[Greenbuilding] A Solar Hot Tub

[Nick Pine]

dantonioli at earthlink.net writes:

> Heating a hot with solar is easy! If you're on a hill and position the panel below the tub you might even be able to get a thermosyphon effect and eliminate the pump altogether.

But how do we store the heat? The tub water is useless, since most people prefer a certain fixed temperature... 104 F is too cold and 106 is too hot. OTOH, we might store 170 F water from thermosyphoning evacuated tubes in a heavily-insulated EPDM-lined box above them. We could keep a tub with a 5K Btu/day heat loss exactly 105 F for 5 cloudy days in a row with 25K/(170-110) = 420 pounds (50 gallons) of 170 F water that cools to 110 F. 

Kathy Cochran writes:

>I imagine that I would line it with concrete and paint it black or dark blue-green to absorb the sun... I live in Northern CA (Calaveras County in the Gold Country)...

NREL says July is the warmest month in Sacramento, when 2520 Btu/ft^2 of sun falls on the ground and 910 falls on a south wall on an average 75.7 F day with a daily max of 93.2 and a humidity ratio wo = 0.0087 pounds of water per pound of dry air with water vapor pressure Pa = 29.921/(1+0.62198/wo)) = 0.413 "Hg. If 1 ft^2 of hot tub at temp Tw (F) with perfect ground insulation and no cover absorbs 2520 Btu = 24hx100(Pw-Pa), using an ASHRAE swimming pool formula, Pw = 1.463 = e^(17.863-9621/(460+Tw)), which makes Tw = 9621/(17.863-ln(Pw))-460 = 90.3 F on an average day, and cooler on cloudy days. With a clear solar pool cover, it might absorb 0.9x2520 = 2268 Btu = 24h(Tw-75.7)1ft^2/R1, which makes Tw = 170.2 :-)

The worst-case month is December, when 550 Btu/ft^2 falls on the ground and 820 falls on a south wall on an average 45.3 F day with a 52.7 F daily max and an average daytime temp of about 49. With a 2" R15 foil polyiso board cover, the tub might lose 24h(105-45.3)1ft^2/R15 = 96 Btu/day. At 140 F (to allow the use of a $10 plastic pipe heat exchanger) for 6 hours, a draindown cover with an EPDM layer under R2 transparent top with 80% solar transmission could collect 0.8x550-6h(140-49)1ft^2/R2 = 167 Btu, which seems like a comfortable margin. We might heat the 1 ft^2 tub for 5 cloudy days with 5x96/(140-110) = 16 pounds of 140 F water cooling to 110. For more solar gain, we might make a lightweight lift-off transparent cover as an equilateral pentagon hat with 4' struts, about 3' tall and 6' diameter with 2 clear polycarbonate triangles facing south and 3 reflective foil-foamboard triangles in the other directions...   

20 PI=4*ATN(1)
50 DATA 4,8,9,10,12,18
60 FOR I=1 TO 6
70 READ L'strut length
80 ANG=36*PI/180
90 R0=L/(2*SIN(ANG))
100 D0=2*R0
110 D0R=INT(D0+.5)
120 D1=L/TAN(ANG)
130 D1R=INT(D1+.5)
140 FILMWIDTH=L*COS(30*PI/180)
150 FWR=INT(FILMWIDTH+.5)
160 H=FILMWIDTH+SQR(L^2-R0^2)
170 HR=INT(H+.5)
180 FILMLENGTH=8.5*L
190 FLR=INT(FILMLENGTH+.5)
200 FILMAREA=FILMWIDTH*FILMLENGTH
210 FAR=INT(FILMAREA+.5)
220 FCR=INT(.05*FILMAREA+.5)
230 PRINT 500+L;"'";TAB(8);D1R;TAB(16);D0R;TAB(24);HR;TAB(32);FWR;
240 PRINT TAB(40);FLR;TAB(48);FAR;TAB(56);FCR
250 FILMAREA=FILMWIDTH*FILMLENGTH
260 NEXT

L      Di      Do      H       FW      FL      FA      FC

4      6       7       6       3       34      118     6
8      11      14      11      7       68      471     24
9      12      15      13      8       77      596     30
10     14      17      14      9       85      736     37
12     17      20      17      10      102     1060    53
18     25      31      25      16      153     2385    119
 
Corwyn <corwyn at midcoast.com>

> PV panels are going to be VERY expensive way to heat the hot tub.  They are about 10% efficient...

We might collect the other 85% with PVs under a shallow water-filled plastic film pillow under under the pentagon. Plastic and water and PVs have similar refractive indices, so adding the pillow only reduces the PV output by about 6%. Raising the PV temp lowers the output by about 0.5%/C, but we can make up for that by concentrating more sun on the water-cooled PVs.

Nick