Article 31319 of alt.solar.thermal: Path: news.misty.com!not-for-mail From: nicksanspam@ece.villanova.edu Newsgroups: alt.solar.thermal Subject: Re: solar air heater review and a few questions Date: 22 Sep 2008 06:55:19 -0400 Organization: Villanova University Lines: 80 Message-ID: References: <086a02d0-abd8-4190-8020-277753418f09@l64g2000hse.googlegroups.com> <4e280031-1ef1-4404-aa08-b44b10be7490@l43g2000hsh.googlegroups.com> NNTP-Posting-Host: acadia.ece.villanova.edu X-Trace: max.inside.misty.com 1222080921 17931 153.104.44.130 (22 Sep 2008 10:55:21 GMT) X-Complaints-To: abuse@misty.com NNTP-Posting-Date: Mon, 22 Sep 2008 10:55:21 +0000 (UTC) Xref: news.misty.com alt.solar.thermal:31319 wrote: >Jersey John wrote: >> >> I intend on building a solar air heater in the coming weeks... >> Dimensions of collector will be 24 inches wide x 12 foot tall. > >... the inlet and outlet vents should be as large as possible to reduce >airflow resistance, which increase airflow, which removes heat better, >which makes the collector more efficient. So the vents should be the same size as the glazing, as in a window? :-) >... the usual design rule is that the depth of the collector should be >1/15 the height. So, for a 12 ft high collector, the 1/15 th rule would >give you a depth of nearly 10 inches. >... If you went with (say) an 8 inch depth, than the cross section area >would be 24*8 = 192 sq inches, and the inlet vent and outlet vent area >should each be about half this -- around 100 sq inches. Why half? Shouldn't the vent area depend on the glazing area? In full sun (250 Btu/h-ft^2), a 70 F room on a 30 F day with R1 glazing with 90% solar transmission might look like this, viewed in a fixed font: 0.9x250x24ft^2 = 5400 Btu/h I = 16.6(100/144)sqrt(12)(T-70)^1.5 --- --- = 40(T-70)^1.5 Btu/h |---------|-->|-----------------------|-->|----> --- | --- | 1/24 | 30 F ------www-------- T with T = 30 + (5400-40(T-70)^1.5)/24 = 70 + (153-0.6T)^(2/3). Plugging in T = 100 on the right makes T = 90.5 on the left. Repeating makes T = 91.4, then 91.3, with I = 3927 Btu/h and a 65% collection efficiency. If 100 in^2 vents collect 23,562 Btu on a 6-hour sunny December day and lose 18h(70-30)2x100/144/R1 = 1000 at "night," would larger or smaller vents be more efficient? 10 FOR A = 25 TO 250 STEP 25'vent area (in^2) 20 AV=A/144'vent area (ft^2) 30 K=16.6*AV*SQR(12)'chimney constant 40 TI=100'initial temp 50 T=70+(255*24/K-24*TI/K)^(2/3) 60 IF ABS(T-TI)>.1 THEN TI=T:GOTO 50'iterate 70 GAIN=6*K*(T-70)^1.5'(Btu/day) 80 LOSS=18*(70-30)*2*AV'(Btu/day) 90 NET=GAIN-LOSS'(Btu/day) 100 EFF=100*NET/(250*24)/6'(%) 110 PRINT 1000+A;"'",T,NET,EFF 120 NEXT A in^2 T (F) net gain eff % 025 117.761 19521.39 54.22607 050 102.3038 21495.75 59.71041 <-- 1.4% of the glazing 075 95.38025 22226.95 61.74154 100 91.30493 22561.75 62.67152 125 88.56468 22706.79 63.07442 150 86.57139 22744.71 63.17974 175 85.04369 22715.67 63.09907 200 83.82823 22641.51 62.89307 225 82.83368 22535.45 62.59846 250 82.00187 22405.81 62.23837 It looks like the vent area doesn't affect the efficiency much. For more efficient heat storage (eg in a shiny ceiling mass), we might use a smaller vent area with a higher outlet temp. Then again, smaller vents work as well on average vs sunny days, and vents also lose heat on cloudy days... 1% might work well above. Nick