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How many (low-temperature) radiators are enough?


Garald

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2 hours ago, ReedRichards said:
 
That's 2561 Watts at delta T = 50 C and, as you correctly observe, 0.41 times this, 1050 Watts for delta T = 25 C (sorry about the error in my earlier comment). 

 

So, what power do I need to heat a 3.75x9.75m library to 20-21C when it's -7 C outside? (Assume it is well insulated: R of insulation at least 3.8, closer to 4.5 counting the brick wall and an air layer; I'm changing the double glazing, whose thermal perfornace is already high, by better double-glazing.) About 3 kW perhaps, so three of these radiators? Maybe put a fourth radiator in the broad corridor leading to it, since it will also be part of the library? (Put a cushy bench in front of that fourth radiator, especially for those who like to get toasted while reading?)

 

(Anything significantly below -7C in Paris is basically Eurostar-stopping weather. If shove comes to push, well, you can see a working fireplace in the pictures.)

 

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8 hours ago, Garald said:

 

So, what power do I need to heat a 3.75x9.75m library to 20-21C when it's -7 C outside?

 

 

I am fortunate in that my installer did all the heat loss calculations for me (using some software called Heat Engineer).  The biggest "room" in my house has a floor area of 48 m2 and was calculated to require about 2.8 kW of heat when it was -4 C outside.

 

1 hour ago, Garald said:

Also, I guess this means I can take an old radiator that is now insufficient for the library or the main bathroom, say, and put it in a smaller space that needs to be heated (such as a WC), if there is room for it?

Yes indeed; just make sure your installer flushes all the gunk out of it before it is relocated.

 

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9 hours ago, Garald said:

 

So, what power do I need to heat a 3.75x9.75m library to 20-21C when it's -7 C outside? (Assume it is well insulated: R of insulation at least 3.8, closer to 4.5 counting the brick wall and an air layer; I'm changing the double glazing, whose thermal perfornace is already high, by better double-glazing.)

 

You have the spreadsheet, so easy enough to calculate.

 

U value is 1/R, so R of 4.5 is 0.22 U value.

 

Double glazed window U value 1.4, single glazed U value 5.

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5 minutes ago, JohnMo said:

It does LL the white cells.  Just size to suit room.  Assume internal walls have no heat loss, as they are heated on the other side.

 

LL? 

 

The (part-wooden) ceiling is below my (heated) attic, and the floor is mostly above a GP's (but his hours are short and he seems to be often on vacation), except for the part above my bikeshed (unheated, but will insulate the ceiling). 0 roof, 50% floor?

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So LL, meant to type 'all'

 

Ceiling, just use 0.1m2 for area to keep the calculation from throwing a miss calculation.

 

Floor just assume it settles to 8 degrees the same as the ground above the shop.  Then make an alteration to compensate for the area of the bike shed, which will cool to your lowest temperature.

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I altered your wall U value, to 0.22 and it puts the heat loss at -10 degC at 1096W.  Which sounds about right.

 

Some perspective, my house 192m2 single storey, whole house heat loss is 3088W

 

Edited by JohnMo
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Just now, JohnMo said:

The more surface area you have of radiators the lower the temperature needs to be to get the same heat output.  Also water volume is good for heat pumps.

 

Other question is do you need to size the system for -10 degrees?

 

It's prudent to size it to -7 degrees (in that it's still fairly common, in that (I think) it happens most or nearly all years; no bougainvillias in Paris, though that may be changing).

 

Anything less than -10 degrees is Eurostar-stopping, highway-closing. airport-disaster-movie weather. I certainly don't think I need to size the system for *less* than -10 degrees. If that ever happens again (not certain because of global warming), I can just use the chimney.

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(What scientists and math people say of engineers is that they do rather precise computations, with the necessarily imperfect data and models that they have, and then multiply everything by 2. I don't know whether this is an idealization or a slur.)

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I've kept U=0.22 and reduced the wall space to 36.56 (since two of the walls are internal). I get 847W, which is really low.

Even if I set the wall space to 48.30 (because a short wall adjoins the kitchen, which won't be insulated - doing so would require ripping it apart) I get a total heat loss of 916W.

 

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