how big a lake?

[derekmt]

My B in law is looking to do a G source heat pump but he has a small lake/pond, How  big practically does it need to be to heat a  3 bedroom house with UFH?
His lake  is  about 25m by 10m by 2m. You get a lot of heat lowering a few hundred tonnes of water a couple of degrees. If he needs to get a digger on site would he be better enlarging the lake than burying 400m of pipe?

[Jeremy]

My B in law is looking to do a G source heat pump but he has a small lake/pond, How  big practically does it need to be to heat a  3 bedroom house with UFH?
His lake  is  about 25m by 10m by 2m. You get a lot of heat lowering a few hundred tonnes of water a couple of degrees. If he needs to get a digger on site would he be better enlarging the lake than burying 400m of pipe?

Is there water flowing into and out of the lake?  If so, then the chances are it'll be OK.  If not, then you may need to do an in-depth assessment to make sure that there isn't a chance of freezing the lake at depth.  The latter point is key, as there may be a significant risk to flora and fauna in the lake, which may or may not be a legal issue.

In crude terms, if the lake was a rectangular box of the dimensions given and starts the heating season at around 12 deg C and ends the heating season at around 4 deg C, then ignoring the large heat loss that would normally occur to the ground and surrounding air, and ignoring any additional solar input through the heating season, there would be about 16,720 MJ of heat available.  If we assume that about half this will be lost naturally, then there could be around 8,000 MJ of heat energy the pump could use over the heating season.  This is about 2,200 kWh, so not a great deal - enough to supply about 1 kW of heat all day and night for 90 days.  In practice I think the actual heat recovery from a lake of this size might be a bit less than this, as the guesstimates above are pretty rough and crude.

Edited to add:
The rough calculations I used above (which may be in error!) were:

Specific heat of water = 4.18 J/g.K
Volume of lake = 25 x 10 x 2 = 500 m² so mass of lake = 5 x 10^8 g
Energy per deg K in lake = 5 x 10^8 x 4.18 = 2.09 x 10^9 J.K
Energy for an 8 K drop = 2.09 x 10^9 x 8 = 1.672 x 10^10 J = 16,720 MJ = 4,644 kWh

[Baz]

Think of it in terms of the catchment area of a pipe burried only 1m deep as it is losing heat to air faster than when underground. Then research minimum recommended separation of such pipes to indicate the area a pipe collects from. I think a lake captures a lot of heat from below but loses huge amounts to air and by radiation. Rather than digging the best bet would be to cover it with bubble wrap in winter.
Heat capacity is about 600KWh per degree, and freezing it about 80 times that.
You shouldn't be using more than 60kWh per day. Worst case would be when you have say 1cm of ice through natural cooling then you might add 3mm to the ice per day. Nature does it far faster than that, so your effect is rather unimportant.
It would be an idea to start measuring the temperature of the lake top & bottom immediately to understand the natural fluctuations wrt air and general weather.
One idea if you are worried about the pond ecosystem is to put a underwater wall across halfway eg by a line of floating logs dangling a weighted sheet.

[dhaslam]

It could be used for skating  but plants and fish would have to be removed every winter, not really the best use for a small pond.

There is such thing as a  solar pond, it would need to be large  and couldn't be used for  fish or water plants so would be a bit of an eyesore but would be very good as a heat source.   


http://en.wikipedia.org/wiki/Solar_pond

http://www.navitron.org.uk/forum/index.php?topic=3863.15;wap2

[derekmt]

It could be used for skating  but plants and fish would have to be removed every winter, not really the best use for a small pond.

...

what do you base that on?

[derekmt]

.. then ignoring the large heat loss that would normally occur to the ground and surrounding air, and ignoring any additional solar input through the heating season,

dont think you can ignore 250 sq m  of air contact and solar gain  and 350+ sq m of ground contact...

I'm think taking a tiny fraction of the bidirectional heat flow through this 250 Sq m pipe

[Jeremy]

.. then ignoring the large heat loss that would normally occur to the ground and surrounding air, and ignoring any additional solar input through the heating season,

dont think you can ignore 250 sq m  of air contact and solar gain  and 350+ sq m of ground contact...

I'm think taking a tiny fraction of the bidirectional heat flow through this 250 Sq m pipe

I don't think you can ignore it, either, which is why I stated later in that same post that you might only be able to use half of the available heat because of the losses to the air and ground and even then that might be being optimistic.

[dhaslam]

It could be used for skating  but plants and fish would have to be removed every winter, not really the best use for a small pond.

...

what do you base that on?

I have a pond just about that size.  It will freeze on the surface when  the  the air temperature falls  just below freezing but with very thin ice.    However if the water is already  cooled from the bottom  by the  heat pump if would form thicker ice on top.  It is the heat from the lower part of the pond that slows the surface freezing.  

The bottom of a two metre depth pool  will  be  only abot 5C  warmer on average, (see page 17  in the link below).  At night there would be a bigger temperature gradient  but in any case the bottom would easily  cool by 5C  over a month or so.    The heat extraction would only need to be a few hundred kWh per degree for the reduced   effective volume.

http://www.mcquay.com/mcquaybiz/literature/lit_systems/AppGuide/AG_31-008_Geothermal_021607b.pdf  

[Jeremy]

Everything I've read in the past few days (I'm looking at using a water source heat pump with boreholes) suggests that if using a lake it either needs to be pretty large or better have a flow of water through it, even if only from underwater springs.  Even a modest flow of groundwater into a lake from a stream, or better, a spring, will provide a significant amount of energy.

[derekmt]

my thinking is this :
60Kw/day = 2500w (water mass averages it all out)
or 10w/ sq m

if a bad winter gives 300mm of ice at -10c air temp that implies a heat source of 66w/sq m (conductivity of ice)
if this reduced to 56w because I'm taking 10w it needs another 60mm of ice to insulate it.

if a bad winter gives 600mm of ice at -10c air temp that implies a heat source of 33w/sq m (conductivity of ice)
if this reduced to 22w because I'm taking 10w it needs another 300mm of ice to insulate it.

if its really bad the solution is to inject air under the ice.