Hi all, I have a single radiator (central heating type) of size 70cm*75cm, circa
1976, that I need to replace. It is in a small bathroom and was wondering what wattage output it was likely to give. Does anyone know or know a formula to calculate it?Many thanks
Ferret
Couple of hundred watts or so.
No more than 500W anyway.
In article , DUMBO writes
Courtesy of John Stumbles 17 Jan, 00:32
Subject: Radiator output formula and finless rad specs
After a bit of firkling around with Myson & Kudox radiator output specs I found a formula which gives a reasonable approximation so one can calculate output from dimensions (and v.v.)
Heat Output = (Height + 12) * Length * FACTOR (Watts) (cm) (cm)
where FACTOR is (for Delta-T 50°C)
See
According to the Wickes Good Ideas Leaflet a 800w x 620h single rad (roughly similar area) is 2,552 BTU's/Hr which is about 750W IMCAC.
The Stelrad catalogue from the late 70s says 605W for one 740h x 640 long, and 756W for 740h x 800 long. Interpolation is left as an exercise for the reader. Other makes won't be very different.
You need to watch out for the average temperature which is claimed to generate the specified power, and then correct for real values (or if comparing, ensure the two datasheets are using the same value for average temperature). When I was designing my system, some of the manufacturer data sheets had small print saying values were at
80C average temperature. Not even a conventional system is likely to ever reach that, as that's typically the maximum flow temperature. (They're usually given as the delta, i.e. difference between room temperature and average radiator temperature, so a delta of 60C would typically be a room temperature of 20C + 60C delta = 80C average radiator temperature, which just isn't realistic.)
Exactly. Which is why my gut feel is it would be in the 200W - 500W range, depending on boiler temps.
Yes - I should have added that these are for average radiator temperature of 170F and room temp of 70F, so delta T of 55.6C. The correction factor to delta T=50C is given as 0.87 +/- 0.02.
Assuming it has no fins on the back, a reasonable rule of thumb would be
1280 watts /M^2 at a Delta-T of 60 degC.For the size quoted, this would equate to about 670 watts.
If your system is running at a lower Delta-T than this, the output will need to be downrated. For example, at a Delta-T of 50 degC it would be about 530 watts.
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