Posted by: @jamespa

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@derek-m Surely that depends on what you are trying to calculate.  Degree days will tell you demand (load) whatever the heating technology, which is a vital parameter of system design. 

I grant its not much good for simulations of COP, for that you ideally need hourly temp data which, so far, I cant find a free source of.  I suspect however you could get a pretty reasonable simulation of cop using a combo of degree days and max/min temps, making some assumptions about the rate of change of temp etc/times of min or max.

The problem with heat pumps is that they are not supplied from a single energy source, and this can be affected quite considerably by changes in outside temperature.

Degree days use the daily mean temperature, so a day where the temperature is a constant 9C throughout the day could give the same value as one where the temperature varies from 4C to 14C, with the mean or average being 9C.

The COP of a heat pump at 9C could be 3, and therefore the thermal energy output can be calculated from the electrical energy input. The COP of the same heat pump at 4C could be 2.5, and at 14C could be 3.5, but the overall COP will not be 3. In both examples the heat demand throughout the day may be the same, but in the second situation a greater portion of the heat demand will be with the heat pump operating at a lower efficiency, so the overall daily efficiency will be lower.

@cathoderay

On the first graph, how did you arrive at the weekly energy output for the oil fired system? Did you record the oil consumption each week?

Posted by: @cathoderay

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My calculated heat loss, done for the heat pump installation, is around 12.3kW, but that is also at minus 2, and most of the time the ambient temp is not minus 2, so the actual total kWh heat loss (and so demand) over a heating season will be much smaller

Be careful not to confuse kW (power) and kWh (energy).  12.3kW is the (calculated) peak power demand at the design temperature.  To get to kWh/year you divide this by the temp diff at the design temp (Indoor-Outdoor) and multiply by degree days for your location and then by 24 (because there are 24 hours in a day).  Tables for these appear in the MCS spreadsheet.  There are various refinements to the calculation which will make 20% difference to the calculation in most cases, except for houses where the DHW demand is significant compared to the heating demand.

Posted by: @cathoderay

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I, and I am sure others, would be interested to know how you get from annual MWh use to a (kW) demand (the reverse is also potentially of interest, given a kW heat loss figure, how do you derive an annual MWh use)? Is this where degree days come in? It might also throw some light (good or bad) on EPC annual use figures.   

To get to peak power demand given total heating energy per year you reverse the calculation.  So in your case the calculation I did, using very rough/assumed figures and without any of the refinements, was 11.4MWh/2000*24/22 = 6.5kW.  This assumes that the house was previously at or close to design temp (assumed to be 20) most of the time.  If the house has a long thermal time constant (12 hrs or more) it should still work even if the house temp is allowed to fall significantly for parts of the day.  However if the house has a short thermal time constant and the house temp was allowed to fall significantly during the daytime/at night, then the calculation will underestimate the peak demand 

Its rough and ready but a good sense check on the room by room calculations.  For me

• the MCS room by room calculations get to 16.5kW (two separate surveyors),
• my calculations using MCS assumptions but the correct u values where the fabric has been upgraded, got to 11kW,
• the actual measured demand when it was -2 outside for 5 days in a row was 7.5kW,
• and the degree day calculation based on actual usage to 8.5kW.  

so actually its a whole lot closer to actual than the MCS calculation, or even mine.

Do you have any idea how quickly your house cools down if you switch the heating off?

Posted by: @derek-m

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@cathoderay

On the first graph, how did you arrive at the weekly energy output for the oil fired system? Did you record the oil consumption each week?

No, as I said in the post, I applied a curve (distribution) derived from recent weekly readings to the annual oil use. Note that the first chart is energy in, not energy out.

The other figures:

The heat pump kWh in come from weekly external dedicated heat pump meter readings.

For the second chart, the oil energy out came from the input (determined as above) multiplied by an efficiency factor (80%)

The heat pump kWh out comes from the Midea app/modbus data.

Posted by: @derek-m

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Posted by: @jamespa

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@derek-m Surely that depends on what you are trying to calculate.  Degree days will tell you demand (load) whatever the heating technology, which is a vital parameter of system design. 

