Taking a tall narrow white radiator / towel rail with heat output 3150 BTU, which equates to circa 900W, how hot would it get with a 600W towel rail element? Is there an easy calculation?
(Element is a MEG thermostatic, I know 600W will fit, not sure if 800W will).
No. It will get as hot as it needs to get in order to dissipate 600 watts into the surroundings. It will get a lot hotter if you wrap it in towels in a hot room than it will if open to the air in a cold room.
But the OP says it has a thermostat so it will get as hot at that will allow?
Bob
Well, it won't get any hotter than the thermostat setting. But it may not necessarily reach the stat setting if it can dissipate 600 watts at a lower temperature - the element will be on all the time, and the stat will never operate.
Not necessarily.
Radiator is rated 3150BTU / 900W.
- Supply it with 900W and it will get to (say) 68oC under a set ambient.
- Supply it with 600W and it will get to (say) 52oC under the same ambient.
So even if the radiator is set to 60oC, a 600W element will never get it to 60oC (unless room ambient were 40oC in which case I do not think I would want any further heat input, heh-heh :-)
So... I will simply photoshop the wall with both old & new radiators, and arrange the tiles to suit either (the radiator will be partly recessed).
Yes, but there's a bit of info missing. The 900W heat output will be at a specific delta-T which is the difference between room temp and the average water temp. This is commonly stated as
50C or 60C in radiator datasheets - I'll assume 50C.So with 600W, it will reach 600/900 x 50 = 33C above room temperature.
This all assumes no towels on towel rail. It will get much hotter (until the thermostat clicks off) if covered in towels. It also assumes that the whole radiator works just as well when heated electrically as when heated with water. That's not always the case.
Not an easy calculation[1] but you might be able to find a curve showing heat output versus flow temperature for the rad. In which cas you can use that backwards to work out the surface temp that equates to 600W of output. However it will only be approximate since the airflow in its vicinity will affect the results, as would putting a towel on it.
[1] The actual heat loss rate will be a combination of convected losses- probably proportional to the temperature. Minimal losses via conduction - again prop to temperature. The radiated losses - proportional to the emissivity of the surface, and probably the forth power of the absolute temperature!
From the website too,
If delta-T were 60C it would be 40C rise, but safer to assume delta-T
50C in absence of info the contrary.Indeed, pumped flow compared to convective.
The radiator in question is a Serrenti White Flat Vertical Radiator
1200x420mm, an interesting slim design. The MEG 600W element is 37cm, I doubt the 800W element will beHomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.