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Heat pumps and on/off working - starting to quantify it

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(@jamespa)
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There has been much discussion on on/off working/nighttime setback in relation to heat pumps, but not much that is quantitative.  @fazel and @derek-m have made some helpful comments in another thread and I have now started to think about it more quantitatively.  I present an example of how one might assess the effect, it needs development and discussion but is a 'starter for 10'

Worked example:

Lets consider a 10kW heat pump fitted to a house which consumes 8kW when the OAT is 0C and the target IDT is 20C.  Lets also assume that the emitters are radiators with a FT of 45C at 0C, and furthermore that we wish save energy by setting back/switching off for a maximum period of 9 hours (overnight) - eg from 10pm-7am.  Lets consider what happens when the OAT is 10C.  Lets further assume that the temperature drops by 5C during an 'off' period of 4.5 hours, when we switch it back on to get the house warmed up in time for 7am.  Once we have worked through this example it will be relatively easy to change the assumptions.

At OAT of 10C and IDT of 20C the load will be 4kW and the flow temperature required to deliver this will be 34.7C (applying the rule that emission from radiators is proportional to (deltaT rad-room)^1.3.  At this FT and OAT we can typically expect to achieve a COP of 4 (as will become evident later it doesn't matter what the exact COP is for the calculation, but it does matter roughly what it is).  

During the on/off cycle the average IDT will be approximately 17.5C, so instead of losing 4kW the house will lose (on average) 3kW.  Thus over the 9 hour cycle the house will lose 9x3=27kWh instead of 9*4=36kWh had we left the heat pump on.  More importantly during the 4.5hour cool down period we will lose approximately half of this amount, 13.5kWh , and this is lost from the fabric so will have to be replaced to return to the state we started with.  We will return to this shortly.

However we don't pay for energy lost from the house, we pay for energy consumed by the heat pump.  This is where it gets interesting as the amount of energy consumed by the heat pump depends on the control strategy during the recovery period.

 

Control strategy 1 - forced recovery

In this strategy we attempt, during the recovery period, to get the house back to the state it was when we first switched off the heat pump.  This means bringing the house back up to 20C and replacing the 12.5kW we have lost from the fabric.  To do this we will have to supply to the house, in addition to the heat loss during the recovery period (average 3kW), an additional 13.5kWh to replace the heat lost from the fabric.  thus we need to supply on average 3+13.5/4.5 = 6kW to the house.  The heat pump is perfectly capable of doing this, but the only way the radiators can emit that amount of energy is by raising the flow temperature during the recovery period to about 40C (40.04C to be precise).  A heat pump that has a fixed WC curve wont do this but one that has room compensation might.  This causes the efficiency of the heat pump to drop by about 10% (2% per degree is a good rule of thumb).  So a heat pump which was operating at a COP of 4 will instead operate, during the recovery period, at a COP of 3.6.  

If we work through all of these numbers we find that the heat pump would have consumed 9kWh to deliver the 36kWh that would have been lost had we left the heating on, and 7.5kWh to deliver the 27kWh that is lost if we switch the heat pump off, a saving of only 16% in energy consumed as opposed to the saving of 25% in energy delivered/lost. 

With these figures its still a saving so still worthwhile, but with other figures, particularly as the OAT drops, the saving becomes much smaller.  Some exploration of the 'space' with an excel model is indicated (my next step).

 

Control Strategy 2 - natural recovery

In this strategy we don't alter the flow temperature during the recovery, it remains at 34.7C.  Heat pumps operating pure WC will do this.  However the rads will still deliver to the house more energy than that needed to maintain the steady state and this the temperature will still recover, albeit more slowly because

a) the rad-room delta T is higher (but gets smaller as the house starts to heat up) and 

b) the house to OAT delta T is lower (but gets larger as the house starts to heat up)

I dont have a formula for this (I think its logarithmic/exponential) but its easy to work out in a spreadsheet, which I have done.  Putting all of the above assumptions in the house will reach about 18.9C after 4.5 hours, and will get to 19.8 after 10 hrs.  The saving in energy is 27/4 vs 36/4, 25% but at the expense of a bit of a loss in comfort, because the house is roughly 1C under temp at the end of the recovery period.  This may be tolerable or alternatively after experiencing it for some days the owner might nudge the WC curve up to compensate, increasing FT by about 1-1.5C, causing a 2-3% penalty on efficiency for the entirety of the day.  However thats a relatively small penalty compared to strategy 1.  There is still a comfort cost however as it means that the house temp will be too low for part of the day, and too high at other times (eg just before the set back period).  This may or may not matter.

