All of which is very fair.  Im not trying to persuade anyone on this particular matter, just identifying (as much for me as for anyone else) the trade offs.

I cant remember the last time we had a power outage lasting more than and hour, and that was because the DNO was changing our main fuse in prep for our HP.  Mostly they last a few minutes.  Having said that I would not be happy without a backup heating solution which, for us, is a log burner.  The most worrying scenario for me is going away during a cold snap and returning to a failed pump but we have cats which means that, even when we are away, someone generally visits the house each day.  Thus we would know about it in time to call a plumber and ask them to drain the system - which triggers the thought that the drain cock needs to be well labelled and easily accessible.  Perhaps it should even be a drain tap that, in extremis, even a cat sitter could operate.  Hmm...

Until we have more hard evidence I would say that its definitely the case that glycol vs antifreeze valves is horses for courses.  Its a pity we dont have info from Scandanavia or somewhere similar that must have the data!

@jamespa

I think that I remember reading that in much colder climes the preference would be for split systems rather than mono, or possibly GSHP's.

@cathoderay 

On  December 15th , 2022 my neighbour chapped my door asking "are you all right?".

He pointed out that my rear garden was covered by a large cloud , some 2 metres deep , covering an area of 64 square metres.

My Heat Pump was indeed  pouring out a metre diameter "tube" of Water vapour.

Freezing Heat Pumps are embarrassing, and consume a lot of energy. !

Glycol OR Anti-Freeze Valves?

Glycol Pros

The addition of Glycol to the Water May NOT reduce the Efficiency but will reduce the output power ?

The use of a Heat exchanger would be justified in protecting my house from toxic Glycol?

Glycol Cons

Glycol is Toxic to both humans and animals.

Glycol is an additional annual expense

Anti-Freeze Valve Pros

A one OFF payment of circa £220.

Anti-freeze Valve Cons

What happens after the anti-freeze valve fires?

What protects the Heat Pump when the Anti-freeze valve fires?

Modelling?

Given that the addition of Glycol to the Pipe Water May NOT reduce the Efficiency but will reduce the output power,

do we need a system model , calculating the complete system performance?

With a model , as one parameter is improved , the effect upon other parameters could be assessed.

ian

@jamespa 

Samsung recommendation

Given that live in Scotland , Samsung recommend that my Pipe Water Temperature, set with the Samsung Water Law , be set to +35 C when the outside Temperature is  -5 C.

 On several December nights  in 2022 my outdoor Temperature fell to -5 C / -6 C at 3/4 AM with the Heat Pump consuming up to  3.5 kWh of Energy.

MCS Buffer Warning

The MCS issued a warning that all Buffers consume Energy , and, consequently lower the Heat Pump  efficiency.

The Energy stored, and dumped into the Heat Pump Fins during Freezing Temperatures should be checked.

All buffers, and all Freezing consumes Energy, and lowers the Heat Pump Efficiency.

ian

@iantelescope

When trying to ascertain what is actually happening within a system, I often find it is useful to put values into the equations to see how the variables change under varying conditions.

Lets assume that a hypothetical house has a calculated heat loss of 12 kW, with an indoor temperature of 21C at an outside temperature of -3C.

If the outside temperature is a constant 9C, with no further effects due to wind, rain or Sun, then the heat loss would be approximately 6 kW. The heat emitters would therefore need to dissipate 6 kW of thermal energy to maintain the indoor temperature at 21C.

Assuming that the system is filled with water, and that the Delta T across the heat emitters is 5C, then the required flow rate would be given by:-

6 kWh / 60 / 5 * 3600000 / 4186 = 17.2 lpm.

If anti-freeze at 10% by volume is now added to the system, with the flow rate and Delta T remaining the same, then the thermal energy transferred would now be given by:-

4036 / 3600000 * 5 * 60 * 17.2 = 5.785 kWh.

If no further changes were made, then it is likely that the indoor temperature would fall from 21C to 20.57C.

