Extractor fan advice - do humidistat fans work well?

[JohnW2]

So we agree on the principle of dilution.

Of course we agree on the principle. The reason that at least some parts of the house will get colder as soon as the extractor is turned on is because the previously-heated air (in at least some parts of the house) will be 'diluted' by incoming cold air.

First of all - the heating system will more than easily make up for any loss ...

As I said, of course it will. However, that is the very issue underlying the reason for this discussion - that turning on the extractor will result in the need to expend ('costly') energy to compensate for the effects of the extraction.

... and secondly dilution is a very major factor, in not extracting as much heat as you would seem to assume.

That's the bit which I don't understand. If you leave the extractor running, and have no heating running to compensate for the heat loss, then, eventually, (some) heat from the house will continue to be lost until the inside temp is equal to the outside temp. The rate of heat loss will gradually reduce, as the previously-heated air in the house gets progressively more and more diluted by incoming cold air (i.e. the air in the house gets colder) but, if one waited long enough (with no heating switched on), the indoor temp would theoretically eventually fall to roughly the outdoor temp.

I would not expect any noticeable difference in the cost of running my heating system, or the indoor temperature whether my bathroom fan ran continuously or not.

As above, basic physics theory indicates that there will be "a difference" (in cost of heating and/or temp inside parts of house). However, you may be saying that difference might not be enough for you to "notice". Although, as below, I somehow doubt this, it's possible that you are right that the difference may not be "all that large" - but with fuel prices as they are, some people would be concerned about any increase in energy usage, even if not all that large - let's face it, we're seeing people getting concerned even about the energy cost of leaving their TVs on 'standby'.

Extract fans are relatively tiny, compared to the air volume in a building, which is why the need to be located close to what ever they are intended to extract.

Approved Doc F suggests minimum extraction rates of 15 L/s (54 m³/h) for a kitchen bathroom, and I think that a typical small house has a volume of around 280 m³. That means that about 19% air in the house would be extracted in 1 hour, and, if (as you mention above) the extractor were run continuously at that rate, it would extract an amount equal to the house's volume in just over 5 hours - or some 4.6 times the house's air volume per day.

It much worse with kitchens, since Approved Doc F then suggests minimum extraction rates of 30-60 L/s (108 - 216 m³/h). That means that, at the upper end of that guidance range, about 77% of the air in the house would be extracted in 1 hour, and, if it were run continuously, something like 18.5 times the volume of volume of air in the house every day.

I would therefore suggest that extraction rates (at least 'minimum recommended' ones) are far from as trivial in comparison with house volumes as you appear to be suggesting.

Don't forget that, regardless of what is being extracted ('diluted' or not, moist or not) if, say 54 m³/h (of something) is being extracted, then 54 m³/h of cold air will enter the house from outside to replace it.

Kind Regards, John
Edit: typo corrected

 

[Harry Bloomfield]

Approved Doc F suggests minimum extraction rates of 15 L/s (54 m³/h) for a kitchen, and I think that a typical small house has a volume of around 280 m³. That means that about 19% air in the house would be extracted in 1 hour, and, if (as you mention above) the extractor were run continuously at that rate, it would extract an amount equal to the house's volume in just over 5 hours - or some 4.6 times the house's air volume per day.

No it would not - dilution would mean mean it would your 5 hours would turn into a much larger figure = 10x - 40x for the entire volume to be changed. Natural ventilation would be much higher.

Don't forget, you have already now agreed to the principle of dilution

 

[JohnW2]

No it would not - dilution would mean mean it would your 5 hours would turn into a much larger figure = 10x - 40x for the entire volume to be changed. Natural ventilation would be much higher.

I never said anything about "changing the entire volume". I merely wrote "... it would extract an amount equal to the house's volume in just over 5 hours" (which also means that an amount of cold air equal to the volume of the house will have been drawn in to replace it). I accept that, because of what you call 'dilution' that doesn't mean that all the heated air will have been extracted, but ...

... Don't forget, you have already now agreed to the principle of dilution

... and, in turn, don't forget that, as I said, in the most common real-world situation, the majority of the ingress points will usually be fairly remote from the bathroom, often with closed doors in the path. That means that, in that situation, for quite a while after the extractor is switched on, the air being extracted will be largely 'undiluted heated air', so a fair bit of the heated air in the house will have been extracted before much 'diluted heated air' gets to the bathroom.

Although I agree with the concepts of mixing/dilution, they do not occur instantly, or even rapidly, over relatively long distances or through closed doors.

