Heat pump radiators

[Dillsue]

52 minutes ago, Michael_S said:

How thick is your 'backbone/main loop' though - ours is rather long due to the house having bene extended and the whole upstairs is 15mm. 

28mm x 7metres from the HP to the centre of the house then 3 x 22mm x 6 metres down the centre of the house, 1 for upstairs and 2 running the length of the house for downstairs. Mainly 10mm drops off the 22mm for each rad typically 6 metre run with 4 drops of 8mm in the original part of the house around 4 metres each. Distance are one way so double for flow and return.

 

15mm for the whole of upstairs sounds a bit constraining!

[Michael_S]

1 hour ago, JohnMo said:

Why do you need rads doing that output with a heat pump system? If you room has that heat loss you must be living in an open shed. Flow rate just sets dT, it is the mean radiator flow temp that sets the output.

 

Why would you cycle the system on and off, or do you mean something else?

With the gas it is tempting to do a overnight setback and of course there is turning down to min heat during holidays and this is when the furthest rads take a while to warm up and then warm the room back up.  That being an issue with the gas it is likely to be even more so with a heat pump.  the issue not being total power output of the heat pump but how quickly that energy can be transferred to the the rads (whatever their size) due to the pipework constraint.  Obviously with an HP I wouldn't use such a big set back.

[JohnMo]

45 minutes ago, Michael_S said:

With the gas it is tempting to do a overnight setback and of course there is turning down to min heat during holidays and this is when the furthest rads take a while to warm up and then warm the room back up.  That being an issue with the gas it is likely to be even more so with a heat pump.  the issue not being total power output of the heat pump but how quickly that energy can be transferred to the the rads (whatever their size) due to the pipework constraint.  Obviously with an HP I wouldn't use such a big set back.

Set backs are generally a waste of time, you still need to replenish the heat back in to be building structure you lost during your setback.

 

If running suitably low temperature for weather compensation, the flow temperature should be equal to supplying only the heat lost no more. If you are running the correct temps you never really recover. So instead, you end up running hotter temps than you really need and bouncing of the thermostats.

 

Setbacks are a false energy saving sold by thermostat manufacturers to sell products.

 

There is a handy calculator here

https://protonsforbreakfast.wordpress.com/2022/12/19/setback-should-you-lower-heating-overnight/

 

So you can see for yourself.

[Michael_S]

11 minutes ago, JohnMo said:

Set backs are generally a waste of time, you still need to replenish the heat back in to be building structure you lost during your setback.

 

If running suitably low temperature for weather compensation, the flow temperature should be equal to supplying only the heat lost no more. If you are running the correct temps you never really recover. So instead, you end up running hotter temps than you really need and bouncing of the thermostats.

 

Setbacks are a false energy saving sold by thermostat manufacturers to sell products.

 

There is a handy calculator here

https://protonsforbreakfast.wordpress.com/2022/12/19/setback-should-you-lower-heating-overnight/

 

So you can see for yourself.

Agree with you re efficiency but it is more abut sleep comfort - I guess it doesn't take long to get used to warner nights as we do it every summer.

[JohnMo]

Just tune the heating to be cooler. Or design radiator to match the lower room design temp 

[marshian]

1 hour ago, JohnMo said:

Set backs are generally a waste of time, you still need to replenish the heat back in to be building structure you lost during your setback.

 

If running suitably low temperature for weather compensation, the flow temperature should be equal to supplying only the heat lost no more. If you are running the correct temps you never really recover. So instead, you end up running hotter temps than you really need and bouncing of the thermostats.

 

Setbacks are a false energy saving sold by thermostat manufacturers to sell products.

 

There is a handy calculator here

https://protonsforbreakfast.wordpress.com/2022/12/19/setback-should-you-lower-heating-overnight/

 

So you can see for yourself.

Absolutely what I found - need a higher temp to recover than what is needed to maintain the house at the desired temp

[marshian]

On 20/08/2025 at 20:45, JamesPa said:

To give an example of the calculation:

 

Radiator output is normally (but not universally) quoted for a average deltaT (rad to room) of 50C.

