```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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