Hybrid Car – More Fun with Less Gas

Re: Another Deployable Doubt Dispeller - Page 2

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

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
 
 

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
  

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
 
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
 

*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?


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>

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?


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


You started this thread with 100% solar.  Are you now suggesting some
fraction less is acceptable?


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...


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?



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>

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?


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?


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


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