 

Control Strategy 2a - natural recovery with fancoil

An interesting variant is the above strategy occurs if the emitters are fancoils.  These can (and do) increase their output without increasing the flow temperature, simply by turning up the fan.  This will increase the deltaT (flow - return) and the HP will make this up by simply delivering more energy at the same flow temperature.  thus you get the benefits of the rapid recovery of strategy 1, but without the disadvantage of requiring a higher FT.  

 

Comments

  • Obviously this is only one example.  There are approximations, most notably that a linear decay in temp is assumed for much of the calculation (but note that a linear recovery is not assumed for recovery strategy 2a).  However its likely to be good enough to give a feel.  Some other examples can be explored using the excel.
  • The calculation for control strategy 2a requires one to derive the house heat capacity from the cool down rate.  That has other uses.
  • No account is taken of the fact that rads first heat the air then heat the house which means that the house may feel warmer more quickly than is indicated in the case of strategy 2.  

The excel attached is very crude, but does some of the calculations referred to above.  It needs developing to make it robust, something I will attempt over the next couple of weeks.  However its good enough to play a bit.  Im calling it v0.1 because its too easy to get silly results if you push it beyond its (undeclared) limits!

This topic was modified 8 months ago by JamesPa

   
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(@derek-m)
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Posted by: @jamespa

There has been much discussion on on/off working/nighttime setback in relation to heat pumps, but not much that is quantitative.  @fazel and @derek-m have made some helpful comments in another thread and I have now started to think about it more quantitatively.  I present an example of how one might assess the effect, it needs development and discussion but is a 'starter for 10'

Worked example:

Lets consider a 10kW heat pump fitted to a house which consumes 8kW when the OAT is 0C and the target IDT is 20C.  Lets also assume that the emitters are radiators with a FT of 45C at 0C, and furthermore that we wish save energy by setting back/switching off for a maximum period of 9 hours (overnight) - eg from 10pm-7am.  Lets consider what happens when the OAT is 10C.  Lets further assume that the temperature drops by 5C during an 'off' period of 4.5 hours, when we switch it back on to get the house warmed up in time for 7am.  Once we have worked through this example it will be relatively easy to change the assumptions.

At OAT of 10C and IDT of 20C the load will be 4kW and the flow temperature required to deliver this will be 34.7C (applying the rule that emission from radiators is proportional to (deltaT rad-room)^1.3.  At this FT and OAT we can typically expect to achieve a COP of 4 (as will become evident later it doesn't matter what the exact COP is for the calculation, but it does matter roughly what it is).  

During the on/off cycle the average IDT will be approximately 17.5C, so instead of losing 4kW the house will lose (on average) 3kW.  Thus over the 9 hour cycle the house will lose 9x3=27kWh instead of 9*4=36kWh had we left the heat pump on.  More importantly during the 4.5hour cool down period we will lose approximately half of this amount, 13.5kWh , and this is lost from the fabric so will have to be replaced to return to the state we started with.  We will return to this shortly.

However we don't pay for energy lost from the house, we pay for energy consumed by the heat pump.  This is where it gets interesting as the amount of energy consumed by the heat pump depends on the control strategy during the recovery period.

 

Control strategy 1 - forced recovery

In this strategy we attempt, during the recovery period, to get the house back to the state it was when we first switched off the heat pump.  This means bringing the house back up to 20C and replacing the 12.5kW we have lost from the fabric.  To do this we will have to supply to the house, in addition to the heat loss during the recovery period (average 3kW), an additional 13.5kWh to replace the heat lost from the fabric.  thus we need to supply on average 3+13.5/4.5 = 6kW to the house.  The heat pump is perfectly capable of doing this, but the only way the radiators can emit that amount of energy is by raising the flow temperature during the recovery period to about 40C (40.04C to be precise).  A heat pump that has a fixed WC curve wont do this but one that has room compensation might.  This causes the efficiency of the heat pump to drop by about 10% (2% per degree is a good rule of thumb).  So a heat pump which was operating at a COP of 4 will instead operate, during the recovery period, at a COP of 3.6.  