If anti-freeze at 20% by volume is now added to the system, with the flow rate and Delta T remaining the same, then the thermal energy transferred would now be given by:-

3886 / 3600000 * 5 * 60 * 17.2 = 5.57 kWh.

If no further changes were made, then it is likely that the indoor temperature would fall from 21C to 20.14C.

So to maintain the indoor temperature at 21C, it would be necessary to increase the flow rate or increase the Delta T, but if the water pump is a fixed speed then the only alternative would be to increase the Delta T.

Increasing the Delta T by lowering the RWT is not really an option, since this would lower the thermal energy transfer from the heat emitters and hence would not increase the indoor temperature. The only alternative is therefore to increase the LWT and thereby increase the thermal energy transfer from the heat emitters.

Now considering the possible effect at the heat pump end. It is generally accepted that increasing the LWT reduces the heat pump efficiency, though the degree by which it is reduced will very much depend upon the operating conditions. Because there is a PHEX within the heat pump system, then having anti-freeze within the water will probably also reduce the thermal energy transfer, and require a greater flow rate and/or a higher Delta T to achieve the required thermal energy output.

I therefore suspect that adding anti-freeze into the system will have a detrimental effect on overall efficiency, which will no doubt vary with varying operating conditions.

It would appear that manufacturers control the LWT by varying the speed of the compressor, and some also control the Delta T by varying the speed of the water pump. Both of which can help keep the heat pump operating in a more consistent manner.

Posted by: @derek-m

↑

@iantelescope

When trying to ascertain what is actually happening within a system, I often find it is useful to put values into the equations to see how the variables change under varying conditions.

Lets assume that a hypothetical house has a calculated heat loss of 12 kW, with an indoor temperature of 21C at an outside temperature of -3C.

If the outside temperature is a constant 9C, with no further effects due to wind, rain or Sun, then the heat loss would be approximately 6 kW. The heat emitters would therefore need to dissipate 6 kW of thermal energy to maintain the indoor temperature at 21C.

Assuming that the system is filled with water, and that the Delta T across the heat emitters is 5C, then the required flow rate would be given by:-

6 kWh / 60 / 5 * 3600000 / 4186 = 17.2 lpm.

If anti-freeze at 10% by volume is now added to the system, with the flow rate and Delta T remaining the same, then the thermal energy transferred would now be given by:-

4036 / 3600000 * 5 * 60 * 17.2 = 5.785 kWh.

If no further changes were made, then it is likely that the indoor temperature would fall from 21C to 20.57C.

If anti-freeze at 20% by volume is now added to the system, with the flow rate and Delta T remaining the same, then the thermal energy transferred would now be given by:-

3886 / 3600000 * 5 * 60 * 17.2 = 5.57 kWh.

If no further changes were made, then it is likely that the indoor temperature would fall from 21C to 20.14C.

So to maintain the indoor temperature at 21C, it would be necessary to increase the flow rate or increase the Delta T, but if the water pump is a fixed speed then the only alternative would be to increase the Delta T.

Increasing the Delta T by lowering the RWT is not really an option, since this would lower the thermal energy transfer from the heat emitters and hence would not increase the indoor temperature. The only alternative is therefore to increase the LWT and thereby increase the thermal energy transfer from the heat emitters.

Now considering the possible effect at the heat pump end. It is generally accepted that increasing the LWT reduces the heat pump efficiency, though the degree by which it is reduced will very much depend upon the operating conditions. Because there is a PHEX within the heat pump system, then having anti-freeze within the water will probably also reduce the thermal energy transfer, and require a greater flow rate and/or a higher Delta T to achieve the required thermal energy output.

I therefore suspect that adding anti-freeze into the system will have a detrimental effect on overall efficiency, which will no doubt vary with varying operating conditions.

It would appear that manufacturers control the LWT by varying the speed of the compressor, and some also control the Delta T by varying the speed of the water pump. Both of which can help keep the heat pump operating in a more consistent manner.

