# A Spencerator

• posted on February 28, 2005, 8:30 pm
A fridge can be a large winter load for an off-grid PV house...
NREL says the average daily November temp in Massena NY is 35.3 F, with an average daily min of 27.7. December has 20.2 F and an 11.7 min. January has 14.3 and 4.3. February has 16.5 and 6.5. March has 28.3 and a 19.1 F min.
A 2'x2'x8' tall R20 34 F fridge in a 70 F room needs (70-34)80ft^2/R20 = 136 Btu/h or 3456 Btu/day of coolth, or 17,280 Btu for 5 days, which might come from at least 17280/144 = 120 pounds of ice in an outdoor box, with a large water tray with a copper pipe in the freezer compartment, with free airflow to the fridge below and a liquid thermosyphoning loop to move coolth in from the box. Can the loop stop working before it freezes the fridge contents?
The ice box might be 2'x2'x8' tall, with a 55 gallon drum packed with 100 2 liter water bottles surrounded by an antifreeze solution thermosyphoning through an old 800 Btu/h-F auto radiator above with an 8' chimney above. Cold air might enter the radiator and flow out at the top of the chimney.
With 800 Btu/h-F of conductance from 32 F water to outdoor air at temp T, (32-T)800 = 3456 Btu makes T = 32-3456/800 = 27.7 for 1-hour per day cooling and T = 32-3456/1600 = 29.8 for 2-hour cooling, and so on.
The crude TMY2 (Typical Meteorological Year) simulation below indicates that a fridge like this could work for 241 days (2/3) of a typical year in Massena, from September 23 through May 22...
Nick
This program makes a "fridge" prep file with the month, day, hour, and dry bulb temp...
10 SCREEN 9:KEY OFF:CLS 40 LINE (0,0)-(639,349),,B:XDF=.073:YDF=2.92 50 FOR TR= 20 TO 40 STEP 10'temp ref lines 60 LINE (0,349-YDF*(TR+20))-(639,349-YDF*(TR+20)):NEXT 70 OPEN "94725.tm2" FOR INPUT AS #1'NREL TMY2 file name (Massena) 80 OPEN "fridge" FOR OUTPUT AS #2 90 LINE INPUT#1,S\$'read header 100 CITY\$=MID\$(S\$,8,25) 110 LAT=VAL(MID\$(S\$,40,2))+VAL(MID\$(S\$,43,2))/60 120 LON=VAL(MID\$(S\$,48,3))+VAL(MID\$(S\$,52,2))/60 130 PRINT#2,CITY\$,LAT,LON 140 FOR H=1 TO 8760'hour of year 150 LINE INPUT#1,S\$ 160 MONTH=VAL(MID\$(S\$,4,2))'month of year (1-12) 170 DAY=VAL(MID\$(S\$,6,2))'day of month 180 HOUR=VAL(MID\$(S\$,8,2))-.5'hour of day 190 N=1+H/24'day of year (1 to 365) 200 TDB=VAL(MID\$(S\$,68,4))*.18+32'dry bulb temp (F) 230 PSET(XDF*H,349-YDF*(TDB+20)) 470 PRINT#2,MONTH;DAY;HOUR;TDB 480 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months 490 NEXT H
Here's the start of the fridge file...
MASSENA NY 44.93334 74.85 1 1 .5 24.98 1 1 1.5 24.98 1 1 2.5 24.98 1 1 3.5 26.06 1 1 4.5 26.06 1 1 5.5 19.94 1 1 6.5 15.98 1 1 7.5 10.94 1 1 8.5 8.959999 1 1 9.5 6.979999 1 1 10.5 6.979999 1 1 11.5 6.079999 1 1 12.5 6.079999 1 1 13.5 6.079999 1 1 14.5 6.979999 1 1 15.5 6.979999 1 1 16.5 6.079999 1 1 17.5 1.939999 1 1 18.5 -2.020001 1 1 19.5 -2.920002 1 1 20.5 -4.000002 1 1 21.5 -4.000002 1 1 22.5 -.9400013 1 1 23.5 -.9400013 ....
This program simulates a typical year...
10 SCREEN 9:KEY OFF:CLS 20 DAYSTART=0'display start time (days) 30 DSÚYSTART*24'display start time (hours) 40 RANGE‡60'display range (hours) 50 LINE (0,0)-(639,349),,B:XDFd0/RANGE:YDF=2.69 60 FOR TR=-30 TO 30 STEP 10'temp ref lines 70 LINE (0,349-YDF*(TR+30))-(639,349-YDF*(TR+30)):NEXT 80 OPEN "fridge" FOR INPUT AS #1:LINE INPUT#1,H\$ 90 FOR H=1 TO 8760'hours of typical (TMY2) year 100 INPUT#1,MONTH,DAY,HOUR,TDB'read fridge file 110 IF TDB<TMIN THEN TMIN=TDB'find min temp (F) 120 PSET(XDF*(H-DS),349-YDF*(TDB+30))'plot outdoor temp (F) 130 IF TDB<32 THEN ICE=ICE+(32-TDB)*800 ELSE ICE=ICE-136'ice amount (Btu) 140 IF ICE<0 THEN ICE=0'lower ice limit (Btu) 150 IF ICE>144*440 THEN ICE4*440'upper ice limit (Btu) 160 IF ICE>0 THEN LINE (XDF*(H-DS),340)-(XDF*(H-DS),330):ICEH=ICEH+1'ice hours 170 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*(H-DS),349)-(XDF*(H-DS),345)'months 180 NEXT H 190 CLOSE #1 200 PRINT TMIN,ICEH 210 END
Min yearly Hours temp (F) with ice
-25.96 5468
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<%-name%>
• posted on February 28, 2005, 11:59 pm
Do you ever get tired of fantasizing your incredibly stupid and incorrect ideas of how reality works , Nick ? And inventing numbers to 'prove' your lunacies ?
On 28 Feb 2005 15:30:51 -0500, snipped-for-privacy@ece.villanova.edu wrote:

