Posted by Morris Dovey on February 19, 2009, 2:08 pm
A number of people have asked me about heat storage within the context
of air-heating solar panels, and I'm trying to develop an reasonable
understanding - and to grasp some of the physics (thermodynamics?) involved.
Jeff's question elsethread prompted me to try to work up some numbers
and methods - and I'd appreciate some critical review of my logic.
I've been thinking of temperature as a measure of energy per unit mass,
so I knew I wanted to be able to determine the mass of the air in a 30'
x 40' x 10' structure. I did a couple of web searches and found:
density of air at sea level = 0.0767 lb/ft^3 [1.229 kg/m^3]
and I calculated
volume of air = 30 ft * 40 ft * 10 ft = 12000 ft^3 [339.8 m^3]
which allows me to figure
mass of air = 12000 ft^3 * 0.0767 lb/ft^3 = 920.4 lb [417.486 kg]
(Yes, I know that a pound isn't a mass unit - but bear with me or
validate with the proper metric units. Is there a name for a "pound"
when we want to use it as a unit of mass?)
I was wondering what mass of concrete it would take (at the same
temperature) to store however much heat there was in the 12000 ft^3 of
air, and I seemed that it would take the exact same mass.
mass of concrete = mass of air
My next question was to what depth would I need to warm the concrete
slab in order to store that much heat? I recognized that unless I
restricted myself to a non-dynamic context I'd get lost, so let me
stipulate a purely static context.
I googled again to come up with
mass of concrete = 149.8 lb/ft^3 [2400 kg/m^3]
so I could calculate
volume of concrete = 920.4 lb / 149.8 lb/ft^3
= 6.144 ft^3 [0.174 m^3]
Since the area of the floor is 1200 ft^2, if the heat were distributed
evenly over the entire floor, then the depth of the concrete needed to
store that heat would be
depth = 6.144 ft^3 /1200 ft^2
= 0.005120 ft = 0.06144 inch = 1.56 mm
At this point I leaned back in my chair and asked myself if the numbers
seemed reasonable, and had to confess that I just don't know - and it
seems like a really good time to get some knowledgeable help?
If I screwed up, please point out where and how...
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/
Posted by Morris Dovey on February 19, 2009, 2:13 pm
Morris Dovey wrote:
> I googled again to come up with
>
> mass of concrete = 149.8 lb/ft^3 [2400 kg/m^3]
Oops - should have been
density of concrete = 149.8 lb/ft^3 [2400 kg/m^3]
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/
Posted by Ecnerwal on February 19, 2009, 5:12 pm
Morris,
You need a property/number called specific heat.
In english (mostly american these days) the mass unit is lbm, or the
godawful slug, btw - lbm is more common.
A btu is the amount of heat to raise one lb water 1 degree F.
Air is easier to heat than water - its specific heat is 0.24, per my
notes as incorporated into my spreadsheet where I figure out how much
heat I need. I have a slightly different (not hugely) mass figure of
13.9 cubic feet of air per pound in my notes, so I'd have 863 lbs of air
for your 12000 cubic feet.
Thus, to heat 863 lbs of air one degree F takes only 207 btu.
863 lbs of water would take 863 btu.
Concrete, per a quick web search, seems to be a specific heat of 0.18
(be sure to get the "Btu specific heat" - one of the metric measures is
wildly different, one seems to be about the same)
863 lbs of concrete takes (or gives, if storing) 155 btu.
So, you need more mass of concrete than mass of air for the same heat
storage/release.
If in reasonable containers, water is a nice thermal storage medium,
physical size to number of btus stored. Concrete, not so much. 1970's
tech was heavy into concrete, and did not work well.
Using the floor adds other problems _ it can't get too hot, or the place
is uninhabitable. It may not be well insulated, so you may lose lots of
heat. To a large extent, you need to think about "heat to raise the
whole slab so many degrees" / heat given off by the whole slab cooling
so many degrees" rather than heating a thin layer of it - which is
connected to the rest of it...
One of the best approaches to hot air heat storage is the thermal
capacitor which Nick Pine used to go on an on about - it's another guys
design/patent (probably expired by now) - Norman ...?