 

I dont think this is quite right.  It is definitely useful to put numbers in, but you have also think about what drives what.  In heating, the driving forces are temperature differences, starting with the house to outside.  the Heat pump resp0onds to the effects of those temperature differences.  So...

Posted by: @derek-m

↑

So to maintain the indoor temperature at 21C, it would be necessary to increase the flow rate or increase the Delta T, but if the water pump is a fixed speed then the only alternative would be to increase the Delta T.

If the flow temp is kept constant then the heat lost from the radiators will still be 6kW because the radiators don't know about the fluid inside them, only its temperature.  Thus, unless the flow rate is increased by the heat pump, the delta T (flow-return) will in fact increase, because to lose 6kW from the glycol mix requires a greater temperature drop than to lose 6kW from pure water.  There will be a second order effect that, because the deltaT flow-return has increased and thus the average radiator temperature decreased, the emitter loss will be very marginally less than 6kW.  Ignoring for now the second order effect, the heat pump will (or at least) should simply respond by dumping 6kW into the glycol water mix to bring it back up to the original design flow temperature.  Thus the effect overall (ignoring second order effects) will be an increase in deltaT (flow-return) unless the heat pump changes the flow rate to maintain both deltaT and flow temp (which it might).

The second order effect is pretty small.  If 6kW is dumped at a deltaT of 5 with water, then the temperature drop with glycol will be 5*4200/3886 = 5.4C.  Thus the average emitter temp will drop by 0.2C.  If the emitter was originally at an average temp of 42.5 its new average temp will be 42.3.  So the second order effect will require the flow temp/wc curve to be nudged up by this amount to maintain the same house temp, causing an efficiency penalty of about 0.4% (using the rule of thumb that 1C rise in flow temp reduces efficiency by 2%)

Almost certainly the more significant effect is the HEX/PHE (choose your acronym).  

Posted by: @derek-m

↑

Now considering the possible effect at the heat pump end. It is generally accepted that increasing the LWT reduces the heat pump efficiency, though the degree by which it is reduced will very much depend upon the operating conditions. Because there is a PHEX within the heat pump system, then having anti-freeze within the water will probably also reduce the thermal energy transfer, and require a greater flow rate and/or a higher Delta T to achieve the required thermal energy output.

If the PHEX is lossless, or the losses are within the thermal envelope, then the HP will adjust to maintain the heat transfer without needing to deliver more heat to the house.  The problem is the temperature drop across the PHEX from heat pump side to CH side.  For each C drop you lose 2% efficiency, and the temp drop may be as much as 5C (I believe a well designed and correctly plumbed HEC can achieve as low as 1C).  So thats 2-10% efficiency loss.   

Hence my summary upthread that the main efficiency loss is from the PHEX

@jamespa 

NIC Visit

The NIC will visit to asses my Heat Pump Installation, this some 18 months since the Heat Pump was "installed".

I am preparing a folder  showing Dailly plots of the

1) Actual Measured water power as  Measured by my "Sharky" water powered meter at the beginning of the Radiator circuit.

2)  Samsung "used " and "Generated" Power  at the Water output of the Heat Pump.

3)  Pipe Water Temperature at the motor input using external K type Temperature Sensors.  

4)  Pipe Temperature at the Beginning of the Radiator circuit using external K Type  Temperature Sensors.

5)  Samsung COP and the Actual measured COP.

6) Cycle times with , and without the 50 L buffer.

Is this , enough , or too much for the NIC Engineer?

I could easily intimidate  the NIC engineer with gigabytes of data !

This time, my "installer " cannot " lose"  the documentation , or fire their engineer !!

Measuring House Power requirements

I have previously Measured the Actual Power Required to Heat my House using the following procedure.

1) I put a 900 Watt oil filled domestic heater in the room to be tested.

2) I attached a commercial power meter to the Heater.

3) I measured both the room Temperature, T_Room  AND the outdoor Temperature, T_Outside.