Paul ( pjm @ pobox . com ) - remove spaces to email me 'Some days, it's just not worth chewing through the restraints.'
HVAC/R program for Palm PDA's Free demo now available online http://pmilligan.net/palm / Free Temperature / Pressure charts for 38 Ref's http://pmilligan.net/pmtherm /
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<%-name%>
• posted on March 1, 2005, 2:49 pm
p j m of the pulsating rectum rites:

Nope. What's wrong with this one, in your opinion?

A couple of fans and pumps could make it more compact and easier to design, with more power and cost and less reliability. Still practical, but less fun.
Nick
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<%-name%>
• posted on March 1, 2005, 7:12 pm
What is so stupid and incorrect? Or is it you just don't understand it.....
Steve Spence Dir., Green Trust http://www.green-trust.org
p j m@see _my _sig _for_address.com wrote:

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<%-name%>
• posted on March 1, 2005, 11:22 pm

You may want to look at that data again. I found that as early as April 7, there are very few hours that get down to 32 F. I counted only 58 hours between 4/7 and 5/22 (46 days) where temperature was below 32, for a total of only 231.5 'fridge-degrees hours' (#hours * (32-T)).
Similarly in the fall, from September 23 to October 31, I counted only 68 hours below 32 for a total of 390. 'fridge-degree-hours'.
Frankly, living not far from Massena (I'm ~hour north of Syracuse), I find it incredible to think that anyone could use this to freeze water reliably from mid March through to May or from late September through mid November. Indian-summers, freakish warm spells and all, just make it pretty iffy. Heck, in mid January we had temperatures in the 50's and 60's for several days. A 'green Christmas' is a real threat every year.
Your program seems to take credit for any temperature below 32, no matter how slight. It doesn't consider that at least a slight temperature difference must exist to circulate the antifreeze. There must be a temperature difference to create a thermosyphon. And without circulation you cannot move any btu/h to the barrel.
Using convection, to carry 800 BTU/h from the radiator to the barrel, the temperature drop across the radiator would have to be 3.2 F if the thermosyphon develops 0.5 gpm. If you don't get that much flow, it would take even more delta T to transport 800 BTU/h. So for an *infinite* radiator, it would still have to be <28.8 at 0.5 gpm of flow to carry 800 btu/h to a 32F sink.
A 50/50 mix of ethylene glycol and water has a density of about 68.27 - 0.01881*T lbm/ft^3 (where T is temperature in F). So a ten foot high thermosyphon with a 5 degree temperature difference between the two legs would have 5*0.01881 * 10 = 0.9405 lbf/ft^2 driving head (a mere 0.0065 psi, or 3/16 inch of water gage). Raise the radiator another 10 feet (20 ft total) and you can double the driving head for the thermosyphon (but it doubles the length of piping).
How much flow can you get through 40 ft of large diameter tubing/piping with 0.013 psi driving head? It would have to be about 0.3 gpm to get 800 BTU/h heat transfer. If not, you can't make '800 BTU' of ice per hour even at 27 F.
In the 'dead of winter' there is plenty of surplus 'fridge-degree days' to make it work. But the spring and fall become pretty dicey as to just how long you could keep the fridge cold with changing weather patterns.
daestrom
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<%-name%>
• posted on March 2, 2005, 12:11 pm

Averages can help predict fuel consumption, altho some days and months and years are not average...

