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workshop on ohm's law for heatflow
17 oct 2003
1. rich komp, ohm, and newton 

rich komp (who is still alive) says heat moves by conduction (a hot frying
pan handle), convection (including air movement), radiation (the sun), and
phase change (it takes 144 btu to melt a pound of ice and about 1000 btu
to evaporate a pound of water.) rich also says "good engineering is just
the beginning in nicaragua. the social process is very important." tom smith
(is he still alive?) said 

    it's a snap to save energy in this country. as soon as more people
    become involved in the basic math of heat transfer and get a gut-level,
    as well as intellectual, grasp on how a house works, solution after
    solution will appear.

about 300 years ago, isaac newton said the amount of heat that flows through
a wall is proportional to its area and the temperature difference from one
side to the other and its thermal conductance. about 100 years later, georg
ohm said the same about electricity: e = ir, ie a current in amps times a
resistance in ohms (the inverse of a conductance in "mhos," which is ohms
spelled backwards) makes a voltage difference instead of a temperature
difference.  for example, if 6 amps flow through a 2 ohm resistor, we'll see
e = ir = 6ax2ohms = 12 volts across it. another example: 120 volts across
48 ohms makes i = e/r = 120v/48ohms = 2.5 amps flow. the electrical power
p = ie = 120vx2.5a = 300 watts. 

2. power and energy

power is a rate of energy flow over time. energy is the stuff we pay for,
measured in joules or watt-hours or kilowatt-hours or "british thermal units"
(btu), which are no longer used in britain :-) a btu is a quantity of heat,
about the same as the energy in a kitchen match. one btu can heat one pound
(16 ounces) of water one degree f. how many btu are needed to heat 8 ounces
of water from 50 to 200 f to make a cup of tea? (0.5btu/f(200f-50f) = 75btu.)

one watt-hour of energy is equivalent to 3.41 btu, like a number-swapped pi.
how long would it take to heat that tea water with a 300 w immersion heater?
(75x60m/h/(300x3.41) = 4.4 minutes. we might check this with a 300 w 120 v
immersion heater and a watch and a $100 raytek ir thermometer. hobos from
onset computer corp are also useful in heating and cooling experiments. one
$85 version is about the size of a small matchbox and records time samples
of its own temperature and rh, and has jacks for two more temperature probes
or other devices on cables. it can record about 64k samples at intervals
ranging from seconds to hours and dump them into a pc for spreadsheet or
other processing.

people often confuse power and energy, saying things like "my house uses
lots of power" (vs energy) or "my furnace capacity is 50,000 btu," vs btu/h.
power is a rate of energy transfer, just a number. unlike energy, it can't
be used or consumed.

some people confuse heat and temperature, too. a bathtub full of hot water
contains a lot of heat, compared to a candle, but the candle is hotter.
temperature is a measure of heat intensity.

3. thermal ohms?

ohm's law for heatflow (aka newton's law of cooling) uses a temperature vs
a voltage difference, and heatflow is measured in units of power, in watts
or btu per hour, and there's no such thing as a thermal "ohm." the closest
thing is the us "r-value" stamped on insulation boards and batts in hardware
stores. beadboard (expanded white polystyrene coffee cup material) has an
r-value of 4 (ft^2-f-h/btu) per inch. blue or pink or green styrofoam board
is 5 per inch. so is air, for downward heatflow. the air spaces near a single
layer of glass have a combined r-value of about 1. a smooth surface in slow-
moving air loses about 1.5 btu/h-f-ft^2, with an r2/3 airfilm resistance.
a rough surface in v mph air loses about 2+v/2 btu/h-f-ft^2.

tiny cold soap bubbles (1/16" at 50 f) have an r-value of about r3 per inch.
bill sturm's calgary greenhouse filled the space between two polyethylene
film covers with air during the day and soap bubble foam insulation at night. 
he thinks this might make an effective refugee shelter in a cold climate.

fiberglass is r3.5/inch, or half that, if it contains 2% moisture, or even
less, if air flows around or through it. to find the thermal resistance of
a wall, we need to divide the r-value by the wall area. an 8'x10' r20 wall
has a resistance of r20/(8'x10') = 0.2 f-h/btu. we might call this "0.2 fhubs"
or "5 buhfs." we can add buhfs in parallel for several kinds of house walls
and windows, or add fhubs in series for series layers of wall insulation.

if it's 70 f indoors and 30 f outdoors (70f-30f)/0.2f-h/btu = 200 btu/h
of heat power will flow through 1 ft^2 of r5 wall. walls with wooden studs
(r1/inch "thermal bridging") have a low resistance in parallel with the
insulation resistance, which lowers their overall r-value. structural
insulated panels (sips, glued plywood-foam-plywood sandwiches) have less
thermal bridging and fewer air leaks. 

metric (european, canadian, australian...) u-values are 5.68 times bigger
than us u-values. a metric u1 window has 1 w/m^2c of thermal conductance, so
it's equivalent to a low-loss us r5.68 window. a 1 m^2 metric u1 window with
20 c air on one side and 0 c air on the other would pass (20c-0c)1m^2x1w/m^2c
= 10 watts of heat power. 

a 3'x4' us u0.25 (btu/h-f-ft^2) window with 70 f air inside and 30 f outside
would pass (70f-30f)3'x4'x0.25btu/h-f-ft^2 = 120 btu/h of heat, no? an 8'x24'
r24 6" sip wall with a 40 f temperature difference lets (70f-30f)8'x24'/r24
= 320 btu/h of heatflow. how much heat flows through a 24'x32' r64 ceiling
with a 40 f temperature difference? (40fx24'x32'/r64 = 480 btu/h.)

