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re: adjustment of dd base temperature for superinsulated structures
7 feb 1998
erich knight  wrote:
> is my understanding that for superinsulated designs, a lower base
>temperature should be used, relative to carry-over thermal performance.

probably. "the superinsulated home book" by j.d. ned nisson and gautam dutt
suggests using a structure's "balance-point temperature" (bpt) as a base,
ie the internal temperature it would have if the heating system were turned
off, which depends on its effective thermal conductance (u) and internal
heat gain (g) and solar heat gain through windows (i). the bpt usually
varies with time. bpt = ti - (g+i)/u, where ti is the indoor temperature.

for instance, a 32' cube surrounded by 2" (r10) styrofoam insulation with
no air infiltration and an r2 transparent south wall has u = 2560ft^2/r10
+ 512ft^2/r2 = 512k btu/h-f. if the south wall admits an average of 512k
btu/day ie i = 21k btu/h of solar heat. if the monthly internal electrical
consumption is 3,000 kwh, ie 100 kwh/day or 14k btu/h, and 8 people live
inside, and they contribute 8x250 = 2k btu/h, then g = 16k btu/h. if the
indoor temp is 65 f, bpt = 65 - (16k+21k)/512 = -7 f, so this cube wouldn't
need any extra heat until the outdoor temperature reached 7 degrees below
zero. on a cloudy day, the balance point would be 65 - 16k/512 = 34 f,
using "ohm's law for heatflow."

if the south glazing were a low-thermal mass sunspace over an r10 wall,
u = 307 btu/h-f and bpt = 13 f on a cloudy day, ie the cube stays about
52 f warmer than the outdoors, because of its internal heat dissipation. 
on an average 30 f phila december day, the sunspace might lose about
6h(80f-30f)512ft^2/r2 = 77k btu of heat, ie 3k btu/h, so without any
backup heating system, the cube has a theoretical indoor temperature
ti = 30 + (16k+21k-3k)/307 = 141 f. if it's vented to keep it 70 f on
an average winter day, and it has 8" solid masonry walls inside the foam
with c = 4x512ft^2x16btu/f-ft^2 = 32k btu/f and rc = 32kbtu/f/(307btu/h-f)
= 104 hours, and the outdoor temp is, say -10 f, the indoor temp will
slowly drop to about -10 + 52 = 42 f. over 2 such cloudy days, the
indoor temp would drop to 42 + (70-42)exp(-48/104) = 60 f (or 68 f,
if we use more insulation, or keep bombing iraq to keep the price of
heating oil low.)

>  any tables available to convert the base 75 f cooling degree days?

the table in the back of the superinsulated home book says philadelphia
has 5,300 (f) heating degree-days at a base temperature of 65 f, and
3,213 at 55 f. you can also calculate this based on monthly average
temperatures as explained on pages 28.5 and 28.6 of the 1993 ashrae hof:

10 dim ta(12),d(12),sigm(12)'monthly average temps and # days/mo
20 data 30.4,33,42.4,52.4,62.9,71.8,76.7,75.5,68.2,56.4,46.4,35.4
30 data 31,28,31,30,31,30,31,31,30,31,30,31
40 for m=1 to 12'read average temp for each month of year
50 read ta(m)
60 tyear=tyear+ta(m)
70 next
80 tyear=tyear/12'average yearly temp (f)
90 print "avg yearly temp (f):",tyear
100 for m=1 to 12'read days/month for each month of year
110 read d(m)
120 sigsum=sigsum+(ta(m)-tyear)^2
130 next
140 sigyear=sqr(sigsum/12)'standard deviation of monthly temps
150 print "standard dev (f):",sigyear
160 tbal=55'balance temperature (f)
170 print "balance temp (f):",tbal
180 print
190 print" month"," theta"," dd"
200 for m=1 to 12
210 sigm=3.54-.029*ta(m)+.0664*sigyear
220 thetam=(tbal-ta(m))/(sigm*sqr(d(m)))
230 am=1.698*sqr(d(m))
240 arg=am*thetam
250 larg=exp(-arg)+exp(arg)
260 fac=thetam/2+log(larg)/(2*am)
270 ddm=sigm*d(m)^1.5*fac
280 print m,thetam,ddm
290 ddmt=ddmt+ddm
300 next m
330 print "total degree days (dd):",ddmt

avg yearly temp (f):         54.29168
standard dev (f):            16.12133
balance temp (f):            55

 month         theta         dd

 1             1.184892      762.5998
 2             1.137994      616.0001
 3             .6693649      390.5999
 4             .1535796      79.5254
 5            -.509225        0.00166
 6            -1.213187       0
 7            -1.633353       0
 8            -1.520849       0
 9            -.9154169      0.000013
 10           -0.084524      5.002172
 11            .4809209      258.0036
 12            .9822557      607.5998

total degree days (dd):     2719.333, 

using newer average weather data with 4,954 hdd at 65 f.
this calculation works for cooling degree days too.

>also if iaq is maintained at year round 50% rh, how much may i reduce my
>cooling load requirements?

i'm not sure what you mean by that...


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