Well, 231.5x800 = 185,200 Btu, 16% more than the 46 day cooling requirement of 46x3456 = 158,976 Btu, and this period may have begun with 144x440/3456 = 18 day's worth of perfectly-insulated ice storage.

And ignores coolth collected from air between 32 and 36 F...

How do low concentrations of antifreeze or salt water thermosyphon when cold?
I figured cold water weighs about 62.67 - 0.003T lb/ft^3, with T in degrees F. The density difference caused by the temperature difference between up and down pipes causes a pressure difference proportional to the height of the water column, making a flow through the resistance of the pipe loop. With 8' of height and a dT temperature difference, dP = 0.024dT lb/ft^2.
A pipe with radius r and length L in feet and pressure diff dP has laminar flow Q = Pir^4dP/(8MuL) ft^3/s. Viscosity Mu is 6x10^-7 lb-s/ft^2 at 32 F. With 16' of 1/2" pipe, 8' up and down, Q = Pi(0.25/12)^4dP/(8x6x10^-7x16') = 0.0077dP ft^3/s or 1729dT lb/h, which might move 1729dT^2 Btu/h. We might collect 3917 Btu of coolth in 2 hours with a dT = sqrt(1958/1729) = 1.1 F thermosyphoning temperature difference, which might be ignored, given the radiator conductance.
Someone who'd like to try this might do a more accurate simulation or use a couple of pumps and fans to start with...
Nick
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<%-name%>
• posted on March 2, 2005, 1:42 pm
snipped-for-privacy@ece.villanova.edu wrote:

This looks remarkably like what my brother-in-law was suggesting on Saturday. But considering the results _last_ time I let him experiment on my off-grid home, I think I'll pass :-)
--
derek

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<%-name%>
• posted on March 2, 2005, 10:41 pm
wrote:

You mentioned glycol, so assuming a 50/50 mix, the density change is more like 0.01884 lb/ft^3-F. This helps since water actually gets denser as it warms up in this temperature range (you can *not* thermosyphon plain water from 32 to 39 F). Even weak solutions aren't very helpful, using just 10% glycol would give you about 0.001 lb/ft^3-F. A bit of a trade off, higher mixtures give better density change per degree, but hurt viscosity and heat capacitance. Probably for a given operating temperture, an 'optimum' mixture could be determined.

More like dP = 0.01884dT * H, for dP = 0.15072dT for 50/50 glycol mix running 8'.

No Way!!!
Check some references, viscosity of water is more like 3.7x10^-5 lbf-s/ft^2 at 32 F (about 1.8 centipoise). Since your number is about 1/32 of the right value, I suspect your's is missing a 'g sub c' division or something (slugs??) (damned imperial units ;-).
50/50 glycol is much worse, about 1.88x10^-3 lbf-s/ft^2 (about 9 centipoise). For laminar flow in straight pipe, I use Q dP*(d^4)/(.000273*mu*L) (Q in gpm, dP in psi, d in inches, mu in centipoise and L in ft) (Crane TP 410 pg 3-2). That gives Q = (0.15072/144 * dT) *(.5^4)/ (.000273*9*16) for about .001664dT gpm for 50/50 glycol.
50/50 glycol is about 0.746 BTU/lbm-F so you would get 0.001664*dT*9*0.746*60 BTU/hr or 0.67 BTU/hr /F. Replace the 1/2 pipe with 2" (always good idea to use large diameter pipe with thermosyphoning :-) and you can get about 171 BTU/hr/F. At that, you would need about 4.7 F delta T to get your target 800 BTU/hr.
daestrom
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<%-name%>
• posted on March 3, 2005, 9:52 am

With a radiator pump and fan and a differential thermostat and a fridge pump and a fridge thermostat and a compact ice box with the drum on the ground. All low power stuff, but can we omit some? What's a good compromise?
The radiator might have a chimney, and it might stay primed for a long time, with both supply and return pipes underwater, and the fridge loop might stop working before it freezes the fridge...

Hey, 6X better for thermosyphoning :-)

...50/50 sounds good, if the viscosity doesn't increase much. We might not care much what happens much below 32 F, when all the water bottles are frozen. Do we need water bottles, vs slush?

