DT is a phrase used too much in HP discussions, one needs to be clear about between what points the difference is being measured. Its normal that DT is used to refer to the difference between flowT and returnT. In your case, DT on the left side is 5C.   (39-34).  DT on the right side is also 5C (40-35).  DT times flow rate gives you the heat energy transfer rate, which has to be the same each side - there is no input or output of energy in the LLH. If the flow rates are different on each side then DT would be different each side, but the energy transfer would still be (MUST BE) the same.  The LLH isn't "losing" energy (it can't, its not an emitter - assuming its well insulated). 

But what the LLH does, is force mixing to happen, which MUST reduce the flowT on the emitter side.  this reduces the efficiency of the system as a whole, once you consider how the energy input device (the HP) works. To get an emitter temp of 39C, you have to run your HP at 40C. This costs you x% in HP efficiency - its guaranteed, for any HP, the higher the HP flowT, the less efficient it is. (as per jamespa, probably around 3% for 1C). If your drop in FlowT is worse, this % gets higher (= worse).

If you have a poorly balanced LLH  - widely differing flowT's - then the drop in flowT would be worse, thus increasing the need to run the HP at an even higher temperature OR leading to too-cold emitters that don't warm the house - thus giving an even less efficient system with an even worse COP.

If you've already got it down to only a 1 degree drop in flowT across the LLH, by tweaking your secondary pump, thats pretty good - implies you've got it down the least amount of mixing that you can (whilst still keeping the LLH). the only way to get better will be to repipe without it. additional benefit of that is you only need 1 pump, and you can also potentially use a variable circulation speed (either a CP or PP pump , or a PWM pump controlled by the HP, or a DT-targeted pump), starting point for that discussion should be what is the pump within your ASHP capable of.

thanks James

Added

(We tried a number of secondary pump settings until we got the lowest change in DT but we couldn’t do much better than 1degC. At least we are dealing with constant flow rates…. )

Ive already built my secondary pump bypass….

yes there is a drop in flow temp across the LLH but there is a more or less equal rise in return temp across the LLH. These were tested with a separate thermistor…

I don’t know if there is a way of working out a quantifiable energy cost for the increased generated DT to create the desired DT at the emitters but perhaps it isn’t as big as I’m thinking it might be?

thanks for your insights on this 

Assuming I understand the question correctly (which I may not) then yes there is.  Plot COP vs flow temperature, then work out the COP (at some representative OAT eg 7C) for the two 'alternative' flow temperatures  required to achieve the same FT at the emitters (ie with and without the LLH, or more simpl the original design FT and that plus the reduction in FT across the LLH).  Then use that to work out the change in energy consumption due to requiring a higher FT from the ASHP.

So for example with a Mitsubishi R32 11.2kW the COP at 7/40 is 4.55 and its 5.11 at 35/40.  That translates to a 12% increase in consumption for a 5C rise in Flow temp, so about 4% for a 1.5C rise.

Here is a quick table for the 11.2kW Mitsi

This accounts only for the reduction in thermodynamic efficiency of the heat pump, not for any energy losses in the LLH (which, assuming its well insulated, should be negligible).  It also assumes that the pump speeds (l/s) both sides of the LLH are approximately equal (if they aren't then youare forcing mixing which is a bad idea)

DT is the same both sides,

hi Ian,

here is my sketch again what do you mean that the DT Is the same both sides? The sketch shows a 1degC drop in one direction and a 1degC rise in the other direction…. Is that what you mean?

the Heat Geek info looks interesting

Will look in due course… thanks

yes there is a drop in flow temp across the LLH but there is a more or less equal rise in return temp across the LLH. 

doesn't matter. check the datasheet 2 posts up. the higher the flowT, the worse the COP. same for any heat pump.

also you are talking about loss of DT, but in your sketch, DT is the same both sides, which it would be if flow rate is the same. heatgeeks explanation is pretty good as they usually are https://www.heatgeek.com/low-loss-headers/

yes there is a drop in flow temp across the LLH but there is a more or less equal rise in return temp across the LLH. These were tested with a separate thermistor…

I don’t know if there is a way of working out a quantifiable energy cost for the increased generated DT to create the desired DT at the emitters but perhaps it isn’t as big as I’m thinking it might be?

thanks for your insights on this 

however, don't forget that one of the major causes of energy loss with an LLH, is the fact that the flowT must drop across it. That means that to achieve a given flowT on the radiator side (which you need to get the emitters providing the required heat output to your house), you need a higher flowT on the source side. Then where the source is a HP, that higher flowT on the source side means the HP is running less efficiently.

to put it another way, for the same DT and flow rate (= heat energy output) at the HP, it costs more in input (electrical) energy , the higher the absolute values of the flowT (and returnT) . 