4. a three-dog house?

now we can talk about superinsulated houses. can a person heat her own house
with the help of a few dogs? people at rest generate about 300 btu/h, like
100 w light bulbs. how big can an l foot r10 cube be, if it's 70 f inside and
30 f outside, with six outdoor faces? if 300 btu/h = (70f-30f)6l^2/r10, l
= sqrt(300/24) = 3.54 feet. too small. with r20 walls, l = sqrt(300/12) = 5'.
better, but cramped. a 100 btu/h dog would make l = sqrt(400/12) = 5.8'.
two dogs make l = sqrt(500/12) = 6.5'. three make l = 7.1'.

a frugal 100 kwh per month of indoor electrical use (vs the us average of
833 kwh/mo) would add 100kwhx3412/(30dx24h) = 474 btu/h, and 474+300 btu/h
= (70f-30f)6l^2/r20 makes l = sqrt(774/12) = 8.03 feet. this is becoming
a house :-) but some people define a "solar house" as "one with no other
form of heat," not even electrical usage or creatures-in-residence. 

5. the sun, and weather

suppose the 8' cube has an a ft^2 r2 window that admits 200 btu/h-ft^2 of
sun... 200a = (70f-30f)(a/2+(6x8'x8'-a)/r20) = 20a+12x64-2a makes a = 4 ft^2,
eg a 2'x2' window, but the temperature instantly drops to 30 f when night
falls, unless the cube contains some thermal mass...

keeping it warm at night also requires a larger south window. the national
renewable energy lab (nrel) solar radiation data manual for buildings (the
"blue book") implies that january is the worst-case month for solar house
heating in boulder, co, when 1370 btu/ft^2 of solar heat falls on a south
wall on an average 29.7 f day, and 1370/(65-29.7) = 38.8. pe norman saunders
says the "worst-case month" is the one with the least solar heat per degree
day, on average. this usually turns out to be december or january.

december in boulder is warmer, with less sun, and 1330/(65-31) = 39.1, which
is more than 38.8, so january is the worst-case month. december is the worst-
case month for solar house heating in seattle, where solar heating is harder,
with 420/(65-40.5) = 17.1. january is the worst-case month in albuquerque,
where solar heating is easier, with 1640/(65-34.2) = 53.2. people often fool
themselves about passive solar performance, saying things like "we only burn
one or two cords of wood per year." if a house can keep itself warm in the
worst-case month, it should do fine in other months.

nrel's "blue book" at http://rredc.nrel.gov contains long-term monthly average
solar weather data for 239 us locations. nrel's typical meteorological year
(tmy2) hourly data files and 30-year hourly data files for each location
can be useful for simulations. 

we might keep our 8' cube at 70 f (eg 75 at dusk and 65 at dawn) over an
average december day with an a ft^2 r2 window with 80% solar transmission that
admits 0.8x1370 = 1096 btu/ft^2. if 1096a/24h = (70-30)(a/2+(6x64-a)/r20)
= 18a+768, a = 28 ft^2. a 4'x8' window would do, with a cube conductance of
32ft^2/r2+(6x64-32)/r20 = 34 btu/h-f, about half window and half walls.

6. thermal mass and direct gain, aka "direct loss"

water is cheap, and easily moved. if it moves, it has a low thermal resistance
to heatflow. a square foot of 1" drywall (a board foot) stores 1 btu/f, like a
pound of water. a 95 lb cubic foot of dry sand with a 0.191 btu/f-lb specific
heat can store 0.191x95 = 18 btu/f. a cubic foot of concrete stores about 25
btu/f, vs 64 for a cubic foot of water or steel, but concrete has significant
resistance to heatflow (about r0.2 per inch, vs r0.4 sand), so it's hard to
move heat into or out of thick concrete. a 32 pound 8"x8"x16" hollow concrete
block with a specific heat of 0.16 btu/f-lb stores about 0.16x32 = 5 btu/f. 

an r fhub cube outside c btu/f of thermal capacitance has a "time constant"
rc = c/g in hours. this is the time it takes the difference between the indoor
and outdoor temps to decrease to about 1/3 (e^-1) of its initial value. for
example, if rc = 24 hours, t(h) = 30+(75-30)e^(-h/24) after h 30 f hours and
t(d) = 30+(75-30)e^(-d) after d days, where e^x is the inverse of the natural
log ln(x) function on an $8 casio fx-260 solar-powered calculator. t starts at
75 f when e^(-0) = 1 and ends up at 30 f (the outdoor temp) much later, when
e^(-oo) = 0. between those times, the exponential factor gradually squashes
the initial temperature difference of (75-30) = 45 degrees.

