@william1066 oxygen ingress that isn't something I had even considered as a specific thing tbh! although I do generally prefer all metal mag-clean on durability grounds. I have Fernox 62252 28mm on the boiler now. Claims 50L/min but the pressure loss is a bit worse than the Inta one you found, about 1.7mh2o at 45l/min. thoughts on the parallel approach?
@derek-m I am back on this topic. So I looked at Grahams post on the carnot cycle, and some other stuff. And flow rate does not come into the efficiency formulas that I can find. Using Grahams post on theoretical efficiency I get the following. It seems to me the focus for efficiency is ultimately a focus on the LWT [direct system, with no mixing, and low temp emitters etc]. Of course as you increase your flow rate, you can reduce [to a point] your LWT to meet your heat load. (assuming constant dT and suitable emitters). Hence my obsession with flow rate. I think that the delta T is really dictated by your emitters as it does not seem to be a parameter wrt efficiency calculations, at least not directly.
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The quantity of heat energy produced by a heat pump is given by the formula:-
Q = mcDT
Where Q = Heat Energy, m = Mass, c = Specific Heat Capacity and DT = Change in Temperature.
Since the volume of 1kg of water is approximately 1 litre around the temperature of a central heating system, it is accurate enough to assume that 1kg of water occupies a volume of 1 litre.
The above formula could therefore be re-written:-
Q = lcDT where l = litres.
The Specific Heat Capacity of water in the temperature range for central heating is approximately 4184 Joules, which equates to approximately 1.16 Watts of heat energy required to raise the temperature of 1kg of water by 1C.
If a heat pump is receiving water at a RWT of 30C, and is heating it by 5C to a LWT of 35C, it will require 1.16 x 5 = ~ 5.8 Watts per litre.
The same would be true if the RWT is 40C and the LWT is 45C.
So using the above formula, the determining factor is the volume of water being heated.
If the flow rate is 10 litres per minute, then the volume of water each hour will be 600 litres.
So to heat that quantity of water will require 600 x 5.8 Watts of energy, which is approximately 3.487kWh.
If the flow rate is now increased to 20 litres per minute, the quantity of energy required is now 1200 x 5.8, which is approximately 6.973kWh.
So from the point of view of heat energy produced, the water flow rate is a prime determining factor.
Heat pump efficiency is quite a simple calculation, since it is 'energy out' divided by 'energy in', but in the real World it is a little more complex.
Whilst 'energy in' can be obtained with reasonable accuracy using electricity power meters, 'energy out' is more problematic, since it requires reasonably accurate measurement of LWT, RWT and water flow rate.
Although, as stated above, it is the DT which determines the quantity of heat energy produced, rather than the actual LWT, the LWT determines how much 'energy in' is required to produce that quantity of 'energy out'. Because a heat pump has to work harder to provide higher LWT, it must therefore increase the 'energy in'. This increase in 'energy in' is not directly proportional to 'energy out', but actually increases as the LWT is increased.
Increased LWT being supplied to a home, will increase the quantity of heat energy being transmitted by the heat emitters, which in turn will require the heat pump to produce more heat energy, which is normally achieved by increasing the water flow rate.
If the heat loss, and hence heat demand, of a home is reasonable well known, it may be possible to obtain an approximation of heat energy produced by a heat pump, utilising reasonably accurate measurement of indoor temperature, outside temperature and LWT, though I doubt that this would be as accurate as measuring the actual 'energy out'.
@derek-m Thanks very much for this explanation. You have confirmed something that I have been wondering about - that the heat energy produced by a heat pump doesn't vary with LWT in the situation where Delta T and flow rate are kept the same. This isn't what one might expect, but is true. Consequently I have often wondered how my Samsung can vary its heat energy production as it almost invariably shows the same delta T of 5 and the same flow rate (I have a non PWM primary pump). According to my heat meter the heat production does vary from day to day, but not by very much. I assume that this is due to small changes in Delta T that I cannot see on my system as it rounds the flow temperatures to integers.
What I have struggled to understand is the interaction between the heat energy produced by the heat pump and the heat energy transmitted by the heat emitters. These have to be the same at equilibrium, so I presumed that the heat pump would determine the value. You seem to be saying that if the LWT is raised from say 35 deg C to 50 deg C, the additional heat energy transmitted by the emitters will cause the heat pump to produce more heat energy until they are once more in balance. This suggests that the emitters have an equal say in determining the heat pump energy production, which I found harder to understand. In cases where the flow rate is fixed as in most Samsung installs, this presumably can only be achieved by an increase in delta T.
I must admit that I have not looked in detail at Samsung heat pumps, but I struggle to understand how it can control the quantity of heat energy produced, without varying either the water flow rate or the DeltaT. The only other way would be to switch the compressor on and off from the controller, or via indoor thermostats.
Thinking about the problem, this may partially explain what could be happening within your system.
If the required LWT is increased, the refrigerant gas flow through the compressor will need to be increase to raise the pressure and temperature of the gas entering the condenser. This in turn will probably absorb more heat energy from the outside air, but also additional energy from the electricity supply.
The hotter gas going through the condenser will increase the temperature of the LWT, which will in turn cause the heat emitters to transfer more heat energy into your home. The temperature within your home will start to increase, which in turn will increase the heat loss through the fabric of your building. At some point equilibrium will be achieved.
What I am not 100% certain about is how this will affect the RWT.
With a fixed water flow rate one would expect an increase in LWT to also cause an increase in RWT, but since the heat emitters are now absorbing more heat energy from the water, will the RWT in fact decrease?