I then recorded the Energy and Power consumption against the Temperature Difference , T_Outside -T_Room.

On Dividing the Power consumed by the Temperature Difference I defined a " Power per Degree Parameter" kW/C.

My Maximum power consumption occurs when the outside Temperature is -5 and the required room Temperature is 21 C or 26 C.

Therefore , my Living Room giving 78 Watts per degree C when the Temperature difference is 26 is 2.1 kW.

Measure the room conductance ....do not guess it.

I , like Derek , and yourself , James, am a firm fan of facts from figures.

However, I still think that an overall system Model is required  !

@derek-m 

Hi Derek,

From the Water Power Equation:

Water_Power ( kW)  =  Specific_Heat_of Water ( kJ/(litres x T)  x  Delta_T  x Flow_Rate (litres/sec)

it follows that for a given Power the multiple of Delta_T and Flow_rate is a constant.

On my system , I can set the Flow rate using the Pulse-Width_Modulated input on the Radiator Grundfos Motor.

With Delta_T X Flow_Rate a constant , I can set the Flow_Rate, thus control Delta_T.

Is this a fair summary of your argument ?

Applying your argument to my current system  I have a Delta_T of 3 C , a flow rate or 10.2 lpm at a load of 2.2 kW.

Posted by: @iantelescope

↑

Is this , enough , or too much for the NIC Engineer?

 

I could easily intimidate  the NIC engineer with gigabytes of data !

I suspect it depends on the engineer.  I also suspect that they will be predisposed to disbelieve anything that the customer tell them unless they can verify it themselves, and that they might have an aversion to customers that they perceive think themselves smart (apologies for my cynicism here!).  

It might be a good idea to feed them with data slowly, starting with pointing out the mis plumbing of the volumiser which is easily verifiable and manifestly bad.  You could play dumb and point out that it seems to be plumbed differently that what Samsung recommended.  So far as I can recall from the thread this is the only thing we have identified which is definitely wrong.  Other things may be sub optimally adjusted, but this is just wrong.  

The failed pwm control/absence of pump control is also worth highlighting, a definite decision is needed on how this should be fixed/adjusted and then appropriate action taken.

You could then, depending on how he reacts, start feeding him more data/info about the install.

Hopefully he will quickly recommend that the manifestly wrong things are fixed and then the system completely recommissioned.

Personally id play this one a little 'by ear', its the outcome that matters not whether the NIC guy gets to see all the evidence to get to the outcome.

Obviously if it looks like its heading towards the wrong outcome, you can use the data to persuade him otherwise.

In this post the masculine includes the feminine and the singular includes the plural!

Posted by: @iantelescope

↑

@jamespa

 

Measuring House Power requirements

 

I have previously Measured the Actual Power Required to Heat my House using the following procedure.

 

1) I put a 900 Watt oil filled domestic heater in the room to be tested.

2) I attached a commercial power meter to the Heater.

3) I measured both the room Temperature, T_Room  AND the outdoor Temperature, T_Outside.

 

I then recorded the Energy and Power consumption against the Temperature Difference , T_Outside -T_Room.

On Dividing the Power consumed by the Temperature Difference I defined a " Power per Degree Parameter" kW/C.

 

My Maximum power consumption occurs when the outside Temperature is -5 and the required room Temperature is 21 C or 26 C.

Therefore , my Living Room giving 78 Watts per degree C when the Temperature difference is 26 is 2.1 kW.

 

Measure the room conductance ....do not guess it.

 

I , like Derek , and yourself , James, am a firm fan of facts from figures.

However, I still think that an overall system Model is required  !

 

ian

 

 

 

 

 

 

 

 

It's a good experiment but how did you isolate one room from the rest of the house?

Posted by: @iantelescope

↑

@derek-m 

Hi Derek,

From the Water Power Equation:

 

Water_Power ( kW)  =  Specific_Heat_of Water ( kJ/(litres x T)  x  Delta_T  x Flow_Rate (litres/sec)

it follows that for a given Power the multiple of Delta_T and Flow_rate is a constant.

