|
|
re: question about heating a home
14 dec 1995
paul milligan exemplifies:
> let's create an example - take 1 lb of water, in a container with 0 r
>value. heat it up from 50 to 100 degrees. that takes 50 btu's, right ?
i guess you would have to put the whole container in an oven to heat it up,
if it had 0 r-value... heat sinks have close to 0 r-value... the r-value
of a slowly moving air film near a smooth surface is about 2/3, in english
units, which would be most of the overall r-value of the container, if it
were made of a thin piece of metal, or a paper cup.
> with no setback, there will be an identifiable rate of heat loss to the
>surrounding 50 degree room.
yes...
>you will need to continuously add 'x' btu's to that water to keep it at 100.
where x is infinity :-) breeng zee beeeg torch...
suppose the container is a cubical box with a lid, with an edge length of
s = 4", ie 1/3 foot. if the cube is sitting in midair, its resistance
will be about 6 s^2/(2/3) = 1 "ohm," so we need to add 50 btu/hour to
keep it at 100 f in the 50 f room.
>let's say it wants to cool down at 25 degrees per hour.
that would make it want to be slightly smaller, with a thermal resistance
of 2 ohms and an edge length of 2.83". (a cube like that would only hold
0.8 pounds of water, but let's ignore that.)
> _with_ setback, there will also be an identifiable rate of heat loss.
>it will start at 25 degrees / hr. one hour later, with the water at 75 in
>your 50 degree room, your heat loss rate is cut in half. one more hour,
>and your heat loss rate is zero, as the water and the room are both at 50.
the time constant of this heat battery is 2 ohms x 1 lb = 2 hours :-) after
time t has passed, the water temp will be t(t) = 50 + (100-50) exp(-t/rc).
after 1 hour, the water temp will be t(1) = 50 + (100-50) exp(-1/2) = 80.3 f.
after 2 hours, the water temp will be t(2) = 50 + 50 exp(-1) = 68.4 f.
if we make the cube an epdm-rubber-lined plywood box, 8' on a side, with a
foot of insulation, ie r-40 sides, with a 6' cube of water weighing 13,500
pounds inside, then rc = 40/(6x8x8)x13500 = 1872 hours or 78 days. if the
water began at 100 f, after 1 hour in a 50 f room it would have a temp of
50 + 50 exp(-1/1872) = 50 + 50 x 0.99947 = 99.97 f. it would cool to 68 f
in t = -rc ln((68-50)/(100-50)) = 1913 hours or 80 days...
we might make the south face of this cube dark and glaze it with thin plastic
over an air gap, and add a $100 10" x 10" honeywell 64ls damper and 6161b1000
motor that opens to the inside of the cube when the sun shines, and replace
the epdm rubber tank with 18 sealed plastic 55 gallon drums full of water...
the average outdoor temp in december in philadelphia is about 32 f. on an
average december day, 1000 btu/ft^2 of sun falls on a south wall, in about
6 hours. how hot will the cube get, sitting outdoors, or inside the house,
behind a window?
i agree with you, paul, setbacks do save energy.
nick
|
|