If we work through all of these numbers we find that the heat pump would have consumed 9kWh to deliver the 36kWh that would have been lost had we left the heating on, and 7.5kWh to deliver the 27kWh that is lost if we switch the heat pump off, a saving of only 16% in energy consumed as opposed to the saving of 25% in energy delivered/lost. 

With these figures its still a saving so still worthwhile, but with other figures, particularly as the OAT drops, the saving becomes much smaller.  Some exploration of the 'space' with an excel model is indicated (my next step).

 

Control Strategy 2 - natural recovery

In this strategy we don't alter the flow temperature during the recovery, it remains at 34.7C.  Heat pumps operating pure WC will do this.  However the rads will still deliver to the house more energy than that needed to maintain the steady state and this the temperature will still recover, albeit more slowly because

a) the rad-room delta T is higher (but gets smaller as the house starts to heat up) and 

b) the house to OAT delta T is lower (but gets larger as the house starts to heat up)

I dont have a formula for this (I think its logarithmic/exponential) but its easy to work out in a spreadsheet, which I have done.  Putting all of the above assumptions in the house will reach about 18.9C after 4.5 hours, and will get to 19.8 after 10 hrs.  The saving in energy is 27/4 vs 36/4, 25% but at the expense of a bit of a loss in comfort, because the house is roughly 1C under temp at the end of the recovery period.  This may be tolerable or alternatively after experiencing it for some days the owner might nudge the WC curve up to compensate, increasing FT by about 1-1.5C, causing a 2-3% penalty on efficiency for the entirety of the day.  However thats a relatively small penalty compared to strategy 1.  There is still a comfort cost however as it means that the house temp will be too low for part of the day, and too high at other times (eg just before the set back period).  This may or may not matter.

 

Control Strategy 2a - natural recovery with fancoil

An interesting variant is the above strategy occurs if the emitters are fancoils.  These can (and do) increase their output without increasing the flow temperature, simply by turning up the fan.  This will increase the deltaT (flow - return) and the HP will make this up by simply delivering more energy at the same flow temperature.  thus you get the benefits of the rapid recovery of strategy 1, but without the disadvantage of requiring a higher FT.  

 

Comments

  • Obviously this is only one example.  There are approximations, most notably that a linear decay in temp is assumed for much of the calculation (but note that a linear recovery is not assumed for recovery strategy 2a).  However its likely to be good enough to give a feel.  Some other examples can be explored using the excel.
  • The calculation for control strategy 2a requires one to derive the house heat capacity from the cool down rate.  That has other uses.
  • No account is taken of the fact that rads first heat the air then heat the house which means that the house may feel warmer more quickly than is indicated in the case of strategy 2.  

The excel attached is very crude, but does some of the calculations referred to above.  It needs developing to make it robust, something I will attempt over the next couple of weeks.  However its good enough to play a bit.  Im calling it v0.1 because its too easy to get silly results if you push it beyond its (undeclared) limits!

A good first attempt but there are quite a number of parameters that have not been considered.

You assume a constant OAT of 10C, which is very unlikely to occur in the real World, it is often lowest in the 6am to 7am period, hence higher LWT in WC mode with reduced efficiency.

Even if the OAT remains constant at 10C, with a constant LWT of 34.7C, the heat emitters will be extracting more energy, as you have stated, but this energy must come from somewhere. This will have the effect of increasing the DT between LWT and RWT, which will need to be restored to the nominal 5C by either increasing the flow rate or raising  the LWT. Both of which will cause the heat pump to work harder and draw more power from the supply.

 


   
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(@jamespa)
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Posted by: @derek-m

You assume a constant OAT of 10C, which is very unlikely to occur in the real World, it is often lowest in the 6am to 7am period, hence higher LWT in WC mode with reduced efficiency.

