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General: What constitutes a low carbon heat source?

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    • CommentAuthorJonti
    • CommentTimeAug 29th 2012
     
    Mike7,

    I realise that my fridge remains cold because I supply it with electricity to keep the temperature down. That is exactly the point!!!! IF YOU KEEP ON REMOVING HEAT HOW WILL THIS BE REPLACED IF THERE IS NO HEAT INPUT???????


    I made the first comment as a off hand verbal note and did not expect it to snowball into such a major topic.

    You are correct in that I am not an expert in this field but what strikes me the more I hear from people who like yourself are more expert is that non of you are able to give a definitive answer as to where the heat source comes from to replace the extracted heat. All the answers are very vague or as in your last sentence make no sense (common or grammatically).

    The point is it will alter the ground temp. and even if this is only by a small amount what effect might this have.


    Jonti
  2.  
    I think this is one of those disagreements looking for a generalised yes or no answer where there is none. The answer will depend on local conditions, depth of abstraction, duration and intensity of abstraction etc etc.
    • CommentAuthortony
    • CommentTimeAug 29th 2012
     
    the thing is it is a small effect in tiny spots on the land surface.

    The heat taken out is replaced by warmth from the sun.
    • CommentAuthorJonti
    • CommentTimeAug 29th 2012
     
    Posted By: tonythe thing is it is a small effect in tiny spots on the land surface.

    The heat taken out is replaced by warmth from the sun.


    But Tony is it? I agree with 'bot de paille' about the individual case. Your post however Tony shows you have either not read all the posts or have not thought about what was in them.

    SO simple question Tony, how does the warmth of the sun replace the heat extracted from under a building? Surely if it could my cellar would be as warm as any other shaded area but it isn't.

    Jonti
    • CommentAuthorbarney
    • CommentTimeAug 29th 2012
     
    It's cold inside your fridge because you continue to abstract heat and reject it outside the box - whereupon it returns to your fridge via the insulation - you're just stirring the heat and coolth about by adding energy.

    The cellar temp would vary - you just couldn't measure it accurately enough

    The heat source that equalises the temperature is solar radiation ?

    I'm not suggesting that the ground temp will not reduce - just that it will also recover - just expand the time frame and it's neutral excepting the energy you supplied to "pump" out the heat and raise it to a useable temperature

    regards

    Barney
    • CommentAuthorbarney
    • CommentTimeAug 29th 2012
     
    Is there any trace of damp in the cellar - are you effectively seeing evaporative cooling going on

    is the latent heat in that evaporating process broadly equal to the rate of heat inflow from surrounding ground - or is your cellar broadly the same temperature as the surrounding ground - so no discernible heat flow going on ?

    regards

    Barney
    • CommentAuthorJonti
    • CommentTimeAug 29th 2012
     
    Barney,

    my cellar is dry and it is similar in temp. to the ground it sits in. Its typical though that you answer the temp. point with that I must have not been able to measure it properly. I have been able to measure it very accurately.

    Rubbishing someone else's point is poor form.

    Jonti
    • CommentAuthorbarney
    • CommentTimeAug 29th 2012
     
    I certainly wasn't rubbishing your point - are you saying that the cellar temp was totally constant and close to ground temperature despite having a slated roof over and that the locale had discernable diurnal swings.

    If that's the case then isn't that proving to you that the cellar was both receiving heat from and rejecting heat to the "ground" locally depending on if it was leading or lagging ground temperature - which you must have seen vary to some degree

    Which aspect of temperature were you measuring out of interest - air temperature in the cellar, temperature of the floor or wall - did you measure humidity at the same time

    Regards

    Barney
    • CommentAuthormike7
    • CommentTimeAug 29th 2012
     
    Posted By: tony
    The heat taken out is replaced by warmth from the sun.


    Not all of it - splitting hairs perhaps but all the heat taken out plus the energy imported from elsewhere to drive the pump will escape from the building one way or another and fractionally warm the local environment, thus transferring more heat to the ground than ....

    Paul, how deep are your boreholes?
  10.  
    <blockquote><cite>Posted By: mike7</cite>Paul, how deep are your boreholes?</blockquote>

    One is 420 feet, the other is 500, so still technically "shallow enough" that the loop temperature is about at the annual average air temperature (just over 6C). Once they're running, the entering water temperature is a bit less than that, but that's as much to do with the rate of transfer of heat to the (chilled - in heating mode) circulating fluid than a measure of the actual temperature of the ground. There's programs available to calculate the right length of loop for a given amount of heat extraction (bot rate and total annual amount) for various soil conditions.

