Doug Kalmer writes:
>Do you really believe that when EVERY ONE of the 19 entries in this years Solar Decathlon uses Direct Gain passive solar just for the "visual appeal"?
I wonder what that means. There were 20 teams and 19 entries this year, including...
ASU/UNM, in which
>Phase-change material distributed throughout the house stores thermal energy and buffers temperature fluctuations by redistributing the energy as passive heating and cooling.
>The capillary radiant system in the ceiling, working with the phase-change materials in the floors, passively charges at night while the ceiling mechanics shift the peak load.
Las Vegas, in which
>Solar thermal collectors provide radiant floor and water heating.
Missouri, in which
>A mixed-mode residential HVAC system marries the automation system and the house HVAC system.
>A radiant heating system with tubes beneath the concrete floor circulates water to heat the interior space from the bottom up.
Norwich, in which
>A mini-split heat pump HVAC system with a single supply diffuser provides widely available, compact, and easily serviceable heating and cooling
Santa Clara, in which
>The radiant heating and cooling system embedded in the ceiling drywall uses radiant panels to heat the house with hot water or cool the house with cold water—ensuring a uniform environment.
>As part of the water heating and storage system, a solar thermal panel supplies heat to a tank containing organic phase-change material.
Sci-Arc/Caltech, in which
>The HVAC system uses state-of-the-art solar water heating technology with solar thermal evacuated tube collectors to maintain a comfortable interior temperature.
Stanford, in which
>A heat-recovery ventilator works with an efficient heating/cooling system, automated windows, phase-change materials, energy-efficient ceiling fans, and a tri-zone ductless mini-split system
Austria, in which
>A heat-recovery ventilator works with an efficient heating/cooling system, automated windows, phase-change materials, energy-efficient ceiling fans, and a tri-zone ductless mini-split system
DC, in which
>A unique under-floor heating and cooling distribution system supplies air at floor-level from ductwork connected to a central air handler
Ontario, in which
>An integrated mechanical system provides space heating, cooling, dehumidification, and domestic hot water through a single system.
Texas, in which
>A ductless radiant heating and cooling system circulates fully as a closed loop of radiant heat.
So Cal, in which
>A combination heat pump system provides heating, cooling, and domestic hot water, and
West Virginia, in which
>The climate-control system enables room-by-room temperature and lighting adjustments. Through smart HVAC technology, users can set different zoning preferences without disturbing the settings of other rooms.
Oodles of PVs, of course. Oddly enough, none of the entries touted simple direct gain solar heating with its refreshing sub-freezing indoor temperatures after 2 cloudy days in a row.
Nick
Tuesday, October 15, 2013
Sunday, October 13, 2013
A direct gain rant
John Canivan writes:
>Dug What happens if you have a week without sun in December?
If it's a direct gain house with sun shining through windows onto a massy floor, you freeze or wave your hands and say it's really warm inside when it isn't, or wear sweaters and arctic army pants and sit in one small room with an electric space heater. Direct gain house owners and builders and designers often fool themselves and others.
What's the solar heating fraction of the Britton's house?
>Every day is sunny for George and Charlotte Britton of Lafayette Hill. The Britton's 2,900-square-foot house is blessed with energy bills 20 percent lower than one of comparable size... The design of the house incorporates "passive" solar principles. There are large double pane windows and sliding glass doors on the south side. Inside, tile floors and a Trombe wall absorb the sun's heat during the day and radiate it at night... A stone fireplace on the south wall of the living area provides additional heat during colder months. Britton said "We have a fire every day of the winter."
Direct gain (aka "direct loss") houses in cold cloudy places rarely have solar heat fractions greater than 50%, according to Passive Solar Institute (aka Sustainable Building Institute) Guidelines... 30% is more common, ie the sun only provides 30% of the house heating over an entire year, but indirect gain houses can have solar heating fractions of 90% or more in the month of January.
An 8' 70 F direct gain cube with R20 walls and ceiling and floor and A ft^2 of R4 south windows with 50% solar transmission needs 24h(70-30) (A/4+(384-A)/20) Btu of heat on an average 30 F January day in Phila with 1000 Btu/ft^2 of sun on a south wall... 0.5x1000A = 24h(70-30)(A/4+(384-A)/20) makes A = 41 ft^2, with a total cube conductance G = 41/4 + (384-41)/20 = 27.4 Btu/h-F, right? (Or would you like to argue about that? :-)
With a 4" 533 Btu/F concrete floorslab, time constant RC = C/G = 19.5 hours. With no sun, 300-year-old high-school physics says the direct gain cube temp drops to 30+(70-30)e^(-24/19.5) = 41.7 F after 24 hours and 30+(70-30)e^(-48/19.5) = 33.4 after 48 hours, and so on. That's MISERABLE solar house heating performance! (Would you like to argue about that? :-)
OTOH, an 8' 70 F indirect gain cube with R20 walls and ceiling and floor with a 384/30 = 19.2 Btu/h-F conductance needs (70-30)19.2 = 768 Btu/h or 18.4K Btu/day. A T (F) 64 ft^2 radiant ceiling heated by thermosyphoning sun-warmed air (as in the Barra system) with a 1.5x64 = 96 Btu/h-F slow-moving airfilm conductance can keep the cube 70 F if (T-70)96 = (70-30)320/20, ie T = 77 F. If the cube has 8'x8' of R2 air heater glazing with 80% solar transmission (eg 2 $42 4'x8' sheets of twinwall polycarbonate) over the R20 south wall,
= 6h(T-30)64ft^2/R2 for the air heater during the day
+ 18h(70-30)64ft^2/R22 for the south wall at night
+ 24h(T-30)64ft^2/R20 for the ceiling all day
+ 24h(70-30)4x64ft^2/R20 for the other walls
+ 18h(70-30)64ft^2/R22 for the south wall at night
+ 24h(T-30)64ft^2/R20 for the ceiling all day
+ 24h(70-30)4x64ft^2/R20 for the other walls
makes T = 167 F.
We can keep the cube exactly 70 F for 5 cloudy days in a row with a slow ceiling fan and a room temp thermostat and 5x18.4K/(167-77)/62.33 = 16.4 ft^3 of water cooling from 167 to 77 F, with an approximate 1-2^-5 = 0.97 solar heating fraction. But that's an overestimate, since the ceiling can be 77 vs 167 F most of the time.
