Subject: MAKE BIG MONEY FAST!!!

From: [email-masked] (Nick Pine)
Date: 19 May 1996 22:05:47 -0400

Here's another litmus test for an energy expert: what will the average water
  temperature be inside a 4' cube full of water surrounded by 5 R20 foam walls,
  sitting outside in December, when the air has an average 24 hour temperature
  of 36 F and the sun puts 1,000 Btu/ft^2/day of heat into the R1 glazed side
  of the box? And how will that change over time if we shade the sunny wall?
  Anyone care to answer that question? I'll offer a $10 reward to the first
  person who answers it correctly.
 
Still only 3 numerical answers to this simple question so far. Two people got
part 1 correct, and one has almost solved part 2 using simple arithmetic, but
there's still a chance for the right person to earn $10 in 2 minutes with
a couple of back of the envelope calculations... This might also be a rare
opportunity for someone else to get paid to learn :-)

Here's a big hint: If a "solar house" has a heat loss of 150 Btu/hr-F and
a thermal capacity of about 8,000 Btu/F, so RC = 8,000/150 = 53 hours, and
the house starts out with an interior temperature of 68 F when it's -10 F
outside, with no electric or wood heating, and no sun, the interior temp after
a couple of days will be approximately - 10 + (68-(-10))exp(-48/53) = 31.5 F. 

Another slightly less accurate way to do this is to find the energy lost
during the first hour, assuming the interior temperature is constant over
the hour, find out how much the thermal mass has to cool to supply that
heat loss, find the new interior temperature, find the energy lost in the
second hour, and so on...

Nick

It's a snap to save energy in this country. As soon as more people become
involved in the basic math of heat transfer and get a gut-level, as well as
intellectual, grasp on how a house works, solution after solution will appear.
 
 					  Tom Smith

---

From: [email-masked] (Tooie)
Date: 20 May 1996 17:22:48 GMT

Nick Pine ([email-masked]) wrote:
:   Here's another litmus test for an energy expert: what will the average water
:   temperature be inside a 4' cube full of water surrounded by 5 R20 foam walls,
:   sitting outside in December, when the air has an average 24 hour temperature
:   of 36 F and the sun puts 1,000 Btu/ft^2/day of heat into the R1 glazed side
:   of the box? And how will that change over time if we shade the sunny wall?
:   Anyone care to answer that question? I'll offer a $10 reward to the first
:   person who answers it correctly.
:  
: Still only 3 numerical answers to this simple question so far. Two people got
: part 1 correct, and one has almost solved part 2 using simple arithmetic, but
: there's still a chance for the right person to earn $10 in 2 minutes with
: a couple of back of the envelope calculations... This might also be a rare
: opportunity for someone else to get paid to learn :-)

: Here's a big hint: If a "solar house" has a heat loss of 150 Btu/hr-F and
: a thermal capacity of about 8,000 Btu/F, so RC = 8,000/150 = 53 hours, and
: the house starts out with an interior temperature of 68 F when it's -10 F
: outside, with no electric or wood heating, and no sun, the interior temp after
: a couple of days will be approximately - 10 + (68-(-10))exp(-48/53) = 31.5 F. 

: Another slightly less accurate way to do this is to find the energy lost
: during the first hour, assuming the interior temperature is constant over
: the hour, find out how much the thermal mass has to cool to supply that
: heat loss, find the new interior temperature, find the energy lost in the
: second hour, and so on...

: Nick

1000 Btu/ft^2-day * 16 ft^2 = 16000 Btu/day

16000 Btu/4000 lbm H2O * 1 Btu/lbm-F = 4 F

assuming  the following

h = 20 Btu/hr ft^2-F
c = 1 Btu/lbm-F
rho = 62.4 lbm
k = .32Btu/hr-ft^2-F

the thermal time constant is;

(62.4 * 1 * 64)/(20 * 16) = 12.5 hours

so if it's dark for 14 hours, at the end of the 14 hours the temp would be

T - 36 = (40 - 36) e^(-14/12.5)

T = 37.3 F

this is highly dependent on the value of h (convective heat transfer 
coefficient) obviously wind (or lack of) would have a majot effect.

Radiative heat transfer would be as follows (assuming blackbody);

q = sigma * Area * (T1^4 - T2^4)

q = 0.1714 E-8 Btu/hf-ft^2-R^4 * 16 ft^2 * ((460+40)^4 - (460+36)^4)

q = 54 Btu/hr therefore neglible