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Workshop on Ohm's law for heatflow

Discussion in 'Home Power and Microgeneration' started by Nick Pine, Oct 17, 2003.

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  1. Nick Pine

    Nick Pine Guest

    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 (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 and 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 = 120 Btu/h of heat, right? 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 about 4 pickup trucks full of bottles, and 16,236 pounds
    of water (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 and divide that
    number 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+360) 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 R1.1.
    At 70 F, the cube needs (70-30)19.2 = 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:

    1/96 1/Q 96Q/(96+Q)
    72 F ---www---www---65 F equivalent to 72------www------65
    -----------> ----------->
    466 Btu/h 466 Btu/h

    where (72-65)(96+Q)/(96Q) = 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(70-40)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, the tubs can supply 768 Btu/h with
    70+768 Btu/h/300Btu/h-F = 72.2 F water.

    With an 8' height difference and 768 Btu/h = 16.6Asqrt(8')(78.9-72.2)^1.5,
    we need 2 vents with A = 0.94 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:

    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
    layers of rubber for more uniform water flow. But then, they pump the warm
    water under a slab beneath lots of sand with a high thermal resistance...

    A Thomason "pancake house" might have a draindown polyethylene film roof pond
    and underfloor 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 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...

    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, and 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 this box. With w = 0.009, T = 72.6 is
    most comfortable. Can we keep our cube in this box 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 "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 loses
    Qe = b(T-Twb)-Qc by evaporation, 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 $8 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 F
    = 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 0.296/4 = 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 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 be used for cloudy day heat storage,
    with a wet basement floor to evaporate water in wintertime, and a LiCl pond
    on the first floor to act 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:

    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
     
  2. On 17 Oct 2003 17:01:11 -0400, (Nick Pine)
    wrote:

    Nick, you really need to get out more :)



    Paul ( pjm @ pobox . com ) - remove spaces to email me

    New HVAC/R program for Palm PDA's !! http://pmilligan.net/palm/
    Free demo now available online !!!!
     
  3. JazzMan

    JazzMan Guest

    Some really good info. Definitely a keeper. :)

    JazzMan
    --
    **********************************************************
    Please reply to jsavage"at"airmail.net.
    Curse those darned bulk e-mailers!
    **********************************************************
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    supply and demand. It is the privilege of human beings to
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  4. George Ghio

    George Ghio Guest


    Ah Nick

    What a lovely piece of work. Don't you just hate the type setting and
    loss of formulas info. May I suggest you have a look at the Tex
    (pronounced "Tech") This is a typesetting program that was developed for
    maths typesetting. It also allows much more than that though.

    The place to start is

    <www.ctan.org>

    I found it best to join the Tex Users Group as this will get you the
    Texlive CD which is most of what one needs to typeset great documents.

    George
     
  5. daestrom

    daestrom Guest

    <snip>

    Nice write-up Nick. But you know me, I have to pick a little 'nit'....

    In 'chapter 10' you inadvertantly state that absolute temperature Rankine is
    degrees Farenheit + 360. I'm sure you *meant* to say Farenheit + 460 and
    the keyboard just got in the way ;-)

    Anyway, a vary nice 'compendium' of information.

    Your link with nrel.gov has me re-thinking about mounting my solar-thermal
    collector on the south wall of my place. Although vertical is not ideal
    from the standpoint of direct solar incidence, I found a correction to
    their data for ground albedo (they use a value for green vegatation).
    http://rredc.nrel.gov/solar/pubs/bluebook/appendix.html#calcincrad

    Since I'm up in the 'snow belt', I'm thinking that ground albedo is much
    higher than 0.2 when it is covered with snow (maybe something like 0.8).
    And in *that* circumstance, a vertical collector might be better than one
    tilted to aim directly at the noon-day sun in winter time. Since this would
    be for home-heating, the winter input is the important factor. And
    certainly wall-mounting is simple ;-)

    daestrom
     
  6. JazzMan

    JazzMan Guest


    My computer is smoking and shaking like an old washing machine now. :)