OK.
Hmmm.
Thanks for the correction. How many legs in a centipoise?

Where do we get a Crane TP 410 pg 3-2?

A gallon of 50/50 glycol weighs 9 pounds?
The flow is 0.898dT pounds per hour, right? And each pound moves 0.746dT Btu, so the flow moves 0.898dTx0.746dT = 0.67dT^2 Btu/h.

Thanks for the improvement. The radiator might have 2" hoses...

...171dT^2 Btu/h?

Or maybe sqrt(800/171) = 2.2. Then again, maybe it's -26 F outdoors.
Nick
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<%-name%>
• posted on March 3, 2005, 9:31 pm
wrote:

1/5 X number of toes, and the same number as feet ;-)

A very useful booklet. Lots of practical formulae for fluid flow. The appendices alone are worth it. Comes in either a metric or imperial version (I have imperial version that I use at work).
http://www.cranevalve.com/tech.htm

That's what my figures show... Here's nice paper about glycol. You can calculate the polynomials, or just guesstimate from the graphs. http://www.mrc-eng.com/Downloads/Brine%20Properties.pdf
Then to convert to just about *any* units under the sun (including the paper's Pascal-Second to lbf-s/ft^2), I recommend this page... http://www.processassociates.com/process/convert/cf_all.htm
(great for metric vs. imperial 'stuff' too)

Hmm... you're right. Doubling the temperature difference would double the dP and hence the flow in a laminar system. And of course doubling the flow *and* the delta T would quadruple the heat transfer. Missed that one, was busy calculating for 1 degree but forgot the second dT in heat rate.
Yes, in the 'dead of winter', the excessive number of 'fridge-degree hours' make it a slam dunk. But just how far into the spring and fall it can work gets a bit more dicey as performance drops. And how many people want to live in Massena, NY besides Steve ;-) Could probably get even better performance in International Falls, MN, but then, who wants to live where they break records for cold all the time ;-)
daestrom
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<%-name%>
• posted on March 2, 2005, 12:02 pm
snipped-for-privacy@ece.villanova.edu wrote:

What I don't understand is why someone short of power would not simply use a window box or its modern counterpart to reduce power loads. I have a dead freezer on my porch that I have little trouble maintaining a nice refrigerator temperature just by opening & closing the lid at appropriate times, with a lot of water thermal ballast.
--
Free men own guns, slaves don't
www.geocities.com/CapitolHill/5357/
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<%-name%>
• posted on March 2, 2005, 12:35 pm

Too small, no light, poor temp control, no coolth storage, hard to clean, and a desire to use the fridge they already own for summertime?

I did that with a stone kitchen with exterior insulation, opening the door to the rest of the house to keep it about 36 F in the winter. The fridge rarely ran. My barn fridge gets unplugged from October through March, with an EH38 "Easy Heat" socket (\$10.99 at Lowe's) in series with a 100 W trouble light in a bottom drawer to keep the carrots and apples from freezing.
Nick
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<%-name%>
• posted on March 2, 2005, 1:48 pm
snipped-for-privacy@ece.villanova.edu wrote:

What's the actual power draw on an Easy Heat socket (whatever that is) in series with a 100W bulb? It sounds like at best a zero-sum game - a quick look at Canadian Energuide ratings shows small refrigerators to be using less than 1.5 kWh daily, so your heating appliance needs to be using less than 50W.
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derek

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<%-name%>
• posted on March 2, 2005, 2:49 pm

Unlike water, they don't actually freeze at 32 F...

The socket is a thermostat that turns on the 100 W bulb when the air near it is less than 38 F, so it uses 0 W when off and 100 W when on. I've never measured the energy usage.