    I just found a report about using GSHPs in Alaska (!) and there is discussion of permanently freezing the ground:

    http://www.uaf.edu/files/acep/Ground-Source-Heat-Pumps-in-Cold-Climates.pdf

    Paul in Montreal.
    • CommentAuthormike7
    • CommentTimeAug 29th 2012
     
    Thanks Paul. Thinking of Tom's point about deep borehles, I've just made this calculation - so crude an approximation I'm not sure how useful it is, but here goes:-

    Suppose you have a house on a plot of 1000m^2 extracting 10,000kWh/yr from a borehole to supply a heatpump, and you are surrounded by neighbours doing just the same. There's no transverse underground water flow, so I'll take that to mean that all the heatflows must only be vertical. What might the themal gradient look like for the surface temperature to conduct enough heat down to sustain the supply of heat to be extracted from the depths?

    To make it simple enough lets have a crude approximation that the gradient is the same all the way down to say 50m, and all the heat travels down to that depth. Yep, that's crude.

    Let's say the ground is something like granite, having a themal conductivity of 4W/mK.

    The heat flux taken as constant through the year (which it would be below a few metres from the surface) would be :-

    10,000kWh/ 9000hrs per year = 1.1kW over the whole plot
    or 1.1W per m^2

    That would give a temperature difference from 50m down to the surface of:-

    Delta T = 1.1W x 50m divided by 4W/mK

    or 14 degrees C

    which is interestingly in the range of credibility, if a bit chilly. Adjust the figures I've chosen up or down and the answer will change pro rata.
  12.  
    Interesting calculation, Mike. But I think you need to consider the specific heat capacity of that cylinder of granite from which you're extracting heat out of the middle of. I calculated that I have about 44500 tons (of limestone in my case) that lies within my property - and this had more than enough heat stored to supply all my needs (approximately 36,000kWh)

    Paul in Montreal
    • CommentAuthorEd Davies
    • CommentTimeAug 29th 2012
     
    Suppose the plot's 100 m x 100 m and the bulk of the extraction happens from a volume with a depth of 100 m. The density of the material is 2.5 tonnes/m³ and it has a specific heat capacity of 1 kJ/(kg·K). Multiplying those all up gives a heat capacity of 1e6 m³ * 2.5e3 kg/m³ * 1e3 J/(kg·K) = 2.5e12 J/K.

    1 kWh is 3.6 MJ so Paul's extraction of 36'000 kWh/year is 3.6e6 * 36e3 J/year = 129.6e9 J/year.

    Therefore the ground cools at a rate of 129.6e9 J/year / 2.5e12 J/K = 0.05184 K/year. You might start to notice that after 20 years or more.

    Of course, in the real world the cooling isn't evenly distributed. That's a bit trickier to work out.
    • CommentAuthortony
    • CommentTimeAug 29th 2012
     
    and even more difficult to notice
  15.  
    Posted By: Ed Davies1 kWh is 3.6 MJ so Paul's extraction of 36'000 kWh/year is 3.6e6 * 36e3 J/year = 129.6e9 J/year.


    I assumed I could only take heat out from the cylinder of rock under my property - so my figures were a bit more pessimistic - but I found the insolation data for that area to give me 7x the heat that I took out and I didn't account for the heat I put back in the summer (as it's only about 1/10 of the heat I take out). I felt confident, though, that I wasn't creating a massive subterranean iceberg!

    Also, I guesstimated my heat extraction by taking my yearly electricity bill and assuming that about 12000kWh of it was for running the heating at a COP of about 3 - the true COP might be closer to 2.7, but that depends on how much auxiliary heating I need in a given year - this past winter was next to nothing due to various house improvements and a mild winter. But it puts me in the right ballpark of about 30-36,000kWh.

    Paul in Montreal.
    • CommentAuthorJonti
    • CommentTimeAug 29th 2012
     
    Just tried to post a long eplanation but the d+~#*d post was lost when the site did not allow me to add.

