Replacement bathroom towel rail

I believe about 30% of the heat output of a radiator is by radiation. Hold your hand in front of a hot one, where there will be no conduction or convection.

Chrome is certainly the worst possible surface in terms of radiation, though black should be better than white. I suspect the figures quoted are not measured for each finish, but someone assumed that a black one would be the same as a white one of the same size.

Chrome is shiny chrome on both the outside and inside face, it reflects heats back, acting like an insulator. First google link:

Then you have piping to the rads which makes around 5% difference in output. Hot in at the top, out at the opposite bottom corner is the most efficient. Having both pipes at the bottom loses around 5% of the possible heat output. Add towels or a cover to the rad, and you can lose between 5 and 10% of output. So, painted white rads, flow of hot water in the top, flow out at the opposite corner, with nothing covering the rad will give the most output from a radiator.

While true that convection will usually be the dominant heat loss method, it can be unfair to say that emitted via radiation is minimal - it can still be a noticeable component of the total. It is very temperature dependant though.

The emissivity of polished chrome could be as low as around 0.1 (where perfect black body would be 1, and a painted surface in the .9 to .95 range typically)

So if we say that a towel radiator has a total surface area of say 1^2m (might be a bit high for a ladder towel rail - but it keeps the sums easy), and the average surface temperature is 60 deg C (333K), we can plug the numbers into the Stefan-Boltzmann equation to get the energy emitted per unit time:

Q/t = sigma * e * A * T^4

where sigma (Boltzmann's constant) = 5.67 x 10^-8 W m2, e = emissivity of the surface, A is the area, and T is the absolute temperature.

So for a chromed surface:

Q/t = 5.67 x 10^-8 x 0.1 * 1 * 333^4 = 70 J/sec (Watts)

For a painted one:

Q/t = 5.67 x 10^-8 x 0.9 * 1 * 333^4 = 627 J/sec (Watts)

Reworded it a bit...

(not sure if you meant this John, or the one who wrote it!)

-- Cheers,

John.

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It might be worth noting that the emissivity figure for a chromed surface is probably from one that has been chemically cleaned or vapour deposited.

In real life such surfaces are contaminated with a variety of compounds. The IR radiation comes mostly from these compounds, which, usually having an organic-chemical source, raise the emissivity by a significant amount. ISTR that a 2 micron layer of oil on an otherwise clean surface raises the emissivity from circa 0.2 to circa 0.8.

It’s the reason why black bathroom radiators radiate much like white ones: it isn’t the pigment that’s doing the radiating, it’s the paint’s organic[1] binder.

My error there - I was going to re-work the figures for half a square meter - but did not actually do it - so that should read 1 m^2

If you have a IR thermometer, you can get a feel for the difference the coating/contamination is making in more tangible terms. Normally they are calibrated assuming an emissivity appropriate for "normal" surfaces, and so will often under read the temperature of metallic ones.

So stick a bit of masking tape on your chrome rad, and take a reading of the temp from that, and again from the chrome - the reading from the metal will show it at a lower temperature as the thermometer is seeing less IR radiation and assumes that is because it is cooler, not because it has a lower emissivity.

It will give some idea of how much difference the chrome is making in a real world case.

I happened to notice the original article was quite "mature". I had assumed that you had written it. Or had you just edited it?

Thanks for all the comments.

I'm coming to the conclusion that the emissivity results can be "massaged" somewhat depending on the method(s) of measurement. It looked pretty clear from Leslie (Leslie's?) Cube experiments what to expect but a quick search suggested some were getting anomalous results. Anyway, I do wonder if looking at chrome/white/black emissivity is the right thing to do when much of the radiator will be covered with towels of varying colours anyway!

In my time in power electronics we used heatsinks of many colors, passive and force cooled.

In short painting them black was often routinely done because they were aluminium for best heat transfer, and so needed anodising, and black is very easy to dye the result with, however the dissipation was so unaffected as to suggets that there was almost no heat loss by radiation itself, but by conduction and convection. Adding a fan could improve great loss by up to a factor of 5 or more.

In short 'radiators' do not radiate to any great extent. The colour is supremely irrelevant at the temperature they are designed to operate.

The only time colour matters in thermal situations is objects subject to the radiation *input* from a very hot body like the sun.

I remember an amplifier with a black heatsink, hooked up to a thermometer, showing 85°C in the midday Johannesburg sun BEFORE I EVEN SWITCHED IT ON.

