[Greenbuilding] Cooling in Tulsa

[nick pine]

Leslie Moyer <unschooler at lrec.org> writes:

>... It's not the aesthetics I'm after (though that would be okay--most. 
>boring. house. ever.), it's relief from this incessant heat-gain.

NREL says 2110 Btu/ft^2 of sun falls on the ground on a 83.3 F average July 
day in Tulsa with an 93.7 and 72.8 high and low and average 88.5 and 78 day 
and night temps and w = 0.0152 pounds of water per pound of dry air and an 
9.5 mph windspeed.

An 82 F house with a 200 Btu/h-F thermal conductance that uses 300 kWh/month 
of electricity indoors would need 24h(83.3-82)200+34.1K = 40.3K Btu/day of 
cooling. With 30 cfm of natural air leakage and 2 gallons per day of 
evaporation from occupants and a 39% indoor RH, ie Pa = 
0.39e^(17.863-9621/(460+82)) and w = 0.62198/(29.921/Pa-1) = 0.0092, inside 
ASHRAE's 55-2004 standard comfort zone, we need to remove 
24hx60x30x0.075(0.0152-w)+16.6 = 36 pounds of water vapor per day.

Fig. 5 of 
http://www.agmcontainer.com/desiccantcity/pdfs/Desiccant%20performance.pdf 
says the maximum moisture content of bentonite clay (eg clumping kitty 
litter) is 15% at 77 F and 5% at 113 F, ie MC = 0.364-0.00278T(F), eg 0.136 
at 82 F. If the clay is solar heated and dried at 113 F during the day and 
cooled to 82 F at night, the house needs 36/(0.136-0.05) = 418 pounds (11.3 
cubic feet) of clay. If air from a 1000 Btu/h-F car radiator cools the 
solar-heated clay bed at night, the house needs (40.3K+36K)/24h = 3179 Btu/h 
of cooling, with an 82-3179/1000 = 78.8 F temp water supply.

The average vapor pressure of the July air Pa = 29.921/(1+0.62198/w) = 0.714 
"Hg and the dew point Tdp = 9621/(17.863-ln(Pa))-460 = 68.6 F and the wet 
bulb Twb(R)= 9621/(22.47-ln(78+460+100Pa-Twb)) = 531.2 R (71.2 F.) Phil 
Niles says a 78.8 F shaded pond in 78 F air with a 68.6 F dew point and a 
71.2 F wet bulb temp loses Qr = 1.63x10^-9((78.8+460)^4-a(78+460)^4) = 17.4 
Btu/h-ft^2 by radiation, where a = 0.002056xTdp+0.7378. Qc = 
(0.74+0.3x9.5mph)(78.8-78) = 2.9 Btu/h-ft^2 by convection, and Qe = 
b(78.8-71.2)-Qc = 104.5 by evaporation, where b = 
3.01(0.74+0.3x9.5)((78.8+71.2)/65-1), totaling 124.8, so it looks like a 
3179/124.8 = 25.5 ft^2 pond could provide the 78.8 F water.

If A ft^2 of R1 glazing with 90% solar transmission evaporates 36 pounds of 
water at 113 F and 0.9x2110A = 36K + 8h(113-88.5)A, 21 ft^2 of pond glazing 
works, eg a 4'x8' clay bed with 4.2" of cat litter. More cat litter can 
store more dryness for non-average days and more glazing in an equilateral 
A-frame with an insulated north wall can do winter air heating.

Nick

50 CLO = .5'clothing insulation (clo)
60 MET=1.1'metabolic rate (met)
70 WME=0'external work (met)
80 TA=(82-32)/1.8'air temp (C)
90 TR=TA'mean radiant temp (C)
100 VEL=.2'air velocity
120 RH=39'relative humidity (%)
130 PA=0'water vapor pressure
140 DEF FNPS(T)=EXP(16.6536-4030.183/(TA+235))'sat vapor pressure, kPa
150 IF PA=0 THEN PA=RH*10*FNPS(TA)'water vapor pressure, Pa
160 ICL=.155*CLO'clothing resistance (m^2K/W)
170 M=MET*58.15'metabolic rate (W/m^2)
180 W=WME*58.15'external work in (W/m^2)
190 MW=M-W'internal heat production
200 IF ICL<.078 THEN FCL=1+1.29*ICL ELSE FCL=1.05+.645*ICL'clothing factor
210 HCF=12.1*SQR(VEL)'forced convection conductance
220 TAA=TA+273'air temp (K)
230 TRA=TR+273'mean radiant temp (K)
250 TCLA=TAA+(35.5-TA)/(3.5*(6.45*ICL+.1))'est clothing temp
260 P1=ICL*FCL:P2=P1*3.96:P3=P1*100:P4=P1*TAA'intermediate values
300 P5=308.7-.028*MW+P2*(TRA/100)^4
310 XN=TCLA/100
320 XF=XN
330 N=0'number of iterations
340 EPS=.00015'stop iteration when met
350 XF=(XF+XN)/2'natural convection conductance
360 HCN=2.38*ABS(100*XF-TAA)^.25
370 IF HCF>HCN THEN HC=HCF ELSE HC=HCN
380 XN=(P5+P4*HC-P2*XF^4)/(100+P3*HC)
390 N=N+1
400 IF N>150 GOTO 550
410 IF ABS(XN-XF)>EPS GOTO 350
420 TCL=100*XN-273'clothing surface temp (C)
440 HL1=.00305*(5733-6.99*MW-PA)'heat loss diff through skin
450 IF MW>58.15 THEN HL2=.42*(MW-58.15) ELSE HL2=0'heat loss by sweating
460 HL3=.000017*M*(5867-PA)'latent respiration heat loss
470 HL4=.0014*M*(34-TA)'dry respiration heat loss
480 HL5=3.96*FCL*(XN^4-(TRA/100)^4)'heat loss by radiation
490 HL6=FCL*HC*(TCL-TA)'heat loss by convection
510 TS=.303*EXP(-.036*M)+.028'thermal sensation transfer coefficient
520 PMV=TS*(MW-HL1-HL2-HL3-HL4-HL5-HL6)'predicted mean vote
530 PPD=100-95*EXP(-.03353*PMV^4-.2179*PMV^2)'predicted % dissatisfied
540 GOTO 580
550 PMV=99999!:PPD=100
580 PRINT TA,RH,CLO,PMV,PPD

27.77778    39     .5     .4958241    10.13743