a liter of water weighs 2.2 pounds, so it stores 2.2 btu/f. n 2-liter bottles
inside the cube would make rc = 4.4n/34/24 = 0.00539n days. starting at 75 f,
t(d) = 30+(75-30)e^(-d/0.00539n) = 30+(75-30)e^(-185d/n) d days later. when
65 = 30+(75-30)e^(-185d/n), ln((65-30)/(75-30)) = -185d/n, and n = 738d.

storing heat for 1 day requires 738 2-liter bottles, 2 days takes 1496, and
so on. if clear and cloudy days were like coin flips, storing heat for 1 day
would make the cube's solar heating fraction 50%, with 75% for 2 days, 88%
for 3, 94% for 4, and 97% for 5. the chance of 5 cloudy days in a row would
be like the chance of 5 tails in a row, ie 2^-5 = 0.03. 

but 738x5 = 3,690 is 4 pickup trucks full of bottles, and 16,236 pounds of
water is a good ballast foundation :-) and pet bottle walls leak water vapor.
they might need topping up once a year. we can fit about 9 4" diameter by 12"
long bottles into a cubic foot, so they would occupy 3690/9 = 410 ft^3 of the
cube's 8^3 = 512 ft^3, ie 80% of the living space, rather intrusively. this
fraction shrinks with larger cubes with larger potential heat-storage volume
to heat-losing surface ratios.

7. indirect gain

why look out a black window at night? with an r20 wall between the cube and
a "low-thermal-mass sunspace" containing the window, we can circulate warm air
between the sunspace and living space during the day and stop air circulation
at night, so we can have the daytime gain of the window without nighttime and
cloudy-day heat loss, and the window and thermal mass can be smaller, but it's
hard to store solar heat from warm air, compared to mass in direct sun. we can
stop air circulation at night with a one-way passive plastic film damper hung
over a vent hole in the r20 wall, with a screen that only allows the film to
swing open in one direction. doug kelbaugh invented this "7-cent solution" in
princeton in 1973. 

for room temperature control, the damper might be in series with an automatic
foundation vent like leslie-locke's $12 8"x16" afv-1b. its louvers open as
air temperature rises, but the bimetallic coil spring that opens them can be
reversed to close the louvers as air temp rises, and we can adjust its soft
threshold temperarture by turning the spring mounting screw. nasa satellites
use more expensive versions of this device as "deep-space coolers" that open
to radiate heat as needed. a very low power motorized damper and thermostat
might control the room air temp more accurately. honeywell's $50 6161b1000
damper motor uses 2 watts when moving and 0 watts when stopped. if it runs
for 1 minute per day, that's 0.03 wh. we might power it with a 10 milliwatt
pv cell and a tiny rechargeable battery and some low-power electronics. 

8. airflow and heatflow

one btu can raise the temperature of 55 cubic feet of air 1 f, so 1 cubic
foot per minute (cfm) of airflow with a temperature difference of 1 degree
moves about 1 (60/55) btu per hour of heat. the american society of heating,
refrigeration and air conditioning engineers (ashrae) say a person needs 15
cfm of outdoor air to stay healthy. if this air just leaks through a house,
that adds about 15 btu/h-f to the house thermal conductance, unless the house
has some sort of air-air heat exchanger that preheats incoming cold air with
outgoing warm air. a small periodically-reversing fan in a partition that
divides the house into two parts might turn all the cracks and crevices in
the house envelope into efficient low-rate bidirectional heat exchangers. 

most us houses leak a lot more than 15 cfm/occupant. an old house might leak
2 air changes per hour (ach), eg 2x2400x8/60 = 640 cfm for a 2400 ft^2 one-
story house. a new us house might leak 1 house volume per hour. i've heard
the swedish standard for new houses is 0.025 ach. how do they do that? people
measure airleakiness with blower door tests. they pressurize and depressurize
houses to 50 pascals (0.00725 psi) and measure the airflow in cfm and divide
that by 20 to estimate the natural air leakage in wintertime.

one empirical formula says an h foot chimney with an a ft^2 vents at the top
and bottom and an average temp ti (f) inside the chimney with an outside temp
to has q = 16.6asqrt(hdt) cfm of airflow, where dt = ti-to. the heatflow in
the airstream is qdt = 16.6asqrt(h)dt^1.5.

a square foot of r1 sunspace glazing with 90% transmission might gain 0.9x1370
= 1233 btu over 6 hours on an average december day. it could be one layer of
clear flat replex polycarbonate plastic, which comes in 0.020"x49"x50' rolls
and costs about $1.50 per square foot, or one layer of corrugated dynaglas
"solar siding," which costs about $1/ft^2. both last at least 10 years. we
might make an 8' long x 8' radius quarter-cylindrical sunspace with 3 $2 1x3
beams on 4' centers. i've made these beams by bending 2 12' 1x3s into an 8'
radius, with 1x3 spacer blocks every 2', and deck screws to hold them together.

