Maker Pro
Maker Pro

Natural cooling

NREL says the average May temp in Phoenix is 88.2 F, with an average
daily min and max of 73 and 105 and humidity ratio w = 0.0056 pounds
of water per pound of dry air and a 41.5 F dewpoint. Night ventilation
won't quite do, even with lots of house thermal mass and surface...

10 TN=73'night temp (F)
20 TD=105'day temp (F)
30 GH=200'house conductance (Btu/h-F)
40 GV=2000'fan conductance (cfm)
50 GM=4000'thermal mass surface conductance (Btu/h)
60 FOR C=2000 TO 8000 STEP 2000'thermal capacitance (Btu/F)
70 RCN=(1/(GH+GV)+1/GM)*C'night time constant (hours)
80 RCD=(1/GH+1/GM)*C'day time constant (hours)
90 NUM=TD+(TN-TD)*EXP(-12/RCD)-TN*EXP(-12/RCN)*EXP(-12/RCD)
100 DEN=1-EXP(-12/RCN)*EXP(-12/RCD)
110 TCD=NUM/DEN'dusk temp (F)
120 TCN=TN+(TCD-TN)*EXP(-12/RCN)
130 PRINT C,TCN,TCD
140 NEXT C

Mass dawn dusk
(Btu/F) temp (F) temp (F)

2000 73.00436 94.79639
4000 73.19868 87.04121
6000 73.61773 83.55931
8000 74.03886 81.73341

More mass helps. The min temp creeps up, but the max decreases faster.

Wet surfaces are better. At 80 F, this house needs 12h(105-80)200 = 60K Btu
of cooling during the day, which might come from 60K/(75-45) = 2,000 pounds
of water warming from 45 to 75 F. That's 31.25 ft^3 of water, like a
4'x8'x2' deep sandbox full of water and rocks with an indoor fan-coil,
eg a $35 used 1984 Dodge Omni radiator with a $12 20" window box fan
that adds no indoor humidity.

The water might be 6" below the rocks during the day and sprinkled over
the rocks at night. With ambient vapor pressure Pa = 29.921/(1+0.62198/w)
= 0.267 "Hg and Pw = e^(17.863-9621/(460+60)) = 0.528 "Hg, dumping 60K Btu
of heat over 12 hours, 60K = 12hx100A(Pw-Pa), so we need A = 192 ft^2 of
rock surface in 4x8x0.5 = 16ft^3, ie N = 16/(4/3Pir^3) = 12/(Pir^3) r'
diameter rocks, with 192 = N4Pir^2 = 48/r, so r = 0.25', or 3", ie 6"
or smaller diameter rocks would do. The rocks protect the water from sun
and children during the day and allow the water to cool in the dewpoint
vs wet-bulb direction.

Nick
 
...At 80 F, this house needs 12h(105-80)200 = 60K Btu of cooling during
the day, which might come from 60K/(75-45) = 2,000 pounds of water warming
from 45 to 75 F. That's 31.25 ft^3 of water, like a 4'x8'x2' deep sandbox
full of water and rocks with an indoor fan-coil...
The water might be 6" below the rocks during the day and sprinkled over
the rocks at night. With ambient vapor pressure Pa = 29.921/(1+0.62198/w)
= 0.267 "Hg and Pw = e^(17.863-9621/(460+60)) = 0.528 "Hg, dumping 60K Btu
of heat over 12 hours, 60K = 12hx100A(Pw-Pa), so we need A = 192 ft^2 of
rock surface in 4x8x0.5 = 16ft^3, ie N = 16/(4/3Pir^3) = 12/(Pir^3) r'
diameter rocks, with 192 = N4Pir^2 = 48/r, so r = 0.25', or 3", ie 6"
or smaller diameter rocks would do.

A 4'x4'x2.5' tall box with 8 ft^3 of 3" diameter rocks would also do,
with 2 12x12 vertical layers of 2 liter bottles under the rocks...

Nick
 
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