Posted by *nicksanspam* on December 7, 2006, 1:34 pm

*>>Can you do better than coin flip for modeling the frequency and run-length*

*>>of cloudy days in Philly, by using some weather/insolation database?*

*>Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.*

Here's a direct gain version:

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)

20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88

30 FOR TR= 60 TO 80 STEP 10'temp ref lines

40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT

50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)

60 LINE INPUT#1,S$'read header

70 CITY$=MID$(S$,8,25)

80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60

90 L=PI*LAT/180'Phila latitude (radians)

100 RHOG=.6'ground reflectance

110 CCUBEB700!'cube thermal capacitance (Btu/F)

120 GCW=4*64/15.5'cube wall and ceiling conductance (Btu/h-F)

130 TCp'initialize cube temp (F)

140 TMIN0'initialize min cube temp (F)

150 FOR H=1 TO 8760'hour of year

160 LINE INPUT#1,S$

170 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)

180 DAY=VAL(MID$(S$,6,2))'day of month

190 HOUR=VAL(MID$(S$,8,2))-.5'hour of day

200 N=1+H/24'day of year (1 to 365)

210 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)

220 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp

230 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months

240 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)

250 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)

260 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)

270 T=HOUR'solar time (EST)

280 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)

290 W=2*PI*(T-12)/24'hour angle (radians)

300 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)

310 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)

320 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall

330 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)

340 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)

350 WLOSS=(TC-TDB)*GCW'cube wall loss (Btu)

360 TC=TC+(SGAIN-WLOSS)/CCUBE'new cube temp (F)

370 IF TC>80 THEN TC'limit max cube temp (F)

380 PSET(XDF*H,349-YDF*(TC-10))'plot cube temp

390 IF MONTH AND TC<TMIN THEN TMIN=TC

400 NEXT H

410 PRINT TMIN

70.01348 (F)

It needs 42,700 pounds of water (685 ft^3, vs 512 for an 8' cube :-)

to stay at least 70 F in December. How much would it need with

indirect gain?

Nick

Posted by *nicksanspam* on December 7, 2006, 5:55 pm

*>>>Can you do better than coin flip for modeling the frequency and run-length*

*>>>of cloudy days in Philly, by using some weather/insolation database?*

*>>*

*>>Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.*

*>Here's a direct gain version:*

*>110 CCUBEB700!'cube thermal capacitance (Btu/F)...*

*>It needs 42,700 pounds of water (685 ft^3, vs 512 for an 8' cube :-)*

*>to stay at least 70 F in December. How much would it need with *

*>indirect gain?*

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)

20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88

30 FOR TR= 60 TO 80 STEP 10'temp ref lines

40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT

50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)

60 LINE INPUT#1,S$'read header

70 CITY$=MID$(S$,8,25)

80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60

90 L=PI*LAT/180'Phila latitude (radians)

100 RHOG=.6'ground reflectance

110 CCUBEU00!'cube thermal capacitance (Btu/F)

120 OWG=4*64/15.5'non-south wall and ceiling conductance (Btu/h-F)

130 SWGd/(15.5+1/.58)'non-south wall and ceiling conductance (Btu/h-F)

140 TCp'initialize cube temp (F)

150 TMIN0'initialize min cube temp (F)

160 FOR H=1 TO 8760'hour of year

170 LINE INPUT#1,S$

180 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)

190 DAY=VAL(MID$(S$,6,2))'day of month

200 HOUR=VAL(MID$(S$,8,2))-.5'hour of day

210 N=1+H/24'day of year (1 to 365)

220 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)

230 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp

240 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months

250 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)

260 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)

270 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)

280 T=HOUR'solar time (EST)

290 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)

300 W=2*PI*(T-12)/24'hour angle (radians)

310 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)

320 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)

330 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall

340 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)

350 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)

360 IF SGAIN<0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG

370 WLOSS=(TC-TDB)*GCUBE'cube loss (Btu)

380 TC=TC+(SGAIN-WLOSS)/CCUBE'new cube temp (F)

390 IF TC>80 THEN TC'limit max cube temp (F)

400 PSET(XDF*H,349-YDF*(TC-10))'plot cube temp

410 IF MONTH AND TC<TMIN THEN TMIN=TC

420 NEXT H

430 PRINT TMIN

70.01386

About 8X less, which would actually fit into the 8' cube :-)

How much for the ceiling mass version with a higher temp swing?