I grant its not much good for simulations of COP, for that you ideally need hourly temp data which, so far, I cant find a free source of.  I suspect however you could get a pretty reasonable simulation of cop using a combo of degree days and max/min temps, making some assumptions about the rate of change of temp etc/times of min or max.

The problem with heat pumps is that they are not supplied from a single energy source, and this can be affected quite considerably by changes in outside temperature.

Degree days use the daily mean temperature, so a day where the temperature is a constant 9C throughout the day could give the same value as one where the temperature varies from 4C to 14C, with the mean or average being 9C.

The COP of a heat pump at 9C could be 3, and therefore the thermal energy output can be calculated from the electrical energy input. The COP of the same heat pump at 4C could be 2.5, and at 14C could be 3.5, but the overall COP will not be 3. In both examples the heat demand throughout the day may be the same, but in the second situation a greater portion of the heat demand will be with the heat pump operating at a lower efficiency, so the overall daily efficiency will be lower.

 

As I said it depends on what you are trying to calculate.  If you are trying to calculate heating load (ie what the house needs) then degree-days are good.  if you are trying to calculate energy consumption by the heat pump (heating load/COP) then degree days don't help too much.  Horses for courses.

Posted by: @cathoderay

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Posted by: @derek-m

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@cathoderay

On the first graph, how did you arrive at the weekly energy output for the oil fired system? Did you record the oil consumption each week?

No, as I said in the post, I applied a curve (distribution) derived from recent weekly readings to the annual oil use. Note that the first chart is energy in, not energy out.

The other figures:

The heat pump kWh in come from weekly external dedicated heat pump meter readings.

For the second chart, the oil energy out came from the input (determined as above) multiplied by an efficiency factor (80%)

The heat pump kWh out comes from the Midea app/modbus data.

 

I assume that the vertical axis is kWh/week (ie energy consumed/delivered in any one week), consistent with the title, can you confirm.  If so oil peaks at about 500kWh/week, an average of about 3kW suggesting a peak of perhaps 6kW, not so far from the degree day calculation based on annual demand above. 

That still doesn't tell us whether this low figure is due to lower than calculated peak demand, or the on-off nature of your heating.  Some handle on the answer to the question 'Do you have any idea how quickly your house cools down if you switch the heating off?'  might.

Posted by: @jamespa

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Be careful not to confuse kW (power) and kWh (energy). 

I like to imagine I get the distinction most of the time, but the use of power (kW) for heat loss and energy (kWh) for energy use does leave me easily confused. I think I am right in saying that if you have a heat loss of 12kW at - 2, then in one hour you will use 12kWh to heat the house to design temps. And so on, 288kWh over 24 hours at - 2 etc. The problem is the outside ambient varies constantly, which is where it seems the degree days come in.

Thanks very much for the explanations, I think I can make sense of them, will go over them again latter as I am busy this afternoon. The variations in heat loss demonstrated by the different ways of getting it are interesting and illuminating.

Posted by: @jamespa

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Do you have any idea how quickly your house cools down if you switch the heating off?

Slowly, I think, by which I mean it takes hours to drop a few degrees, though I haven't specifically tested this (as not inclined to turn the heating off when it's cold outside, only when it is warm!). As I recall, it didn't get really cold overnight with the oil boiler, which was off overnight. It is also very slow to heat up with the heat pump (many hours to recover a few degrees, but as I recall the oil boiler could heat the house from cold (say 10 degrees or so) to warm on an hour or so. This seems consistent with the hourly output idea, an oil boiler can put out a lot more heat per hour than a heat pump (but that's OK, because we run heat pumps differently, slow and steady, so they get their in the end). 

Posted by: @jamespa

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I assume that the vertical axis is kWh/week (ie energy consumed/delivered in any one week), consistent with the title, can you confirm. 

You're right, I should have labelled the axis, I hoped the title was enough, and it allowed a slightly larger plot area. Yes, it is kWh/week.

Posted by: @jamespa

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Some handle on the answer to the question 'Do you have any idea how quickly your house cools down if you switch the heating off?'  might.

I will think more about this, see if perhaps I have at some point accidentally collected data on this (its possible, we had a power cut of several hours not so long ago). The problem is it will vary, and for obvious reasons noted above, I don't usually turn the heating off when it is cold outside!