Nick
Monday, September 30, 2013
Ridesharing
If 100 people live on a circle 5 miles from church and they each drive 10 miles per week to and from church, that's 1000 miles per week, ie 50 gallons of gas at 20 mpg, ie $150/week at $3/gallon.
The circle circumference is 10Pi = 31.4 miles. If 20 cars move 5 people each, efficiently, they move 20x10 = 200 miles per week in and out + 2x20/100x31.4 circular miles, in a pie-shaped wedge, ie 451 miles per week, using 451/20 = 22.6 gallons of gas worth $67.70.
The weekly difference is $150-$67.70 = $82.30, ie $4280 per year, not counting reduced wear and tear on the cars or environmental benefits or increased socialization opportunities.
The basic economics make sense, but this needs social engineering, eg considerations of fairness, flexibility, privacy, safety, personality conflicts, and so on.
Why doesn't everyone car pool? How can computers help? Who travels with whom, and in what order? How can we avoid George feeling burdened, or personality conflicts with Alice? What happens if someone decides to skip church one week or only comes once a month or isn't ready to go when the driver shows up?
Poland had a national Autostop (hitchhiking) system in the 70s. A person by the side of a road would hold up a book with a target symbol on it, and a driver would stop and give her a ride, in exchange for a coupon torn from the book. The coupons were national lottery tickets redeemable by the driver. The trips were also insured for safety by the government.
At one point, the Israeli army mostly traveled by hitchhiking...
>... In Cuba, picking up hitchhikers is mandatory by government vehicles, if passenger space is available. Hitchhiking is encouraged, as there are few cars, and designated hitchhiking spots are used. Waiting riders are picked up on a first come first go basis. http://en.wikipedia.org/wiki/Hitchhiking
>In 2009 carpooling represented 43.5% of all trips in the United States[1] and 10% of commute trips.[2] The majority of carpool commutes (over 60%) are "fam-pools" with family members.
>Carpooling, or car sharing as it is called in British English, is promoted by a national UK charity, Carplus, whose mission is to promote responsible car use in order to alleviate financial, environmental and social costs of motoring today, and encourage new approaches to car dependency in the UK. Carplus is supported by Transport For London, the British government initiative to reduce congestion and parking pressure and contribute to relieving the burden on the environment and to the reduction of traffic related air-pollution, in London.
>Challenges for carpooling
>Flexibility - Carpooling can struggle to be flexible enough to accommodate en route stops or changes to working times/patterns. One survey identified this as the most common reason for not carpooling. To counter this some schemes offer 'sweeper services' with later running options, or a 'guaranteed ride home' arrangement with a local taxi company.
>Reliability - If a carpooling network lacks a "critical mass" of participants, it may be difficult to find a match for certain trips. In addition, the parties may not necessarily follow through on the agreed-upon ride. Several internet carpooling marketplaces are addressing this concern by implementing online paid passenger reservation, billed even if passengers do not turn up.
>Riding with strangers - Concerns over security have been an obstacle to sharing a vehicle with strangers, though in reality the risk of crime is small.[12] One remedy used by internet carpooling schemes is reputation systems that flag problematic users and allow responsible users to build up trust capital, such systems greatly increase the value of the website for the user community.
http://en.wikipedia.org/wiki/Carpool
>Real-time ridesharing (also known as instant ridesharing, dynamic ridesharing, ad-hoc ridesharing, or dynamic carpooling) is a service that arranges one-time shared rides on very short notice. This type of carpooling generally makes use of three recent technological advances:
>GPS navigation devices to determine a driver's route and arrange the shared ride
>Smartphones for a traveler to request a ride from wherever they happen to be
>Social networks to establish trust and accountability between drivers and passengers
>These elements are coordinated through a network service, which can instantaneously handle the driver payments and match rides using an optimization algorithm.
>Real-time ridesharing is promoted as a way to better utilize the empty seats in most passenger cars, thus lowering fuel usage and transport costs. It can serve areas not covered by a public transit system and act as a transit feeder service. It is also capable of serving one-time trips, not only recurrent commute trips.Furthermore, it can serve to limit the volume of car traffic, thereby reducing congestion and mitigating traffic's environmental impact.
>One potential drawback may be economic harm to the auto industry due to sharing; however, some auto companies such as Daimler are quite supportive of real-time ridesharing research. Opposition may also come from taxi companies and public transit operators.
>Implementation
>Early real-time ridesharing projects began in the 1990s, but they faced obstacles such as the need to develop a user network and a convenient means of communication. Gradually the means of arranging the ride shifted from telephone to internet, email, and smartphone; and user networks were developed around major employers and universities. As of 2006, the goal of taxi-like responsiveness still generally eluded the industry; "next day" responsiveness was considered the state of the art.
>Most instant ridesharing services are still in their early stages. Successful pilot projects have been completed, but no real-time ridesharing company seems to have yet reached a critical mass of users.
>Two dynamic ridesharing pilots in Norway received government funds from Transnova in 2011. One pilot in Bergen had 31 passenger in private cars during one day. Thirty-nine users acted as drivers or passengers between June 30 and September 15 with four ridesharing episodes or more. The phone apps that was used was Avego Driver and HentMEG.no cell client, a prototype developed for the NPRA of Norway. The other pilot is run by the company Sharepool.
>In France, real-time ridesharing is provided by Geocar from Villefluide, which focuses on the commute market and utilizes a cluster model and algorithms.
http://en.wikipedia.org/wiki/Real-time_ridesharing
>Nous sommes spécialisés depuis 2008 dans les problématiques de déplacements liés au travail. Grâce à notre expertise, nous sommes capables de penser à l’échelle d’organisations afin de proposer à chaque employé la meilleure solution personnalisée de mobilité. Plans de mobilité, déménagements d’entreprises, covoiturage domicile travail, inter-modalité, VilleFluide est un acteur majeur de la mobilité 2.0.
"Notre vision est celle d’un monde où les déplacements du futur seront plus responsables, plus partagés et plus économiques. Notre ambition est à la fois de marier le meilleur des technologies à un concept entièrement innovant de réseau local de transport pour apporter des solutions de partage aux trajets récurrents, quotidiens, en relation avec tous les autres moyens de déplacements.