with a ft^2 of sunspace glazing, we can keep the cube 70 f cube an average day
if 1233a = (70-30)(6a/2+18a/20+24(6x24^2-a)/r20), so a = 18.3 ft^2. a 4'x5'
window would do, with a cloudy-day cube conductance of 6x64/20 = 19.2 btu/h-f.

if 0.9x250 = 225 btu/h-ft^2 of peak sun enters a ft^2 of r1 sunspace glazing
and 70 f air near the glazing (on the south side of a dark screen north of the
glazing, with warmer air north of the screen) loses (70-30)1ft^2/r1 = 40 btu/h,
the net gain is 185 btu/h-ft^2, or 3.7k btu/h (1.1 kw) for 20 ft^2 of glazing.
if 70 f room air enters the sunspace through the lower vent and exits into the
house at 120 f through the upper vent, the average temp inside the hot part
of the sunspace is 97.5 f, and 3.7k btu/h = 16.6asqrt(8')(97.5-70)^1.5 makes
a = 0.55 ft^2. we might use 1 ft^2 vents with an 8' height difference. 

putting n 2-liter water bottles inside the cube makes rc 4.4n/19.2/24
= 0.00955n days... 65 = 30+(75-30)e^(-d/0.00955n) makes n = 417d, eg 2084
bottles (9170 pounds of water and 45% of the floorspace) for 5 cloudy days...
45% is better than 80%, no?

9. less mass with more swing?

we might warm ceiling mass with 120 f air from a sunspace. on an average day,
we need to store 18h(70-30)19.2 = 13.8k btu of overnight heat. if the ceiling
mass temp t hardly varies over a day and its conductance to slow-moving air
below is 1.5x64 = 96 btu/h-f, 6h(120-t)96 = 13.8k makes t = 96 f. after 5
cloudy days, the cube needs (65-30)19.2 = 672 btu/h of heat. the ceiling mass
must be at least 65+672/96 = 72 f to provide this. over 5 cloudy days, the
cube needs 5x24h(70-30)19.2 = 92.2k btu of heat, so we need 92.2k/(96-72)
= 3840 btu/f of ceiling mass (vs 4x more for "direct loss"), ie 3840/64
= 60 pounds (11.25") of water per square foot of ceiling. water tends to
stratify, with warmer water on top (ever swim in the sun in a muddy lake?)
and an r0.25/inch resistance to downward heatflow, so we'd have to stir it 
somehow to get the heat out. 

keeping the heat in the ceiling allows the cube to be cooler when vacant, so
stored solar heat can last longer, provided we reduce the ceiling's downward
heatflow by radiation...

10. radiation

a surface emits se(t^4) of heat flux by radiation, where t is an absolute
temperature in rankine (f+460) or kelvin (c+273) degrees, and s is the stefan-
boltzman constant, 0.1714x10^-8 btu/ft^2-r^4 or 5.660x10^-8 w/m^2-k^4, and 
e is the surface's "emissivity," which varies from 0 (shiny) to 1 according
to shininess. most natural surfaces are close to 1, but mirrorlike surfaces
have emissivities close to 0. the net heatflow from a surface at t1 degrees
surrounded by a t2 degree surface is se(t1^4-t2^4). for example, a 50 f pane
of window glass (e = 0.88) exposed to a 30 f outdoors loses 0.1714x10^-8x0.88
((50+460)^4-(30+460)^4) = 15 btu/h-ft^2 by radiation.

if tb is their average temperature, the "linearized radiation conductance"
between two surfaces is approximately 4setb^3. a 96 f ceiling exposed to
a 70 f room with tb = 543 r has 4setb^3 = 1.097 btu/h-f-ft^2, roughly r0.9.
at 70 f, the cube needs (70f-30f)19.2btu/h-f = 768 btu/h, and e(96-70)64x1.097
= 768 makes e = 0.42. we might make about 60% of the ceiling a low-e (0.05)
foil surface and the rest an ordinary surface and let radiation warm the room
on an average day and use a slow ceiling fan and a thermostat to bring warm
air down on cloudy days. a 72 f ceiling would supply 0.42(72-65)64ft^2x1.097
= 206 btu/h of radiant heat. the other 672-206 = 466 btu/h might come from
q cfm of airflow, as in diagram 1, viewed in a fixed font: 

        1/96  1/q                                   (96+q)/(96q)
72 f ---www---www---65 f        equivalent to   72------www------65
       ----------->                                 ----------->
        466 btu/h                                    466 btu/h

where (72-65)96q/(96+q) = 466, so q = 217 cfm. grainger's $120 48" 315 rpm
86 w 21k cfm 4c853 ceiling fan might move 217 cfm at 217/21kx315 = 3.3 rpm
with 86(217/21k)^3 = 100 microwatts, according to the fan laws :-) large slow
fans can be very efficient and quiet... 

