that the market for high quality panel radiators is squeezed in the middle between cheap imported panel radiators made in eastern Europe for rental properties, refurbs, and homeowners who are more interested in saving money on radiators than on heating bills, and designer radiators for homeowners who want style. The central heating installer then mentioned how designer radiators are designed to look good rather than radiate efficiently, where many are mediocre in operation or have lower outputs for their physical size than an average panel radiator.


The tooling for the Barlo Delta Compact and other panel radiators is in the now closed down Quinn factory in Newport but whether anybody wants to buy it to resume production remains to be seen.


The way the radiator market has moved in recent years could have a detrimental effect on fuel consumption and CO2 emissions. Much attention is devoted to the efficiency of boilers but far less so to the efficiency of radiators.

Simon Barker:

I'm sure that, given a bit of time, a physicist could give you an near-exact formula to calculate the size of radiator you would need to maintain the temperature of a given room.


The trouble is, it would have many variables that you don't have the exact values for.  Plugging in wild guesses would give you a result no more accurate than the plumber's rule of thumb.


Off the top of my head, you would need to know:

The temperature of the water in the radiator (and do you want to account for the temperature drop from the inlet to the outlet?).

The thermal conductivity of the steel and paint of the radiator.

. . .

What temperature that air is at.

 

I could add a few more things to this list, which could be endless.

• What about the insulation on the pipes feeding the radiators, or lack thereof? (I was appalled to find in my house many uninsulated water pipes below floor level.)

• Is there any reflective material on the wall behind the radiator. (Does this really make much difference? I suspect not.)

• What about factoring in the pipes gradually becoming furred up over the years?

• What about people who constantly alter the thermostat? (Not for bad reason. People want the room air warmer when they feel cold and vice versa.)

A formula needs to be straightforward, in view of all the unknowns, even though it may be based on rigorous thermodynamic analysis.


Let me give an analogy. An electrical engineer working for a electricity supply network is to design a substation to supply a new, large office block. How should he assess the MVA rating of the transformer? I don't think he would seek a detailed inventory of all the lights and appliances expected to be used, tot these up and apply a diversity factor. Over the years reliable statistical information has been produced from which one can assess the volt-ampéres required per square metre of floor area. It would depend, of course, on the type and purpose of the building.


In a similar manner, a plumbing-cum-heating engineer can assess the rating of a boiler for a house of given size, and judge the size of radiator needed according to the size of the room. If house insulation level is judged to be poor then I suggest this is where attention is required rather than just boost radiator sizes to compensate.

Arran Cameron:

. . .

It's notable that the central heating trade stubbornly sticks with BTU/h (more often than not erroneously quoted as just BTU) but not all central heating installers know the definition of a BTU or that it is convertible into watts, and vice versa.

 

I am not sure the trade is as stubborn as Arran suggests. I have seen in boiler specifications ratings in the dated BThU/h, usually as a bracketed alternative to kilowatts. When we had our central heating boiler replaced last year, the conversation about the rating required took place in kilowatts. Some installers are moving with the times. In the end it was a case of installing a boiler just slightly larger than the previous one, which served well while it worked.


In answer to Arran's original question, I think the current range of calcuating aids plus others suggested by contributors to this conversation are about as good as he will get.