Traditionally-styled low temp radiators?

[YodhrinForge]

I'm aware you can't use actual traditional radiators with a low temp heatpump driven system, but I'm wondering if anyone knows if there's such a thing as a low temp radiator that is *styled* at least somewhat like old Victorian cast iron units? Searching myself has yielded only either modern looking low temp units or actual Victorian salvage, but these days who can tell if that's because what you're looking for doesn't exist or just search engine enshittification.

[Garald]

1 minute ago, YodhrinForge said:

I'm aware you can't use actual traditional radiators with a low temp heatpump driven system, but I'm wondering if anyone knows if there's such a thing as a low temp radiator that is *styled* at least somewhat like old Victorian cast iron units? Searching myself has yielded only either modern looking low temp units or actual Victorian salvage, but these days who can tell if that's because what you're looking for doesn't exist or just search engine enshittification.

 

How low is low? My cast-iron radiators work just fine at 40C - that's their usual working temperature.

For truly low temp (25C, say) I don't think it makes sense to use any radiators, traditional or untraditional - they would need to be huge. That's why people use underfloor heating (basically huge radiators hidden by the floor.

 

Is there a real difference between traditional radiators and radiators that are sold as "low-temperature"? I was taught here that there was not, and followed the advice, which turned out to be good.

 

[JohnMo]

20 minutes ago, YodhrinForge said:

anyone knows if there's such a thing as a low temp radiator

A radiator is a radiator, takes its output at T50/T60 from manufacturer data and then use the correct correction factor, based on actual flow temperature you intend to use. 

[Garald]

2 hours ago, JohnMo said:

A radiator is a radiator, takes its output at T50/T60 from manufacturer data and then use the correct correction factor, based on actual flow temperature you intend to use. 

 

Right. For a back-of-the-envelope calculation, I just use Newton's law - radiators' function as literal radiant bodies is a thing but it's a second-order term (and at any rate it varies less with temperature in the range we are talking about). What correction factor do you use?

[John Carroll]

3 hours ago, Garald said:

 

How low is low? My cast-iron radiators work just fine at 40C - that's their usual working temperature.

For truly low temp (25C, say) I don't think it makes sense to use any radiators, traditional or untraditional - they would need to be huge. That's why people use underfloor heating (basically huge radiators hidden by the floor.

 

Is there a real difference between traditional radiators and radiators that are sold as "low-temperature"? I was taught here that there was not, and followed the advice, which turned out to be good.

 

There's a big difference in output between running a rad with a flow temp of 40C and a T40 rated rad 

A T50 rated rad (now the standard) often has 75C/65C/20C printed on the top of its spec sheet, is the mean rad temp - the required room temp,

the mean rad temp is the (flowtemp+returntemp)/2 and 20C is the most often quoted required room temp, so the T50 rad above is ((75+65)/2) -20, 70-20, 50C, a T50 rad. A dT of 10C will require a flowrate of 1.43LPM/kW

If the flowrate is kept the same then a T40 rated rad will have, flowtemp/returntemp/dT/output, 63.73C/56.25C/7.48C/74.8% but if that rad is run with a flowtemp of 40C (same flowrate) then these numbers become 40.0C/37.23C/2.77C/25%

A T25 rated rad numbers are 47.03C/42.97C/4.06C/41% but if run with a flowtemp of 25C then these numbers are 25.0C/24.53C/0.47C/4.7%.

 

One might thik that a T25 rad would give 50% output, (25/50)x 100, but it doesn't, it gives (25/50)^1.3 x 100, 41%, similarly a T40 rad will not emit 80% of a T50 rated rad, it gives, (40/50)^1.3 x 100, 74.8%

Edited February 22 by John Carroll

[John Carroll]

4 minutes ago, Garald said:

 

Right. For a back-of-the-envelope calculation, I just use Newton's law - radiators' function as literal radiant bodies is a thing but it's a second-order term (and at any rate it varies less with temperature in the range we are talking about). What correction factor do you use?

 

To the power of 1.3, see above.

Edited February 22 by John Carroll

[Garald]

10 minutes ago, John Carroll said:

 

To the power of 1.3, see above.

 

What is the physical reason for that?

[Garald]

The document https://www.delta-q.de/wp-content/uploads/heizflaechenarten_recknagel_sprenger.pdf

gives an exponent of 1.3 as typical for flat panel radiators; is the exponent an empirical approximation?

 

It does not give a value for what would be a typical exponent for legacy cast-iron radiators (what I have).

