In article ,
Stan Brown wrote:
My understanding is that the furnace uses less fuel overall to keep
the water in the hating pipes at a constant temperature than to let
it cool down by 10 or 15 degrees in the day time (when I'm at work)
and the night (when I'm in bed) and then reheat it.

My understanding is that this is true for hot-water heat but not for
forced-air, since it takes much less energy to heat air than water.

In general, if you have something that you need to be above ambient
temperature at time 1 and at time 2, it takes more energy to keep it at
that temperature than it does to let it cool after time 1, and heat it
back to the desired temperature at time 2.

Here's how to see that. You need to know two facts of physics first.

(1) Imagine the object surrounded by something that can measure the
heat energy entering or exiting the object. You'll find that when heat
energy leaves the object, the temperature goes down. When heat energy
enters, the temperature goes up. You'll also find that there is a
conservation law at work here. If the temperature is at a given
temperature, and a given amount of heat energy leaves, then to get the
object back to the original temperature, you have to put that amount of
heat energy back.

Basically, the temperature of the object is the integral of the heat
flow over time.

The important point here is that the temperature just depends on the net
change in heat energy of the object.

(2) The rate the object loses heat to its surroundings goes up as the
temperature difference goes up. If the ambient temperature is, say, 50,
and the object is 70, it will lose more heat energy per second than it
would if the object temperature were 60.

Putting these two together, let's do a thought experiment. We have two
objects, both at, say, 70. The ambient temperature is 50.

Object 1 we keep at a constant 70. Object 2 we allow to cool, until
just before we need to use it again, and then we heat it back to 70.

For object 1, it is at a constant 20 above ambient, so is losing heat at
a constant rate. So, the total energy lost is the amount it loses per
second at 70 times the number of seconds between time 1 and time 2.
That's how much total energy our furnace has to put into the object to
keep it at a constant 70 from time 1 to time 2.

For object 2, it starts out at 20 above ambient, so in the first second
it loses about as much energy as the first object. But we are letting
it cool, so it gets colder. That slows the rate of heat loss slightly.
In the second second, it loses slightly less energy than the first
object. The advantage grows as time goes on. Finally, time 2
approaches, and we have to use the furnace to heat the object. The
amount of heat we have to supply is exactly the amount it has lost since
time 1, which is LESS than object 1 has lost.

So, strictly from the viewpoint of energy required to have an object at,
say, 70 at time 1 and at time 2, with an ambient temperature of 50, much
less energy is required to let the object cool between time 1 and 2 and
then heat it back to 70 at time 2, than just keeping it at a constant 70.

However, it is possible that there could be other considerations in
practice. If you had some kind of furnace that takes a while to reach
full efficiency after startup, that could change things, depending on
how much time is between time 1 and time 2.


--
--Tim Smith