Posted by: @majordennisbloodnok

that gives me headaches when I try putting them into practice

Indeed. I have read some accounts that do a sort of vanishing trick, where everything gets divided by itself, and all disappears into oblivion, though the authors seem to think they have gone full circle.

From my analysis above:

A joule is an amount, a quantum, of energy (and energy is the capacity to do work, ie move things, heat something, whatever).

These quanta, amounts, of energy can be delivered at a certain rate, and that is what we use watts for: 1 watt = 1 joule per second.

It's always per single time unit, because its a rate. This is why this applies:

Posted by: @majordennisbloodnok

I also get that a flow of 1 Joule per second over 3 seconds will deliver the same number of Joules (i.e. the same total energy) as a flow of 3 Joules per second over a time period of just one second.

In case one, 1 watt for 3 secs = 3 watt seconds (a quantity, like kWh), in case two 3 watts for one sec = 3 watt seconds. The second  3 watt case means the rate of delivery is three times as fast, so the same amount of energy is delivered in a third of the time (and, over the same time interval, will deliver three times as much energy).

Posted by: @majordennisbloodnok

the numbers only dictate how quickly or slowly that total energy is delivered.

Exactly, we are always, or almost always, using rates (kWs) as our everyday working units, describing how quickly or slowly things are happening.

Posted by: @majordennisbloodnok

What I struggle with, however, is that it looks to me like both kW and kWh are units of power i.e. dependent on time. It's the fact one is per second and one is per second every second for an hour that gives me headaches when I try putting them into practice.

The key thing, I think is that although both involve time, they use time differently. Power, kW, is a rate: so much energy per given interval of time. The crucial thing is there is a denominator, per unit of time. Energy, on the other hand, isn't in and of itself dependent on time, and so doesn't have a denominator, but it can be delivered at different rates. To get the total amount of energy delivered/consumed, we have to multiply the rate (in kW) by the time is has been delivered/consumed for. Rate (X quanta of energy delivered at Y quanta per unit of time) multiplied by Y length of time = Z total quanta delivered/consumed.  10kW (a rate) for one hour (a time interval) delivers 10kWh (a quantity) of energy, likewise 5kW for 2 hours delivers the same amount (10kW) but takes twice as long because the rate is half as fast.

The key thing I think, to keep things simple, is we only really ever talk about rates, ie kW. Amounts (kWh) only get considered very rarely (on bills, total energy use per annum etc). The counter-intuitive bit, at least for me, is I always thought I was talking about amounts: my rad emits this much or that much amount of heat energy, but in fact I have only ever been talking about the rate at which it emits energy. As long as everything is in the same units (using rates) that's fine. My home's heat loss happens at this rate (in kW), my heating system add heat in at this rate (kW) etc. I suppose it's a bit like two racing cars, we are more interested in their relative speed (ie rate of travel, mph) than the total mileage covered (miles)...

If I can find the article that did the full circle trick (made kW and kWh disappear up each others backside) I'll post it here. When I read it I felt I was close to knowing the answer to everything (and that my heat pump's output was 42), but the feeling quickly dissipated once I moved on.    

One final very simple, I hope, example: two convector heaters, one 1.5kW and one 3kW. The 3kW one delivers heat energy at twice the rate of the 1.5kW one, and that's all we normally care about, and talk about. It's twice as powerful, it has twice the power, and when we go shopping for a heater, a primary interest along with cost and quality is how powerful it is. We only worry about the amount of energy it uses when we consider our energy bill, and to find that out, we need to multiply the rate by the length of use, to get kilowatt hours.

Power (rate) is so many quanta of energy delivered per (divided by) unit of time, total consumption (amount) is rate (watts, joules per second) times (multiplied by) duration.

If I have got any of this wrong, please please do say so! I would hate to mislead people...

Posted by: @batalto

You should be able to estimate flow rates using a bit of maths.

Interesting. I will see if anyone comments further on your 'ramblings' (which look very erudite to me), in the meantime, to clarify:

Posted by: @batalto

kg/s = kW / (specific heat capacity * temp difference)

ie kg/s = l/s (because its water) = heat pump capacity (in kW from Midea controller) x 0.9 (to account for losses)/ specific heat capacity x secondary flow return temp diff at PHE

However, for example, using hypothetical but in range data:

l/s = (6 x 0.9)/(4200 x 10)  = 0.000128571l/s = 0.462857143l/h = 0.000462857 cu metres per hour = something not quite right...