    Barney,

    I have never said that temperature does not radiate through the ground just that if you remove heat and there is no compensating input then the ground temperature must drop at and close to the point of extraction. So if you remove through a heat pump and do not give back because no A/C needed where is the compensation coming from?

    Also can you define 'locale' for me as this is the village pub in certain places I have lived though it would experience daily changes in temperature depending on the footy results I doubt this is what you meant:wink:

    Jonti
    • CommentAuthormike7
    • CommentTimeAug 29th 2012 edited
     
    Thanks Paul - I did consider vol sp ht but left it out to keep it simple - I was also really only looking for an idea of the vertical thermal gradient once equilibrium is reached. Ed's calc sugggsts that might take some time even for my smaller plot size.

    From Jonti's point of view the interesting bit might be that the temperature gradient I get is of the order of one third of a degree C per metre different than it would be without the heat extraction. Note that's with a conductivity of 4W/mK. Wet soil is probably less than that so the gradient there would be proportionately higher. Town gardeners please note.
    •  
      CommentAuthorSteamyTea
    • CommentTimeAug 30th 2012
     
    Heating and cooling is not linear, look up Newton and one of his most important laws, then look up Fourier for a quick way to apply it in practice.
    • CommentAuthorEd Davies
    • CommentTimeAug 30th 2012
     
    Posted By: JontiJust tried to post a long eplanation but the d+~#*d post was lost when the site did not allow me to add.

    I often see that. I think there's a timeout; probably it's an anti-spam thing. Just press “Add your comments” again and it should work.
    • CommentAuthorEd Davies
    • CommentTimeAug 30th 2012
     
    Posted By: SteamyTeaHeating and cooling is not linear, look up Newton and one of his most important laws,

    But here there's a simplification of assuming a constant temperature so it is linear.
    • CommentAuthorbarney
    • CommentTimeAug 30th 2012
     
    Well jonti, I think the assumption is that the input is solar irradiation - which need not be above the point of extraction as heat will flow 3 dimensionally. I'm not sure if I can find the article now, but a reverse of this effect was experienced when the original billingsgate fish market closed. Many years of ice storage had frozen the ground (to some depth) - once the ice was removed, the ground thawed - you don't get a lot of bare earth around billingsgate - so the heat flow came from somewhere - it had to because a temperature gradient existed

    For locale, i guess your general surroundings - ie the pub was still there from summer to winter and that summer and winter had discernable temperature differences

    Regards

    Barney
    • CommentAuthorJonti
    • CommentTimeAug 30th 2012 edited
     
    Thanks Ed, will try that next time.

    Barney,

    I do not know if I am not making myself clear or you are just being contrary. Using your example of the Billingsgate Fish Market where the ground froze because of the ice which backs up my point. The ground did not thaw until the ice was taken away. You seem to agree with this and this is what I have been saying (removing heat from the ground will cause a lowering of the ground temp).

    Jonti
    • CommentAuthormike7
    • CommentTimeSep 2nd 2012 edited
     
    <blockquote><cite>Posted By: barney</cite>Well jonti, I think the assumption is that the input is solar irradiation - </blockquote>

    That's what I thought, too. But on reflection perhaps the main source of replacement heat is via the temperature of the immediate surroundings - the air temperature and the local radiant temperature, ie the local weather and climate. (Of course these temperatures are what they are ultimately because they're at a level where Earth's radiation out balances the Sun's radiation in, but overall that's a steady state (or was!). Small changes in absorbtion and re-radiation of solar heat in the immediate cooled area may play a part, but generally a minor one IMO.

    PS. Not sure I've managed to put this at all well. If you can see what I'm getting at and can put it better, do have a go.
    • CommentAuthorJonti
    • CommentTimeSep 2nd 2012
     
    I would agree with your assessment Mike7 but as of yet nobody has addressed my main point which is what is the effect on the ground temperature when heat is extracted at a faster rate than it is replaced. My main gripe with the point of view of people such as Barney is they seem to me to basing their entire argument on what happens when the heat extraction stops. The point being as far as I can tell the nett extraction year on year exceeds the replenishment of heat.

    A drop of a small amount in the ground temp. might seem of little importance but what effect would it have on micro-life in the ground?