If you have no LLH, the flowT of the HP and the flowT of the radiators are by definition the same, thus your HP is as efficient as it can be given whatever other conditions are in play.

Just to recap and summarise on earlier posts:

We had seen a small loss of temperature across the Flow Pipes as the water passed through the Low Loss Header tank, into the central heating circuit. This loss was approx 1 to1.5c. This was thought to be a small amount of mixing in the Low loss header tank as the water passed through. 

So to check this we tested the return temperatures and confirmed that the Primary Return water was exiting the Low Loss Header Tank slightly hotter than the incoming Central Heating Return water. This increase in Primary Return temperature was also about 1c. This more or less matched the Flow pipe temp. loss, suggesting that a small amount of mixing as might be expected. This mixing could be considerably more if flow rates created by each pump were different. 

The test limitation was that the thermistor readings only measure in 1c increments which give a wide ish margin of accuracy.

Can anyone help with some maths on estimating percentage of energy loss of DT….?

After a summer break we are now instigating some changes to our HP by making the 2 circuit (DHW and secondary heating circuits) layout into a single circuit, driven off a single pump. 

But before I discus this with our installer it would help to clarify in my own mind what potential losses are occurring through the LLH.

The above extract describes a  1c to 1.5c loss of Flow temperature as the LWT crosses the LLH and exits into the secondary circuit becoming the Secondary Flow.

Previously I didn’t think this was a major loss. Since the flow temperature was 37 or 40c. However I’m doubting that thought now:-

My thought now is that the loss is a continuous 11/2 degrees loss of the DT5c.

If this were a true loss this equates to a potential 33% reduction of the DT due to water mixing within the LLH.

HOWEVER I realise the energy isn’t lost, it is recycled by passing back into the Primary Return pipe and so the return temperature is raised by an equivalent 11/2c. 
Also it would appear that the HP is having to raise the DT by 6 or 7 degrees in order to provide only a 5C DT at the emitters…. so there must be some dynamic losses in this situation which I can’t quantify?

So my question is how do I calculate the actual energy inefficiency as a result of mixing through the LLH. 

I’ve attached a sketch of the pipes through the LLH with a simplified 1c loss of DT

Thanks in advance…

Can anyone explain.

This might be a daft question but I’m curious about how the different outputs are selected? Between Max, Nominal, Mid and Min. Are these four output groups selected automatically by the FTC or are they somehow derived by any of the settings?

It is not possible to select a particular output, they are included within the test data to indicate the likely power consumption and efficiency when operating at different levels of loading. The notes below indicate how they should be interpreted. When operating the heat pump will vary the compressor speed (frequency) to achieve the required output energy.

Definition of terms
Max :Performance at Maximum compressor frequency
Nominal :Performance at Nominal compressor frequency
Mid :Performance at Medium compressor frequency (80% of Nominal)
Min :Performance at Minimum compressor frequency
:This icon means injection circuit is active.
NOTES:
• The reference data at water outlet temperatures of 35°C,40°C,45°C,50°C,55°C and 60°C are shown.
• Gray highlighted data means integrated data including defrost operation.
• Actual performance may vary depending on operating conditions.
• These data are measured based on EN14511-2013.
Nominal heating capacity
A7W35: Heating outside air DB 7°C/WB 6°C, Water outlet temperature 35°C (ΔT=5°C)
A7W45: Heating outside air DB 7°C/WB 6°C, Water outlet temperature 45°C (ΔT=5°C)
A2W35: Heating outside air DB 2°C/WB 1°C, Water outlet temperature 35°C ( ΔT=5°C)

Can anyone explain.

This might be a daft question but I’m curious about how the different outputs are selected? Between Max, Nominal, Mid and Min. Are these four output groups selected automatically by the FTC or are they somehow derived by any of the settings?