Matty D Posted October 3 Share Posted October 3 (edited) Hi wise forum people, I'm in the process of getting quotes to replace an oil boiler with an ASHP for my farm house. I've got two quotes now in place, one of which was nice and detailed with all the heat loss metrics, and the other which just listed suggested radiator replacements. Despite this, after some discussion, the radiators they both suggest to replace match fairly well, except for the lounge. After a bit of questioning, I got some figures for the less detailed quote and am now having difficulty comparing the two. Company A (detailed quote) say the heat loss of the lounge is 2800 watts, and company B stated 4700 watts. I could understand a little variation, but this seem like a large difference. Further confusion comes in when I compare what each company suggests the CURRENT radiator will produce: Company A: 782 watts Company B: 1732 watts. The both correctly identified the rad as a K2 1000 x 600 mm, and I'm assuming they MUST have to use the same MCS methodology and flow rate and target temperature. There is clearly something funny going on. Does anyone have any ideas what, and if there are other methodologies that could be at play? Thanks. Edited October 3 by Matty D Add tags Link to comment Share on other sites More sharing options...
JohnMo Posted October 3 Share Posted October 3 Different flow temperatures give different radiator outputs. Perhaps the one quoting the lowest radiator output runs the system a much lower temperature? If so this is good Link to comment Share on other sites More sharing options...
ReedRichards Posted October 3 Share Posted October 3 13 minutes ago, Matty D said: K2 1000 x 600 mm The Stelrad model of such a radiator has a quoted output of 1672 W at Delta T = 50 C. So the 1732 W figure must be the rated output and not what you will achieve with a heat pump, which will operate at a much lower delta T than 50 C. But it could be roughly what the current radiator produces at your current operating conditions with an oil boiler (80 C flow, 60 C return perhaps?). Either way, if this is the only radiator in the lounge then the calculations suggest it is currently insufficient to keep the lounge warm enough in colder weather. If that's not true then you need to consider why. Link to comment Share on other sites More sharing options...
Matty D Posted October 3 Author Share Posted October 3 Thank for answers. Now I look into rad specs, you are right @ReedRichards, the rad figures provided are Delta T = 50C. This is where I get confused, as they have been provided in the context of an ASHP quote which I understand are provided to MCS criteria (the company stated they are MCS certified), AND the heat loss for the room is the heat loss for the room. i.e., I don't see how you can use delta T50 rad figures, when the system will run with a flow rate of 50C and a design temp of 20C, and then compare it to a heat loss figure which is calculated from U values, air changes etc... Could they be somehow taking the heat loss figures and converting them to be comparable to delta T50 rad specs (and if so why)? I've asked for clarification and will post back, but any further insight would be appreciated. Link to comment Share on other sites More sharing options...
ectoplasmosis Posted October 3 Share Posted October 3 50degC design temp is too high, you should be aiming for a maximum of 45degC. Our MCS heat loss calculation document and associated emitter schedule had radiator Watt/BTU ratings listed at Delta-T 50 for ease of comparison to manufacturer's specs. Link to comment Share on other sites More sharing options...
JamesPa Posted October 3 Share Posted October 3 (edited) 36 minutes ago, Matty D said: Thank for answers. Now I look into rad specs, you are right @ReedRichards, the rad figures provided are Delta T = 50C. This is where I get confused, as they have been provided in the context of an ASHP quote which I understand are provided to MCS criteria (the company stated they are MCS certified), AND the heat loss for the room is the heat loss for the room. i.e., I don't see how you can use delta T50 rad figures, when the system will run with a flow rate of 50C and a design temp of 20C, and then compare it to a heat loss figure which is calculated from U values, air changes etc.. Radiators are normally specified for deltaT50. Many heat pump designers will work out the heat loss and from that (and the flow temperature) work out the DT50 rating that this would require, because its then easy to pick the radiator. Or they will produce a table that includes both. For a most radiators output is proportional to deltaT^1.3. Another way installers do the calculation is to work up the 'oversize factor' = (50/(actual deltaT)^1.3. Again an easy way to work when looking through radiator catalogues. Take another look at the quote, they have probably done one of these. Edited October 3 by JamesPa Link to comment Share on other sites More sharing options...
PhilT Posted October 3 Share Posted October 3 1 hour ago, JamesPa said: (50/(actual deltaT)^1.3. Again an easy way to work when looking through radiator catalogues. ( actual delta T / 50 ) ^ 1.3 ? Link to comment Share on other sites More sharing options...