Whatever, I think you are now just arguing about the magnitude of the energy needed to compensate for the extraction, not the concept (since your 'dilution' can only reduce, not eliminate, the heat loss) - and, as I've said, in current times a lot of people are concerned about even small increases in their energy usage (and cost).

Kind Regards, John

 

[JohnD]

Approved Doc F suggests minimum extraction rates of 15 L/s (54 m³/h) for a kitchen...

...It much worse with kitchens, since Approved Doc F then suggests minimum extraction rates of 30-60 L/s (108 - 216 m³/h).

Did you mean "kitchen?"

 

[JohnW2]

Did you mean "kitchen?"

No, I obviously didn't - now corrected. Thanks for noticing!

Kind Regards, John

 

[JohnW2]

..... dilution of cold air with warm air becomes a major factor, so not as great a heat loss as you suggest.

You are of course wrong, because of dilution of the air in the bathroom, moist air will be drawn out along with it. You seem to have an problem understanding that air does mix, and mix very easily.

No it would not - dilution would mean mean it would your 5 hours would turn into a much larger figure = 10x - 40x for the entire volume to be changed.

Don't forget, you have already now agreed to the principle of dilution

Harry, per the above snippets, my understanding of your view is that you believe I am over-estimating/overstating the heat energy loss (hence reduction in indoor temp, in the absence of compensatory heating) resulting from extraction of air - because the incoming cold air (replacing that which has been extracted) will rapidly and completely mix with the air within the building, such that the incoming cold air will be 'diluted' by the heated air within the building Is that correct?

If so, as you are aware, whilst I accept the 'dilution' concept, it does not alter my view about the amount of heat/temp loss when an extractor is running - particularly in the fairly common situation of a bathroom (with an extractor) fairly remote from the main routes of ingress of outside air into the building.

However, the fact that someone else has a different view from me does not stop me considering and exploring their viewpoint. I have therefore undertaken simulations on the assumption of immediate and complete mixing (hence 'dilution') of incoming air with heated air within the building.

I have considered two rooms with extractors. Firstly a 2.5m x 2.5m x 2.5m kitchen with an extraction rate of 15 mL/sec (minimum recommended by App Doc F) and a 2.5m x 4m x 2.5m kitchen, with an extraction rate of 45 L/sec (mid-range of App Doc F minimum recommendation).

For both rooms, I have considered two scenarios, at the 'extremes'. Firstly a fairly unrealistic case of the room being 'sealed' from the rest of the building, with a ventilator within the room to outside air. Secondly, at the other extreme, a situation in which the room with the extractor is completely open to the rest of the building, such that (per the/your assumption) incoming outside air was rapidly and completely mixed with (hence 'diluted by') all of the heated air within the building. For the latter, I have assumed a total house volume of 280 m³.

In all cases I have assumed an outside air temp of 5 °C and an indoor temp of 20 °C when the extractor was first switched on. I have also assumed there is no heating in operation within the building and that, in the initial state, the entirety of air in the building is at the same temp. I have also assumed that the extraction is continuous for the periods I have indicated.

The results of my simulations are shown in the graphs below, which are hopefully fairly self-explanatory, so I will provide only minimal description/commentary, an almost no 'comment'. In general, the graphs show the period up to when the indoor temperature (of room or whole house) falls to, or near to, the outside air temperature.

In the (fairly unrealistic) scenario of a bathroom or kitchen totally 'sealed' from the rest of the house, with its own ventilation to outside air, following switching on of the extractor, the temperature of air in the room fairly rapidly falls to the outside sir temperature. However, once that has happened there will have been only a very modest loss of heat energy, because there is relatively little air in the room, hence relatively little heated air mixing with, and 'diluting', the incoming cold air (or vice versa).

In contrast, when the room with the extractor is open to the whole of the building, a very much larger volume of heated air is involved, and if one assumes that the incoming cold air rapidly and completely mixes with (hence is 'diluted by') all of that heated air, the fall in indoor temperature (now of the whole house, because of the assumed 'mixing') is considerably lower, but the loss of heat energy is considerably greater. By the time the indoor temperature has fallen to close to the outdoor temperature, about 6 kWh of heat energy has been lost - so that is roughly the amount of energy (gas, electricity, oil or whatever) that would be required to restore the indoor temperature to its original level.

The real-world situation will be somewhere between the two extreme scenarios depicted below, depending primarily on the locations of air ingress points relative to the extractor and the speed at which mixing/'dilution' of incoming and indoor air actually occurs.