Essential idea is to have a insulated high temperature heat store that
is insulated from the living space, and run that up to 150-180 degrees,
if enough heat is available. When heat is called for in the space, run
air through the store if the collectors are not producing heat. Put your
hot water heater in it and get solar hot water form it as well...
--
Cats, coffee, chocolate...vices to live by
Posted by Morris Dovey on February 19, 2009, 6:56 pm
Ecnerwal wrote:
> Morris,
>
> You need a property/number called specific heat.
>
> In english (mostly american these days) the mass unit is lbm, or the
> godawful slug, btw - lbm is more common.
Thank you - lack of vocabulary can be a problem for me. I'll note the
"slug" and /use/ "lbm" in the future. :)
> A btu is the amount of heat to raise one lb water 1 degree F.
Yuppers. I was kinda hoping to sidestep around energy units, but see
that a nod in that direction is necessary to ensure that I'm working
with only one "flavor" of specific heat.
> Air is easier to heat than water - its specific heat is 0.24, per my
> notes as incorporated into my spreadsheet where I figure out how much
> heat I need. I have a slightly different (not hugely) mass figure of
> 13.9 cubic feet of air per pound in my notes, so I'd have 863 lbs of air
> for your 12000 cubic feet.
Your number may be better - I just grabbed the first I found, and ran
with it.
> Thus, to heat 863 lbs of air one degree F takes only 207 btu.
>
> 863 lbs of water would take 863 btu.
>
> Concrete, per a quick web search, seems to be a specific heat of 0.18
> (be sure to get the "Btu specific heat" - one of the metric measures is
> wildly different, one seems to be about the same)
>
> 863 lbs of concrete takes (or gives, if storing) 155 btu.
>
> So, you need more mass of concrete than mass of air for the same heat
> storage/release.
Aha! Then I care about the /ratio/ of the materials' specific heat
values. That was absent from my calculation (but won't be in the
future!) Thank you!
> If in reasonable containers, water is a nice thermal storage medium,
> physical size to number of btus stored. Concrete, not so much. 1970's
> tech was heavy into concrete, and did not work well.
>
> Using the floor adds other problems _ it can't get too hot, or the place
> is uninhabitable. It may not be well insulated, so you may lose lots of
> heat. To a large extent, you need to think about "heat to raise the
> whole slab so many degrees" / heat given off by the whole slab cooling
> so many degrees" rather than heating a thin layer of it - which is
> connected to the rest of it...
Right. This is the dynamic aspect that I wanted to overlook until I got
a better handle on the static one. As a non-physicist I feel somewhat
constrained to "baby steps" (and I'm procrastinating on my DiffEq review
- it's been a /long/ time...)
> One of the best approaches to hot air heat storage is the thermal
> capacitor which Nick Pine used to go on an on about - it's another guys
> design/patent (probably expired by now) - Norman ...?
>
> Essential idea is to have a insulated high temperature heat store that
> is insulated from the living space, and run that up to 150-180 degrees,
> if enough heat is available. When heat is called for in the space, run
> air through the store if the collectors are not producing heat. Put your
> hot water heater in it and get solar hot water form it as well...
I've been working on a passive water-based storage system (much as you
described), but it's not ready for "prime time" yet - resources are
already stretched to the limit.
Nick and I seem to come at this stuff from opposite directions. He seems
to focus on calculation, with (perhaps) practical results down the road;
and I focus on practical results and end up wondering why things work so
well. If we could ever stand to work together we'd probably be
dangerous. :-D
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/
Posted by daestrom on February 19, 2009, 6:08 pm
> Jeff's question elsethread prompted me to try to work up some numbers
> and methods - and I'd appreciate some critical review of my logic.
> I've been thinking of temperature as a measure of energy per unit
> mass,
As long as your mass doesn't undergo a phase change (liquid->solid or
liquid->gas), this is a very good measure of the amount of heat stored in a
mass.