...1.5kWh/24h = 62.5 W, but it probably uses less, on a continuous basis. The fridge might have 80 ft^2/R10 = 8 Btu/h-F of conductance. January is the coldest month in Phila, averaging 30.4 F. The unheated barn tends to average daily temp swings. The trouble light is in a drawer with a few 1 liter water bottles under some fiberglass insulation, so it keeps the fridge just above freezing. In January, it might supply 24h(32-30.4)8 = 307 Btu/day, using 307/3412 = 0.09 kWh, ie 3.75 W on a continuous basis, ie 17X less than a fridge in a warm room. It seldom comes on in February (average 33.0) or March (average 42.4.)
With some rewiring, the fridge might use its own light near the top of the box for heat, with the fan running without the compressor whenever the light is on to distribute warmer air near the lower part of the box.
Nick
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<%-name%>
• posted on March 3, 2005, 4:42 am
I seem to get the impression of quite a Rube Goldberg type of heat engine, apparently intended to transfer outdoor coolth into a kitchen fridge, with the intent of saving the energy that would otherwise be used to run the fridge's own compressor.
Why not simply run a pair of insulated air ducts from the interior of the fridge to the outdoors, and equip the exhaust duct with a muffin fan. One could either tap into the fridge's own thermostat, or simply provide another. You could either manually select which system would provide the coolth, or have it automated by means of an outdoor temperature sensor.
Again, the question arises, would the energy saved be worth the trouble? Will you have to burn more fuel to keep the interior of the house at the same temperature, without the heat rejected from the fridge?
I think, if I were to design and build a home from scratch, that I would incorporate a cold box in the outside wall of the kitchen, and equip both with cooling coils so it could act as a refrigerator in the warm months, and with an air-exchange system to draw in outdoor coolth in the cold months. I would also ensure that this cold box was extremely well-insulated, and that it had a number of small doors, rather than one large one, so that a minimum amount of cold air would spill into the room with each item fetched.
Gordon Richmond
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<%-name%>
• posted on March 3, 2005, 12:03 pm

Yes...
Lower performance with less coolth storage, and ducts are larger than pipes, but that's not a bad idea. Grainger's \$31 2.6 watt 24 VDC 44 cfm 6KD73 fan might run 100x163/(44(32-20)) = 31% of the time to keep a fridge cold on 20 F days from December through March in Concord NH, using 19 Wh/day from a PV battery bank vs 1 kWh from an inverter. A 4" round 0.087 ft^2 duct with a one-way plastic film damper would make V = 44/A = 504 lfm, without much reduction in airflow.

We might fill the freezer compartment with 15x16x24/(5x11x1.5) = 70 1-pound ice cube trays and turn on the fan when it's less than 32 F outdoors and the fridge is warmer than 36 F, and turn off the fridge or set it for 40, or use the fridge fan and thermostat to circulate air between the freezer and fridge.

My Vermont friends have lots of wood, but only 240 W of PV. They might double that for \$1K, with free labor...
Photovoltaics are hideously expensive. Nick
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<%-name%>
• posted on March 3, 2005, 1:58 pm

That knocks the natural cooling hours down from 5468 to 4000 hours (46%) of a TMY2 year in Massena NY, but it still covers December through March, the difficult PV period.
We could squeeze 44 vertical 1-liter water bottles into a 16x24x15" deep GE freezer compartment, with 14 ice cube trays on top, ie 104 pounds of water. There's no need to run the fan when the return temp is much less than 32 F, which indicates that the ice is all frozen.
10 SCREEN 9:KEY OFF:CLS 20 DAYSTART=0'display start time (days) 30 DSÚYSTART*24'display start time (hours) 40 RANGE‡60'display range (hours) 50 LINE (0,0)-(639,349),,B:XDFd0/RANGE:YDF=2.69 60 FOR TR=-30 TO 30 STEP 10'temp ref lines 70 LINE (0,349-YDF*(TR+30))-(639,349-YDF*(TR+30)):NEXT 75 ICEMAX4*104'max ice storage (Btu) 80 OPEN "fridge" FOR INPUT AS #1:LINE INPUT#1,H\$ 90 FOR H=1 TO 8760'hours of typical (TMY2) year 100 INPUT#1,MONTH,DAY,HOUR,TDB 110 IF TDB<TMIN THEN TMIN=TDB'find min temp (F) 120 PSET(XDF*(H-DS),349-YDF*(TDB+30))'plot outdoor temp (F) 130 IF TDB<32 AND ICE<ICEMAX THEN ICE=ICE+(32-TDB)*44 ELSE ICE=ICE-136'new ice 150 IF ICE>ICEMAX THEN ICE=ICEMAX 160 IF ICE>0 THEN LINE (XDF*(H-DS),340)-(XDF*(H-DS),330):ICEH=ICEH+1'ice hours 170 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*(H-DS),349)-(XDF*(H-DS),345)'months 180 NEXT H 190 CLOSE #1 200 PRINT TMIN,ICEH 210 END
-25.96 5468 with 440 pounds of ice outdoors
-25.96 4000 with 104 pounds of ice indoors
Nick
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<%-name%>
• posted on March 3, 2005, 3:42 pm