Kind Regards, John

 

[Harry Bloomfield]

Harry, per the above snippets, my understanding of your view is that you believe I am over-estimating/overstating the heat energy loss (hence reduction in indoor temp, in the absence of compensatory heating) resulting from extraction of air - because the incoming cold air (replacing that which has been extracted) will rapidly and completely mix with the air within the building, such that the incoming cold air will be 'diluted' by the heated air within the building Is that correct?

If so, as you are aware, whilst I accept the 'dilution' concept, it does not alter my view about the amount of heat/temp loss when an extractor is running - particularly in the fairly common situation of a bathroom (with an extractor) fairly remote from the main routes of ingress of outside air into the building.

John, I believe you are overthinking it all, and not accepting just how leaky a home is to air flow. Even if you closed and sealed a bathroom door, there are still lots of entry points for air, it is impossible to create even a slight vacuum. To suck even 50% of the heat of a home out, would need a massive amount of through air flow, and that air flow would only manage to cool the air immediately, the cooling the fabric takes much longer.

If you burn your bacon on the stove, with no extract hood, it takes a very long time for the smelly air to be dissolved and replaced. A hood is simply able to help with that, because it sucks the smell laden air out at source. Install the extract fan somewhere other than above where the smell is created and again it takes a long time to clear the smell, because of dilution.

It's the principle of dilution which you are missing, in assuming a fan will take out 100% of the heat. In fact the fan is just taking out a mix of heated air and the cold fresh air.

And with that - I am going to leave this thread..

 

[JohnW2]

John, I believe you are overthinking it all

I don't think so. In fact, I would say that you are the one doing the 'overthinking' by making a very simple situation seem more complicated than it actually is.

, and not accepting just how leaky a home is to air flow. Even if you closed and sealed a bathroom door, there are still lots of entry points for air, it is impossible to create even a slight vacuum.

Exactly. My second pair of simulations (with bathroom/kitchen 'open to house') made absolutely no assumptions about where (or in how many places) the replacement air was entering the house - only that, as you believe, that once it entered (wherever, and through any number of routes) it was immediately and completely 'diluted' by the air within the house.

To suck even 50% of the heat of a home out, would need a massive amount of through air flow, and that air flow would only manage to cool the air immediately,

If there were NO mixing/dilution (i.e. the entering cold air simply 'followed behind' the heated air which was moving towards the extractor}, then to completely remove all of the (heated) air within a house of, say, 280m³ volume would only require 280m³ or so to be extracted, since all of the extracted air would be 'heated air' before any of the replacement cold air got to the extractor. However, if mixing/dilution does occur, then much more has to be extracted before 'all the heated air has been removed', since the extracted heater air is progressively diluted by the cold replacement air that has entered.

Looking at my figures. in the case of the kitchen 'open to the house', it takes about 8 hours (at 45 L/sec extraction) for all of the heated air to be removed (i.e. for indoor temp to get close to outdoor temp) - so involves the extraction of about 1,296 m³ of air from the house . That is some 4.6 times the total volume of air in the house and, as above, that is because, as you acknowledge, the heated air is progressively more and more diluted by the cold replacement air.

However, interesting though it is to look at how long it takes to extract almost all of the heated air (hence energy), people are rarely going to leave extractor fans on for that long, so it is of little more than theoretical interest, much more important being what happens in the much shorter-term real-world situation. If you look at the last of my simulation graphs (kitchen 'open to house') you will see that about 1.5 kWh of energy is lost during the first 30 mins of extraction, and nearly 3 kWh is lost during the first 60 minutes - amounts which are 'not insignificant' at current energy prices.

the cooling the fabric takes much longer.

True, but that is irrelevant to what I've been discussing, in which discussions I have only been considering the air. Although, as you say, the cooling of the fabric of the building takes a lot longer, in the absence of heating it eventually will cool, and to restore the temperature of both air and fabric will require the replacement of all the energy lost by extraction. That's about as basic of physics as one can fins.

If you burn your bacon on the stove, with no extract hood, it takes a very long time for the smelly air to be dissolved and replaced. A hood is simply able to help with that, because it sucks the smell laden air out at source. Install the extract fan somewhere other than above where the smell is created and again it takes a long time to clear the smell, because of dilution.

You now appear to be contradicting yourself because, whilst what you say is true, it is only true because there is not rapid and complete mixing/dilution of (in this case) the cooking smells in the situation you describe. IF those smells were rapidly diluted by mixing with all the air in the house, then you would have to extract vast amounts of air (as above, a lot more than the total volume of air in the house) before those smells were totally removed. However, as you say, if the extraction point is close to the source of the smells (with the air ingress point(s) more remote from the extractor), one will extract most of the smelly air quite quickly - and that is because there has not been significant mixing/dilution before the smelly air gets extracted..