I'm in a similar place to you and was wondering how the numbers can vary so wildly. We'll be replacing a small but powerful double panel radiator with a towel radiator and don't want to lose the comfort from the room. Plus the radiator is on the "always on" route for the boiler so needs a certain output to satisfy the requirement. so the higher the output the better for us especially when factoring in the towels. It seems to be a minefield and two visually identical radiators can have such different reported outputs. One thing to watch out for is the 'delta T' the figure is reported at, as this also has an effect.

I don't know if your situation is sensitive to output but we're considering one like this,

Do you happen to know how much of a rad's output is radiated, or how it's calculated?

I was, ahem, taught that for all practical purposes radiators are convectors, and that should part-infrom where you put them.

Cheers, Rob, Sheffield UK

I think that I'll probably go for chrome, as we've got chrome taps, a shiny SS cupboard and drawers, and a shiny SS medicine cabinet. Chrome will match well.

I hadn't seen that and it is a clever design, but is on the expensive side. As it happens, it wouldn't fit as the bathroom door opens onto the towel rail. I'd thought of using two or three passive rails which clip over the towel rail bars as supports for the towels, but the problem would be the same - the door would hit them when opened fully.

I would fit a small normal single/double convector radiator and a couple of simple towel rails above it. Having plenty of heat to warm a bathroom up quickly seems important to me and the "towel rail" type radiators dont provide much heat for their size.

re finish: several years ago I did fit a floor standing towel rail type rad, when researching I noticed that stainless steel ones had a much higher quoted heat output compared to equivalent Chrome plated ones.

If you click the "view history" link you can see all the edits and who did each. John Stumbles wrote the original in 2008. I only did a few bits on it - mainly reverting edits by spammers! It looks like it had not been changed since 2010.

Alas towel rail rads are often a bit of a compromise... I fitted a bathroom for an Aunt once, and she wanted one of the tall ladder style towel rads. I did mention that she might not find it good for heating the room when it had towels on it. So I suggested we install pipes to allow another conventional rad to be added later if required. As soon as it got to winter, she was pleased we did and decided she wanted the normal rad added after all.

Although slightly clunky looking, when I did my bathroom refits, I went for a hybrid towel rail design that included a rad, and had the (heated) towel rail positioned so that it does not actually cover the main read part directly. Towels are suspended in front of the rad, but there is always a convection path up and through it:

I'm currently doing something similar. I have a ladder type (chrome) towel rail that doesn't give off enough heat in the winter months to make the bathroom comfortable for showers etc. The upstairs and bedrooms in this household are deliberately kept below 18C.

I have a limited space just wide enough for a 500mm radiator plus valves but I'm also limited to just under a metre in height. Above this I need clearance for opening an airing cupboard door. My solution is for a type

The 950W figure is for a flow temperature of 75C and a return of 65C and for the the purposes of radiator specification an average of 70C, and for a notional room temperature of 20C, giving a Delta T of 50. (Ave radiator temperature minus notional room temp)

As others have pointed out this may be a bit unrealistic with recommended condensing boiler settings these days. I run my flow most of the heating season at 60C (Flow 60, Return 50, ave 55, room 20 so Delta T = 35) Conversion factors Delta 45 = Delta 50 specified output x 0.87 Delta 40 = Delta 50 specified output x 0.75 Delta 35 = Delta 50 specified output x 0.63

Table at

The new radiator is sat in junk room awaiting a round tuit.

Those advocating towel rails should also take the above into consideration if this is the only source of heating in a bathroom.

correction Type 22 (double panel)

Erm, yes... from the very post you quoted :-)

You can calculate the gross energy radiated using the following formula:

Q/t = sigma * e * A * T^4

where sigma (Stefan-Boltzmann's constant) = 5.67 x 10^-8 W/m^2 K^4 e = emissivity of the surface A is the area in m^2 and T is the absolute temperature in Kelvins.

So work out the total surface area of the rad, and the surface temperature. Convert that temperature in deg C to an absolute temp in Kelvin (add 273 to it). For a painted rad assume the emissivity is 0.9

So say the rad temp is 60 deg C, that is 333K. If the rad area is say

So plug in the numbers:

Note that is the gross radiation - and that will be omnidirectional. With double panel or heat sink rads, some of that will be radiated and re-absorbed internally between panels and heat sink fins. Some will be absorbed straight into the wall or floor (which will then heat the room by convection). There will also be some absorption from other radiant sources in the room. So you may only have a couple of hundred watt being radiated directly into the room, a fair chunk of the remainder will be heating adjacent surfaces which will then in turn cool by convection.

Convection is the dominant mode of loss - both directly from air contact, and indirectly from air contact with local surfaces that were first heated by radiation. You will still get a proportion of the output directly radiated - particularly at higher flow temperatures.