Nous poursuivons notre effort de R&D afin de toucher un public le plus large possible. En particuliers ceux qui ont un besoin fondamental de faire évoluer leurs pratiques de mobilité, pour des raisons économiques le plus souvent.
En vue d’atteindre cet objectif, nous nouons des partenariats stratégiques, avec des industriels, mais aussi des collectivités locales et territoriales, qui ont compris l’énorme bénéfice qu’elles pouvaient tirer de ces nouvelles solutions de déplacements."
>Wedrive est l’application grand public de VilleFluide. Il s’agit de la première application permettant de covoiturer vraiment chaque jour pour aller travailler.
http://villefluide.fr/
Wedrive is Villefluide's smartphone app, which uses facebook.
A church transportation system with a fixed schedule and one fixed destination and fixed origins might not need GPS or smartphones, but it seems to need a mechanism for fairness, eg the Phoenixville Area Time Bank http://www.pa-timebank.com/
If a driver (eg George) gives a passenger (eg Alice, who is unable to drive a car) a ride to church, and that takes another 0.8 hours, Alice could give George 1.2 hours from her time bank account (earned by babysitting or cooking?) , and George could get a different time bank member to mow his lawn for an hour.
If Alice is not ready to go on one Sunday morning when George arrives at her house, and that makes 4 people 1/2 hour late for church, Alice could donate 0.8+4x0.5 = 2.8 hours to the time bank that week, with an appropriate distribution to the accounts of George and the other 3 passengers. If George's late start makes everyone 1/2 hour late for church, George could donate extra time to the bank...
Nick
The circle circumference is 10Pi = 31.4 miles. If 20 cars move 5 people each, efficiently, they move 20x10 = 200 miles per week in and out + 2x20/100x31.4 circular miles, in a pie-shaped wedge, ie 451 miles per week, using 451/20 = 22.6 gallons of gas worth $67.70.
The weekly difference is $150-$67.70 = $82.30, ie $4280 per year, not counting reduced wear and tear on the cars or environmental benefits or increased socialization opportunities.
The basic economics make sense, but this needs social engineering, eg considerations of fairness, flexibility, privacy, safety, personality conflicts, and so on.
Why doesn't everyone car pool? How can computers help? Who travels with whom, and in what order? How can we avoid George feeling burdened, or personality conflicts with Alice? What happens if someone decides to skip church one week or only comes once a month or isn't ready to go when the driver shows up?
Poland had a national Autostop (hitchhiking) system in the 70s. A person by the side of a road would hold up a book with a target symbol on it, and a driver would stop and give her a ride, in exchange for a coupon torn from the book. The coupons were national lottery tickets redeemable by the driver. The trips were also insured for safety by the government.
At one point, the Israeli army mostly traveled by hitchhiking...
>... In Cuba, picking up hitchhikers is mandatory by government vehicles, if passenger space is available. Hitchhiking is encouraged, as there are few cars, and designated hitchhiking spots are used. Waiting riders are picked up on a first come first go basis. http://en.wikipedia.org/wiki/Hitchhiking
>In 2009 carpooling represented 43.5% of all trips in the United States[1] and 10% of commute trips.[2] The majority of carpool commutes (over 60%) are "fam-pools" with family members.
>Carpooling, or car sharing as it is called in British English, is promoted by a national UK charity, Carplus, whose mission is to promote responsible car use in order to alleviate financial, environmental and social costs of motoring today, and encourage new approaches to car dependency in the UK. Carplus is supported by Transport For London, the British government initiative to reduce congestion and parking pressure and contribute to relieving the burden on the environment and to the reduction of traffic related air-pollution, in London.
>Challenges for carpooling
>Flexibility - Carpooling can struggle to be flexible enough to accommodate en route stops or changes to working times/patterns. One survey identified this as the most common reason for not carpooling. To counter this some schemes offer 'sweeper services' with later running options, or a 'guaranteed ride home' arrangement with a local taxi company.
>Reliability - If a carpooling network lacks a "critical mass" of participants, it may be difficult to find a match for certain trips. In addition, the parties may not necessarily follow through on the agreed-upon ride. Several internet carpooling marketplaces are addressing this concern by implementing online paid passenger reservation, billed even if passengers do not turn up.
>Riding with strangers - Concerns over security have been an obstacle to sharing a vehicle with strangers, though in reality the risk of crime is small.[12] One remedy used by internet carpooling schemes is reputation systems that flag problematic users and allow responsible users to build up trust capital, such systems greatly increase the value of the website for the user community.
http://en.wikipedia.org/wiki/Carpool
>Real-time ridesharing (also known as instant ridesharing, dynamic ridesharing, ad-hoc ridesharing, or dynamic carpooling) is a service that arranges one-time shared rides on very short notice. This type of carpooling generally makes use of three recent technological advances:
>GPS navigation devices to determine a driver's route and arrange the shared ride
>Smartphones for a traveler to request a ride from wherever they happen to be
>Social networks to establish trust and accountability between drivers and passengers
>These elements are coordinated through a network service, which can instantaneously handle the driver payments and match rides using an optimization algorithm.
>Real-time ridesharing is promoted as a way to better utilize the empty seats in most passenger cars, thus lowering fuel usage and transport costs. It can serve areas not covered by a public transit system and act as a transit feeder service. It is also capable of serving one-time trips, not only recurrent commute trips.Furthermore, it can serve to limit the volume of car traffic, thereby reducing congestion and mitigating traffic's environmental impact.
>One potential drawback may be economic harm to the auto industry due to sharing; however, some auto companies such as Daimler are quite supportive of real-time ridesharing research. Opposition may also come from taxi companies and public transit operators.
>Implementation
>Early real-time ridesharing projects began in the 1990s, but they faced obstacles such as the need to develop a user network and a convenient means of communication. Gradually the means of arranging the ride shifted from telephone to internet, email, and smartphone; and user networks were developed around major employers and universities. As of 2006, the goal of taxi-like responsiveness still generally eluded the industry; "next day" responsiveness was considered the state of the art.
>Most instant ridesharing services are still in their early stages. Successful pilot projects have been completed, but no real-time ridesharing company seems to have yet reached a critical mass of users.
>Two dynamic ridesharing pilots in Norway received government funds from Transnova in 2011. One pilot in Bergen had 31 passenger in private cars during one day. Thirty-nine users acted as drivers or passengers between June 30 and September 15 with four ridesharing episodes or more. The phone apps that was used was Avego Driver and HentMEG.no cell client, a prototype developed for the NPRA of Norway. The other pilot is run by the company Sharepool.
