quote:

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> Richard J Kinch <kinch@truetex.com> wrote:
>
>
> That scheme is unlikely to be efficient, with lots of thermal resistance
> (insulation) between the living space and the roof, and it's the wrong
> time of day to collect coolth, but Harry Thomason did something like this
> in the 60's for hundreds of houses, including his own in Washington, DC
> (a bad climate for evaporative cooling) pumping water over the roof at
> night and storing and distributing coolth during the day. He also ran
> an AC at night, when ACing is more efficient.
>
> In "Solar Space Heating and Air Conditioning in the Thomason Home" (Solar
> Energy Journal, Vol 4, No. 4, Oct. 1960, pp 11-19), Dr. Thomason wrote
> "...the water on the north sloping roof flows slowly in an almost perfect
> thin sheet from top to bottom. The granules of sand in the asphalt shingles
> help to spread the water evenly and to retard the rate of flow... On the
> night of 14 July the rate of cooling was 25,700 Btu/h at the peak when
> the sky was overcast, but there was a breeze of about 15 mph. Outside air
> was at 68 F, water to the roof was at 73 F, with a return temp of 65.5 F
> and a flow rate of 5 gallons in 41 seconds. Humidity was about 65%."
>
> Thomason used a 1600 gallon steel tank surrounded by rocks for heat and
> coolth storage and distribution. A homeowner with a basement might cool
> a house by pumping water over a roof at night and collecting it in a
> gutter and letting it flow back into some 4" thinwall PVC pipes or poly
> film water ducts tucked up between basement ceiling joists and circulate
> air between the house and the basement during the day.
>
> NREL says an average 75.6 day July day in Sterling, VA (close to DC) with
> an average daily 64.1 and 87.0 min and max has a humidity ratio w = 0.0136
> pounds of water per pound of dry air and an average V = 6.2 mph windspeed.
>
> With average night temp Ndbt = (75.6+64.1)/2 = 69.9 F and vapor
> pressure Pa = 29.921/(1+.62198/w) = 0.640 "Hg and dew point Tdp
> = 9621/(17.863-ln(Pa))-460 = 65.5 F, the approximate wet bulb temp
> Twb = (Ndbt+Tdp)/2 = 67.7 F. With lots of T = 70 F water flow,
> Phil Niles says A = 0.002056*Tdp+0.7378 = 0.8725 for a radiation loss
> Qr = 1.63E-09*((T+460)^4-A*(Ndbt+460)^4) = 16.5 Btu/h-ft^2 plus
> convection loss Qc = (0.74+0.3*V)*(T-Ndbt) = 0.3 Btu/h-ft^2, with
> B = 3.01*(0.74+0.3*V)*((T+Twb)/65-1) = 8.753 and evaporation loss
> Qe = B*(T-Twb)-Qc = 19.8 Btu/h-ft^2, for a total Q = Qr + Qc + Qe
> = 36.6 Btu/h-ft^2, or 12Q = 440 Btu/ft^2 per night.
>
> A small 1000 ft^2x8' house with 200 Btu/h-F of thermal conductance and
> 0.5 air changes per hour with 600 kWh/mo of indoor electrical use
> that's 80 F with w = 0.012 indoors would need about 70K Btu/day of
> cooling, which might come from 70K/440 = 160 ft^2 of roof.
>
> It would need about 7100 Btu/h of peak cooling. If that comes from 1000
> ft^2 (about $150) of ducts in the basement ceiling, the min duct water
> temp would be about 80-7100/(1000x2x1.5) = 77.6 F. If the house needs
> 350K Btu for a warm humid week, with no significant cooling at night,
> 350K = (77.6-70)C makes C = 45,851 Btu/F, eg 23 tons of water in 8.8" deep
> ducts (a good ballast foundation :-) More roof or basement ceiling surface
> would reduce the water depth.
>
> As an alternative, we could cool static basement ducts without plumbing
> or pumping with a whole house fan on cool nights. Dr. Thomason also used
> a small dehumidifier.
>
> Nick