 

Convector rad output is approximately proportional to deltaT^1.3 (manufacturers do quote exponents very slightly different to 1.3, but they are almost always close to 1.3). 

 

So a convector radiator operating at ft 35 and DT (flow to return) of 5, target room temp 20, will have an average deltaT of 35-5/2-20 = 12.5.  The output will therefore be (12.5/50) ^ 1.3 = 0.16 times the output at deltaT 50.  Thus a radiator quoted as 1kW at deltaT 50 will emit only 160W at deltaT 12.5.

 

You can invert this to get the 'oversize factor', in this case 1/0.16 = 6.25, the figure by which you must multiply the calculated loss to get the radiator output (at deltaT50).  For any given type of radiator (unless it is very short or very tall) the oversize factor is a constant so this is a very convenient way to work out which radiator from a range you need

 

Column radiators may have a different exponent than 1.3, or even a different basic formula (although the latter is unlikely); you need to check with the manufacturer.  There is, however, a fair likelihood that the same formula will apply.

 

Incidentally if you could pump a bit harder and reduce the DT to say 3, the oversize factor reduces to 5.25.

 

I keep coming back to this thread to make sense of the calculations in Bold 

 

I can't make 35-5/2-20 = 12.5 It's driving me a bit cranky

 

@JamesPa

 

My issue is my heat loss calcs were done room by room and in some rooms the rads currently fitted shouldn't provide the heat required at -2.5 Deg OAT if I use the std DT10 factor for T50 Rad so I'm trying to work out what they will actually output based on my own flow, return temps

Flow 35 Return 25 Room 20 DT 10 Correction 0.123

Edited September 25 by marshian

[JamesPa]

18 minutes ago, marshian said:

35-5/2-20 = 12.5 

That's the average delta t from rad to room.

 

If 35C is the ft and the water temp drops by 5C as it flows through the rad, then the average rad temp is 32.5 ie 35-5/2.

 

Thus the average deltat rad to room if the room temp is 20 = 32.5-20 = 12.5.

 

Does that  make sense and if not which bit are you struggling with 

 

Edited September 25 by JamesPa

[JamesPa]

32 minutes ago, marshian said:

if I use the std DT10 factor for T50 Rad so I'm trying to work out what they will actually output based on my own flow, return temps

Flow 35 Return 25 Room 20 DT 10 Correction 0.123

Average rad temp =35-10/2=30.  Dt rad to room = 30-20 = 10. Correction factor for rads  = (10/50)^1.3 = 0.123.

 

 

Why are you designing for DT 10?  Better if you can to design for DT 5 or less and use a heftier water pump if necessary.

Edited September 25 by JamesPa

[marshian]

2 minutes ago, JamesPa said:

Average rad temp =35-10/2=30.  Dt rad to room = 30-20 = 10. Correction factor for rads  = (10/50)^1.3 = 0.123.

 

 

Why are you designing for DT 10?  Better if you can to design for DT 5 or less and use a heftier water pump if necessary.

I've not designed for any DT - I believe the expression is I am where I am with this house

 

I've been using DT 10 compensation factor of 0.123 to work out what the rad output is v the heat loss at -2.5 OAT and two of my rads are coming out as undersize (both bathrooms with vertical column rads)

 

This confuses me because the room temps are actually way closer than they should be (no matter what the OAT and room temp is just 1 deg below what I want it to be 21 although on bath nights Mrs Alien would like it a bit warmer.) so before I upsize them to hit the target temp I want to be absolutely sure of the factor to apply so I don't end up going to large and then restricting the flow to trim the temps.