>In France, real-time ridesharing is provided by Geocar from Villefluide, which focuses on the commute market and utilizes a cluster model and algorithms.
http://en.wikipedia.org/wiki/Real-time_ridesharing
>Nous sommes spécialisés depuis 2008 dans les problématiques de déplacements liés au travail. Grâce à notre expertise, nous sommes capables de penser à l’échelle d’organisations afin de proposer à chaque employé la meilleure solution personnalisée de mobilité. Plans de mobilité, déménagements d’entreprises, covoiturage domicile travail, inter-modalité, VilleFluide est un acteur majeur de la mobilité 2.0.
"Notre vision est celle d’un monde où les déplacements du futur seront plus responsables, plus partagés et plus économiques. Notre ambition est à la fois de marier le meilleur des technologies à un concept entièrement innovant de réseau local de transport pour apporter des solutions de partage aux trajets récurrents, quotidiens, en relation avec tous les autres moyens de déplacements.
Nous poursuivons notre effort de R&D afin de toucher un public le plus large possible. En particuliers ceux qui ont un besoin fondamental de faire évoluer leurs pratiques de mobilité, pour des raisons économiques le plus souvent.
En vue d’atteindre cet objectif, nous nouons des partenariats stratégiques, avec des industriels, mais aussi des collectivités locales et territoriales, qui ont compris l’énorme bénéfice qu’elles pouvaient tirer de ces nouvelles solutions de déplacements."
>Wedrive est l’application grand public de VilleFluide. Il s’agit de la première application permettant de covoiturer vraiment chaque jour pour aller travailler.
http://villefluide.fr/
Wedrive is Villefluide's smartphone app, which uses facebook.
A church transportation system with a fixed schedule and one fixed destination and fixed origins might not need GPS or smartphones, but it seems to need a mechanism for fairness, eg the Phoenixville Area Time Bank http://www.pa-timebank.com/
If a driver (eg George) gives a passenger (eg Alice, who is unable to drive a car) a ride to church, and that takes another 0.8 hours, Alice could give George 1.2 hours from her time bank account (earned by babysitting or cooking?) , and George could get a different time bank member to mow his lawn for an hour.
If Alice is not ready to go on one Sunday morning when George arrives at her house, and that makes 4 people 1/2 hour late for church, Alice could donate 0.8+4x0.5 = 2.8 hours to the time bank that week, with an appropriate distribution to the accounts of George and the other 3 passengers. If George's late start makes everyone 1/2 hour late for church, George could donate extra time to the bank...
Nick
Thursday, September 26, 2013
A combined solar closet in Pittsburgh
Young Ben would like to live inexpensively on a farm near Pittsburgh...
http://pittsburgh.craigslist.org/fod/4062701580.html offers a 12' by 56' 2 bedroom mobile home with a new steel roof, a front kitchen, and a large bath, in very good condition for $3,900.00.
http://pittsburgh.craigslist.org/cto/4039312791.html lists "Antique CLASS C - 22' MOTOR HOME for sale - pictures sent to serious inquires - inspected and ready-to-go anywhere you want. Mechanically VERY RELIABLE, needs TLC (tune-up) - everything works. Great for camping, hunting, or just fix it up for the ultimate tail-gator mobile. Comes with kitchen, refrigerator, stove, oven, microwave, furnace, holding tanks (for gray and black water), full bathroom, overhead bed, couch, chair and dinette table. It has a 350 small block engine (that is very reliable), and transmission also very reliable, no slips or bangs, very smooth. Very easy to drive and is still being driven daily. Selling for $1500, or best offer."
Here's one for $100... http://pittsburgh.craigslist.org/for/4056583831.html "1974 Holly Park mobile home for sale. Must be moved. Serious inquiries only; sold as-is. You move it. Needs repairs and attention, but would make a great hunting cabin."
Ben has also been offered the free use of a pop-up camper. None of these structures have much thermal insulation, but they could be solar-heated if enclosed in a strawbale structure inside an inexpensive plastic film greenhouse. Ben might be happy living inside a strawbale structure without a pop-up camper, with a strawbale bed and a strawbale table and a sawdust toilet.
An 8'x12'x8'-tall room with about 536 ft^2 of R55 dry 2' strawbale walls would have thermal conductance G = 536/55 = 9.7 Btu/h-F... 10 cfm of fresh air would raise the conductance to about 20. Keeping it 70 F on an average 31.5 F December day in Pittsburgh would only require (70-31.5)20 = 770 Btu/h.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 1.7 kW/m^2 of sun falls on a vertical south wall and 1.91 falls on a south wall with a 45 degree slope on an average December day with a 35.1 average daytime temp. Suppose a 6 cent/ft^2 4-year 14'x24' R1 sloped polyethylene film south wall with 90% solar transmission and 100% longwave IR transmission connects the top edge of the south room wall to the ground 10' south of the room wall and transmits 0.9x14'x24'x1.91x317 = 185K Btu and keeps the room 70 F for 24 hours and radiates heat from the ground inside the wall and the south room wall at absolute Rankine temp Ta to an outdoor temp Ts (R) and 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-Ts^4)480ft^2, where V = 1/4 is the approximate view factor of the radiating surfaces. (A more expensive and durable polycarbonate south wall would block longwave IR.)
Duffie and Beckman's 2006 Solar Engineering of Thermal Processes book has equation (3.9.2) for sky temperature Ts = Ta[0.711+0.0056Tdp+0.000073Tdp^2+0.013cos(15t)]^(1/4), where Ts and Ta are in degrees Kelvin and Tdp is the dew point temp in degrees Celsius and t is the number of hours from midnight.
Ta = 35.1+460 = 495.1 degrees R, ie 495.1/1.8 = 275.1 degrees K. NREL says the humidity ratio w = 0.0030 pounds of water per pound of dry air on an average December day in Pittsburgh, which makes the partial pressure of water in air Pa = 29.921/(0.62198/w-1) = 0.145 "Hg, which makes the dew point Tdp = 9621/(17.863-ln(Pa)) = 486 R, ie 486-460 = 26 F, ie -3.3 C. With t = 12 hours (noon), cos(15t) = -1, so Ts = 275.1[0.711-0.0056x3.3+0.000073x3.3^2-0.013)]^(1/4) = 253.2 K, ie 1.8x253.2 = 455.7 R, ie -4.3 F, 39 F degrees less than the air temperature.