Apparently it's sometimes a bit lower (i.e. better) than 1.3; see, e.g., https://www.cinier.com/en/products/voltaire-47/

 

I am a happily heated man, but then I didn't just calculate 2 = (60-40)/(40-20), I also added a big fat safety factor of about 2 on top of that.

Edited February 22 by Garald

[John Carroll]

55 minutes ago, Garald said:

 

What is the physical reason for that?

 

I've no idea really, it was only years ago when I was looking at the manufacturers correction factors that I realized the ^1.3 "fitted" best. Its a bit strange I suppose that while heat loss is linear, rad output isn't, presumably this is why they have a WC "curve".

 

Re your own correctIon factor above, (60-40)/40-20), presume you mean (60-20)/(40-20)? to give 2, you say you added another ~ 2, X 4 in all?.

 

[Garald]

38 minutes ago, John Carroll said:

 

I've no idea really, it was only years ago when I was looking at the manufacturers correction factors that I realized the ^1.3 "fitted" best. Its a bit strange I suppose that while heat loss is linear, rad output isn't, presumably this is why they have a WC "curve".

 

Re your own correctIon factor above, (60-40)/40-20), presume you mean (60-20)/(40-20)? to give 2, you say you added another ~ 2, X 4 in all?.

 

 

4ish; 3 or so minimum. Since it was the first time I was working on heating calculations, I decided to take the joke on engineers ("calculate everything very precisely, optimize, and then multiply by 2") literally.

 

I wasn't really going out and buying oversized radiators - I had bought a place with vintage radiators, which were of course oversized, and became much more oversized when I insulated and made the place more airtight. In rooms where  I figured I didn't have enough of a safety margin, I would move an existing radiator to a smaller room and then buy a particularly tall new radiator. Had to do that only a couple of times.

 

The insulation on the north wall is not as good as I was promised, and I think the insulation in the last floor (attic) really leaves something to be desired (it was mostly done by the owners before the previous ones; it's not quite thick enough (16cm or so) but the bigger problem is that there's no airtight membrane) so having a large correction factor turned out to be the right decision, I'd say.

[Garald]

43 minutes ago, John Carroll said:

 

I've no idea really, it was only years ago when I was looking at the manufacturers correction factors that I realized the ^1.3 "fitted" best. Its a bit strange I suppose that while heat loss is linear, rad output isn't, presumably this is why they have a WC "curve".

 

Right, that's surprising. As I said, radiators are, what, 80% convectors and 20% radiators (more or less, no?), but as radiators their dependency on flow temperature is actually minor, in the range we are discussing: it is proportional to the cube of temperature in Kelvin, and (273+60)^3/(273+40)^3 = about 1.2, which is much less than 2, so you'd expect the power of a radiator at 40C to be a bit *greater* than half the power at 60C.

[SteamyTea]

16 hours ago, YodhrinForge said:

I'm aware you can't use actual traditional radiators with a low temp heatpump driven system

Not the case at all.

As power transfer is, in part, a function of surface area, you need larger surface area radiators.

The material they are made from makes little difference i.e. steel or aluminium.

 

Temperature is not energy.

If you look at the SI units for energy, the joule, you get kg.m².s^-2 [mass times area divided by time squared], no mention of temperature [K for kelvin}.

Edited February 23 by SteamyTea

[John Carroll]

Yes, but if you take the example of a T25 rad (above), this will give 41% output of a T50 but will require a oversize factor of, 1/0.41, X 2.41, to give a T50s rated output and so on.

[Garald]

At any rate, the TL;DR version is that the OP now knows he can get some nice, big iron-cast radiators (used/antique are fine, they just need to be properly refurbished/checked for flow). If the house is well-insulated and the radiators are big enough (likely the case, but he should use the spreadsheet), he'll be fine at T40.

 

What I'm left with is the curiosity for the real reasons why the exponent is greater than 1 (other than "practice is never as nice as theory").

 

I'd also like to know what to expect at T25. I don't imagine any radiators are large enough for that to work in winter. I've set the temperature curve in mine to start at T25 in early autumn/mid-spring temperatures, but the heat-pump seems to want to go up to 40. OTOH my heat-pump is probably a bit oversized, especially in those seasons.

 

[John Carroll]

50 minutes ago, Garald said:

At any rate, the TL;DR version is that the OP now knows he can get some nice, big iron-cast radiators (used/antique are fine, they just need to be properly refurbished/checked for flow). If the house is well-insulated and the radiators are big enough (likely the case, but he should use the spreadsheet), he'll be fine at T40.

