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re: toby in need of help
6 oct 2000
toby spreads more confusion:
>suppose the outside temperature, averages 27f in a typical winter day.
>a 67f, 200, 40 *40 house would lose 24hr*(67-27)f*200 = 192 kbtu/day.
ok. a 40x40x8' tall house with 200 btu/h-f of heat conductance might be
"airtight" with 0.2 air changes per hour, ie 40x40x8x0.2/60 = 43 cfm of
air leaks accounting for about 43 btu/h-f of the 200 btu/h-f; 128 ft^2
(8% of the floorspace) as r4 windows adds 32. an r40 ceiling adds 40,
leaving 85 btu/h-f for 1152 ft^2 of r14 walls, or 4" r16 sip walls...
(but the original poster wanted a 74 f house, no?)
>the 3, 300 btu/day residents, occupying it for 15 hrs each day would
>supply 3*15*300 = 14k btu/day...
they'd be our 300 btu per hour residents, no?
>the fridge and freezer, lights, tv, etc...generating: 29k... thus,
>a thermal cistern would need to supply 192-29 – 14 = ~150k/day.
ok, on a cloudy day...
>the thermal cistern might discharge the heat, lowering it’s temp
>from 82f to 72f.
how will the cistern "discharge the heat"?
and what happened to the 95 to 75 f swing?
>sand, with a moisture content of 10%mc, has a thermal capacity of about
>30 btu/f/cuft...
arguably. maybe it's 30 with 20% moisture...
>let’s make it 2’ thick... so ccist =... 96k btu/f
>the house needs 150k btu/day from the sun, but we need to store more
>than that because it might be totally cloudy one day or several days
>and you might not gather any heat from the sun and you want 5 days
>worth of heat stored up. suppose that, the monthly average, during the
>day, says the sky is “clear” 60% of the time… so, in a 30day month,
>30*0.6 = 18days are sunny and 12 days are cloudy.
ok.
>the 18 sunny days, need to gather enough heat for all 30 days plus
>5 extra days.
i'd say we only need to gather enough heat during the 18 sunny days
for the 30 days, including the 12 cloudy days. we need to distinguish
"gathering" (which requires an energy balance, collecting as much
energy over 18 sunny days as we use over 30 days) from "storing"
(making sure that the thermal store can supply heat for 5 cloudy days
in a row, which requires certain max and min storage temps and thermal
capacitance and heat transfer rate.) "gathering" may be done at a lower
temperature than "storing," and it requires more peak charging power
but less heat capacity, enough for 1 day vs 5.
>letting d be the amount of btu’s to store for 1 day:
let's say "the amount to gather."
> (30+5)days * 150kbtu/day = 18days * d; where d is the heat gathered on
>the sunny day. d = 290k btu/day.
how about 30x150k = 18d, so d = 250k? heat energy that is stored is not
necessarily used. it may not be necessary to gather more energy on a sunny
day in order to store 5 day's worth of heat, especially if the store
is under the house, and it operates at a low temperature.
>actually, some heat will be lost through the floor of the cistern, but
>also, some heat will not be needed because i didn’t count the solar
>radiation entering the house through the windows.
let's count that. say we collect 1000 btu/ft^2x50%x128ft^2 = 64k btu on
a sunny day (vs an average day) via the windows, leaving 186k that must
be collected by other means on a sunny day.
>let’s let the sun shine directly on water-filled black irrigation
>tubing.
ok. maybe it hangs on the north wall of a sunspace, in some sort of
festoon that can drain down at night.
>we might expect the water in it to gain only 50% of the solar radiation, so
>we need to double the size of the solar collector to 2*290k = 580k btu/day.
we might try to calculate the gain and loss, vs just doubling the size.
>suppose 1000 btu/sqft/day of solar radiation shines down on the southwall
>and another 1000 btu on the floor, through glazing, upon the ~80sqft
>surface area of 1000’ of 1” diameter, water-filled, irrigation
>tubing in an enclosure. that’s 2000 btu/sqft of solar radiation.
half on the wall and half on the floor, or maybe all of it on a 45 degree
sloped surface, like sloped big fins in a sunspace, near the glazing...
>suppose the water is able to absorb 50% of this
(0.5*80sqft/tube*notube*2000btu/sqft/day = 580kbtu/day, and the rest
>is re-radiated out or lost back out to the environment.
suppose the tubes absorb about 100% of it through a single layer of r1
glazing with 90% solar transmission, ie they gain 1800 btu/ft^2-day. at
130 f in an 80 f sunspace, they might lose about 6h(130-80)1ft^2xu1.5
= 450 btu/day, gaining 1350 btu/ft^2 or 108k btu/tube on a sunny day.
>then the number of tubes we need (notube) = 580k/( 0.5*80*2000) = 6 tubes.
or maybe 2 tubes, enough to gather 186k btu on a sunny day.