11. even less mass?

on an average day, we might only store 13.8k of overnight heat in c btu/f
in the ceiling, with t(6) = 120+(72-120)e^(-6x96/c) = 120-48e^(-576/c) and
(t(6)-72)c = 18,432, so c = 384/(1-e^(-576/c). c = 384 on the right makes
c = 494 on the left, and plugging that in on the right again makes c = 558,
596, 620, 635, 644, 649, 653, 655, 657, and 657 (whew!), so it looks like we
can store overnight heat with 657/64 = 10.3 psf (about 2") of water above
the ceiling, with t(6) = 120-48e^(-576/657) = 100 f. 

the 8' cube needs 4x24(70-30)19.2 = 73,728 btu for 4 more cloudy days. this
might come from a "solar closet" (see my web page paper) inside a sunspace
(in which the heat lost from the closet air heater glazing efficiently ends
up in warm sunspace air that heats the cube) with about 73,728/(120-70)
= 1475 lb or 184 gal. or 23 ft^3 of water cooling from 120 f to 70 f over 4
days. with 1792 pounds of water in 56 10"x10"x13" 4-gallon ropak plastic tubs,
we can supply 73,728 btu as it cools from 120 to 120-73728/1792 = 78.9 f,
stacking the tubs 7-high and 4-wide and 2-deep in 2'x4'x8' tall closet that's
completely surrounded by insulation, with an air heater with its own closet
vs sunspace glazing over the closet's insulated south wall and one-way dampers
in that wall. with 56x4x10x13/144 = 202 ft^2 of tub surface and 300 btu/h-f
(buhfs) of water-air thermal conductance, we have diagram 1 again with a
78.9-70 = 8.9 f temperature difference and a (300+q)/(300q) resistor, so
q = 121 cfm.

with an 8' height difference and 121 cfm = 16.6asqrt(8'(78.9-70)),
we need 2 vents with a = 0.86 ft^2 for natural airflow into the cube.

as an alternative, the cloudy-day heat might come from 3072 pounds of water
inside a 2'x4'x8' tall "shelfbox" with a 2'x4'x2' tall water tank below 18
2'x4' wire shelves on 4" centers, with 2" of water inside a $20 continuous
piece of poly film duct folded to lay flat on each shelf and a small pump to
circulate tank water up through the duct as needed. the tank might have a
pressurized tank inside to make hot water for showers, with the help of an
efficient external greywater heat exchanger, eg 300' of 1.17" od plastic pipe
pushed into two coils inside a 35"x23.5"" id 55 gallon drum. the outer coil
can be longer, with 27 turns and a 23.5-1.17 = 22.3" diameter and a 5.85'
circumference and 27x5.85 = 157' of pipe. if the inner coil perfectly nested
inside the turns of the outer one, its diameter would be 20.3", with a 5.31'
circumference and 143' of pipe. pushing the pipe inside the drum is awkward
but doable with two people, trying to avoid kinks.

11. greywater heat exchange

i used a new $35 55 gallon lined steel drum with a strong removable lid
(because the drum might end up under 2' of greywater head, with the inlet
and outlet above the lid) and bolt ring and 100 psi/73.4 f pipe from pt
industries (800) 44 endot. their pbj10041010001 1"x300'100psi nsf-certified
pipe is actually tested to 500 psi. the price is $59.99 from any true value
hardware store. lowes sells the rest of the hardware needed, all of which
is installed through the lid, so the drum itself has no holes:

sales        total
#       qty  price   description

25775    1   $5.73   24' of 1.25" sump pump hose (for greywater i/o)
105473   1    1.28   2 ss 1.75" hose clamps (for greywater hose) 
54129    2    3.24   1.25" female adapter (greywater barb inlet and outlet)
23859    2    2.36   1.25x1.5" reducing male adapter (bulkhead fittings)
75912    1    0.51   2 1.25" conduit locknuts (bulkhead fittings)
28299    1    1.53   2 1.25" reducing conduit washers (")
22716    1    1.36   1.5" pvc street elbow (horizontal greywater inlet)
23830    1    2.98   10' 1.5" pvc pipe (for 3' greywater outlet dip tube)

the parts above are greywater plumbing ($18.99.)

75450 9  1    0.29   2 3/4" conduit locknuts (fresh water i/o)
23766    2    0.64   3/4" cpvc male adapter (fresh water i/o)
141830   1    0.42   2 3/4" reducing conduit washers (fresh water i/o)
23813    1    1.39   10' 3/4" cpvc pipe (for 1" fresh water outlet)
23760    2    0.96   3/4" cpvc t (fresh water i/o)
22643    2    0.86   3/4" cpvc street elbow (fresh water i/o)
         4      -    1" 3/4" cpvc pipes (fresh water i/0)
         1      -    3' 3/4" cpvc pipe (fresh water inlet)
23574    4    3.88   3/4" cpvc fip adapters (")
54142    4    3.28   3/4"x1" male adapter barb (fresh water i/o)
22667    2    2.56   2 ss 1.125" hose clamps (fresh water i/o)
219980   1    4.87   10.1 oz dap silicone ultra caulk (bulkhead fittings) 
150887   1    3.94   4 oz primer and 4 oz pvc cement

parts above are fresh water plumbing. subtotal $42.08.