True, but folding in variable oat before developing the basic theory is making it unnecessarily complex at an early stage and getting an idea based on constant OAT will inform, probably more so initially than variable.

Also, since this is ultimately cost Vs comfort, I 'ought' to build in the effect of time of use tarrifs.  I'm definitely not going to do that, this is about quantifying the broad brush principles of how heat pumps behave and testing the assertion that 24*7 operation is THE way to operate, not the minituae.

Posted by: @derek-m

Even if the OAT remains constant at 10C, with a constant LWT of 34.7C, the heat emitters will be extracting more energy, as you have stated, but this energy must come from somewhere. This will have the effect of increasing the DT between LWT and RWT, which will need to be restored to the nominal 5C by either increasing the flow rate or raising  the LWT. Both of which will cause the heat pump to work harder and draw more power from the supply.

That is accounted for in the power emitted/power consumed calculation.  Whether the hp chooses to do this by allowing delta t to increase or increasing flow rate is secondary.  Some pumps modulate water pump speed for constant delta t, others don't/can't.  In the latter case delta t varies anyway as a function of OAT.  There are arguments both ways.  Either way the energy/cop calculation works because it relies only on the principle of conservation of energy.

This post was modified 8 months ago 2 times by JamesPa

   
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(@derek-m)
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@jamespa

I'm afraid that I cannot see how the heat pump could be supplying 5.86kW to the heat emitters rather than the previous 4kW, without working harder and hence drawing more power from the supply.

Please explain how this is achieved.


   
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(@kev-m)
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Posted by: @jamespa

There has been much discussion on on/off working/nighttime setback in relation to heat pumps, but not much that is quantitative.  @fazel and @derek-m have made some helpful comments in another thread and I have now started to think about it more quantitatively.  I present an example of how one might assess the effect, it needs development and discussion but is a 'starter for 10'

Worked example:

Lets consider a 10kW heat pump fitted to a house which consumes 8kW when the OAT is 0C and the target IDT is 20C.  Lets also assume that the emitters are radiators with a FT of 45C at 0C, and furthermore that we wish save energy by setting back/switching off for a maximum period of 9 hours (overnight) - eg from 10pm-7am.  Lets consider what happens when the OAT is 10C.  Lets further assume that the temperature drops by 5C during an 'off' period of 4.5 hours, when we switch it back on to get the house warmed up in time for 7am.  Once we have worked through this example it will be relatively easy to change the assumptions.

At OAT of 10C and IDT of 20C the load will be 4kW and the flow temperature required to deliver this will be 34.7C (applying the rule that emission from radiators is proportional to (deltaT rad-room)^1.3.  At this FT and OAT we can typically expect to achieve a COP of 4 (as will become evident later it doesn't matter what the exact COP is for the calculation, but it does matter roughly what it is).  

During the on/off cycle the average IDT will be approximately 17.5C, so instead of losing 4kW the house will lose (on average) 3kW.  Thus over the 9 hour cycle the house will lose 9x3=27kWh instead of 9*4=36kWh had we left the heat pump on.  More importantly during the 4.5hour cool down period we will lose approximately half of this amount, 13.5kWh , and this is lost from the fabric so will have to be replaced to return to the state we started with.  We will return to this shortly.

However we don't pay for energy lost from the house, we pay for energy consumed by the heat pump.  This is where it gets interesting as the amount of energy consumed by the heat pump depends on the control strategy during the recovery period.

 

Control strategy 1 - forced recovery

In this strategy we attempt, during the recovery period, to get the house back to the state it was when we first switched off the heat pump.  This means bringing the house back up to 20C and replacing the 12.5kW we have lost from the fabric.  To do this we will have to supply to the house, in addition to the heat loss during the recovery period (average 3kW), an additional 13.5kWh to replace the heat lost from the fabric.  thus we need to supply on average 3+13.5/4.5 = 6kW to the house.  The heat pump is perfectly capable of doing this, but the only way the radiators can emit that amount of energy is by raising the flow temperature during the recovery period to about 40C (40.04C to be precise).  A heat pump that has a fixed WC curve wont do this but one that has room compensation might.  This causes the efficiency of the heat pump to drop by about 10% (2% per degree is a good rule of thumb).  So a heat pump which was operating at a COP of 4 will instead operate, during the recovery period, at a COP of 3.6.  