Nick

Posted by *nicksanspam* on December 8, 2006, 12:59 pm

*>>>>Can you do better than coin flip for modeling the frequency and run-length*

*>>>>of cloudy days in Philly, by using some weather/insolation database?*

*>>>*

*>>>Sure. A simulation using NREL's Phila TMY2 hourly weather data file.*

Here's a ceiling mass version:

10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)

20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88

30 FOR TR= 60 TO 80 STEP 10'temp ref lines

40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT

50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)

60 LINE INPUT#1,S$'read header

70 CITY$=MID$(S$,8,25)

80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60

90 L=PI*LAT/180'Phila latitude (radians)

100 RHOG=.6'ground reflectance

110 CCEIL17!'ceiling capacitance (Btu/F)

120 CEGd/15.5'ceiling conductance (Btu/h-F)

130 OWG=3*64/15.5'non-south wall conductance (Btu/h-F)

140 SWGd/(15.5+1/.58)'non-south wall and ceiling conductance (Btu/h-F)

150 TRp'constant room temp (F)

160 TCp'initial ceiling temp (F)

170 TMIN0'initial min ceiling temp (F)

180 FOR H=1 TO 8760'hour of year

190 LINE INPUT#1,S$

200 MONTH=VAL(MID$(S$,4,2))'month of year (1-12)

210 DAY=VAL(MID$(S$,6,2))'day of month

220 HOUR=VAL(MID$(S$,8,2))-.5'hour of day

230 N=1+H/24'day of year (1 to 365)

240 TDB=VAL(MID$(S$,68,4))*.18+32'dry bulb temp (F)

250 'PSET(XDF*H,349-YDF*(TDB-10))'plot dry bulb temp

260 IF DAY=1 AND HOUR=.5 THEN LINE (XDF*H,349)-(XDF*H,345)'tick months

270 IGLOH=VAL(MID$(S$,18,4))*.317'global horizontal radiation (Btu/ft^2)

280 IDIF=VAL(MID$(S$,30,4))*.317'diffuse horizontal radiation (Btu/ft^2)

290 IDIR=VAL(MID$(S$,24,4))*.317'direct normal radiation (Btu/ft^2)

300 T=HOUR'solar time (EST)

310 D=PI*23.45/180*SIN(2*PI*(284+N)/365)'declination (radians)

320 W=2*PI*(T-12)/24'hour angle (radians)

330 X=COS(D)*SIN(L)*COS(W)-SIN(D)*COS(L)

340 THETAI=-ATN(X/SQR(-X*X+1))+PI/2'incidence angle to s wall (radians)

350 IF THETAI>=PI/2 THEN THETAI=PI/2'beam sun behind wall

360 SS=IDIR*COS(THETAI)+IDIF/2+IGLOH*RHOG/2'sun on south wall (Btu/ft^2)

370 SGAIN=.8*64*SS-(TC-TDB)*.58*64'solar gain/loss (Btu/h)

380 IF SGAIN<0 THEN SGAIN=0:GCUBE=OWG+SWG ELSE GCUBE=OWG

390 CLOSS=(TR-TDB)*GCUBE+(TC-TDB)*CEG'cube loss (Btu)

400 TC=TC+(SGAIN-CLOSS)/CCEIL'new ceiling mass temp (F)

410 IF TC>130 THEN TC0'limit max ceiling temp (F)

420 PSET(XDF*H,349-YDF*(TC-10))'plot ceiling temp

430 IF MONTH AND TC<TMIN THEN TMIN=TC

440 NEXT H

450 PRINT TMIN

70.02401

Only 1200 pounds, here assumed all under the ceiling, although putting most

of it under the floor would be more practical and efficient. With about

18h(70-30)20 = 14.4K Btu of overnight heat, the ceiling only needs about

14.4K/(130-70) = 240 pounds of water, or less all around, if the cube is

70 F for 12 hours per day and 50 for the other 12 hours.

Nick

Posted by *nicksanspam* on December 8, 2006, 1:20 pm

*> *

*>>>>>Can you do better than coin flip for modeling the frequency and run-length*

