Combining radiators with UFH - some surprising conclusions

[sharpener]

5 hours ago, Beelbeebub said:

So you could run open loop WC, and just balance the UFH and rad flows to give you the best results. 

 

What about the salus autobalancing trv/actuators? They monitor the flow and return and adjust the flow (they sit on standard trv valves or even ufh loop actuators) so the dT is 7c (or maybe 5c, I forget) 

 

Yes. Salus ABAs look interesting (see also this thread) but address the wrong problem. They would control the recirculating flow in the manifold post-mixer. I need to control the amount of water from the primary flow into the manifold. Also it needs to be different in the two scenarios, by a factor of 4.

 

Interesting prog on Ch 5 as I write, not too bad as they go with good pointers to BUS.

Edited January 29 by sharpener

[Beelbeebub]

11 hours ago, sharpener said:

 

Yes. Salus ABAs look interesting (see also this thread) but address the wrong problem. They would control the recirculating flow in the manifold post-mixer. I need to control the amount of water from the primary flow into the manifold. Also it needs to be different in the two scenarios, by a factor of 4.

 

Interesting prog on Ch 5 as I write, not too bad as they go with good pointers to BUS.

Hang on how is your setup. 

 

Is you ufh via a "standard for boilers" manifold with it's own pump and mixing valve?  Is your flow and return direct from the HP or. Is there a thermal store/buffer? 

[sharpener]

55 minutes ago, Beelbeebub said:

Hang on how is your setup. 

 

Is you ufh via a "standard for boilers" manifold with it's own pump and mixing valve?  Is your flow and return direct from the HP or. Is there a thermal store/buffer? 

 

Yes yes and no. At least not in the sense you mean.

 

Before anyone gives the standard advice for HPs to remove the mixing valve and pump, note that I need them when both sets of emitters are operating in order to dilute the hot water at the temp of the rads down to a suitable temp for circulating in the UFH. Otherwise we get back to the original problem which is the UFH return flow is too much and too cold.

[Beelbeebub]

3 hours ago, sharpener said:

Before anyone gives the standard advice for HPs to remove the mixing valve and pump, note that I need them when both sets of emitters are operating in order to dilute the hot water at the temp of the rads down to a suitable temp for circulating in the UFH. Otherwise we get back to the original problem which is the UFH return flow is too much and too cold.

Thats what the auto actuators would sort out. They would regulate the flow so the dt was 7/5C.

 

The alternative is to manually balance the ufh and rads so the dt was to the HP's liking and run it all (ufh and rads) as one big zone. 

 

From the HP's point of view there isn't any difference between a rad and a UFH. All it sees is a flow rate at a given pump speed and a dT and will adjust it's power to achive the setpoint temp. 

 

If you are hitting the maximum power and are unable to achive the set temp then isn't the problem is your HP is too small?

 

 

 

 

[JohnMo]

3 minutes ago, Beelbeebub said:

the HP's point of view there isn't any difference between a rad and a UFH. All it sees is a flow rate at a given pump speed and a dT and will adjust it's power to achive the setpoint temp

That's the issue doing UFH, the floor can suck the heat away really quickly, this means return temp stays low, as heat pump wants to manage dT first, target temperature second. So if you try to run UFH for short periods, you never achieve a stable operating point. This means UFH can happily absorb most heat given to it, and big flows of return water keeping the return temperature low, never achieving target flow temperature.

 

Run UFH long a long time the return temp increases,  smaller quantities of hot water are taken into the mixer, more water is recycled within the mixer. Return temp increases because, more flow is routed to radiators and less to UFH. Heat pump achieves target.

[Dillsue]

On 29/01/2025 at 21:15, sharpener said:

 

That table was purely to show you how Heat Engineer tabulates the output at various MWTs, to help you scale your rads for a 35C flow temp if you can't get that info from the rad mfr's data sheets. The relative data are what is important, the absolute values less so.

 

After a look at Stelrads conversion factors I think to get the rad size the table needs rejigging to divide the heat loss by the conversion factor. Your bed 2 would need a rad size of 1276/0.396= 3222 Watts at the standard dT 50. Is that roughly what you ended up with??

[sharpener]

Can't recall. It is a 600 x 1200 K22 if that helps. (If BBOE is the quoted output condition then add 5% for TBOE.) It was a trade-off to avoid having a K33 or one so long it would have overlapped the door frame and looked rubbish.

 

Heat Engineer presentation could be better, so could rad mfrs who all have slight variations. Unfortunately there is no standardised operating condition for rads on HP.

If you are using a factor of 1/0.396 that is 2.5x, AFAIR installers used 1.9, can't now remember how that was derived, whether empirically, from rad data tables or raising the ratio of the two mean temps to power of 1.3 which is usually reckoned pretty good and well within other inaccuracies.