ReedRichards Posted October 3 Share Posted October 3 5 hours ago, Matty D said: AND the heat loss for the room is the heat loss for the room. This is absolutely correct so there are two issues: Why do the two companies quote vastly different values for the heat loss? Why do both companies quote heat loss values far higher than your current radiator is capable of matching with its current heat output? Link to comment Share on other sites More sharing options...
ReedRichards Posted October 4 Share Posted October 4 Just a thought but I have an open plan lounge and dining room. This was originally heated by three radiators and I had to add a fourth one we we moved from oil to a heat pump. The heat loss of the two rooms combined was calculated to be 3900 W but the calculation split this into 3 "areas" and assigned a heat loss to each. So the difference between Company A and Company B could be down to a difference of opinion as to what they think constitutes the lounge, if it isn't clearly separated from another area by a door. Link to comment Share on other sites More sharing options...
JamesPa Posted October 4 Share Posted October 4 15 hours ago, PhilT said: ( actual delta T / 50 ) ^ 1.3 ? ((actual delta t)/(delta t at which the rad is specified (ie 50)) to the power 1.3 So if a radiator emits 1kW at delta T 50, then the power output at delta T 20 will be 20/50^1.3 *1000 = 300W. The inverse ratio (50/20)^1.3 = 3.3 is called the oversize factor. This can conveniently be used to work out the dt50 rating required for a particular output at the design dt. So if my room needs 400W and the oversize factor is 3.3 then I need to look for a radiator rated at 1320W. 1 Link to comment Share on other sites More sharing options...
ReedRichards Posted October 4 Share Posted October 4 18 hours ago, JamesPa said: For a most radiators output is proportional to deltaT^1.3. So there are some radiators where it isn't? What differentiates these ones? Link to comment Share on other sites More sharing options...
JamesPa Posted October 4 Share Posted October 4 1 hour ago, ReedRichards said: So there are some radiators where it isn't? What differentiates these ones? I haven't seen any, but then I haven't done a comprehensive trawl. Fan radiators almost certainly don't follow this rule, and I imagine very tall rads may not. Its not going to vary materially from manufacturer to manufacturer, its just the physics of convection. Some rad manufacturers quote the formula in their literature, giving a value for the exponent. Every one I have seen the exponent is 1.3+/-0.02 Link to comment Share on other sites More sharing options...
Beelbeebub Posted October 4 Share Posted October 4 I think the 1.3 holds good for standard designs, but if you had some sort of designer rad, or on maybe a cast alu rad with lots of fins, it might change a bit. Link to comment Share on other sites More sharing options...
PhilT Posted October 4 Share Posted October 4 (edited) On 03/10/2024 at 08:55, Matty D said: Further confusion comes in when I compare what each company suggests the CURRENT radiator will produce: Company A: 782 watts Company B: 1732 watts. So one could be quoting the existing 1732w @ ΔT of 50 with gas And the other 782w @ ΔT of 27 approx. with a heat pump, which makes sense if the heat pump average radiator temp is 47, less target room temp of 20. Not sure about BUS but certainly RHI specified a MAX heating flow temp of 50, not the average of the in & out. So my heat loss calcs all use radiator temp of avg. 50, 44 = 47 maximum target. In practice it operates way below that 99% of the time But no light yet shed on total lounge heat loss. 4700w seems implausible unless there is a large glass area somewhere Edited October 4 by PhilT Link to comment Share on other sites More sharing options...
SteamyTea Posted October 4 Share Posted October 4 Just a point on the maths. 20/50^1.3 is not the same as (20/50)^1.3 Which is it, I suspect the latter. 1 Link to comment Share on other sites More sharing options...
JamesPa Posted October 4 Share Posted October 4 37 minutes ago, SteamyTea said: Just a point on the maths. 20/50^1.3 is not the same as (20/50)^1.3 Which is it, I suspect the latter. The latter. Apologies. 1 Link to comment Share on other sites More sharing options...
Iceverge Posted October 4 Share Posted October 4 (edited) Waiting to take flak again for this. Is an A2W ASHP the correct choice for your farmhouse? Those room heat losses suggest that you have an older building that might be more suitable to intermittent heating. Espically if it's not occupied 24/7. Continuous maintenance of say 21deg will get very expensive unless you're in a super low energy building. Intermittent heating needs a high power heat source. Domestic heat pumps tend to top out at about 16kW while the smallest oil boilers begin at 20kW. You may simply not get enough power for economical operation with a heat pump vs FF. Edited October 4 by Iceverge Link to comment Share on other sites More sharing options...
ReedRichards Posted October 4 Share Posted October 4 That seems like a perfectly valid observation. But if the heat loss is at least 2800 W, how on earth is it currently being heated by a 1700 W (@ Delta T = 50 C) radiator? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now