26371    1    6.83   1500 w electric water heater element
22230    1    2.31   1" galvanized t ("nut" for heating element)
61294    1   11.76   single element thermostat with safety
136343   1    0.56   5 10-24x3/4" machine screws (mount thermostat with 3)
33368    1    0.37   5 #10 ss flat washers (mount thermostat with 3)
198806   1    1.38   10 #0 rubber faucet washers (mount thermostat with 3)
8763     1    0.67   5 10-24 ss nuts (mount thermostat with 3)

the above would make a standalone water heater, if needed. grand total: $65.96.

for 4 10 min showers per day and 20 minutes of dishwashing at 1.25 gpm we
might heat 75 gallons of 55 f water to 110 with 8x75(110-55) = 33k btu with
about 10 kwh worth about $1/day at 10 cents/kwh. if the "heat capacity flow
rate" cmin = cmax = 75gx8/24h = 25 btu/h-f and the pipe coil has a = 300pi/12
= 78.5 ft^2 of surface with u = 10 btu/h-f-ft^2 (for an hdpe pipe wall with
slow-moving warm dirty water outside and 8x300pi(1/2/12)^2 = 13 gallons of
fresh water inside), the "number of heat transfer units" for this counterflow
heat exchanger ntu = au/cmin = 78.5ft^2x10btu/h-f-ft^2/25btu/h-f = 31.4, so 
the "efficiency" e = ntu/(ntu+1) = 97% for hot water usage in bursts of less
than 13 gallons. 

the hazen-williams equation says l' of d" smooth pipe with g gpm flow has a
0.0004227lg^1.852d^-4.871 psi loss. at 1.25 gpm, the pressure drop for 2 150'
coils of 1" pipe might be 0.0004227x150x(1.25/2)^1.852x1^-4.871 = 0.03 psi.

if greywater leaves a shower drain and enters the heat exchanger at 100 f and
fresh water enters at 50 f, the fresh water should leave at 50+0.97(100-50)
= 98.5 f. warming it further to 110 f would take 8x75(110-98.5) = 6.9 btu/day
with 2 kwh worth about 20 cents/at 10 cents/kwh, for a yearly savings of about
($1-0.20)365 = $292. the 1500 w heater might operate 2kwh/1.5kw = 1.3 hours
per day. we might wrap the drum with 3.5" of fiberglass and a 4'x8' piece of
foil-foamboard with 7 4' kerfs (knife cuts partially through the board) to
make an octagon and aluminum foil tape to cover the kerfs and hold it closed. 

12. other promising solar house heating schemes

harry thomason's trickle collectors were used in hundreds of houses. pump
water up to the ridgeline of a metal roof and trickle it down under a glass
cover to a gutter, then down to a large tank on the ground surrounded by
rocks to help with water-air heat distribution. these days, a hydronic floor
with a low water-air thermal resistance in an airtight house with lots of
insulation might be simpler and use less power for heat distribution. this
system needs lots of collector pump power, and it looks like the roof cover
has to be glass, since polycarbonate quickly degrades in warm water vapor.
it can only heat water to 80 or 90 f on cold winter days, so heat storage
and distribution are inefficient. 

zomeworks may soon have a more efficient "double-play" system with plastic
tubes under closely-spaced metal roof standing seams, with a polycarbonate
cover or a selective surface.

donald wright's habitat for humanity houses near safford arizona use large
hydronic solar collectors with fiberglass window screen between two hypalon
rubber layers for more uniform water flow. but then, they pump the warm water
under a slab beneath lots of sand with a high thermal resistance, which seems
like a big mistake, thermally-speaking.

a thomason "pancake house" might have a draindown polyethylene film roof pond
and an underfloor poly film pillow for heat distribution. a "concentrating
solar attic" might have a transparent steep south roof and a north roof that
approximates a parabola with 4 or 5 line segments (we want to avoid line foci)
aimed slightly above the southern horizon to reflect 2-3 suns down into a
4'-wide water trough along the attic floor near the north wall. the trough
might be 30" round polyethylene film greenhouse air duct (about 40 cents per
linear foot) that lays flat to 48", with 1-2" of water inside during the day.
this might lay on top of standard photovoltaic panels on the attic floor.
cooled to 120 f, they should have a long lifetime under 2-3 suns. your
milage may vary.

13. natural cooling

less electrical usage helps. 

and shading. in the northern hemisphere at noon on 12/21, sun elevation emin
= 90-latitude-23.5 degrees. at noon on 6/21, emax = 90-latitude+23.5 degrees.
a horizontal overhang that projects p feet from a south wall d feet above the
top of the glass of an h foot tall window can completely shade the window on
6/21 and admit all the sun on 12/21 if tan(emin) = d/p, tan(emax) = (h+d)/p,
and p = h/(tan(emax)-tan(emin)) and d = ptan(emin). for example, at 40 n. lat,
emin = 26.5 degrees and emax = 73.5, so an h = 8' tall window needs a p = 2.7'
overhang d = 1.2' above the top of the glass. 

shading walls and making them light-colored also helps. plants on trellises
come to mind. walls have more insulation than windows, but they also have
much more surface. a house might have 8% of the floorspace as windows...
attics and sunspaces need venting.

night ventilation can help, if a house has lots of internal thermal mass
and lots of insulation. ventilate with cool night air and button the house
up during the day and let the thermal mass keep it cool. nrel says july is
the warmest month in boulder, with average 73.5 f days and daily lows and
highs of 58.6 and 88.2 and an average humidity ratio w = 0.009. this is the
number of pounds of water vapor per pound of dry air. it is more constant
over a day than the relative humidity (rh), which depends how much water
the air can hold, which depends on the air temperature. 