 

Set up if it helps

 

Viesmann 100-W 16kW "Heat only" range rated to 4kWh on CH with external OAT sensor running WC

 

(DHWP takes care of HW)

 

Heating 23.5/7 (HW cycle once a day 30 mins does what we need 6 out of 7 days 2 x HW on her bath night)

 

So at -2.5 my boiler flow temp is 36 Deg C (that worked fine last winter and the WC curve is nicely dialed in for pretty much all circumstances)

 

DT at the boiler is a pretty stable 7 Deg (regardless of flow temp determined by the WC curve) 

 

Two rads concerned have a DT of 6 Deg between flow and return (Pretty much all the rads run similar DT T22 and K33 are a little bigger but other towel rails are a little narrower  - all the rads are set up with the correct flow to meet the room requirements using Drayton EB4 TRV bodies)

 

Now I understand the calculation (and I take the worst case of -2.5 OAT)

 

36 - 6/2 = 33

33 - 21 = DT 12

13/50 = 0.240

 

So Heat loss of Bathroom is 210 W

Current rad is 800W at DT50

This at 0.123 is 98.4 W (Just under 50% of the output I need based on heat loss calcs)

 

800W @ new correction factor of 0.240 is 192 W 

 

That's much closer and kinda fits with the reality I need to upsize it but not by much ( a 1000 W similar rad with a DT 50 rating should give a little more wiggle room for the Bath night ).

 

Thanks for your help

 

[JamesPa]

12 hours ago, marshian said:

I've been using DT 10 compensation factor of 0.123 to work out what the rad output is v the heat loss at -2.5 OAT and two of my rads are coming out as undersize (both bathrooms with vertical column rads)

 

This confuses me because the room temps are actually way closer than they should be (no matter what the OAT and room temp is just 1 deg below what I want it to be 21 although on bath nights Mrs Alien would like it a bit warmer.) so before I upsize them to hit the target temp I want to be absolutely sure of the factor to apply so I don't end up going to large and then restricting the flow to trim the temps.

Are they gaining heat from other rooms?  My bathroom rad is well undersized but the bathroom temperature is fine because thew adjacent hall is heated....

 

12 hours ago, marshian said:

So Heat loss of Bathroom is 210 W

Current rad is 800W at DT50

This at 0.123 is 98.4 W (Just under 50% of the output I need based on heat loss calcs)

 

100W isnt a lot to gain from elsewhere (if surrounding rooms are warmer of course).  50% makes it seem like a lot but 100W isnt, particularly if you run a bath/shower when you get some instant heat from the water!

 

12 hours ago, marshian said:

That's much closer and kinda fits with the reality I need to upsize it but not by much ( a 1000 W similar rad with a DT 50 rating should give a little more wiggle room for the Bath night ).

 

Sounds like you are more or less sorted, good luck!

Edited September 26 by JamesPa

[SteamyTea]

14 hours ago, marshian said:

35-5/2-20 = 12.5

BIDMAS.

It is a mathematics convention (the grammar).

Learn that, and the laws of indices, and you are halfway to a mathematics A Level.

Don't learn it and you are fully qualified in social sciences and politics.

[MrPotts]

Shouldn’t it be written as (35-(5/2))-20?

[SteamyTea]

Just now, MrPotts said:

Shouldn’t it be written as (35-(5/2))-20?

Totally depends on what was intended.

Personally I prefer to use brackets as it saves confusion, most of the time.

[marshian]

56 minutes ago, JamesPa said:

Are they gaining heat from other rooms?  My bathroom rad is well undersized but the bathroom temperature is fine because thew adjacent hall is heated....

 

100W isnt a lot to gain from elsewhere (if surrounding rooms are warmer of course).  50% makes it seem like a lot but 100W isnt, particularly if you run a bath/shower when you get some instant heat from the water!

 

Sounds like you are more or less sorted, good luck!

Unfortunately from a stealing heat perspective both bathrooms

 

1. share a wall with the stairwell (18 deg set point) 

 

2. share a wall with the landing (same 18 deg) 

 

3. One shares a wall with a heated bedroom (20 deg) the other with a box room (18 deg) this box room has 3 external walls and a much higher heat loss hence the lower target temp

 

4. Both have an external wall and small window (plus extractor fan) 

 

5. Both have 300 - 325mm of loft insulation above

This conversation has stopped me making a big error and doubling the size of the rad based on 50% of wattage required - I just need to either buy slightly taller one 1800 v current 1600 or same height doubled but a reduced width.