The result is so small because the denominator (42,000) is so large... 

Posted by: @batalto

@cathoderay as a note my Midea is the 12kw version, however I doubt it working very hard at these ambients.

You should be able to estimate flow rates using a bit of maths. You know the flow rate and temperature going into one side of the transfer, we can punt on efficiency and call it 90% for transfer of heat and you know the incoming temperature to the exchange and outgoing temperature.

We also know The specific heat capacity of water is 4,200 Joules per kilogram per degree Celsius (J/kg°C)

heat = mass flow * specific heat capacity * temperature difference

kW = kg/s * kJ/kg/°C * °C

kg/s = litres per second (as its water). We'll rearrange the equation

kg/s = kW / (specific heat capacity * temp difference)

Its been a while since I did any thermodynamics, can someone confirm my ramblings? 

Isn't it W rather than kW?  Watts is the SI unit of power; kW is a convenient multiple to make the everyday numbers used more familiar. Like centimetres or kilometres. 

@kev-m you are correct

6000 / 42,000 = 0.14/s

0.14l/s * 60 seconds * 60 minutes = 514 l/h

514 l/h = 0.514 m3/h

That would seem a respectable number for central heating given the pressures around the pipework

Posted by: @cathoderay

....

Posted by: @majordennisbloodnok

What I struggle with, however, is that it looks to me like both kW and kWh are units of power i.e. dependent on time. It's the fact one is per second and one is per second every second for an hour that gives me headaches when I try putting them into practice.

The key thing, I think is that although both involve time, they use time differently. Power, kW, is a rate: so much energy per given interval of time. The crucial thing is there is a denominator, per unit of time. Energy, on the other hand, isn't in and of itself dependent on time, and so doesn't have a denominator, but it can be delivered at different rates. 

...

The key thing I think, to keep things simple, is we only really ever talk about rates, ie kW. Amounts (kWh) only get considered very rarely (on bills, total energy use per annum etc)

... 

OK, so it seems to me you're suggesting a kWh is the amount of energy IF you sustained a flow rate for a whole hour. That makes sense, and links back with what @derek-m was saying. In essence, despite the term kWh using a time unit in its name, it's just a clumsily worded alias for 3.6MJ.

As an analogy, a litre is a unit of volume and a litre-per-second is a flow rate. Therefore a (litre-per-second)hour is still just a measure of volume - how many litres would have been delivered in one hour if a given (litres-per-second) rate was sustained.

Assuming I've got that right, thanks all for clarifying. Once a concept clicks I've got it, but it doesn't necessarily click quickly.

@majordennisbloodnok you are correct. Batteries are measured in kWh e.g. I have 14kwh of storage, but I cannot deliver 14kw of power from them at any one moment (your bucket has 14kWhs in it, but it doesn't mean you can chuck it all out at once, so to speak). My solar panels are measured in kW, hours have nothing to do with them, they can produce 8kw of power.

kWh is a measure of power over time, kW is just a measure of power.

Hi @cathoderay

It would appear that something went slightly wrong with your calculation.

To hopefully explain things a little clearer lets consider Specific Heat Capacity. We know that if energy is added to a quantity of water its temperature will be increased, the degree by which the temperature is increased is dependent upon the amount of energy (Joules) being added, the quantity of water to which the energy is being added, and the specific heat capacity, which for water is in the order of 4184 J per kg per degree C.

So if 10000 J of energy is added to 1 kg of water, it is possible to calculate that the temperature of the water should be increased by:-

10000/4184 = 2.39C

If the 10000 J is added to 2 kg of water then the temperature increase would only be 1.195C.

The rate at which the temperature increase takes place is dependent upon the size of the heat source, if a 10kW heater was used then 1 kg of water could be heated by 2.39C in 1 second, but would take 10 seconds if a 1kW heater was used.

For all intents and purposes it will be assumed that 1 Litre of water has a mass of 1 kg.