"comfort" depends on temperature, humidity, air velocity (faster is cooler),
activity (sleeping vs wrestling) and clothing (three-piece suits vs shorts and
a t-shirt.) the ashrae comfort zone relates temperature and humidity ratio.
experiments have found that people in developed countries are "comfortable"
from about 67 to 81 f, with w = 0.0045 to 0.012. t = 89.4-1867w is a "constant
comfort" line diagonally down through the zone. with w = 0.009, t = 72.6 is 
most comfortable. can we keep our cube in the zone with night ventilation?

we might put n 2-liter water bottles inside the cube and remove the roof
to make a "cold trap" at night with no lower temperature limit, or put
a motorized damper near the top that lets warm air flow out of the cube
every night until the inside air temp drops to 70.6 f.

if we model an average boulder july day as a square wave with 12 hours at
(58.6+73.5)/2 = 66 f followed by 12 hours at (73.5+88.2) = 81 f (not very
close to the real sine wave, but maybe more useful than solar architects'
rules of thumb), we might have something like diagram 1, with c = 4.4n btu/f
in series with its 1.83n thermal conductance to slow moving air and a 19.2
bhuf cube-to-outdoor conductance that's shunted at night with a q cfm airflow
conductance.

as the bottles warm, 74.6 = 81+(70.6-81)e^(-12/rcc) makes the charge time
constant rcc = 24.7 h = 4.4n/19.2, approximately, so n = 108 and c = 475 btu/h.
as the bottles cool, 70.6 = 66+(74.6-66)e^(-12/rcd) makes the discharge time
constant rcd = 19.2 h. each bottle has 1.2 ft^2 of surface with a 1.5x1.2
= 1.8 bhuf conductance, so 108 bottles have a 198 bhuf conductance or a 5.06
millifhub thermal resistance. if rcd = 19.2, the total series resistance r
to outdoors is 19.2/475 = 0.0404 f-h/btu, which makes 1/q = 0.0404-0.00506 
= 0.035 fhub, which corresponds to q = 28.3 cfm = 16.6asqrt(8'(70.6-66)),
so a = 0.28 ft^2 for natural airflow. we could cool 2" of water above the
ceiling with a gable vent or a thermal chimney above a flat roof. 

we could also use a fan. in pablo laroche and murray milne's tiny ucla test
house with some thermal mass, a "smart whole house fan controller" turned on
a fan when outdoor air was cooler than indoor air. "enthalpy economizers" do
this for large buildings, but seldom for houses. with a humidity sensor, we
might heat as well as cool a house by ventilation, and avoid condensation
on mass surfaces, and bias the house temperature into the upper part of the
comfort zone in wintertime and the lower part in summertime in order to store
more heat or coolth in the mass of the house.

we might also cool 2" of water above the ceiling with a roof pond, as in a
zomeworks thermosyphoning "architectural cool cell." phil niles says a t (f)
pond in to (f) air loses 1.63x10^-9((t+460)^4-a(to+460)^4) btu/h-ft^2 by
radiation, where a = 0.002056tdp+0.7378, with a dew point temp tdp (f). with
v mph of wind, it also loses qc = (0.74+0.3v)(t-to) btu/h-ft^2 by convection,
and it evaporates qe = b(t-twb)-qc, where b = 3.01(0.74+0.3v)((t+twb)/65-1),
and twb (f) is the wet bulb temperature.

the pond radiates more heat to the sky if the air contains less moisture,
with a higher dew point, since water vapor is a "greenhouse gas" that blocks
radiation. air at the dew point temperature is saturated with water vapor.
the relative humidity is 100%. we can find the approximate dew point by first
finding the vapor pressure of water in air on an average july day in boulder
with humidity ratio w = 0.009. pa = 29.921/(1+0.62198/w) = 0.427 inches of
mercury ("hg--29.921 "hg is 1 atmosphere.) then we use a clausius-clapeyron
approximation (don't ask) to find the temperature corresponding to that
pressure, at 100% rh. if pa = e^(17.863-9621/tdp), ln(pa) = 17.863-9621/tdp,
so tdp = 9621/(17.863-ln(pa)) = 514 r or 54 f in boulder in july. 

this makes a = 0.7399 in the formula above, so a 70.6 f pond in 58.6 f night
air would lose 1.63x10^-9((70.6+460)^4-0.7399(58.6+460)^4) = 42 btu/h-ft^2.
nrel says the average july windspeed in boulder is 8.1 mph, so a pond would
lose (0.74+0.3x8.1)(70.6-58.6) = 38 btu/h-ft^2 by convection. an 8'x8' roof
pond would lose 8x8(42+38) = 5120 btu/h by radiation and convection, more than
a 5,000 btu/h window air conditioner. keeping the cube 72.6 f on an 81 f day
in boulder only requires 12h(81-72.6)19.2 = 1935 btu, so we don't need to
evaporate water from this roof pond. it might have a polyethylene film cover,
since poly film is essentially transparent to radiation.