 

On my system , I can set the Flow rate using the Pulse-Width_Modulated input on the Radiator Grundfos Motor.

With Delta_T X Flow_Rate a constant , I can set the Flow_Rate, thus control Delta_T.

 

Is this a fair summary of your argument ?

 

Applying your argument to my current system  I have a Delta_T of 3 C , a flow rate or 10.2 lpm at a load of 2.2 kW.

 

 

 

ian

It actually becomes quite complex, even without taking solar gain, wind chill and rain effect into consideration.

Taking the hypothetical example that I posted earlier, with a calculated heat loss of 12 kW per hour, at -3C outside and 21C inside.

With the outside temperature at a constant 9C, the heat loss would be approximately 6 kW, so to maintain the indoor temperature at 21C would require a supply of 6 kW of thermal energy from the heat emitters. If the supply of thermal energy were increased to 6.5 kW, then the indoor temperature would increase to approximately 22C, and if the supply of energy were reduced to 5.5 kW then the indoor temperature would reduce to approximately 20C. So to keep the indoor temperature at a desired level it is a matter of balancing supply to demand.

A degree of complexity starts to appear when considering the heat emitters, since the quantity of thermal energy that they can provide is not only determined by the average water temperature, but also the actual indoor temperature, the total output capacity (normally specified at a Delta T (DT) of 50C) and also the Specific Heating Capacity (SPH) of the liquid within the system.

In the hypothetical example, to supply 12 kW of thermal energy at an average water temperature of 50C, would require heat emitters with a specified capacity of approximately 24.4 kW. For these heat emitters to provide 6 kW of thermal energy would require an average water temperature of 38C. If the DT across the heat emitters is 5C, then the LWT would need to be 40.5C, with a RWT of 35.5C.

If it were possible to introduce a 20% anti-freeze mixture, then the quantity of thermal energy being supplied to the heat emitters would be reduced from 6 kW to 5.57 kW.

The indoor temperature would still be at 21C, so the building heat loss would still be 6 kW, but the quantity of thermal energy being supplied to the heat emitters would now be 5.57 kW.

The heat emitters still require 6 kW of thermal energy to maintain the indoor temperature at 21C, but are now being supplied with only 5.57 kW, with the flow rate being kept constant. The LWT is being kept constant at 40.5C, so the net effect must be for the RWT to be reduced. This in turn will cause the average water temperature to be reduced, which will therefore reduce the amount of thermal energy being supplied by the heat emitters, with the subsequent effect that the indoor temperature will start to reduce. This process will continue until the thermal energy supplied by the heat emitters balances the heat loss from the property, at a value calculated to be 20.14C.

I have attached an Excel spreadsheet with a table showing the probable effect.

If the flow rate remains the same, then the only way to bring the indoor temperature back to the desired 21C is to increase the average water temperature. This can only be achieved by increasing the LWT.

As you have probably gathered by now the process is not exactly straightforward, since increasing the LWT will probably cause an increase in the RWT, with the pair of them causing the average water temperature to increase. This has the effect of transferring more energy into the property and hence increasing the indoor temperature, which in turn will increase the heat loss from the property. Balance should be achieved when the thermal energy supply and demand are approximately equal.

I have not yet developed formula's to calculate the change in RWT when the indoor temperature is brought back to the desired 21C, which will therefore affect the required LWT.

The effect at the heat pump will be a slight reduction in efficiency due to the higher LWT.

@iantelescope

With regard to changing the flow rate by varying the pump speeds, I do believe that some manufactures use this method to keep the Delta T (DT) across the heat pump fairly constant.

In your particular system you could try setting the secondary pump speed to provide a DT of 5C across your heat emitters and then adjust the primary pump speed to give a DT of 5C between LWT and RWT. You may have to re-adjust several times since one may affect the other.

Then monitor the effect at your PHE.