If we work through all of these numbers we find that the heat pump would have consumed 9kWh to deliver the 36kWh that would have been lost had we left the heating on, and 7.5kWh to deliver the 27kWh that is lost if we switch the heat pump off, a saving of only 16% in energy consumed as opposed to the saving of 25% in energy delivered/lost. 

With these figures its still a saving so still worthwhile, but with other figures, particularly as the OAT drops, the saving becomes much smaller.  Some exploration of the 'space' with an excel model is indicated (my next step).

 

Control Strategy 2 - natural recovery

In this strategy we don't alter the flow temperature during the recovery, it remains at 34.7C.  Heat pumps operating pure WC will do this.  However the rads will still deliver to the house more energy than that needed to maintain the steady state and this the temperature will still recover, albeit more slowly because

a) the rad-room delta T is higher (but gets smaller as the house starts to heat up) and 

b) the house to OAT delta T is lower (but gets larger as the house starts to heat up)

I dont have a formula for this (I think its logarithmic/exponential) but its easy to work out in a spreadsheet, which I have done.  Putting all of the above assumptions in the house will reach about 18.9C after 4.5 hours, and will get to 19.8 after 10 hrs.  The saving in energy is 27/4 vs 36/4, 25% but at the expense of a bit of a loss in comfort, because the house is roughly 1C under temp at the end of the recovery period.  This may be tolerable or alternatively after experiencing it for some days the owner might nudge the WC curve up to compensate, increasing FT by about 1-1.5C, causing a 2-3% penalty on efficiency for the entirety of the day.  However thats a relatively small penalty compared to strategy 1.  There is still a comfort cost however as it means that the house temp will be too low for part of the day, and too high at other times (eg just before the set back period).  This may or may not matter.

 

Control Strategy 2a - natural recovery with fancoil

An interesting variant is the above strategy occurs if the emitters are fancoils.  These can (and do) increase their output without increasing the flow temperature, simply by turning up the fan.  This will increase the deltaT (flow - return) and the HP will make this up by simply delivering more energy at the same flow temperature.  thus you get the benefits of the rapid recovery of strategy 1, but without the disadvantage of requiring a higher FT.  

 

Comments

  • Obviously this is only one example.  There are approximations, most notably that a linear decay in temp is assumed for much of the calculation (but note that a linear recovery is not assumed for recovery strategy 2a).  However its likely to be good enough to give a feel.  Some other examples can be explored using the excel.
  • The calculation for control strategy 2a requires one to derive the house heat capacity from the cool down rate.  That has other uses.
  • No account is taken of the fact that rads first heat the air then heat the house which means that the house may feel warmer more quickly than is indicated in the case of strategy 2.  

The excel attached is very crude, but does some of the calculations referred to above.  It needs developing to make it robust, something I will attempt over the next couple of weeks.  However its good enough to play a bit.  Im calling it v0.1 because its too easy to get silly results if you push it beyond its (undeclared) limits!

I've tried both (forced and natural) and measured the effects on consumption.  My measured results are broadly in line with your modelled ones.  I used WC and switched off 12-5am.  That saves me about 15% over leaving it on 24/7.  I don't need the house to be as warm in the morning so this suits us.  I now use Ecodan AA but I limit the maximum flow; it bumps it up a little to warm up the cooler house but not a lot.  It still takes 3-4 hours.  Savings are about the same, or close enough not to be able to measure the difference. 

EDIT - and this looked in detail at a cold start and how much extra energy it took to warm up the radiators (not the house) from cold.

https://renewableheatinghub.co.uk/forums/renewable-heating-air-source-heap-pumps-ashps/heat-pump-truth-or-myth1-keeping-it-running-24-7-uses-less-energy#post-12704

This post was modified 8 months ago by Kev M

   
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(@jamespa)
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Posted by: @derek-m

@jamespa

I'm afraid that I cannot see how the heat pump could be supplying 5.86kW to the heat emitters rather than the previous 4kW, without working harder and hence drawing more power from the supply.

Please explain how this is achieved.