It would be an interesting test to perform.
As I have said previously a home central heating system is not as straightforward as one would think, since there are quite a number of parameters which can be continually balancing and re-balancing.
Whilst a system with a fixed water flow rate may indeed keep the indoor temperature reasonably constant, when correctly optimised, I suspect that it will do so with a higher LWT, and hence be less efficient than a system with a variable speed water pump.
@derek-m Thanks very much for this explanation. You have confirmed something that I have been wondering about - that the heat energy produced by a heat pump doesn't vary with LWT in the situation where Delta T and flow rate are kept the same. This isn't what one might expect, but is true. Consequently I have often wondered how my Samsung can vary its heat energy production as it almost invariably shows the same delta T of 5 and the same flow rate (I have a non PWM primary pump). According to my heat meter the heat production does vary from day to day, but not by very much. I assume that this is due to small changes in Delta T that I cannot see on my system as it rounds the flow temperatures to integers.
This is really interesting. Thinking "out loud", I think that assumption is likely - you would only see a significant change in Delta T if your whole house was losing significantly more heat energy (c. 30-40% for a 1 degC difference in DT). Taking my house as an example, my heat loss Q right now (at 0 degC outside give or take, and 21 degC inside) is about 4.2kW, and my DT is about 3 degC (sound low? that's why I'm lurking in this thread! it's usually around 2!). That puts the flow rate at about 20 L/min. Ok, so, assuming that flow rate is fixed, what heat loss would I need in order to see a DT of 4 degC? Plugging the numbers back in to the equation, my house would need to lose 5.6kW of heat! That's a 33% increase in heat loss - huge! That corresponds with a drop in outdoor temperature of 7 degC, i.e. if it went from 0 degC to -7 degC, my DT would increase from 3 degC to 4 degC. You probably wouldn't notice a difference looking day to day as temps rarely vary that much, but if you compared a day when it's -7 degC outside to a day when it's +7 degC outside, you might see a 2 degC difference in DT.
With a fixed water flow rate one would expect an increase in LWT to also cause an increase in RWT, but since the heat emitters are now absorbing more heat energy from the water, will the RWT in fact decrease?
I suspect RWT will increase but to a slightly lower level than the increase in LWT, meaning a slightly larger DT that corresponds (via Q = m.c.DT) with the slightly increased heat loss from the house at the higher equilibrium temperature.
I don't know the accuracy of the temp sensor in my pump. Assuming it is precise, but not accurate, and we are only interested in the delta then the below deviation (0.7 degC) results in a 1.4 kW delta. My flow rate is rock solid and not changing at all.
I don't know the accuracy of the temp sensor in my pump. Assuming it is precise, but not accurate, and we are only interested in the delta then the below deviation (0.7 degC) results in a 1.4 kW delta. My flow rate is rock solid and not changing at all.
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Forgive me if I'm misunderstanding something but that graph appears to be the Leaving Water Temperature (LWT)? If so you need to also know the Return Water Temperature (RWT), and then subtract that from the LWT to find the Delta T that is being referenced.
For example, taking the trough first, the LWT is 34.3. Let's say the RWT is 30.0. That gives a Delta T of 4.3 degC (i.e. 34.3 - 30.0).
Now, let's take the peak LWT of 34.8. If the RWT has gone up to 30.5, then the Delta T is still 4.3 degC; in theory, the emitters won't be emitting any more heat, and the heat pump won't be using any more power. On the other hand, if the RWT only goes up to say 30.3, then the DeltaT is now 4.5 degC. The difference between 4.3 and 4.5 is 0.2, so we know from the equation Derek posted that at 30L/min flow rate (i.e. 0.5 kg/s): Q = 0.5 * 4.2 * 0.2 = 0.42 kW -- i.e. the heat pump is producing 0.42 kW more than it did before.
My point is that it is unlikely the heat pump's sensors, recording instruments, or the app they let you view those things on, will be accurate enough to see a 0.2 degC difference in Delta T. There's an error margin on the LWT, an error margin on the RWT, and the limitations in the accuracy of what the app exposes to end users. Those things will all blow away a 0.2 difference in Delta T. There's also the problem that LWT and RWT are themselves changing, so there is a lag between the LWT increasing and the RWT increasing (the curves will be slightly out of phase) - it's hard to line them up just by eyeballing.
When I get some time I'll post some of my own data for comparison. I personally don't notice a big change in Delta T unless the outdoor temperature changes significantly.
I don't know the accuracy of the temp sensor in my pump. Assuming it is precise, but not accurate, and we are only interested in the delta then the below deviation (0.7 degC) results in a 1.4 kW delta. My flow rate is rock solid and not changing at all.
The indicated value of flow rate may be 'rock solid', I doubt that the actual flow rate is.
@derek-m Yes, the flow rate "meter" in the pump is done via fancy calculations by the pump electronics. So probably some "fancy" filtering going on as well :-). I plan to read the Sika sensor data directly at some point, if I can find a way of safely doing that. Then I will report back as to how good the grundfos flow estimation is.
If so you need to also know the Return Water Temperature (RWT),
Of course, I should have been clear, I need to accurately know the dT with much more accurate sensors, but it was more to illustrate the point that minor fluctuations in delta T, would translate to, not so minor, changes in kW especially over time. Though you could argue that the plus and minus deviations may cancel out. I need more data and more accurate data, for sure.
There is a reason the "proper" heat meters cost lots of wonga.
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