    Jonti
    • CommentAuthormike7
    • CommentTimeSep 3rd 2012 edited
     
    Continuing my crude calculations a few days ago, where the figures I chose resulted in a heat flow into the ground of 1.1W/m^2:-

    If the replacement heat is entirely by radiation, the surface temperature drop needed would be about 0.22 C.
    If entirely by conduction across a boundary layer of air 1 cm thick it would be about 0.45 C

    (For the radiation, deltaT = 1.1W divided by (4 times the cube of the temperature in Kelvin times the Stefan-Bollzmann constant)

    For conduction, deltaT = 1.1 W times the boundary layer thickness divided by the conductivity of air.)

    .....I think....

    Edit : Boundary layer mre likely to be 1mm - see later post.
    • CommentAuthorSeret
    • CommentTimeSep 3rd 2012 edited
     
    Posted By: JontiI would agree with your assessment Mike7 but as of yet nobody has addressed my main point which is what is the effect on the ground temperature when heat is extracted at a faster rate than it is replaced.


    Are you talking about small-scale boreholes, or large geothermal power? The latter will be drawing heat from quite deep, and AFAIK there's not a lot going on down there microbially (although there is some). Geothermal plants do often extract heat faster than the rocks can replace it, which is why it's incorrect to class geothermal as a truly sustainable source much of the time. The economics have traditionally encouraged extraction at excessive rates, although they do seem to be behaving themselves a bit better.

    Basically the ability of a geothermal system to replace extracted heat is determined by the thermal and hydrodynamic properties of the rocks below, while extraction near the surface benefits from solar too. Apples and oranges.

    Adding to the confusion is that in some places like NZ it's quite possible to have a small scale borehole in your garden that's 100% geothermal. Although the authorities try to discourage that, as it depletes the field in an uncontrolled way.
    • CommentAuthorJonti
    • CommentTimeSep 3rd 2012
     
    Seret,

    I am talking more about the smaller type though your example of excessive extraction seems to prove my point. As with most things it is my belief that a certain density of GSHP are sustainable but in an urban environment not for all households.

    Jonti
    •  
      CommentAuthorSteamyTea
    • CommentTimeSep 3rd 2012 edited
     
    Many of the geothermal installation that were depleted in the USA used naturally occurring steam, it was the steam that got used up (taking too much water out of the rocks). This water eventually gets replaced, turns to steam, and the turbines can start up again.
    There really does have to be a distinction between all the different technologies that are clumped together. GSHP as we know them (horizontal or vertical collection) are really just using solar, Hot Rocks (not that it works) is using nuclear decay near the surface (lots around here but it does not make it warmer). You have to put a scale onto it to get a better picture. You are not comparing just two surface areas (absorption area and emitter area).
    But the physics says that if you extract heat from one place it will cool. If at the same time you are inputting energy into the system then it will cool at a different rate (slower) and find a new equilibrium point. This is not a linear equation but an exponential one (think radiators in a cold room the delta T may start of at 70 °C but end up at 50 °C as the room warms).
    If you want to test this at home, get an electric kettle a thermometer and stopwatch. Leave the lid of the kettle open, turn the kettle on and not how long it takes for each degree of temperature rise. You know that when it boils it is 100 °C, but because the lid is open it will not turn off, so energy is going in but the temperature is not rising. That is the new equilibrium.
    If you want to get really clever, log the cooling as well, then take one from the other and plot that.
    • CommentAuthorJonti
    • CommentTimeSep 3rd 2012
     
    SteamyTea,

    your analogy is a good one but seems to me to deal with inputting in to a system. What happens when you remove at a higher rate. Take a reservoir where you are removing water faster than water flowing into it. At some point it will be empty. What would be required in this case would be to only extract at the same rate as you input.

    Jonti
    • CommentAuthorSeret
    • CommentTimeSep 3rd 2012
     
    The reservoir analogy isn't quite accurate, because when you're extracting heat the more of a temperature delta you build up, the faster the replacement heat flows in. Different rocks provide a different amount of resistance to heat flow, so if you extract quickly from low permeability rock the temperature will decrease quickly. Eventually however it'll reach a point where the temp delta is enough that the heat flows in at the same rate you pull it out. At that point your temperature should be pretty stable. That's backed up by measurements in actual boreholes that do show an initial reduction in temperature, then they stabilise.

    That's a bit simplistic, because there's all sorts of stuff going on under the ground that geologists get very excited about, but you get the idea.

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