 

E.g. installers forgot B1 and B2 have vaulted ceilings, when I pointed this out they fixed one but not the other, in the end I gave up trying to get all the small errors fixed. Was at the time more concerned about getting the electrics installed in a way which integrated with the battery system, met the regs and kept their spark happy, which was not easy to arrive at.

 

As to the main thread am planning some experiments over the w/e and will report back.

[Dillsue]

16 hours ago, sharpener said:

Can't recall. It is a 600 x 1200 K22 if that helps. (If BBOE is the quoted output condition then add 5% for TBOE.) It was a trade-off to avoid having a K33 or one so long it would have overlapped the door frame and looked rubbish.

From stelrad tables a 600x1200 k2/type 22 puts out about 2100watts at dT50. That's a bit short of the calculated loss so how's things in the real world..... does that rad keep the room up to temp when it's freezing outside??

[Beelbeebub]

From the stelrad catalog. 

 

Remember it's the dT of the mean temp ie DT50 would be something like a flow of 80C and return of 60C (assuming room temp of 20C) 

 

So a typical HP with a flow of 45C and return of 40C would be midway between DT20 and dt25 so about 1/3 the rated output. 

[sharpener]

This has all drifted a bit off topic, but the formula is quite well known and quoted above.

 

(65/50) ^ 1.3 = 1.406

 

(20/50) ^ 1.3 = 0.304

 

so it is fairly clear that is what they have used.

 

In my case the design is for a bedroom @18C with a MWT of 47.5 so the correct factor is (29.5/50) ^ 1.3 = 0.504 which is pretty close to the installers' 1/1.9. Also they have not taken a/c of the heat flux from the kitchen below.

 

9 hours ago, Dillsue said:

does that rad keep the room up to temp when it's freezing outside?

 

The major problem is distribution not production. The cold comes from the window wall which is the only external wall, the heat comes from the rad on the opposite side of the room. No, a fan of any kind would not be acceptable to OH (whose study this is) bc of the noise.

 

[Beelbeebub]

2 hours ago, sharpener said:

This has all drifted a bit off topic, but the formula is quite well known and quoted above.

 

(65/50) ^ 1.3 = 1.406

 

(20/50) ^ 1.3 = 0.304

 

so it is fairly clear that is what they have used.

 

In my case the design is for a bedroom @18C with a MWT of 47.5 so the correct factor is (29.5/50) ^ 1.3 = 0.504 which is pretty close to the installers' 1/1.9. Also they have not taken a/c of the heat flux from the kitchen below.

 

 

The major problem is distribution not production. The cold comes from the window wall which is the only external wall, the heat comes from the rad on the opposite side of the room. No, a fan of any kind would not be acceptable to OH (whose study this is) bc of the noise.

 

You're always going to be a bit buggered if the external wall is opposite the heat source. You'll just get a steeper gradient if you up the flow temp. 

 

Is there any prospect of moving the rad to the outside wall? Your only other alternative would be some sort of forced air movement - there are fans that are pretty much silent, you don't need alot of air movement to make a big difference. 

 

Maybe a large diameter, low speed ceiling fan? 

[Dillsue]

On 01/02/2025 at 18:28, sharpener said:

In my case the design is for a bedroom @18C with a MWT of 47.5 so the correct factor is (29.5/50) ^ 1.3 = 0.504 which is pretty close to the installers' 1/1.9. Also they have not taken a/c of the heat flux from the kitchen below. 

Using 0.504 gives a dT50 equivalent 2530 watts radiator for bed 2 with the rad fitted seemingly around 2100 watts output. Given the gain/loss from the room below wasn't accounted for and some rads might not be positioned optimally, I think I'd be tempted to revisit all the heat loss calcs and verify rad sizing at least meets the heat loss.

 

What you do now will likely be in place for decades and if you're upping flow temps to deal with undersized rads, you've potentially got decades of throttled COP!!

[sharpener]

On 01/02/2025 at 20:54, Beelbeebub said:

Is there any prospect of moving the rad to the outside wall?

 

No, none whatsoever. The rad sizes and positions are now set in stone. This philosophical discussion was all a fork from the original thread caused  by this enquiry:

 

On 29/01/2025 at 14:52, Dillsue said:

Did you do heat loss calcs for each room to size the rads and factor in the the massive reduction in output for low temp flow?? I'm working my way through all our rooms/rads just now to resize everything for 35 degree flow and intrigued to know the approach you took??

 

The only changes I am wanting to make now are to do with stopping the UFH from preventing the rads running at their design flow temp.

Today I have been making a series of measurements but paradoxically the OAT has been 8.5C so the o/p of the HP is more like 15 or 16 kW and the problem is hard to replicate under these conditions.