In raising the temperature of water it is possible to store energy, the amount being stored equates to the quantity of water and the increases in temperature. Increasing the temperature of 10L of water from 25C to 35C will require the same amount of energy as that required to heat the same 10L of water from 35C to 45C, but the latter will contain more stored energy.

Actually I do believe that I got my knickers in a twist the other day and gave some incorrect information, for which I have suitably smacked my legs.

Energy is a quantity and can be measured in Joules or kWh, whilst Power is a rate or level and is measured in kW or Horse Power etc. Sorry for the confusion.

In CathodeRay's situation the required amount of heat demand can be approximated from the attached spreadsheet at various outdoor air temperatures and specified indoor air temperature.

Based on the data supplied, at an outdoor air temperature of 5C and a desired indoor air temperature of 20C, the estimated heating demand would be approximately 8kW. To obtain this amount of heat energy from the radiators would require an average water temperature of approximately 41C. Dependent upon the water flow rate around the secondary circuit and the delta T across the radiators, the outlet water temperature from the PHE would need to be in the region of 43C to 46C, to meet the demand.

For adequate energy transfer within the PHE the primary circuit inlet temperature needs to be probably 5C higher than the secondary circuit outlet temperature, which would indicate that LWT from the heat pump would need to be approximately 51C.

I would therefore suggest that you try running your heat pump for a period of time with a fixed LWT of 50C, and monitor the temperatures and flow rate of your system under varying outdoor air temperatures. This may hopefully indicate where any problem areas may be within your system.

Posted by: @majordennisbloodnok

That makes sense, and links back with what @derek-m was saying. In essence, despite the term kWh using a time unit in its name, it's just a clumsily worded alias for 3.6MJ.

Exactly, it is the unfortunate use of a unit of time in the name that makes it seem like a rate, when in fact it is a quantity (that can then be produced, delivered or consumed at a certain rate, for which name is power, and the unit is kW).

Posted by: @batalto

514 l/h = 0.514 m3/h

That would seem a respectable number for central heating given the pressures around the pipework

Actual figures right now, including the 90% efficiency correction: 

capacity: 5.1 - 5.6 (say 5.35)kW

secondary circuit PHE flow/return: 39 - 40/30 - 31 degrees

This was over a few mins observation. At the same time, primary flow rate 1.4 - 1.5M3/H, Primary LWT/RWT 42/38 degrees. Ambient 12 degrees.

5350 x 0.9/ 4200 x 9 = 0.13l/s = 468l/h = 0.468M3/H

suggesting the secondary flow rate is around one third of the primary flow rate. 

Isn't this a bit on the low side?

Posted by: @derek-m

Dependent upon the water flow rate around the secondary circuit and the delta T across the radiators, the outlet water temperature from the PHE would need to be in the region of 43C to 46C, to meet the demand.

@derek-m - my PHE numbers above aren't a million miles out, secondary flow from the PHE is 39 - 40, but what about the average delta t across all the rads and secondary pipework as represented by PHE flow/return (9 degrees) and the secondary flow rate estimate above, bearing in mind the current ambient is 12, not 5 degrees.   

@cathoderay I have no idea the efficiency of the exchange, so I'd check if I was you. On the flow rate, best to ask someone else, but as the pipes are smaller, there is a longer distance and the pump is far less powerful I can see that number making sense.

Hi @cathoderay

Based on the data in the spreadsheet, which I believe is reasonably accurate, at an outdoor temperature of 12C, the heating demand of your home should be approximately 4.3kW. This would require an average radiator temperature of 33C, assuming the system is fully balanced. I suspect that the system is not fully balanced, and also the fact that the operational delta T is 9C, then the probable PHE outlet temperature to balance the heating demand would be:-

33C + 4.5C + ?C (maybe 1C or 2C of imbalance) = 38.5C to 39.5C (This appears to match fairly closely to your measured values).

There would appear to be approximately a 3C temperature loss through the PHE from primary to secondary, which would therefore equate to a required LWT in the order of 41.5C to 42.5C.

One item that has not been considered is the concentration of anti-freeze within the system, which has the effect of reducing the overall specific heat capacity of the primary water circuit, and thereby reducing the energy transporting capacity of the system.