... 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-455.7^4)480 makes 166.5K = 1.234E-6(Ta^4-455.7^4), so Ta = 649.5 R, ie 190 F. A quarter-cylindrical R1 film wall with a 10' radius and a 24' length and 377 ft^2 of surface with 190 F air inside and 35.1 F (495.1 R) air outside would lose 6h(190-35.1)377ft^2 = 350.4K Btu/day to the outdoor air, which is more than the solar input, so 190 is too hot.
If 185K = 6h(T-495.1)377 + 1.234E-6(Ta^4-455.7^4), ie T = 576.9 - 5.46E-10(Ta^4-455.7^4). Plugging in T = 540 R on the right makes T = 554.0 on the left. Repeating makes T = 549.0, 550.8, and 550.2 R, ie 90.2 F, warm enough to solar heat the room on an average day.
If the room is 70 F at dusk and 60 at dawn, 18 hours later, and 60 = 31.5+(70-31.5)e^(-18/RC), time constant RC = -18/ln((60-31.5)/(70-31.5)) = 60 hours = C/G, and C = 60hx20Btu/h-F = 1200 Btu/F of room temp thermal mass with lots of surface can store average-day heat, eg 1200/5 = 240 8"x8"x16" hollow concrete blocks, but they would take up a lot of space in the room. Putting 2 750 ml glass bottles of Pellegrino water into the holes in each block would raise its capacitance to about 9 Btu/F and lower the number of blocks to 1200/9 = 133.
But it seems more interesting to combine daily and cloudy-day heat storage in a single higher-temp glazed Solar Closet http://www.ece.villanova.edu/~nick/solar/solar.html with 8'x8' of R1 vertical polycarbonate south air heater glazing over an insulated south wall. If the room is 70 F for 16 hours per day and 60 F for 8 hours and we count Ben's 300 Btu/h body heat for 8 hours and ignore the south room wall heat gain from the 90.2 sunspace air during the day, we need to store 8h((60-31.5)20-300)+16h(70-31.5)20 = 14.5K Btu of average-day heat. With lots of surface, we can store this heat in C Btu/F of daily mass cooling from T (F) to 60, with (T-60)C = 14.5K, ie T = 60+14.5K/C.
If the air heater glazing transmits a constant 0.9^2x1.7x317/6h = 73 Btu/h-ft^2 of sun with a 90.2+73xR1 = 163 F Thevenin equivalent temperature and the daily mass warms to T = 163-(60-163)e^(-6/RC) F in 6 hours of sun, and we replace T with 60+14.5K/C, 60+14.5K/C = 163-(60-163)e^(-6/RC), ie 14.5K/C = 103(1-e^(-6/RC), ie 141/C = 1-e(-6/RC), and an R = 1/64 glazing resistance makes 141/C = 1-e(-384/C), ie C = 141/(1-e^(-384 /C)). Plugging in C = 160 on the right makes C = 155.1. Repeating makes C = 153.9, then 153.7 Btu/F, eg 31 hollow sparsely-stacked concrete blocks cooling from 60+14.5K/155 = 154 to 60 F on an average night.
But wait! An 8"x8"x16" hollow concrete block with 2 4"x4"x8" holes has 384 in^2 for the solid faces + 192 in^2 for the two faces with holes + 256 in^2 for the holes, totaling 832 in^2, ie 5.78 ft^2. With a 1.5 Btu/h-F-ft^2 slow-moving airfilm conductance, one block's airfilm conductance is 8.67 Btu/h-F, and 31 blocks have a 269 Btu/h-F film conductance.
If the blocks absorb 14.5K/6h = 2417 Btu/h from hot air, the air will be 2417/269 = 9 F warmer than the blocks. And the air in the air heater has to be warmer than the air near the blocks for thermosyphoning to occur. One empirical chimney formula says cfm = 16.6Asqrt(HdT), where A is the area of the chimney opening in square feet and H is the chimney height in feet and dT (F) is the temperature difference between the chimney and outdoor air. With 6"x8' slots at the top and bottom of an 8' tall air heater, A = 4 ft^2 and H = 8'. Heatflow in Btu/h is cfmxdT, approximately, so Btu/h = 2417 = 16.6x4xsqrt(8)dT^(3/2) makes dT = 12.9^(2/3) = 5.5 F. With 2 1 ft^2 vents to the room and an 8' height difference, 770 = 16.6x1xsqrt(8)dT^(3/2) makes dT = 16.4^(2/3) = 6.4 F.
Given these effects, the daily heat store has a smaller temp swing, so it needs more capacitance... If T = 66.4+14.5K/C = 148.5-(66.4-148.5)e^(-384/C), ie 14.5K/C = 82.1(1-e^(-384/C), ie C = 177/((1-e^(-384/C)). Plugging in C = 200 on the right makes C = 207 on the left. Repeating makes C = 210.0, then 210.9 Btu/F, eg 42 hollow concrete blocks cooling from 135.4 to 66.4 F. Or fewer blocks, since 42 would have more surface than 31, with smaller charging and discharging temp drops.
Heating the room directly for 6 hours with 90.2 F air from the sunspace instead of air from the daily store would also reduce the number of blocks required in the daily store. And a thermosyphoning air-air heat exchanger could help. If 31.5 F air falls down through the corrugations of 64 ft^2 of 8 mm Coroplast and 10 cfm of 65 F air rises up between the Coroplast faces with possible condensation and freezing, NTU = AU/Cmin = 128ft^2x0.75Btu/h-F-ft^2/10Btu/h-F = 9.6, and E = 9.6/10.6 = 0.906, and Tco = 31.5+E(65-31.5) = 61.9 F, and Tho = 65-E(65-31.5) = 34.6 F, with 46.7 and 49.8 F average air temps in the hot and cold chimneys and a 3.1 F difference between the averages. Not much, for thermosyphoning air. So maybe this heat exchanger needs 2 small DC fans that only run when the indoor RH or CO2 concentration exceed 60% or 1000 ppm, with an Arduino controller that can also shut off the cold air fan if condensation in the outgoing air passage begins to freeze, eg http://www.youtube.com/watch?v=wexdNx_StRc
Without these refinements, the room needs 5x14.5K = 72.5K Btu for 5 cloudy days in a row, with an approximate solar heating fraction of 1-2^-5 = 0.97. This could come from 72.5K/(135-70) = 1108 Btu/F of cloudy-day mass with lots of surface cooling from 135 to 70F, eg 2 vertically-stacked 450 Btu/F steel 55 gallon drums with plastic film liners surrounded by a layer of rocks inside a 3'-diameter x 6' tall cylindrical welded-wire ag fence gabion.
