Subject: thermal calcs cleaned up & expanded
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Date: Thu, 11 Sep 2003 13:12:49 -0700

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thermal calcs cleaned up & expanded
by
Darel Henman
23 October 2002 07:33 UTC

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This is the newest latest version. I added an example of how to use the
engineering RC constant
to calculate resulting temperature after so many hours at such and such
a temperature differential.

Also some people had some trouble getting some special graphic
characters I used
in the formulas to display. I took all of those out and tried to make
it
easier to follow:

-----------
Here's an example that I found right off using real figures.
Simplified a bit by ignoring air film resistance etc.

I use the symbols below:

Tr ... for thermal resistance ( defined below )
Tc ... for thermal capacitance (")
RC ... just used this for Tr * Tc
Te ... for thermal energy (btus)
Tg ... for thermal gradient = delta temperature (T_out - T_in)
Sh ... specific heat coefficient for a material
hr ... for hours
deg_F ... for temperature in degrees fahrenheit
ft ... for feet
ft^2 ... for square feet (area)
ft^3 ... for cubed feet (volume )
btu ... for british thermal unit
lb ... for pound (where 32 ounces is one pound )
t ... time in hours
e ... natural log number equal to 2.718281828....... so whenever
you see 'e' use 2.7183

--------------------------
EXAMPLE: Thermal Mass RC example: (thanks to Nicholas Pine)

Given that:
* Water has a Sh capacity of 1 btu / (lb per deg_F)
* 6" fiberglass insulation has a thermal blocking value
(R-Value) of 19. units are: (hr per ft^2 per deg_f )/btu

If you made a big box out of plywood, say a cube, 8' x 8' x 8' (with a
nice low surface to volume ratio), and surrounded it by the fiberglass
insulation mentioned above, and lined it with a big sheet of rubber
roofing material folded up like a Chinese take-out box, you would
have a thermal battery with a time constant of...

Volume of water = 8' x 8' x 8' = 512 ft^3
Its Thermal Capacitance=
Tc = Density * Volume * Sh
= ( density * volume ) * Sh
= (62.4 lb/ft^3 * 512 ft^3) * 1 btu/lb * deg_F
= 31,948.8 btu / deg_F

This figure means that about 32,000 btus can be absorbed or sourced
per each degree F change in the thermal storage medium, in this
case water.

Its Thermal Mass Resistance,
Tr = R-Value / Area_through_which_energy_is_flowing
= (19 hr per ft^2 per deg_F/btu) / (6 sides of 8' * 8')
= (19 hr per ft^2 per deg_F/btu) / ( 384 ft^2 )
= 0.04948 hr per deg_F/ btu

This 8' cube of water would have an RC time constant of:
RC = Tr * Tc from above we have:
= (0.04948 hr per deg_F / btu) * (31,948.8 btu / deg_F)
= 1580.83 hours or about 65.88 days

This RC value is what we would use in the charging and discharging
formulas below:

Charging Thermal Mass Capacitors: (R and C are thermal resistance (Tr)
and
thermal capacitance (Tc) defined above. This is in hours so use hours
time units below:

Thermal_masses_temp_F = Charging_deg_F * (1-e^(-t/RC) )
Resulting temperataure= (F_high - F_low)* (1-e^(-t/RC) )

----

If you put this thermal energy storage box in a constant temperature 0
degree_F room with the water at 100 degrees_F , then in a one day period
about:

Te = ((Tg) * hrs ) / Tr which has units of
= ((100 degrees_F - 0 degrees_F) * 24 hrs) / (0.04948 hr per deg_F /
btu )
= 48,504 btus would transfer from the water into the cold room.

So about 48,504 buts of heat would leak out of the box, to heat up the
room. This seems like a lot, but, because water has such a high
thermal mass, the water inside the box would only cool down by,

Box_water_delta_deg_F = Te / Tc
= (48,504 btu) / (31,948.8 btu / deg_F)
= 1.52 deg_F

So after one day the thermal battery's temperature would be:
100 deg_F - 1.52 deg_F or 98.48 deg_F.

------

Or we might have used the discharging forumla to calculate what
temperature your mass will be
after t-hours. In this example 24 hours. given static conditions. We
can use the RC time constant discharing formula to check this. For
example:

Discharging Thermal Mass Capacitors:
Capacitor's_F = Discharging_deg_F * e^(-t/RC)
resulting deg_F = (F_high - F_low) * e^(-t/RC)
= ( 100F - 0F) * 2.7183^(-24hr/1580.83 hours)
= 98.49 deg_F pretty much the same as we got
above.
Note: I used the spreadsheet formula =(100-0) *
EXP(-24/1580.83) for this.

Just change the amount of time in hours to see what temperature your
thermal storage will end up with.
I suppose it's possible to for the storage to be colder than the
surround room, but I assume that the sign will show you which direction
energy is flowing. In this example I used F_high assuming that the
thermal storage medium, the water was hotter than the room that it was
sitting in. If the reverse were true the water would be sucking up heat
from the room.

This looks good, but the storage has to get charged up some way.

Sorry I've been meaning to get around to sending this. I converted it
into normal ascii-ish characters, but if any strange character gets
through let me know.

For the earth people we just need to somehow come to terms with some
assumptions about boundaries. People who use underground umbrellas have
some kind of R-figure for above but what to do about the sides and
underneath... Some assumptions should be given then tested to get some
kind of usable for other folk modeling information, that doesn't
require
corporate expense accounts to give us some kind of rough answer.
It's always better than picking numbers out of the air. And we can
always modify it to make it more and more accurate.

Happy Reckoning,
Darel

---------------

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