Posted by: @cathoderay

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I like to imagine I get the distinction most of the time, but the use of power (kW) for heat loss and energy (kWh) for energy use does leave me easily confused. I think I am right in saying that if you have a heat loss of 12kW at - 2, then in one hour you will use 12kWh to heat the house to design temps. And so on, 288kWh over 24 hours at - 2 etc. The problem is the outside ambient varies constantly, which is where it seems the degree days come in.

Its easily done, I had to edit my response to get everything right.  Yes if you have 12kW heat loss at -2 then you will need to inject 12kWh into the house each hour to maintain the temperature, thats 288kWh/day.  How much you use is of course dependent on COP/efficiency. 

As you rightly say OAT varies constantly, which is where degree days come in to work out how much you have to inject into the house.  For fossil fuels they will also give you a good estimate of how much fuel you need but not for HPs because the COP varies steeply with temperature, so you need to know (or have a model of) both the temperature and the variation in temperature. 

Posted by: @cathoderay

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I will think more about this, see if perhaps I have at some point accidentally collected data on this. The problem is it will vary, and for obvious reasons noted above, I don't usually turn the heating off when it is cold outside!

You could make the measurement when its mild but not freezing outside, so long as you know the outside temperature and the OAT is fairly constant.  This will give you a figure for degrees per hour (lost) per degree (temp diff inside-outside).  It wont be accurate though if the OAT varies a lot from night to day (so best on a cloudy day & night when diurnal variations are usually smaller) nor if there is large solar gain (so best on a cloudy day when solar gain is less).

Posted by: @cathoderay

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I like to imagine I get the distinction most of the time, but the use of power (kW) for heat loss and energy (kWh) for energy use does leave me easily confused. I think I am right in saying that if you have a heat loss of 12kW at - 2, then in one hour you will use 12kWh to heat the house to design temps. And so on, 288kWh over 24 hours at - 2 etc. The problem is the outside ambient varies constantly, which is where it seems the degree days come in.

Its easily done, I had to edit my response to get everything right.  Yes if you have 12kW heat loss at -2 then you will need to inject 12kWh into the house each hour to maintain the temperature, thats 288kWh/day.  How much you use is of course dependent on COP/efficiency. 

As you rightly say OAT varies constantly, which is where degree days come in to work out how much you have to inject into the house.  For fossil fuels they will also give you a good estimate of how much fuel you need but not for HPs because the COP varies steeply with temperature, so you need to know (or have a model of) both the temperature and the variation in temperature. 

Posted by: @cathoderay

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I will think more about this, see if perhaps I have at some point accidentally collected data on this. The problem is it will vary, and for obvious reasons noted above, I don't usually turn the heating off when it is cold outside!

You could make the measurement when its mild but not freezing outside, so long as you know the outside temperature and the OAT is fairly constant.  This will give you a figure for degrees per hour (lost) per degree (temp diff inside-outside).  It wont be accurate though if the OAT varies a lot from night to day (so best on a cloudy day & night when diurnal variations are usually smaller) nor if there is large solar gain (so best on a cloudy day when solar gain is less).

Posted by: @cathoderay

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I think my Midea can get down to about 25% of it's nominal output, then it starts cycling, need to check.

 

According to the databook the minimum output of the Midea unit at LWT55 OAT 7C is 9kW.  So its cycling much of the time even if the 12.6kW calculated load at -2 is correct.  If the actual load at -2 is much less than the calculated 12.6kW it will be cycling almost all of the time.  Depending on your system volume and allowed hysteresis, this could have a rather negative effect on performance.  Any idea what the cycle time/frequency is?

A few ideas to consider:

The ashp is outside - no doubt it's measuring produced power at the unit, so will include any pipe losses that the oil boiler didn't have.

Does the ashp controller "know" it has glycol in it? - it makes a noticeable change to calculated heatflow.

How does the ashp actually calculate heatflow?  I know it seems obvious that it will be based on temperature measurements and flowrate, but are we really sure it doesn't just base it on electricity used?  I'm betting there's no accuracy spec on the heat produced.  Colour me cynical.

I bet the ashp has a immersion in it.  Any chance it's stuck on?  

For all the above, I recommend an independent, mid certified, heat meter.  I would put it where the pipes come into the house.