If the drums are well-insulated with strawbales, trickle-charging them hot won't require much daily overflow hot air from the daily air heater, which could first heat the daily store to 135 with thermosyphoning air and a passive one-way plastic film damper, then heat the cloudy store behind the daily store with more thermosyphoning air and another film damper.
Nick
http://pittsburgh.craigslist.org/fod/4062701580.html offers a 12' by 56' 2 bedroom mobile home with a new steel roof, a front kitchen, and a large bath, in very good condition for $3,900.00.
http://pittsburgh.craigslist.org/cto/4039312791.html lists "Antique CLASS C - 22' MOTOR HOME for sale - pictures sent to serious inquires - inspected and ready-to-go anywhere you want. Mechanically VERY RELIABLE, needs TLC (tune-up) - everything works. Great for camping, hunting, or just fix it up for the ultimate tail-gator mobile. Comes with kitchen, refrigerator, stove, oven, microwave, furnace, holding tanks (for gray and black water), full bathroom, overhead bed, couch, chair and dinette table. It has a 350 small block engine (that is very reliable), and transmission also very reliable, no slips or bangs, very smooth. Very easy to drive and is still being driven daily. Selling for $1500, or best offer."
Here's one for $100... http://pittsburgh.craigslist.org/for/4056583831.html "1974 Holly Park mobile home for sale. Must be moved. Serious inquiries only; sold as-is. You move it. Needs repairs and attention, but would make a great hunting cabin."
Ben has also been offered the free use of a pop-up camper. None of these structures have much thermal insulation, but they could be solar-heated if enclosed in a strawbale structure inside an inexpensive plastic film greenhouse. Ben might be happy living inside a strawbale structure without a pop-up camper, with a strawbale bed and a strawbale table and a sawdust toilet.
An 8'x12'x8'-tall room with about 536 ft^2 of R55 dry 2' strawbale walls would have thermal conductance G = 536/55 = 9.7 Btu/h-F... 10 cfm of fresh air would raise the conductance to about 20. Keeping it 70 F on an average 31.5 F December day in Pittsburgh would only require (70-31.5)20 = 770 Btu/h.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 1.7 kW/m^2 of sun falls on a vertical south wall and 1.91 falls on a south wall with a 45 degree slope on an average December day with a 35.1 average daytime temp. Suppose a 6 cent/ft^2 4-year 14'x24' R1 sloped polyethylene film south wall with 90% solar transmission and 100% longwave IR transmission connects the top edge of the south room wall to the ground 10' south of the room wall and transmits 0.9x14'x24'x1.91x317 = 185K Btu and keeps the room 70 F for 24 hours and radiates heat from the ground inside the wall and the south room wall at absolute Rankine temp Ta to an outdoor temp Ts (R) and 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-Ts^4)480ft^2, where V = 1/4 is the approximate view factor of the radiating surfaces. (A more expensive and durable polycarbonate south wall would block longwave IR.)
Duffie and Beckman's 2006 Solar Engineering of Thermal Processes book has equation (3.9.2) for sky temperature Ts = Ta[0.711+0.0056Tdp+0.000073Tdp^2+0.013cos(15t)]^(1/4), where Ts and Ta are in degrees Kelvin and Tdp is the dew point temp in degrees Celsius and t is the number of hours from midnight.
Ta = 35.1+460 = 495.1 degrees R, ie 495.1/1.8 = 275.1 degrees K. NREL says the humidity ratio w = 0.0030 pounds of water per pound of dry air on an average December day in Pittsburgh, which makes the partial pressure of water in air Pa = 29.921/(0.62198/w-1) = 0.145 "Hg, which makes the dew point Tdp = 9621/(17.863-ln(Pa)) = 486 R, ie 486-460 = 26 F, ie -3.3 C. With t = 12 hours (noon), cos(15t) = -1, so Ts = 275.1[0.711-0.0056x3.3+0.000073x3.3^2-0.013)]^(1/4) = 253.2 K, ie 1.8x253.2 = 455.7 R, ie -4.3 F, 39 F degrees less than the air temperature.
... 185K = 24h(70-31.5)20 + 6hx0.1714E-8V(Ta^4-455.7^4)480 makes 166.5K = 1.234E-6(Ta^4-455.7^4), so Ta = 649.5 R, ie 190 F. A quarter-cylindrical R1 film wall with a 10' radius and a 24' length and 377 ft^2 of surface with 190 F air inside and 35.1 F (495.1 R) air outside would lose 6h(190-35.1)377ft^2 = 350.4K Btu/day to the outdoor air, which is more than the solar input, so 190 is too hot.
If 185K = 6h(T-495.1)377 + 1.234E-6(Ta^4-455.7^4), ie T = 576.9 - 5.46E-10(Ta^4-455.7^4). Plugging in T = 540 R on the right makes T = 554.0 on the left. Repeating makes T = 549.0, 550.8, and 550.2 R, ie 90.2 F, warm enough to solar heat the room on an average day.
If the room is 70 F at dusk and 60 at dawn, 18 hours later, and 60 = 31.5+(70-31.5)e^(-18/RC), time constant RC = -18/ln((60-31.5)/(70-31.5)) = 60 hours = C/G, and C = 60hx20Btu/h-F = 1200 Btu/F of room temp thermal mass with lots of surface can store average-day heat, eg 1200/5 = 240 8"x8"x16" hollow concrete blocks, but they would take up a lot of space in the room. Putting 2 750 ml glass bottles of Pellegrino water into the holes in each block would raise its capacitance to about 9 Btu/F and lower the number of blocks to 1200/9 = 133.
