System boiler modulation with low loss header LLH

[John.g]

Yes, regarding the UFH heating requirements and boiler output, both will (have to) be the same. With your system if you want to run at a manifold flowtemp of 40C then you would have to reduce the boiler flowtemp to 59C to give the original 27C boiler dT, normally, you would have a separate UFH manifold recirc pump on each UFH system, then the boiler dT can be controlled to whatever you require irrespective of boiler or manifold temperatures but if your system is working OK then why worry as it certainly wont require more energy and is actually more efficient with those very low return temperatures.
See below schematic of a system with a LLH and UFH manifold recirc pump but with a boiler flowtemp of 65C and UFH flowtemp of 40C and (if) a required boiler dT of 20C, the boiler return will now be 45C.

 

[HRP123]

Yes, regarding the UFH heating requirements and boiler output, both will (have to) be the same. With your system if you want to run at a manifold flowtemp of 40C then you would have to reduce the boiler flowtemp to 59C to give the original 27C boiler dT, normally, you would have a separate UFH manifold recirc pump on each UFH system, then the boiler dT can be controlled to whatever you require irrespective of boiler or manifold temperatures but if your system is working OK then why worry as it certainly wont require more energy and is actually more efficient with those very low return temperatures.
See below schematic of a system with a LLH and UFH manifold recirc pump but with a boiler flowtemp of 65C and UFH flowtemp of 40C and (if) a required boiler dT of 20C, the boiler return will now be 45C.

View attachment 82194

Thanks for this calculation, appreciate it. Sorry for these potentially naive questions, but how do you come up with the LPM rate of 6.45 going into the header and 3.91 going out of the header into the UF circuit? And then the pump LPM rate of 16.13?

 

[John.g]

Start at the UFH requirement of 9.0kw, UFH loops might typically have a dT of 6C to 10C so a dT of 8C might be a reasonable average, now, LPMX60XdT/860 = kw, LPM =9X860/60/8, 16.13LPM. You require a manifold flowtemperature of 40C so the return temperature is, 40-8, 32C. You must mix x LPM at 65C with y LPM at 32C. let LPM32 = flowrate required at 32C, and LPM65 = flowrate required at 65C, LPM65 = 16.13-LPM32, so (16.13x40)=(LPM32X32)+(16.13-LPM32X)X65, solve the equation for LPM32 and you get LPm32=12.22LPM so LPM65=16.13-12.22, 3.91. So you require 3.91LPM at 65C mixed with 12.22LPM at 32C to give you 16.13LPM at 40C and 9kw. You now know that the boiler must suppy that 3.91LPM at 65C and that 3.91LPM at 32C returns to the boiler to give a dT of 65-32, 33C but you might require a dT of 20C so again you use the same method to calculate that you require 2.54LPM at 65C mixed with 3.91LPM at 32C to give you 6.45LPM at 45C to give a dT of 20C and 9kw boiler output.
I do these numbers in a spreadsheet so takes the donkey work out of it, You can see below the LLH part of the first schematic. In the LLH 2.54LPM at 65C is short circuiting in the primary and mixing with 3.91LPM at 32C coming from the secondary to give 6.45LPM at 45C.

 

[HRP123]

Start at the UFH requirement of 9.0kw, UFH loops might typically have a dT of 6C to 10C so a dT of 8C might be a reasonable average, now, LPMX60XdT/860 = kw, LPM =9X860/60/8, 16.13LPM. You require a manifold flowtemperature of 40C so the return temperature is, 40-8, 32C. You must mix x LPM at 65C with y LPM at 32C. let LPM32 = flowrate required at 32C, and LPM65 = flowrate required at 65C, LPM65 = 16.13-LPM32, so (16.13x40)=(LPM32X32)+(16.13-LPM32X)X65, solve the equation for LPM32 and you get LPm32=12.22LPM so LPM65=16.13-12.22, 3.91. So you require 3.91LPM at 65C mixed with 12.22LPM at 32C to give you 16.13LPM at 40C and 9kw. You now know that the boiler must suppy that 3.91LPM at 65C and that 3.91LPM at 32C returns to the boiler to give a dT of 65-32, 33C but you might require a dT of 20C so again you use the same method to calculate that you require 2.54LPM at 65C mixed with 3.91LPM at 32C to give you 6.45LPM at 45C to give a dT of 20C and 9kw boiler output.
I do these numbers in a spreadsheet so takes the donkey work out of it, You can see below the LLH part of the first schematic. In the LLH 2.54LPM at 65C is short circuiting in the primary and mixing with 3.91LPM at 32C coming from the secondary to give 6.45LPM at 45C.

View attachment 82196

Thanks for taking the time to explain all this and share the spreadsheet. Interesting to see how you work out the math above. So how does one work out what the boiler is actually capable of producing in LPM at 65 degrees?

