Formulas to calculate the required power output of a radiator for a room of a specific size. Do they exist?

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

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