But it seems more interesting to combine daily and cloudy-day heat storage in a single higher-temp glazed Solar Closet http://www.ece.villanova.edu/~nick/solar/solar.html with 8'x8' of R1 vertical polycarbonate south air heater glazing over an insulated south wall. If the room is 70 F for 16 hours per day and 60 F for 8 hours and we count Ben's 300 Btu/h body heat for 8 hours and ignore the south room wall heat gain from the 90.2 sunspace air during the day, we need to store 8h((60-31.5)20-300)+16h(70-31.5)20 = 14.5K Btu of average-day heat. With lots of surface, we can store this heat in C Btu/F of daily mass cooling from T (F) to 60, with (T-60)C = 14.5K, ie T = 60+14.5K/C.
If the air heater glazing transmits a constant 0.9^2x1.7x317/6h = 73 Btu/h-ft^2 of sun with a 90.2+73xR1 = 163 F Thevenin equivalent temperature and the daily mass warms to T = 163-(60-163)e^(-6/RC) F in 6 hours of sun, and we replace T with 60+14.5K/C, 60+14.5K/C = 163-(60-163)e^(-6/RC), ie 14.5K/C = 103(1-e^(-6/RC), ie 141/C = 1-e(-6/RC), and an R = 1/64 glazing resistance makes 141/C = 1-e(-384/C), ie C = 141/(1-e^(-384 /C)). Plugging in C = 160 on the right makes C = 155.1. Repeating makes C = 153.9, then 153.7 Btu/F, eg 31 hollow sparsely-stacked concrete blocks cooling from 60+14.5K/155 = 154 to 60 F on an average night.
But wait! An 8"x8"x16" hollow concrete block with 2 4"x4"x8" holes has 384 in^2 for the solid faces + 192 in^2 for the two faces with holes + 256 in^2 for the holes, totaling 832 in^2, ie 5.78 ft^2. With a 1.5 Btu/h-F-ft^2 slow-moving airfilm conductance, one block's airfilm conductance is 8.67 Btu/h-F, and 31 blocks have a 269 Btu/h-F film conductance.
If the blocks absorb 14.5K/6h = 2417 Btu/h from hot air, the air will be 2417/269 = 9 F warmer than the blocks. And the air in the air heater has to be warmer than the air near the blocks for thermosyphoning to occur. One empirical chimney formula says cfm = 16.6Asqrt(HdT), where A is the area of the chimney opening in square feet and H is the chimney height in feet and dT (F) is the temperature difference between the chimney and outdoor air. With 6"x8' slots at the top and bottom of an 8' tall air heater, A = 4 ft^2 and H = 8'. Heatflow in Btu/h is cfmxdT, approximately, so Btu/h = 2417 = 16.6x4xsqrt(8)dT^(3/2) makes dT = 12.9^(2/3) = 5.5 F. With 2 1 ft^2 vents to the room and an 8' height difference, 770 = 16.6x1xsqrt(8)dT^(3/2) makes dT = 16.4^(2/3) = 6.4 F.
Given these effects, the daily heat store has a smaller temp swing, so it needs more capacitance... If T = 66.4+14.5K/C = 148.5-(66.4-148.5)e^(-384/C), ie 14.5K/C = 82.1(1-e^(-384/C), ie C = 177/((1-e^(-384/C)). Plugging in C = 200 on the right makes C = 207 on the left. Repeating makes C = 210.0, then 210.9 Btu/F, eg 42 hollow concrete blocks cooling from 135.4 to 66.4 F. Or fewer blocks, since 42 would have more surface than 31, with smaller charging and discharging temp drops.
Heating the room directly for 6 hours with 90.2 F air from the sunspace instead of air from the daily store would also reduce the number of blocks required in the daily store. And a thermosyphoning air-air heat exchanger could help. If 31.5 F air falls down through the corrugations of 64 ft^2 of 8 mm Coroplast and 10 cfm of 65 F air rises up between the Coroplast faces with possible condensation and freezing, NTU = AU/Cmin = 128ft^2x0.75Btu/h-F-ft^2/10Btu/h-F = 9.6, and E = 9.6/10.6 = 0.906, and Tco = 31.5+E(65-31.5) = 61.9 F, and Tho = 65-E(65-31.5) = 34.6 F, with 46.7 and 49.8 F average air temps in the hot and cold chimneys and a 3.1 F difference between the averages. Not much, for thermosyphoning air. So maybe this heat exchanger needs 2 small DC fans that only run when the indoor RH or CO2 concentration exceed 60% or 1000 ppm, with an Arduino controller that can also shut off the cold air fan if condensation in the outgoing air passage begins to freeze, eg http://www.youtube.com/watch?v=wexdNx_StRc
Without these refinements, the room needs 5x14.5K = 72.5K Btu for 5 cloudy days in a row, with an approximate solar heating fraction of 1-2^-5 = 0.97. This could come from 72.5K/(135-70) = 1108 Btu/F of cloudy-day mass with lots of surface cooling from 135 to 70F, eg 2 vertically-stacked 450 Btu/F steel 55 gallon drums with plastic film liners surrounded by a layer of rocks inside a 3'-diameter x 6' tall cylindrical welded-wire ag fence gabion.
If the drums are well-insulated with strawbales, trickle-charging them hot won't require much daily overflow hot air from the daily air heater, which could first heat the daily store to 135 with thermosyphoning air and a passive one-way plastic film damper, then heat the cloudy store behind the daily store with more thermosyphoning air and another film damper.
Nick
Friday, September 20, 2013
A large streamwater heat exchanger
Drew asks about a 2 million Btu/h heat exchanger (big enough to heat 40 houses):
>a heat pump loop cooled by gravity flow stream thru a shell and tube htx...
>200 gpm 90 degrees on tube side heat pump loop
>600 gpm 70 degrees gravity flow on shell side. 8 inch pipe stream water
So the heat capacity rate ratio Z = Cmin/Cmax = 1/3.
>It is now able to get the leaving loop water temp down to 75
So the heat exchanger effectiveness E = (90-75)/(90-70) = 0.75
= (1-e(-NTU(1-Z)))/(1-Ze(-NTU(1-Z))), ie
0.75-0.75x0.33e^(-NTU(1-1/3)) = 1-e^(-NTU(1-1/3)), ie
0.75-0.25e(-NTU2/3) = 1-e^(-NTU2/3), ie
0.75e^(-NTU2/3) = 0.25, ie
e^(-NTU2/3) = 1/3, ie
-NTU2/3 = -1.099, ie
NTU = 1.648 = AU/Cmin, ie
AU = 1.648x200x500 = 165K Btu/h-F.