That is the very argument I presented in terms of water-laden (rather than smell-laden) air in a bathroom -but you rejected that view.

It's the principle of dilution which you are missing, in assuming a fan will take out 100% of the heat. In faact the fan is just taking out a mix of heated air and the cold fresh air.

I cannot be 'missing' it, because the (only) basis of the simulations I've now presented is the (very simple) "principle of dilution".

And with that - I am going to leave this thread..

That's obviously your decision. As far as I am concerned, the view I have been expressing, and which I have now illustrated by simulation, is very basic physics based almost entirely on 'the principle of dilution'. Hence, if you do not present a convincing argument against my application of the physics, I will have to assume that you now agree with what I am saying.

Kind Regards, John

 

[Harry Bloomfield]

If there were NO mixing/dilution (i.e. the entering cold air simply 'followed behind' the heated air which was moving towards the extractor}, then to completely remove all of the (heated) air within a house of, say, 280m³ volume would only require 280m³ or so to be extracted, since all of the extracted air would be 'heated air' before any of the replacement cold air got to the extractor.

I said I wouldn't, but... Your suggestion that no mixing will occur is incredibly wild of the mark, because lots of mixing and dilution will take place. No matter how you configure the intake and extract, other than a wind tunnel, extracting 280m³ of air, will not and cannot extract 280m³ of the room temperature air - air flow simply will not work that way.

Back to the cup of tea, with 1/2" left in the bottom and you want to rinse it. There are two ways to rinse the remaining tea out...

1. You tip the remaining tea out, pour a tiny bit of water in swill in around and pour it out - that comes very close to leaving the cup clear of tea.

2. You leave the 1/2" of tea in put it under the running tap. That method requires many times more than water volume used, than in method 1.

Tea = the heat in the house. Method 2, is similar to what will happen in the house, you need to extract an incredibly greater amount than the 280m³ volume of the house, in order to suck all of the warm air out and replace it entirely with fresh cold air.

However, as you say, if the extraction point is close to the source of the smells (with the air ingress point(s) more remote from the extractor), one will extract most of the smelly air quite quickly - and that is because there has not been significant mixing/dilution before the smelly air gets extracted..

Exactly, which also serves to minimise the amount heat which is extracted too.

 

[JohnW2]

... Your suggestion that no mixing will occur is incredibly wild of the mark, because lots of mixing and dilution will take place.

I never suggested that there was no mixing. You have quoted my statement about what would be the situation IF there were no mixing- but you failed to quote my next sentence which says what (more realistically, and very different) happens when there IS mixing.

There will always be some mixing. The real-world situation will be somewhere between the two extreme situations we are talking about - 'no mixing' and 'complete and immediate mixing'.

No matter how you configure the intake and extract, other than a wind tunnel, extracting 280m³ of air, will not and cannot extract 280m³ of the room temperature air - air flow simply will not work that way.

I've never claimed otherwise. As I illustrated, with a kitchen open to the rest of the house (probably with outside air leaking into it all over the place), a 45 L/sec extractor would have to extract about 1,296 m³ of air (mixed heated and incoming cold) from a 280 m³ house before virtually all of the heated air had been removed.

Back to the cup of tea, with 1/2" left in the bottom and you want to rinse it.

I really don't think that your 'analogies', which are not really analagous to the situation we are discussing, are helpful, since it is far simpler and easier to just talk about the actual situation of interest.

It seems that you may have some difficulty in grasping the fact that if heated air is 'diluted' (due to mixing with cold incoming air) to a certain extent, then moisture or smells will also be diluted (due to mixing with incoming air without moisture or smells) to the same extent. Hence, if one wants to extract 'most' of the moisture/smells, one also has to extract 'most' of the heat that was contained in that moist/smelly air, regardless of how much or how little 'dilution' has taken place.

So, what I'm really reminding you of is that, in the case of, say, a bathroom full of water-laden, and also heated, air, then 'heated air' and 'water-laden air' are the same thing. Hence, if one wants to extract a certain proportion of the moisture (be that 'most', 'all', 50% or whatever), then one has to extract the same proportion of the heat. Put another way, if one doesn't extract 'much of' the heat, nor will one extract 'much of' the moisture or smells.