*>>>>>of cloudy days in Philly, by using some weather/insolation database?*

*>>>>*

*>>>>Sure. A simulation using NREL's Phila TMY2 hourly weather data file.*

*>Here's a ceiling mass version:*

*>10 SCREEN 9:KEY OFF:CLS:PI=4*ATN(1)*

*>20 LINE (0,0)-(639,349),,B:XDF=.073:YDF=3.88*

*>30 FOR TR= 60 TO 80 STEP 10'temp ref lines*

*>40 LINE (0,349-YDF*(TR-10))-(639,349-YDF*(TR-10)):NEXT*

*>50 OPEN "13739.tm2" FOR INPUT AS #1'NREL TMY2 file name (Phila)*

*>60 LINE INPUT#1,S$'read header*

*>70 CITY$=MID$(S$,8,25)*

*>80 LAT=VAL(MID$(S$,40,2))+VAL(MID$(S$,43,2))/60*

*>90 L=PI*LAT/180'Phila latitude (radians)*

*>100 RHOG=.6'ground reflectance*

*>110 CCEIL17!'ceiling capacitance (Btu/F)...*

*>Only 1200 pounds, here assumed all under the ceiling, although putting most*

*>of it under the floor would be more practical and efficient. With about*

*>18h(70-30)20 = 14.4K Btu of overnight heat, the ceiling only needs about*

*>14.4K/(130-70) = 240 pounds of water, or less all around, if the cube is*

*>70 F for 12 hours per day and 50 for the other 12 hours.*

Only 879 pounds altogether, with the night setback.

110 CCEIL9!'ceiling capacitance (Btu/F)...

390 IF T>6 AND T<18 THEN TRp ELSE TRP'night setback temp

400 CLOSS=(TR-TDB)*GCUBE+(TC-TDB)*CEG'cube loss (Btu)...

460 PRINT TMIN

70.00746

Nick

Posted by *daestrom* on December 8, 2006, 7:03 pm

*>>nicksanspam@ece.villanova.edu writes:*

*>>*

*>>>*

*>>> >>>You don't seem to account for any heat to 'charge' the under-floor *

*>>> >>>tank.*

*>>> >>*

*>>> >> I figure that's already accounted for, since most of the heat that *

*>>> >> leaks*

*>>> >> from the tank moves up through the floor...*

*>>> >*

*>>> >What I was referring to is that how does the tank get heated in the *

*>>> >first*

*>>> >place? If it's from circulating water from the overhead storage, then *

*>>> >your*

*>>> >calcs for the TS (temperature of the sun space) don't consider this.*

*>>>*

*>>> I disagree. The tank loss heats the cube. Try mentally drawing a box *

*>>> around*

*>>> the cube and balancing the solar energy that enters with the heat energy*

*>>> that leaves.*

*>>*

*>>Nick, you keep talking about tank heat loss.*

*>>Daestrom keeps asking about tank heat gain...*

*> The tank won't need much heat, since it only needs to provide heat on*

*> cloudy days, ie it is rarely used.*

*THAT* makes no sense. You've been saying the ratio of cloudy/sunny days is

1:1, yet you say the storage tank '..is rarely used'. Care to explain that

a bit?

If the *average* solar gain is 1000 BTU/ft^2-day, then if you assume 0

BTU/ft^2-day for cloudy days, you must also be *assuming* 2000 BTU/ft^2-day

on 'sunny' days. How hot does the sunspace get on such days, and how much

are the losses increased by this higher temperature?

If you're not storing some of that heat for the 'cloudy' day, then how cold

does the d-cube get on a cloudy day? Using your original code, with zero

solar gain for just 24 hours it gets pretty chilly in your d-cube. Your

obvious reply is to use the heat stored in the under-floor tank, but you

haven't put any heat *into* the tank. Besides, now we hear that it's only

'rarely used'. Meaning not on a cloudy day that happened to occur after a

sunny day?

*> We might pump tank water up through*

*> the ceiling to keep the tank hot. Or heat it with Big Fins or fin-tube*

*> in a sunspace, or use a closet with some overnight heat storage mass.*

Yes, you *might* do that. But your model doesn't take that into account, so

as far as dispelling any doubt, it hasn't done much.

<snip>

*>>> Sure, if you wanted to recharge the tank completely in 5 days,*

*>>> after 5 cloudy days in a row, but is that a requirement?*

*>>*

*>>Yes.*

*> Why?*

Well, you haven't provided *ANY* recharging of the tank, so what are we to

expect? That Maxwell's daemons do it for you?

If cloudy/sunny days are 50/50, it seems quite reasonable that a 5-day

cloudy storage should be recoverable in 5 sunny days. But you haven't shown

this a possibility in your model. You haven't shown *any* recovery, so how

long would it take in your design?

*>>There has to be some recharge requirement,*

*> Sure.*

*>>otherwise, we can assume the heat tank is charged at the beginning*

*>>of the winter, and once we have 5 cloudy days, the heat tank*

*>>is discharged and useless for the rest of the year.*

*> If it takes 2 average weeks to recharge,*

Would you have evidence for this article of faith? Will the tank recharge

at all?

*> how much does that lower*

*> the maximum solar heating fraction?*

You started this thread with 100% solar. Are you now suggesting some

fraction less is acceptable?