>we want to circulate the water in these tubes into the sandbox, so
>let’s put 6 tubes in the sanbox as well. let’s bundle them together and
>zigzag the tubing, in the middle depth and throughout the sandbox so
>that it forms about 40 rows each row 40 foot long, so that the heat travels
>from the pipe thru 1 foot of dirt, before it hits the insulated walls.
this will work better with the tubes unbundled, say 6000' (wasn't this
3000' before) of tubing in n 40' rows, ie 150 40' rows on 0.27' (3.2")
centers. now the heat only needs to travel 1.2" horizontally, but it
still needs to travel a foot vertically. let's vary the tube depth too.
let's put the tubes on an equal xy grid looking horizontally, d inches
on center; 150(d/12)^2 = 40x2', so d = 8.8", 3 tubes vertically, and
50 horizontally. now the heat only has to travel 4.4" max. what happens
if one of these tubes kinks or starts to leak?
>...since there are 6 tubes, the velocity through each tube is... 16ft/sec
if you say so.
>the 6 tubes in the solar collector hold:
> 6 tubes*1000’/tube*[pi*(1/2”*1’/12”)^2]sqft * 8gal/cuft = 260 gal
so what?
>since the water is circulating at 4gal/min, the water is exposed to the
>sun for: 260gal/(4g/min) * 1hr/60min = ~1 hour. in a 6 hour day, the
>solar collector gathers 580 kbtu, so if 50% of this energy enters the
>water, the 260 gal of water will heat up in 1 hour by dtw:
>1btu/lb/f * dtw*260gal/hr * 8lb/gal = 50% * 580k btu/day * 1day/6hrs
here's where your logic starts to go screwy...
>dtw=0.5*580k/6/260/8 = 23f/hr
so what? the water isn't being heated in batches. this way of looking
at things won't work. the water is losing heat to the sandbox at the
same time that it's gaining heat from the collector...
>let’s say the sandbox and the water in it start out (in the morning) at
>80f, at the end of the 1 hour in which the water is exposed to the sun
>in the collector, the temperature of the water entering the sandbox is:
>80+23= 103f=tw
nonono. the water is losing heat to the sandbox during that hour.
>according to “earth-coupled heat transfer” by hart and couvillion –
>especially around page 59-61: the convective water resistance,
>
>rconv = 1/(h*pi*di); where: di = inside pipe diameter in inches, and h
>is ashrae’s heat transfer coefficient and given by:
>
>h = [150*(1+0.011tw)v^0.8]/[di^0.2];
>where tw = the water temperature,f =103f;
>v=the velocity in the tube,ft/sec=8ft/sec;
didn't you say 8 ft/s?
>di=the inside diameter of the pipe,inch = 1”
>
>h = [150*(1+0.011*103)16^0.8]/[(1” * 1’/12”)^0.2] = 2940
it's hard to tell with these ascii codes, but the numerator looks like
150*(1+0.011*103)16^0.8 = 2940.2 to me, and the denominator looks like
(1/12)^0.2 = 0.608, which makes h = 4833(?)
>rconv = 1/(2940 *pi * 1” ) = 0.000108 f*sqft*hr/btuh/ft of each pipe
>(i believe this is the resistance in a 1 foot piece of pipe.)
the resistance from where to where?
>heat is flowing from the 103f water to the 80f sand. the heatflow thru
>each pipe is very roughly: (103-80)/rconv...
nonsense. the sand has resistance, so the part near the pipe will be warmer.
>i am not sure why the book has “btuh”. normally resistance is in
>f*sqft*h/btu. anyways, to be safe, since the water is in the pipe for 1
>hour, i’m going to multiply this twice to get the right units….
>rws=1hr*1hrs*0.007518 f*sqft/btuh/ft = 0. 007518 f*ft*h/btu
hell, let's multiply by pi^2, just for kicks.
>the thermal capacitance of the water in a 1foot section...
>cw = 6 tubes * 1’*[pi*(1/2”*1’/12”)^2]sqft*64lb/cuft*1 btu/lb/f
>= 2 btu/f/ft
who cares?
>thus, the thermal time constant is...
irrelevant.
>now, mr. pine believes we ought to include the resistance of the sand..
yes.
>part of the heat travels up into the 67 house through r20 insulation
>into the house...
how can the sandbox keep the house warm on a cloudy day with r20 insulation
above it? the house needs about 150k btu/day ie 6250 btu/h (not much), but
it looks like the floor can only supply this if the sand temperature is
67f+6250xr20/1600ft^2 = 145 f.
>and part travels down through r20 insulation into the 67f ground.
sounds like half of the sand's heat is wasted.
>...if we use mr. pine's numbers of r0.44/foot...
that was r0.44 per inch, according to the 1993 ashrae hof.
>...it looks like the system works fairly well...
perhaps you should look again.
nick
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