It does work harder, I have never said otherwise, and wouldn't do so because it would violate the laws of physics.  I calculate the total energy supplied to the emitters (which must equal the energy supplied by the heat pump) and divide it by the COP to get the total energy consumed by the heat pump.  

I think the confusion may arise because it is not necessary to know whether the extra energy is delivered by an increase in flow rate or an increase in deltaT (flow-return) or by a combo of both as neither of these materially affects the COP.  Thus I don't bother to consider these factors explicitly, I can just calculate total energy. 

It is however necessary to know what the flow temperature is because that does materially affect the COP.  In the 'forced' recovery model the flow temperature is increased so that the emitters emit sufficient energy in the allotted recovery time, and thus not only does the heat pump have to work harder to supply more energy, it is less efficient while doing so.  In the natural recovery model the flow temperature is maintained, but the heat pump still supplies more energy due to the fact that the radiators emit more because deltaT rad-room is larger (due to the fact that the room is colder until recovery is complete).  COP is not affected, but the extra energy supplied will be reflected in the energy consumed.  My model and explanation does this.  There isn't yet a line in the spreadsheet for this (comparable to A32, A33 in the case of the forced recovery model)) but in this case the energy consumed 'on' will be 9kW as before, and the energy consumed 'on-off' will be 27/4=6.75kW, the saving being the difference in energy loss over the cycle (9kW) in the two cases divided by the constant COP (4.0) = 2.25kW

 

This post was modified 8 months ago by JamesPa

   
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Also, since this is ultimately cost Vs comfort, I 'ought' to build in the effect of time of use tarrifs.  I'm definitely not going to do that,

With the new Cosy tariff (and various EV tariffs ) available, it would be great to do this..... I understand your calculations are about the energy needed, but our off-peak rate is about 5x cheaper than on-peak so it's no longer about efficiency and energy use, for us it's about cost.

So would be great to get some modelling on whether we can take advantage of these tariffs. For example our off-peak is 23:30 - 5:30 which is pretty much when we'd have a set-back so does it make any sense to run hard between say 04:30 - 05:30? Probably, but be good to model...

250sqm house. 30kWh Sunsynk/Pylontech battery system. 14kWp solar. Ecodan 14kW. BMW iX.


   
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(@derek-m)
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Posted by: @jamespa

Posted by: @derek-m

@jamespa

I'm afraid that I cannot see how the heat pump could be supplying 5.86kW to the heat emitters rather than the previous 4kW, without working harder and hence drawing more power from the supply.

Please explain how this is achieved.

It does work harder, I have never said otherwise, and wouldn't do so because it would violate the laws of physics.  I calculate the total energy supplied to the emitters (which must equal the energy supplied by the heat pump) and divide it by the COP to get the total energy consumed by the heat pump.  

I think the confusion may arise because it is not necessary to know whether the extra energy is delivered by an increase in flow rate or an increase in deltaT (flow-return) or by a combo of both as neither of these materially affects the COP.  Thus I don't bother to consider these factors explicitly, I can just calculate total energy. 

It is however necessary to know what the flow temperature is because that does materially affect the COP.  In the 'forced' recovery model the flow temperature is increased so that the emitters emit sufficient energy in the allotted recovery time, and thus not only does the heat pump have to work harder to supply more energy, it is less efficient while doing so.  In the natural recovery model the flow temperature is maintained, but the heat pump still supplies more energy due to the fact that the radiators emit more because deltaT rad-room is larger (due to the fact that the room is colder until recovery is complete).  COP is not affected, but the extra energy supplied will be reflected in the energy consumed.  My model and explanation does this.  There isn't yet a line in the spreadsheet for this (comparable to A32, A33 in the case of the forced recovery model)) but in this case the energy consumed 'on' will be 9kW as before, and the energy consumed 'on-off' will be 27/4=6.75kW, the saving being the difference in energy loss over the cycle (9kW) in the two cases divided by the constant COP (4.0) = 2.25kW

 

Are you assuming a constant COP of 4 and an indoor temperature drop of 5C over a 4.5 hour period? Please clarify.