Furthermore, if there is substantial dilution of heated air within the house by cold air entering from outside of the house, then there will obviously be a consequential substantial reduction in temp inside the house - something which you don't seem to think could/would happen, despite your view that 'dilution' (presumably 'substantial dilution', for your argument to work) is a crucial factor. Conversely, if there is not much reduction in indoor temp, even in the absence of heating (which seems to be your belief) then there cannot have been much dilution with cold incoming air - again, despite your belief in the importance of that dilution.

In terms of the extent/.impact of 'dilution', a 15 L/sec running for, say 30 minutes, will extract 27 m³ - which is less than 10% of the air in a 280 m³ house. That means that, even after 30 mins of extraction, and assuming 'complete mixing', the amount o heat in the air being extracted will have fallen (due to dilution) by less than 10%.

I suspect that one reason why people tend not to be aware of these effects of extraction are that, if it is cold outside, people will not often have extractors running when the house is not being heated. They will therefore probably not experience any fall in indoor temp, even though their heating system is having to utilise energy in order to compensate for the heat loss due to extraction.

Kind Regards, John

 

[Harry Bloomfield]

I really don't think that your 'analogies', which are not really analagous to the situation we are discussing, are helpful, since it is far simpler and easier to just talk about the actual situation of interest.

It works fine for me as a reasonable analogy, and we only differ on the extent of the heat loss due to air extraction, so as said earlier I I think it best to just drop the thread.

 

[JohnW2]

It works fine for me as a reasonable analogy, ...

The problem with your 'Method 2' is that it completely ignores the 'extraction'. A closer analogy would be to have a tube sucking liquid out from the very bottom of the cup whilst water trickled into the cup from a tap at the same rate as liquid was being sucked from the bottom of the cup.

and we only differ on the extent of the heat loss due to air extraction,

That's a very big 'only', given this whole discussion has been about the 'extent' of heat loss due to extraction

In fact, to be more accurate, the discussion has been implicitly about the rate (not 'extent') of heat loss - since, in the absence of heating, with any scenario heat will continue be lost so long as the extractor is running and, if the extractor runs for long enough, the indoor temp will eventually approach that of the outdoor temp.

so as said earlier I I think it best to just drop the thread.

As I said, that's up to you.

Unless I've done something wrong, I've demonstrated what happens with 'immediate complete mixing' - and if mixing is any less than that, then heat will be lost more rapidly than I have illustrated. Hence, unless you can identify some flaw in what I've done, I think one has to accept that what I've illustrated is the lowest possible rate of heat loss.

Kind Regards, John

 

[Harry Bloomfield]

The problem with your 'Method 2' is that it completely ignores the 'extraction'. A closer analogy would be to have a tube sucking liquid out from the very bottom of the cup whilst water trickled into the cup from a tap at the same rate as liquid was being sucked from the bottom of the cup.

The cup will overflow, which is equal to the extraction.

 

[JohnW2]

The cup will overflow, which is equal to the extraction.

OK - but 'so what?'

IF one assumes 'rapid and complete mixing/dilution' (which you seem to do, in both cases), then what you describe would be a reasonable analogy of the extractor/air situation - but that (your 'Method 2') is 'how it is' (i.e. "a fact of life/physics") - and there is no alternative I can think of.

In the case of an extractor extracting air, there is (IF rapid/substantial mixing/dilution is occurring) no equivalent to your 'Method 1' (first 'tip out' the great majority of the tea, before finishing the removal of the small tea remnants by dilution (and 'extraction') with tap water.

However, if you abandon your belief that mixing/dilution is an important factor, then your 'Method 1' does have a reasonable equivalent with air/extraction. If there is little/no mixing/dilution, and the point of extraction is close to the source of the moisture/smells/whatever, then nearly all of the initial extraction will be of undiluted moist/smelly (and also 'heated') air (analogous to starting by 'tipping out' most of the tea) - which is the most desirable/efficient situation.

What one cannot escape from is the fact that moist/smelly air is heated moist/smelly air. Hence, if one extracts X litres of moist/smelly air, one has also extracted X litres of heated air (to be replaced by cold outside air entering the house somewhere).

I suppose everyone else got bored stiff with all this very early, but I do wish some others would participate, since all we have had so far is a dialogue between the two of us, who seem to hold very different views of the matter! Anyone?

Kind Regards, John

 

[Harry Bloomfield]

I suppose everyone else got bored stiff with all this very early, but I do wish some others would participate, since all we have had so far is a dialogue between the two of us, who seem to hold very different views of the matter! Anyone?

I suspect anyone with any sense, will have gone off to make a cup of tea long ago