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Posted

Would it be possible for someone with knowledge of this to confirm these figures, please, in relation to an aluminium vertical radiator in a bathroom:

 

Water temperature going into rad [flow in] is 40C and water exiting [flow out] is 35C.

Average temp in rad is 37.5C.
Target temperature of bathroom is 22C.
DeltaT is 37.5-22= 15.5C
 
DeltaT correction factor is 0.179 [from manufacturer]
Radiator Output at DeltaT 60C is 1661 Watts [from manufacturer]
 
Corrected rad output is 1661 x 0.179= 297.32Watts
 
Many thanks.
Posted (edited)

What  I have all ways used based on the mauufacturers correction tables for mild steel rads at any rate is the exponential 1.3.

A 15.5C rad will have a correction factor of (15.5/60)^1.3, 0.17212 to give 285.894 watts from a 1661 watt (T60)  rad, or 96.16% of your figure of 297.32 watts.

Edited by John Carroll
Posted
20 hours ago, John Carroll said:

What  I have all ways used based on the mauufacturers correction tables for mild steel rads at any rate is the exponential 1.3.

A 15.5C rad will have a correction factor of (15.5/60)^1.3, 0.17212 to give 285.894 watts from a 1661 watt (T60)  rad, or 96.16% of your figure of 297.32 watts.

If the rad is producing 285 watts and the calculated total heat loss for the bathroom is 643 watts, does that mean the rad has been incorrectly spec’d? Or someone has not accounted for the corrected output.

IMG_6291.jpeg

Posted
1 hour ago, John Carroll said:

Simply put, yes, the specified T60 rad output should, IMO, be 643/0.17212, 3736 watts.

 

Don't know what the corrected loss of 2074 watts means though.

Thanks for that, John.

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