    JazzMan
    --
    **********************************************************
    Please reply to jsavage"at"airmail.net.
    Curse those darned bulk e-mailers!
    **********************************************************
    "Rats and roaches live by competition under the laws of
    supply and demand. It is the privilege of human beings to
    live under the laws of justice and mercy." - Wendell Berry
    **********************************************************
     
  7. Ecnerwal

    Ecnerwal Guest

    This is, in fact, a standard bit of "proven by experiment" lore in the
    snowbelt (Maine, anyway). I don't know offhand of a weblink to it, but
    this was checked out by the University of Maine (decades ago) and they
    did find that vertical windows with snow were actually better than
    angled collectors. Nick has in the past also mentioned using either a
    pond or even a hunk of white EDPM rubber (without actually making it a
    pond) to artifically increase the albedo when snow is not present.
     
  8. Ecnerwal

    Ecnerwal Guest

    Collectors, not windows, insufficient coffee.
     
  9. Simon

    Simon Guest

    Wow, a 39K quotation!
     
  10. Simon

    Simon Guest

    Power is not just a number, it has units, and it measures our rate of
    consumption.
    The bathtub of hot water contains a lot of internal energy (related to its
    temperature). How can something contain a lot of power?
    No, that depends upon the temperature *difference* and the thermal
    conductivity.
    temperature vs voltage,
    temperature difference vs voltage difference
    There is for a steady state. The difficulty comes when you apply a
    temperature difference to a conductor that it takes time to reach this
    steady state, and its internal energy will rise or fall in doing so.
    A resource normally wasted.
     
  11. Guest

    I think of this figure of merit as "solar heat per degree day."
    "Worst-case design" is an aerospace engineering concept. You can design
    something to work under average conditions or specify that it must work
    under the most strenuous conditions, in this case, the month that combines
    minimum sun and maximum cold.
    NREL says January gets 1000 Btu/ft^2 day in Phila, vs 1150 in October.
    I posted an overhang calc...
    The idea is to make sure the heating system performs well under the most
    trying ("worst-case") conditions, in which case it should do well
    in other months.
    One of those months is more trying than others, on average.
    Sure... Ever notice the black plastic latticework over the near-horizontal
    back windows of long low sports cars? Those are "bris-soleil" overhangs
    that allow rear vision while blocking overhead sun.

    Nick
     
  12. If you had a work order to take the resistance measurement of
    a squash, would that order read 'Ohm my gourd' ??

    If there are two resistances separated by a dielectric, is
    that 'Ohm away from Ohm'?

    If they charged for it, would we have to pay Ohmage ?

    Watt does it all mean ?

    Who let that Watt in here ? Kill it !! Kill a Watt !



    Paul ( pjm @ pobox . com ) - remove spaces to email me

    New HVAC/R program for Palm PDA's !! http://pmilligan.net/palm/
    Free demo now available online !!!!
     
  13. this ng would be mostly wasted space without Nick.
     
  14. a gourd is not a squash!!

     
  15. It is if you jump up and down on it a few times.

     
  16. Don K

    Don K Guest


    Ohms was hooked up with Meg Ohms and initially the sparks flew
    between them. Wattever happened, they eventually got totally
    dissipated. Ohms broke off all contact and switched to
    his current friend, Flo.

    After considerable doping, Ohms became a semiconductor in
    later years, and his controlling efforts to have his way
    with Flo, rolled off with frequency.

    Don
     
  17. News

    News Guest

    They are. Both Btu/Hr and kW/Hrs are given. I prefer everything in kWs,
    even car engines, electricity, gas, etc. Direct comparisons are then very
    easy.
     
  18. Nick Pine

    Nick Pine Guest

    Why kW/Hr?

    Nick
     
  19. News

    News Guest

    Doesn't make any difference we quote both equivalents. But I prefer the
    metric system as it's far easier and logical. Strange that the French
    thought of it, as I don't usually associate them with logic (ever seen them
    drive?).
     
  20. Nick Pine

    Nick Pine Guest

    WRONG!!! Please learn the difference between power and energy!

    Nick
     
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