 

I really want to make sure I don’t end up bouncing along the top temp with TRV intervention - currently I have a DAB pump set to lowest speed on constant pressure 

 

watts 9.5

head 1.8 m

m3/hr 0.5 (so 8.5 litres per min flow rate) 

 

in the shoulder seasons with current set up Boiler fires for ~12 mins in every hour and with a nice open circuit all rooms stay nicely at close to target temps

 

again thank you for the help 

 

[saveasteading]

34 minutes ago, MrPotts said:

(35-(5/2))-20?

= 12.5 

 

half of 5 = 2.5.  35 less 2.5 = 32.5. less 20 = 12.5.

 

14 hours ago, JamesPa said:

35-5/2-20 

 = -35

 

ie 35- 5 =30. Then divide by 2 = 15, then subtract 20.

 

from left to right after first doing the stuff in brackets.

If there is officially another protocol or convention in some fields of design then its important we all know. These results are drastically different. Buildings will collapse.

40 minutes ago, SteamyTea said:

Personally I prefer to use brackets as it saves confusion, most of the time.

It's not personal preference, but crucial.

[JamesPa]

1 hour ago, MrPotts said:

Shouldn’t it be written as (35-(5/2))-20?

It can be but it doesn't have to be.  In arithmetic grammar exponentiation takes precedence over multiplication and division which both take precedence over addition.  Look up BIDMAS.

 

 

 

1 hour ago, saveasteading said:

 

16 hours ago, JamesPa said:

35-5/2-20 

 = -35

 

ie 35- 5 =30. Then divide by 2 = 15, then subtract 20.

 

from left to right after first doing the stuff in brackets.

If there is officially another protocol or convention in some fields of design then its important we all know.

 

Sorry thats just wrong, the sum 35-5/2-20 is correctly parsed as 35-2.5-20 = 12.5.  Its not left to right after first doing the stuff in brackets its BIDMAS.

 

There isnt 'another' protocol, there is the globally adopted protocol which we would all have been taught at GCSE or O Level as applicable (KS3 I believe), but may well have forgotten.  Its not helped by the fact that some early calculators didn't implement BIDMAS (maybe some still don't), but that doesn't change the rule! 

 

Incidentally I just checked my Android calculator and it DOES implement BIDMAS, so if you type my formula in exactly as written and without brackets it correctly gives the answer 12.5.

Edited September 26 by JamesPa

[SteamyTea]

Brackets, Indices, Division, Multiplication, Addition, Subtraction.

 

 

[marshian]

27 minutes ago, SteamyTea said:

Brackets, Indices, Division, Multiplication, Addition, Subtraction.

 

 

You’ve replace google for me

[SteamyTea]

18 minutes ago, marshian said:

You’ve replace google for me

Here is a bit about indices.

 

https://www.sofatutor.co.uk/maths/videos/proofs-of-laws-of-indices

 

Half way to calculus now.

[MrPotts]

6 hours ago, JamesPa said:

It can be but it doesn't have to be.  In arithmetic grammar exponentiation takes precedence over multiplication and division which both take precedence over addition.  Look up BIDMAS.

 

 

 

 

Sorry thats just wrong, the sum 35-5/2-20 is correctly parsed as 35-2.5-20 = 12.5.  Its not left to right after first doing the stuff in brackets its BIDMAS.

 

There isnt 'another' protocol, there is the globally adopted protocol which we would all have been taught at GCSE or O Level as applicable (KS3 I believe), but may well have forgotten.  Its not helped by the fact that some early calculators didn't implement BIDMAS (maybe some still don't), but that doesn't change the rule! 

 

Incidentally I just checked my Android calculator and it DOES implement BIDMAS, so if you type my formula in exactly as written and without brackets it correctly gives the answer 12.5.

Add me to the forgotten list! Age is getting the better of me 😩