 


   
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Posted by: @benseb

Also, since this is ultimately cost Vs comfort, I 'ought' to build in the effect of time of use tarrifs.  I'm definitely not going to do that,

With the new Cosy tariff (and various EV tariffs ) available, it would be great to do this..... I understand your calculations are about the energy needed, but our off-peak rate is about 5x cheaper than on-peak so it's no longer about efficiency and energy use, for us it's about cost.

So would be great to get some modelling on whether we can take advantage of these tariffs. For example our off-peak is 23:30 - 5:30 which is pretty much when we'd have a set-back so does it make any sense to run hard between say 04:30 - 05:30? Probably, but be good to model...

I am looking to model the possible energy usage during different hours of the day under different operating conditions, you should then be able to calculate the cost.

 


   
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(@jamespa)
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@derek-m I take account of cop change due to flow temp change.  I don't take account of the (generally second order and due to engineering not physics thus model dependent) change in cop due to changes in output/flow rate etc


   
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Posted by: @jamespa

@derek-m I take account of cop change due to flow temp change.  I don't take account of the (generally second order and due to engineering not physics thus model dependent) change in cop due to changes in output/flow rate etc

Could you please reply in 'English'.

So you assumed a constant COP of 4 and a temperature drop of 5C?

Did you remove your spreadsheet, I don't appear to be able to find it?

 


   
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Posted by: @derek-m

Posted by: @jamespa

@derek-m I take account of cop change due to flow temp change.  I don't take account of the (generally second order and due to engineering not physics thus model dependent) change in cop due to changes in output/flow rate etc

Could you please reply in 'English'.

So you assumed a constant COP of 4 and a temperature drop of 5C?

Did you remove your spreadsheet, I don't appear to be able to find it?

 

I assume a COP of 4 at the starting conditions (but you can adjust the value in the spreadsheet).  In the forced recovery model the FT changes so the COP changes.  In this model I degrade the modelled efficiency of the heat pump by 2% for every 1C rise in FT, by adjusting the COP (If you look at the data on actual performance this is not a bad approximation; this is based on Mitsubishi, but as this is driven directly by the laws of thermodynamics, its likely to be similar for other models).   FT is the only variable in response to which I change COP, I don't change COP based on HP output or OAT, but you can adjust the starting COP for the starting OAT if you want.

I doubt, for the specific question I am trying to answer, that the exact starting COP matters as much as the change in COP.  The specific question I am trying to answer is 'is it really true that 24/7 operation is the most efficient, and if so under roughly what conditions?'

The spreadsheet as posted has a drop of 5C as described upthread, but this is adjustable in the spreadsheet

I did not remove the spreadsheet.

To be clear I'm not particularly interested in modelling the exact performance of any particular model of heat pump.  What I am interested in is understanding general trends which are likely to be applicable across most heat pumps and confirming or debunking 'perceived wisdom'.  Thus I will generally try to reduce anything I do to the basic physics, or model performance differences between various scenarios rather than absolute performance, basing the difference model wherever possible on something that is likely to be universal or at worst quite common.

So the model (like my model of weather compensation posted several months ago) is relatively simple and in this case I am seeking to understand the physics and limitations behind the generally stated view that 24/7 operation is optimal.

So far the very crude model (and the supporting rationale) seems to support the intuitive view that

  • the benefits (in terms of energy consumed) of night time set back are not as much as for a gas boiler
  • that, at least in mild OAT scenarios, there are still benefits (in terms of energy consumed)
  • that the benefits, where they exist, depend significantly on the control strategy during the recovery phase, which becomes a comfort/cost trade off unless you have fancoils (a discovery I was not expecting and triggers an interesting in thinking more about the distinct properties of fancoils)

My next steps are to 'turn the model round' so that house heat capacity, rather than the cycle time is fixed, and then try it for various OATs to see how the conclusion varies with OAT.  I also want to play with the scenario where the house temp is set back for 18 hrs instead of the current 9hrs, which would suit a working day.  I have a hunch that it will transpire that

  • set back makes less sense when the OAT is very low
  • for the 'working day' scenario' the breakfast time recovery would be more efficient with COP1 fan heaters or something else that simply heats air quickly

But these are only hunches at present!

 

 

 

 

 

 


   
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