*>>If the recharge rate is less than the depletion rate,*

*>>then this design does not fulfill the design goal*

*>>in any location where there are more cloudy days than sunny days.*

*>>So a complete recharge in 5 days sounds like the minimum requirement.*

*> You might think harder about this. What is "the depletion rate"? How often*

*> do strings of 5 cloudy days occur? This is like "the gambler's ruin," with*

*> a little gain, vs even odds.*

The 'gambler' is working with perfectly independent events. Each toss of

the coin; roll of the die; or other 'game of chance' is independent of all

previous runs. The idea that a 'streak' of good luck or bad luck happens to

gamblers is false in games of chance. I think we can agree on that.

But weather patterns aren't completely independent events. If a storm front

heads towards Phila from the west and you have your first cloudy day, the

chances of tomorrow *also* being cloudy are much higher than 50/50.

You're trying to apply statistics for indpendent events to weather

prediction. And any fool knows that doesn't work. After all, if it snows

on average 90 days in a year, that doesn't mean the chances of snow on July

4 is 90/365.

Consider this. You say that cloudy days are about 50/50 in your area. So

if a winter storm heads into the area Monday morning, you're methodology

*assumes* that the chances of clouds on Tuesday are completely independent

of Monday's storm. That's a good one, go on, pull the other leg...

*>>What is the ratio of cloudy days to sunny days in Philly?*

*> About 1:1.*

*>>What is the rate at which the heat tank can be recharged on a sunny day?*

*> About twice the average-day rate.*

Have you any evidence of this article of faith? We haven't seen you present

any. Recharging at 'twice the average-day rate'?? And the average-day rate

of charging is zero?

*>>> >So to 'recharge' in five days, your system has to develop '2X' BTU/day.*

*>>> >1X for heating the house, and 1X for dumping into storage.*

*>>>*

*>>> Sure, in the above scenario, which is unlikely to be required.*

*>>*

*>>How do you calculate it is an unlikely requirement?*

*> Cloudy days are like coin flips.*

No, they are not. I've explained this to you in previous threads, yet you

persist in this fantasy. Just because you can't be bothered with the higher

statistics to model cloudy/sunny days more accurately, doesn't make your

gross assumption correct.

<snip>

*>>You missed the point again Nick.*

*>>By this point, I'm beginning to think it is deliberate.*

*> You might think harder about this :-) If sun shines in through a window*

*> and heats a black spot on the floor to 100 F, that does not increase*

*> the amount of heat the cube needs to stay 70 F on a 30 F day...*

*>> On a cloudy day, there is no solar gain, so all the heat comes*

*>>from the tank, right?*

*> Right (in this simple model, with no internal heat gains.)*

*>> On an Average day, the solar gain equals the loss,*

*>>so there is no net change in the heat tank, right?*

*> I think we need a little tank gain on an average day.*

FINALLY! Now, just how much tank gain is needed? Above you suggested 5 to

6 days to recharge the tank 'seems to work well'. So, to recharge the tank

in 5 to 6 days, how much 'little tank gain' is that? More than a 'little'

isn't it?

*>> On a sunny day (only), the heat tank charge can be increased, right?*

*> On a sunnier or warmer than average day, when there is excess solar heat.*

Changing the conditions again. What a joke! Heck Nick, you can gain a lot

in July!! But for every 'sunnier' or 'warmer than average day', there is a

'cloudier' or 'colder than average day'. Consider the definition of

'average'.

Are you now conceding that your idea won't work in 'below average' weather?

*>>Lets check the surface to volume ratio. Ok, thats right.*

*>>Dcube has 6x8x8 surface to 8x8x8 volume = 3:4 = 0.75 ratio*

*>>A 42x42x8 house has 42x42x2+42x8x4= 3108 surface to 42x42x8 vol = 0.22 *

*>>ratio*

*> Counting the floors, which I ignore. How about a 24' cube?*

It's still nothing more than a 'doghouse' with no practical living

potential. Two and a half stories tall? Still no

windows/doors/ventilation?

What about the fact that if you start out with a depleted storage tank and

you have several sunny days in a row, the first day will increase the

storage temperature by X degrees. But the second day the sunspace will be

operating at higher temperatures so it will only raise the storage

temperature by (1-<deltaloss>)*X. And the third day by (1-<deltaloss>^2)*X

and so on. The hotter the tank gets the fewer BTU's that will be collected

and stored.

daestrom

>>Can you do better than coin flip for modeling the frequency and run-length>>of cloudy days in Philly, by using some weather/insolation database?>Sure. A simulation using NREL's Phila TMY2 hourly measured weather data file.