I had a look at the boiler specs but this does not tell me. I'm just trying to work out with lets say multiple circuits on e.g. underfloor and radiators or towel rails or hot water - is the boiler actually capable of producing the total LPM flow from the LLH to these circuits?

Are you a heating engineer yourself? Cheers!

 

[John.g]

If you assume a 30kw boiler and assuming flow/return temps of 65C/45C, dT20C, then this will give it at a flowrate of 21.5LPM (1290LPH) which would be the comfortable flowrate for this output, otherwise the pressure drop across the boiler heat exchanger can get excessive requiring a very high head pump, if your rad demand is say 30kw with flow/return temps of 55C/45C , dT10C, then a flowrate of 43LPM (double the boiler flowrate) is required, the secondary side of a a LLH will achieve this.

No, not a heating engineer. Ex Marine (15 years) and utilities engineer (31 years).

 

[HRP123]

If you assume a 30kw boiler and assuming flow/return temps of 65C/45C, dT20C, then this will give it at a flowrate of 21.5LPM (1290LPH) which would be the comfortable flowrate for this output, otherwise the pressure drop across the boiler heat exchanger can get excessive requiring a very high head pump, if your rad demand is say 30kw with flow/return temps of 55C/45C , dT10C, then a flowrate of 43LPM (double the boiler flowrate) is required, the secondary side of a a LLH will achieve this.

No, not a heating engineer. Ex Marine (15 years) and utilities engineer (31 years).

View attachment 82207

I assume your engineer training has helped with this approach to calculating flow rates etc. Thanks for taking all the time to explain this!

In my case there is an additional pump going from the boiler into the LLH and then from the LLH four heating circuits plumbed off three flow outputs (UF, towel rails, and the HW and CH circuits share one output from the LLH). I don't think the typical radiator demand is anywhere close to 30kw, probably more like 8-9kw typically. HW should be around 8-9kw, UF around 5-10kw (zoned), and TR 6-7kw.

What confuses me though is that in theory the UF demand should be less than 10kw but it typically pulls almost the maximum capacity of the boiler when open (which I have ranged rated down to around 24kw). The heat loss calculations don't suggest this should be the case as the floors and walls are insulated. This is why I wonder if somehow the flow and return (mixed) are distorting something in the LLH ... my very basic noivce understanding is that the UF is pumping out a lot of water from the header (even though only set on pump speed 1) - and/or returning a large flow of cold water - which is reducing the return temp to the boiler for a long period of time, making it think it needs to heat/pump at full capacity (when in reality it could work slower).

 

[John.g]

Most UFH Loops have flow meters thar are adjustable and you can read off the flowrate in LPM, a typical flowrate per loop might be 2LPM with a typical dT of 8C so each loop might emit 2x60x8/860, 1.116kw, say 1.0kw per loop or require a boiler output of 10kw for 10 loops.
Do you have flow readings from yours??, if so and you have flow and return manifold temperature readings then very easy to calculate the total UFH heat output required.

 

[ShaunCorbs]

The ufh will pump out 3 times more flow than the boiler as it’s dt 7 ish

 

[John.g]

What confuses me though is that in theory the UF demand should be less than 10kw but it typically pulls almost the maximum capacity of the boiler when open (which I have ranged rated down to around 24kw). The heat loss calculations don't suggest this should be the case as the floors and walls are insulated. This is why I wonder if somehow the flow and return (mixed) are distorting something in the LLH ... my very basic noivce understanding is that the UF is pumping out a lot of water from the header (even though only set on pump speed 1) - and/or returning a large flow of cold water - which is reducing the return temp to the boiler for a long period of time, making it think it needs to heat/pump at full capacity (when in reality it could work slower).

The LLH only distribute the water in the primary and secondary, they do not lose or gain heat, if the flowrate is higher in one than the other then the dTs will be different to give exactly the same enmergy content.
In the schematic there are three different flowrates and dTs but all have the same energy content of 9.0kw.

The boiler connected with the primary flows 6.45LPm at a dT of 20C, energy content = 6.45x60x20/860 = 9.0kw
The Secondary flows 3.91LPm at a dT of 33C, energy content = 3.91x60x33/860 = 9.0kw
The UFH flows 16.13LPm at a dT of 8C, energy content = 16.13x60x8/860 = 9.0kw

 

[HRP123]

Most UFH Loops have flow meters thar are adjustable and you can read off the flowrate in LPM, a typical flowrate per loop might be 2LPM with a typical dT of 8C so each loop might emit 2x60x8/860, 1.116kw, say 1.0kw per loop or require a boiler output of 10kw for 10 loops.
Do you have flow readings from yours??, if so and you have flow and return manifold temperature readings then very easy to calculate the total UFH heat output required.

I will try and get the flow return temps to the loops as well.