 

What I'm left with is the curiosity for the real reasons why the exponent is greater than 1 (other than "practice is never as nice as theory").

 

I'd also like to know what to expect at T25. I don't imagine any radiators are large enough for that to work in winter. I've set the temperature curve in mine to start at T25 in early autumn/mid-spring temperatures, but the heat-pump seems to want to go up to 40. OTOH my heat-pump is probably a bit oversized, especially in those seasons.

 

 

Well, using the "1.3" exponent, a T25 rad will emit 41% (of a T50) at a flowtemp of 47C, but the rad, a T18.6, will only emit 28% at a flowtemp of 40C (HP max?) and at a flowtemp of 25C, a T4.8 rad,  will only emit 5%.

 

 

 

Condensing Calcs Extract Gar Rev0.xlsx

[Garald]

1 hour ago, John Carroll said:

 

Well, using the "1.3" exponent, a T25 rad will emit 41% (of a T50) at a flowtemp of 47C, but the rad, a T18.6, will only emit 28% at a flowtemp of 40C (HP max?) and at a flowtemp of 25C, a T4.8 rad,  will only emit 5%.

 

 

 

Condensing Calcs Extract Gar Rev0.xlsx 10.98 kB · 2 downloads

 

Well, sure (0.1^1.3 is indeed about 0.05), but I wouldn't expect any empirical exponent to give a good approximation that close to the minimum.

 

 

[marshian]

I’m not actually sure why anyone would think a radiator is more or less suited to low flow temp

 

I’m sure surface area is a factor for heat transfer but the lower the flow temps go the ratio of convection to radiation changes

 

At a higher temp most radiators convect more than radiate

 

At a lower flow temp the radiator may convect less than it radiates 

 

I know that from my perspective I’m more comfortable at a lower room temp with less convection creating air movement but that might just be me!

 

 

[SteamyTea]

4 hours ago, marshian said:

At a higher temp most radiators convect more than radiate

 

At a lower flow temp the radiator may convect less than it radiates 

Other way around I think, until you get to the extremes.

 

 

The OP has never come back to this topic.

Edited February 27 by SteamyTea

[marshian]

3 hours ago, SteamyTea said:

Other way around I think, until you get to the extremes.

 

 

The OP has never come back to this topic.

 

pretty sure I’m right @SteamyTea I can’t feel much convection above a radiator at a 30 deg flow temp and come to think of it does UFH convect or radiate 😉 

 

re the OP happens occasionally 🙂 

[SteamyTea]

7 hours ago, marshian said:

can’t feel much convection above a radiator at a 30 deg flow temp

That is because there is not much temperature difference between the emitter and the surrounding air, so not much density change.

Ideally, the emitter would be very close in temperature to the surrounding air, which is why UFH does not cause updrafts, but emitters under windows do (you can fly a paper plane above one).

Radiative forces are calculated from the Stefan-Boltzmann  constant multiplied by the forth power of absolute temperature. 

Then the area of the emitter and the receiver, as well as the relative angles, and the inverse of the square of the distance MUST be factored in.

And don't forget it is travelling though air, which will absorb some of the energy because of the water content (a reasonably good absorber of IR).

If the radiative component was significant, we would be running IR heating everywhere.

 

(In reality, all thermal energy, as is all energy, is radiative, but over tiny distances i.e. half the diameter of an atom)

[Big Jimbo]

I just love @SteamyTea. I never understand a blinking word he says, but, I always makes me realise that a decent school, was completely wasted on me. I defo bunked off that lesson.

[Big Jimbo]

Always reminds me of being told how many atoms you could fit on a pin head. Which, I might add, I immediately forgot.

[SteamyTea]

4 minutes ago, Big Jimbo said:

Always reminds me of being told how many atoms you could fit on a pin head. Which, I might add, I immediately forgot.

Depends on the material, and how the crystalline structure is arranged.

I never worry about the number of atoms in a structure, it is the phase they are in, or how they are arranged, that is important.

[marshian]

8 minutes ago, Big Jimbo said:

I just love @SteamyTea. I never understand a blinking word he says, but, I always makes me realise that a decent school, was completely wasted on me. I defo bunked off that lesson.

 

His pedantry over units can be a little tiresome but I do admire his persistence

[JohnMo]

12 minutes ago, marshian said:

 

His pedantry over units can be a little tiresome but I do admire his persistence

It is just as easy to kW as it is kw, so why not use the correct units?