> so I knew I wanted to be able to determine the mass of the air in a
> 30' x 40' x 10' structure. I did a couple of web searches and found:
> density of air at sea level = 0.0767 lb/ft^3 [1.229 kg/m^3]
> and I calculated
> volume of air = 30 ft * 40 ft * 10 ft = 12000 ft^3 [339.8 m^3]
> which allows me to figure
> mass of air = 12000 ft^3 * 0.0767 lb/ft^3 = 920.4 lb [417.486 kg]
> (Yes, I know that a pound isn't a mass unit - but bear with me or
> validate with the proper metric units. Is there a name for a "pound"
> when we want to use it as a unit of mass?)
(Actually, in a lot of engineering, we use two 'pound' units. One is the
pound-mass (lbm) and the other is the pound-force (lbf). One pound-mass
resting on your scales in earth-normal gravity (32.2 ft/s^2), results in a
force of one pound-force. So Newton's equation is modified slightly to make
all the math work out F=m*A/g-subc where g-subc is a 'conversion factor'
between pound-force and pound-mass).
But that's besides the issue here :-)
> I was wondering what mass of concrete it would take (at the same
> temperature) to store however much heat there was in the 12000 ft^3 of
> air, and I seemed that it would take the exact same mass.
> mass of concrete = mass of air
Nope. Here you have missed one important point. The amount of heat it
takes to warm up one lbm of air by one degree is different than what it
takes to warm up one lbm of concrete by one degree.
Assuming you aren't containing the air in a fixed volume and causing the
pressure to rise as you heat it, the amount of heat it takes to warm one lbm
of air one degree F is about 0.24 BTU. The amount of heat to warm up a lbm
of concrete one degree is a bit less at about 0.18 BTU.
So to store the same amount of heat, you need
mass-of-concrete * 0.18 = mass-of-air * 0.24
mass-of-concreate = mass-of-air * 0.24 / 0.18
(the values 0.24 and 0.18 are referred to as 'specific heat capacity' for
the concrete and 'specific heat capacity at constant pressure' for air.
With gasses you have to specify whether it is at constant pressure or
constant volume).
> My next question was to what depth would I need to warm the concrete
> slab in order to store that much heat? I recognized that unless I
> restricted myself to a non-dynamic context I'd get lost, so let me
> stipulate a purely static context.
> I googled again to come up with
> mass of concrete = 149.8 lb/ft^3 [2400 kg/m^3]
> so I could calculate
> volume of concrete = 920.4 lb / 149.8 lb/ft^3
> = 6.144 ft^3 [0.174 m^3]
Adjusting that for the difference in specific heats, you would actually
need...
volume of concrete = 6.144 ft^3 * 0.24/0.18 = 8.19 ft^3
> Since the area of the floor is 1200 ft^2, if the heat were distributed
> evenly over the entire floor, then the depth of the concrete needed to
> store that heat would be
> depth = 6.144 ft^3 /1200 ft^2
> = 0.005120 ft = 0.06144 inch = 1.56 mm
> At this point I leaned back in my chair and asked myself if the
> numbers seemed reasonable, and had to confess that I just don't know
> - and it seems like a really good time to get some knowledgeable help?
> If I screwed up, please point out where and how...
Not off by much. This illustrates the fact that 'air' by itself is a lousy
way to store heat. If you had just 120 ft^3 of water (specific heat
capacity in these units for water is 1 BTU/lbm-F and density of about 62.2
lbm/ft^3) you could store...
heat-stored per degree F = 120 ft^3 * 62.2 lbm/ft^3 * 1 BTU/lbm-F = 7464 BTU
/ F
Compared with your volume of air (100 times larger)
heat-stored per degree F = 12000 ft^3 * 0.0767 lbm/ft^3 * 0.24 BTU/lbm-F =
220 BTU / F
It's the very low density of air that really kills you for heat storage.
Now, for more heat storage in a small space, look for phase-changing
systems. Some salts can absorb quite a bit of heat changing from solid to
liquid. Glauber's salt is supposed to absorb something like 99 BTU/lbm at
80 F as it changes from solid to liquid (and give it back up when it
'freezes' back to a solid). (But Glauber's salt has one drawback, it 'wears
out' and won't change from solid-liquid as easily the more you cycle it)
daestrom
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>
> mass of concrete = 149.8 lb/ft^3 [2400 kg/m^3]