Re: OT - Geothermal Heat issue...?

In article <pvvmk310vhhsrc7d51susg9akla7fb6f2s@xxxxxxx>,
Kenneth <usenet@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
On Mon, 26 Nov 2007 16:44:48 -0800, "Lew Hodgett"
<lewhodgett@xxxxxxxxxxxxx> wrote:

In a nutshell, thermal inertia.

Once the system is balanced, it requires minimum energy to maintain
the balance.

Change the set point to a lower level, remain there for a while, then
return to the higher level requires a lot of thermal work.

Heat intensive industries such as steel, refineries, etc, run 24/7 for
just this reason.

Lew


Hi Lew,

I don't deny it... I just don't understand it:

(Though there may be parallels to industrial applications,
I'll stick with home heating for my example.)

For a given outside and inside temperature, the house loses
a constant amount of heat per hour, and that amount must be
replaced if we are to keep the internal temperature
constant.

If the internal temperature of the house is allowed to drop,
two things happen. First, there is the direct energy savings
because it takes fewer BTUs to keep the house at the lower
temp;

_THOSE_ BTUs are just 'deferred spending'. you spend exactly
that amount to raise the temp back to the original setting.

but perhaps less obviously, the rate of heat loss to
the outside environment is decreased. (Because the greater
the temperature differential, the more rapid the rate of
equalization.)

That is the -totality- of the energy savings -- the lowered losses.
from the reduced temperature.

So, for the eight hours or so that the interior temperature
was lowered, there are savings for two reasons: We are
providing less heat to the house, and we are losing less per
hour of what heat we do supply.

FALSE. you are double-counting the same saving there.

to maintain any system 'at equilibrium', all you do is replace
the losses. if you are maintaining a lower equilibrium point,
the 'savings' are exactly equal to the difference in the losses
at the two equilibrium points.

When we decide to go back to the original interior
temperature, at every stage (prior to reaching that temp)
the hourly rate of heat loss is something less than it would
be when we reach the desired internal temperature.

Now, of course, heating up the house those 10 degrees will
take a bushel of BTUs, but (unless I am way off here) that
would have to be fewer than those saved.

I well understand that the efficiency of the system goes
down as the well cools, but it seems to me that the
diminished efficiency, though regrettable, is more than
balanced by the savings at the lower temperatures.

authoritative answer: "it depends".

1) _how_much_ lower are the building thermal losses for the temperature
reduction employed?
2) _how_much_ less efficient is the heat plant as the -rate- of draw
increases?

Depending on the _quantitative_ answers to those two questions the
'savings' can 'net' to either a positive or negative result.

The exact answers to both questions will be specific to a particular
installation.

Getting an answer by 'science' is -very- messy. It's much simpler to
use the 'experimentalist' approach and simply 'measure' what actually
happens.

The building loss rates are relatively easy -- measure the required heat
input at both equilibrium points. It _is_ reasonable to assume that the
delta on the loss rates is the same for both temperature rising and falling,
so the cool-down, and warm-up phases effectively cancel each other.

The changing 'efficiency' of the heat plant is harder. You really need
to have a running monitor on the well-water temperature for that. (with
that you can tell 'when' things have 'recovered' from the excessive
consumption to raise the building back to the higher level.

Failing instrumentation on the water temperature, one can use outside
air temperatures as a -rough- basis for comparison. (it helps greatly if
you have historical power usage data [at stable inside temperature operation]
that you can correlate with 'heating degree days' for various periods)

If you have the above-mentioned historical data, you'll see that 'cost of
operation' goes up as the heat demand increases. both in absolute terms
and on a per unit basis.

Now, run the system for a while in 'set-back' mode. Total the 'heating
degree days', and the cost. See where that 'per unit' cost falls relative
to the same degree-days for stable temperature operation.

NOTE: this is all figuring 'cost' on the basis of "how cold it is outside"
-not- on a "per BTU of heat added" basis, so you have a direct comparison
of the 'efficiency' of the methods, and can reasonably predict what, if any,
the overall savings will be.


Heat pumps are, by their nature, less efficient, the larger the temperature
differential between the 'external' and 'internal' sides. And that efficiency
does degrade significantly with relatively small increases in that
differential.

Things will depend 'a whole lot' on the thermal conductivity of the external
heat reservoir, and how fast stuff in the vicinity of the 'radiator' there
recovers to equilibrium after a draw-down.

W/o extensive geological testing, that's hard to quantify.

I _would_ tend to believe that the designers/installers *DO* know what
they're talking about when they recommend stable (and not 'set back')
operation, counter-intuitive though it may seem.



.

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