If heat losses within the system are reasonably small, then the 4.3kW heating demand should be approximately matched be the heat energy output from the heat pump. If my understanding is correct the controller is indicating a heat energy output of the order of 5.3kW, which I suspect may be reading high due to the concentration of anti-freeze within the system.

Since the specific heat capacity of the mixture is lower than that for water, then to transport the same quantity of heat energy will require a higher flow rate. If the heat energy output calculation has been based upon having only water in the system, then the higher flow require with an anti-freeze mixture, will lead to inaccurate reading which is always higher than the true value.

I would suggest that you monitor the operation of your system over the coming days and record the variations under different operating conditions. It may be that operation could be improved by changes to the weather compensation slope, or if the system cannot meet the heating demand at lower outdoor air temperatures then this would indicate a weakness in the design and/or configuration.

@batalto - you will have seen @derek-m's remarks about the PHE above, 3 degrees loss through the PHE from primary to secondary, which suggests that at the moment is it reasonably efficient (3/40 suggests ~7.5% loss from primary to secondary, or 92.5% efficient). The secondary pipework is fairly standard 28mm at first, then 22mm and finally 15mm to the rads themselves, and it got power flushed so it shouldn't offer extreme resistance. The rads are all new, and the secondary filter only collects small amounts of sludge, so all in all the secondary circuit should function reasonably well.

@derek-m - the balancing has been tricky, the system is volatile. I can get the rads reasonably balanced one day, the next they are out again. There are also one or two rads that have a habit of remaining cool, even with their lockshield valves fully open, and another rad further down the same branch gets warm, ie there is definitely warm water on that branch's pipework, it just doesn't get into that rad. Other rads have a habit of being relatively warmer than others, even with their lockshield valve all but closed (maybe only a quarter to a half turn open). All the TRV heads are currently off, because I have noticed that just screwing down the head in the fully open position can start to close the valve (felt as a sudden increase resistance as the head is screwed on).

Your point about antifreeze is pertinent. My understanding is there is antifreeze in the primary circuit, but I don't know the concentration. It doesn't appear to be in any documentation, so I will have to ask my installer once he is back.

This morning, the system is still under-performing. Ambient 10 degrees, room stat set to 21.0 in the kitchen which has a design temp of 19 degrees, but it is only 18.4 degrees this morning. This means the room stat effectively calls for heat 24 hours a day, but not all the rooms are at (or above) design temp, even with double figure ambients and room stat constantly calling for heat. Bathroom is 19.5 degrees (vs 22 design), bedroom is 18.5 (vs 19 design). The point is that given double figure ambients and a room stat constantly calling for heat, the rooms should easily be at and above design temps, but they are not. If the total heat demand is of the order of 5.4kW at 10 degrees ambient (from your Midea 14kW spreadsheet), then it should be well within the heat pump's capabilities (unless perhaps the weather comp curve is somehow throttling the kW (as opposed to LWT) output???), suggesting the failure to reach temps is perhaps down to weaknesses in the design and/or configuration.  

At 10 ambient, using my one degree down in LWT for every one degree increase in ambient approximation for the standard Midea weather comp curve, LWT should be 43, and that is where it is, 43 degrees, with a RWT of 39 degrees. The secondary flow and return are currently 40 and 30 degrees respectively, and the top middle of the warmest rad (bathroom) is 38 degrees, but most are at 20 something degrees. I have closed the bathroom rad lockshield from all but fully open to only half open, but it feels awkward doing this, because the bathroom currently is 2.5 degrees below design temp, and closing the lockshield valve to half open isn't going to help it warm up!

I will continue to monitor temps over the next few days, and report back. I have also found some open source weather data from the Met Office for my neck of the woods (Hampshire/Sussex border) and plan to do a plot of hourly ambient temps and hourly kitchen temps (I have a data logger in the kitchen) over the last month, which will include both milder spells and the recent colder spell. I think such a plot may be the simplest way of establishing the system isn't performing as expected, eg if the kitchen temp movements mirror the ambient temp movements, then it suggests to me that the system is in effect always flat out, with no reserve, otherwise it would maintain the kitchen at the design temp regardless of ambient temp, unless and until the ambient drops below -2, the design temp.