That knocks the natural cooling hours down from 5468 to 3675 hours (42%) of a TMY2 year in Massena NY, but it still covers November through March, the difficult PV period.
We could squeeze 44 vertical 1-liter water bottles into a 16x24x15" deep GE freezer compartment, with 14 ice cube trays on top, ie 104 pounds of water. There's no need to run the fan when the return temp is much less than 32 F, which indicates that the ice is all frozen.
10 SCREEN 9:KEY OFF:CLS 20 DAYSTART=0'display start time (days) 30 DSÚYSTART*24'display start time (hours) 40 RANGE‡60'display range (hours) 50 LINE (0,0)-(639,349),,B:XDFd0/RANGE:YDF=2.69 60 FOR TR=-30 TO 30 STEP 10'temp ref lines 70 LINE (0,349-YDF*(TR+30))-(639,349-YDF*(TR+30)):NEXT 80 ICEMAX4*104'max ice storage (Btu) 90 OPEN "fridge" FOR INPUT AS #1:LINE INPUT#1,H\$ 100 FOR H=1 TO 8760'hours of typical (TMY2) year 110 INPUT#1,MONTH,DAY,HOUR,TDB 120 IF TDB<TMIN THEN TMIN=TDB'find min temp (F) 130 PSET(XDF*(H-DS),349-YDF*(TDB+30))'plot outdoor temp (F) 140 IF TDB>32 GOTO 160'don't make ice 150 IF ICE<ICEMAX THEN ICE=ICE+(32-TDB)*44:FANúN+1'make ice 160 ICE=ICE-136'subtract fridge cooling (136 Btu/h) 170 IF ICE<0 THEN ICE=0'lower ice limit (Btu) 180 IF ICE>ICEMAX THEN ICE=ICEMAX 190 IF ICE>0 THEN LINE (XDF*(H-DS),340)-(XDF*(H-DS),330):ICEH=ICEH+1'ice hours 200 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*(H-DS),349)-(XDF*(H-DS),345)'months 210 NEXT H 220 CLOSE #1 230 PRINT TMIN,ICEH,2.6*FAN/1000'fan energy (kWh/year)
-25.96 3675 5.0934
Nick
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<%-name%>
• posted on March 5, 2005, 2:53 pm
Anthony suggested an outdoor icehouse. Looks like an 8' cube with R14 walls and 2 108 cfm 6.7 W 24 VDC icemaking fans could keep an indoor fridge cool all year. Could be an interesting lawn sculpture, with another solar-heated cube for house heating on cloudy days.
10 SCREEN 9:KEY OFF:CLS 20 DAYSTART=0'display start time (days) 30 DSÚYSTART*24'display start time (hours) 40 RANGE‡60'display range (hours) 50 LINE (0,0)-(639,349),,B:XDFd0/RANGE:YDF=2.69 60 FOR TR=-30 TO 30 STEP 10'temp ref lines 70 LINE (0,349-YDF*(TR+30))-(639,349-YDF*(TR+30)):NEXT 80 CUBESIZE=8'ice box size (ft) 90 GC=6*CUBESIZE^2/14'box conductance (Btu/h-F) 100 ICEMAX4*62.33*CUBESIZE^3'max ice storage (Btu) 110 OPEN "fridge" FOR INPUT AS #1:LINE INPUT#1,H\$ 120 FOR H=1 TO 8760'hours of typical (TMY2) year 130 INPUT#1,MONTH,DAY,HOUR,TDB 140 IF TDB<TMIN THEN TMIN=TDB'find min temp (F) 150 PSET(XDF*(H-DS),349-YDF*(TDB+30))'plot outdoor temp (F) 160 IF TDB>29 GOTO 180'don't make ice 170 IF ICE<ICEMAX THEN ICE=ICE+(32-TDB)*216:FANúN+1'make ice 180 ICE=ICE-200'subtract fridge cooling (200 Btu/h) 190 ICE=ICE-(TDB-32)*GC'subtract ice cube heating 200 IF ICE<0 THEN ICE=0'lower ice limit (Btu) 210 IF ICE>ICEMAX THEN ICE=ICEMAX 220 IF ICE>0 THEN LINE (XDF*(H-DS),340)-(XDF*(H-DS),330):ICEH=ICEH+1'ice hours 230 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*(H-DS),349)-(XDF*(H-DS),345)'months 240 NEXT H 250 CLOSE #1 260 PRINT TMIN,ICEH,2*6.7*FAN/1000'fan energy (kWh/year)
-25.96 8760 25.4332
Nick