the dew point temperature only depends on the amount of water in an air
sample, vs its temperature. a little water inside a perfectly-insulated cup
might find itself at the dew point temp, as water evaporates from the surface
and water vapor molecules diffuse to the top of the cup, while the air above
the water acts as a good insulator (about r5 per inch for downward heatflow.) 
which water depth will keep it coolest in an r5 3" diameter x 6" tall cup?  
as we fill the cup, the air layer insulates less, but more water evaporates,
according to fick's law, because the concentration gradient increases as the
air layer thins...

when the water gets to the top of the cup, we'd expect to see it at the wet
bulb temperature, when its heat loss by evaporation equals its heat gain by
convection. a perfect swamp cooler would make air at this temperature. in
1926, i.s. bowen said a pond's ratio of heat loss by evaporation to heat gain
by convection equals 100(pp-pa)/(tp-ta), regardless of windspeed. this ratio
is -1 at the wet bulb temp. with pa = 0.009, and ta = 58.6 f, and tw (r),
100(e^(17.863-9621/tw)-0.427) = 58.6+460-tw, so tw = 9621/(22.47-ln(561.3-tw)).

the wet bulb temp is easy to find on a calculator. plugging in tw = 518.6 r 
(58.6 f) on the right of the equation above makes tw = 514 on the left. doing
this again makes tw = 517, 515, 516.2, 515.6, 515.9, 515.7, and 515.8 (55.8 f),
between the dew point and dry bulb temperatures. so an uncovered roof pond
would lose qe = b(t-twb)-qc, where b=3.01(0.74+0.3x8.1)((70.6+55.8)/65-1)
= 9.01 and qe = 9.01(70.6-55.8)-38 = 95 more btu/h-ft^2 by evaporation, like
a 10k btu/h window ac. this could be useful in phoenix. one simple ashrae
swimming pool formula says q = 100(pw-pa) btu/h-ft^2, regardless of air temp.
pw = e^(17.863-9621/(460+70.6)) = 0.764 "hg and pa = 0.427 "hg makes q = 34
btu/h-ft^2.

to make the cube "comfortable" when the outdoor air is 88.2 f (the average
daily max in boulder in july), using a swamp cooler, we might make t = 81 f.
this corresponds to w = 0.0045 on the downward diagonal "constant comfort"
line through the ashrae zone, so we can't get there by adding water to air
with w = 0.009, but it's still within the zone, and slightly "warm," to the
right of the line. but swamp coolers rarely have thermostats, and we might
want to use as little water as possible, rather than flooding the space with
moist air...

to keep the cube 81 f while moving q cfm of outdoor air through it and
evaporating p pounds of water per hour, we need (88.2-81)(19.2+q) = 1000p.
a ft^3 of air weighs 0.075 lb, so p = 60q(0.075)(w-0.009) = 4.5q(w-0.009),
and q = p/(4.5(w-0.009)). substituting this for q in the first equation,
(88.2-81)(19.2+p/(4.5(w-0.009)) = 1000p makes p = (13w-0.117)/(94.3w-1).
we can minimize p within the comfort zone by making w = 0.012, which makes
p = 0.296 pounds of water per hour (less than 1 gallon per day) and q
= 0.296/(4.5(0.012-0.009)) = 22 cfm (not much.) moving more air means
evaporating more water...

we can make 22 cfm move through the cube with a ft^2 vents with 8' of height
difference if 22 = 16.6asqrt(8(88.2-81)), ie a = 0.175 ft^2, eg 5"x5" vents.
the 0.296 pounds of water per hour might come from a solenoid valve and a 0.5
gph mister nozzle with a 100x0.296/4 = 7.4% duty cycle, or some plant leaves
at the 81f/w=0.012 wet bulb temp in an indoor greywater wetland or an a ft^2
81 f indoor "swimming pool" with pa = 29.921/(1+0.62198/0.012)) = 0.483 "hg
and pw = e^(17.863-9621/(460+81)) = 1.082 "hg and a = 0.296/(0.1(1.082-0.483))
= 4.9 ft^2, according to that simple ashrae swimming pool formula.

a more sophisticated cooling system with no water consumption or outdoor air
exchange might have a multi-effect licl solar still on the roof that absorbs
water vapor at night from the pond below and distills water out of the licl
solution during the day. this might also store cloudy day heat, with a wet
basement floor to evaporate water in wintertime, and a licl pond on the first
floor that acts as a "chemical heat pump." 

nick

nicholson l. pine                      system design and consulting
pine associates, ltd.                                (610) 489-1475
821 collegeville road                           fax: (610) 831-9533
collegeville, pa 19426                     email: nick@ece.vill.edu

computer simulation and modeling. high performance solar heating and
cogeneration system design. bsee, msee, sr. member, ieee. registered
us patent agent. web site: http://www.ece.vill.edu/~nick 




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