>1) if we parallel another identical htx my guess is flow would be 100 gpm on each tube side still 600 gpm on shell sides and the leaving loop water would drop to 72.5? what you think?
If the 600 gpm is divided into 2 300 gpm streams for the 2 heat exchangers,
Z = 1/3 and NTU = 165K/(100x500) = 3.3 and
E = (1-e(-3.3x2/3)))/(1-e(-3.3x2/3))/3)
= 0.8889/0.9630 = 0.9231, so
Tho = Thi-E(Thi-Tci)
= 90-0.9231(90-70)
= 71.54 F. Or a bit more, with less water velocity in the tubes.
http://www.gewater.com/handbook/cooling_water_systems/ch_23_heat.jsp and
http://www.gewater.com/handbook/cooling_water_systems/fig23-3.jsp say
a single moving water film has conductance
U = 100+200V Btu/h-F-ft^2, with V in fps, eg
U = 1500 at V = 7 fps.
>2) as an alternate we are thinking of laying some poly pipe directly in the stream. my suggestion is 20 100 foot 1 inch pipes in parallel so the loop water flow in each would be 10 gpm. Stream flows at about 7 fps. what's your guess as to the leaving loop temp?
1" 100 psi NSF HDPE pipe with a 0.07" wall thickness would have a 0.38x6.93/0.07 = 37.6 Btu/h-F-ft^2 conductance. The water velocity inside the pipe would be about 10gpm/60s/m/7.48g/ft ^3/(Pi(1.049"/24)^2) = 3.7 fps with a 100+200x3.7 = 843 Btu/h-F-ft^2 film conductance. With 7 fps water on the outside, U = 1/(1/843+1/37.6+1/1500) = 35.15 Btu/h-F-ft^2. A = 100'xPi1.12"/12 = 29.3 ft^2 makes NTU = AU/Cmin = 29.3x35.15/(10x500) = 0.206. With an infinite stream heat capacity and Z = 0, E = 1-e^-NTU = 0.186, and Tho = Thi-E(Thi-Tci) = 90-0.186(90-70) = 86.3 F.
>3) as another alternate, we are thinking of diverting some drainage water from the exterior of the parking garage to an isolated section of the stream bed and laying the pipes in it. its advantage is it is 50 degrees. There is less flow perhaps 200 gpm. if this were in a 2 inch by 24 inch crossection with the tubes it would flow at about 3 fps. What do you think the exiting loop temp would be?
U = 1/(1/843+1/37.6+1/700) = 34.23 Btu/h-F-ft^2. NTU = 29.3x34.23/(10x500) = 0.201. With Z = 1, E = NTU/(NTU+1) = 0.167, and Tho = 90-0.167(90-50) = 81.6 F.
Nick
>a heat pump loop cooled by gravity flow stream thru a shell and tube htx...
>200 gpm 90 degrees on tube side heat pump loop
>600 gpm 70 degrees gravity flow on shell side. 8 inch pipe stream water
So the heat capacity rate ratio Z = Cmin/Cmax = 1/3.
>It is now able to get the leaving loop water temp down to 75
So the heat exchanger effectiveness E = (90-75)/(90-70) = 0.75
= (1-e(-NTU(1-Z)))/(1-Ze(-NTU(1-Z))), ie
0.75-0.75x0.33e^(-NTU(1-1/3)) = 1-e^(-NTU(1-1/3)), ie
0.75-0.25e(-NTU2/3) = 1-e^(-NTU2/3), ie
0.75e^(-NTU2/3) = 0.25, ie
e^(-NTU2/3) = 1/3, ie
-NTU2/3 = -1.099, ie
NTU = 1.648 = AU/Cmin, ie
AU = 1.648x200x500 = 165K Btu/h-F.
>1) if we parallel another identical htx my guess is flow would be 100 gpm on each tube side still 600 gpm on shell sides and the leaving loop water would drop to 72.5? what you think?
If the 600 gpm is divided into 2 300 gpm streams for the 2 heat exchangers,
Z = 1/3 and NTU = 165K/(100x500) = 3.3 and
E = (1-e(-3.3x2/3)))/(1-e(-3.3x2/3))/3)
= 0.8889/0.9630 = 0.9231, so
Tho = Thi-E(Thi-Tci)
= 90-0.9231(90-70)
= 71.54 F. Or a bit more, with less water velocity in the tubes.
http://www.gewater.com/handbook/cooling_water_systems/ch_23_heat.jsp and
http://www.gewater.com/handbook/cooling_water_systems/fig23-3.jsp say
a single moving water film has conductance
U = 100+200V Btu/h-F-ft^2, with V in fps, eg
U = 1500 at V = 7 fps.
>2) as an alternate we are thinking of laying some poly pipe directly in the stream. my suggestion is 20 100 foot 1 inch pipes in parallel so the loop water flow in each would be 10 gpm. Stream flows at about 7 fps. what's your guess as to the leaving loop temp?
1" 100 psi NSF HDPE pipe with a 0.07" wall thickness would have a 0.38x6.93/0.07 = 37.6 Btu/h-F-ft^2 conductance. The water velocity inside the pipe would be about 10gpm/60s/m/7.48g/ft ^3/(Pi(1.049"/24)^2) = 3.7 fps with a 100+200x3.7 = 843 Btu/h-F-ft^2 film conductance. With 7 fps water on the outside, U = 1/(1/843+1/37.6+1/1500) = 35.15 Btu/h-F-ft^2. A = 100'xPi1.12"/12 = 29.3 ft^2 makes NTU = AU/Cmin = 29.3x35.15/(10x500) = 0.206. With an infinite stream heat capacity and Z = 0, E = 1-e^-NTU = 0.186, and Tho = Thi-E(Thi-Tci) = 90-0.186(90-70) = 86.3 F.
>3) as another alternate, we are thinking of diverting some drainage water from the exterior of the parking garage to an isolated section of the stream bed and laying the pipes in it. its advantage is it is 50 degrees. There is less flow perhaps 200 gpm. if this were in a 2 inch by 24 inch crossection with the tubes it would flow at about 3 fps. What do you think the exiting loop temp would be?
U = 1/(1/843+1/37.6+1/700) = 34.23 Btu/h-F-ft^2. NTU = 29.3x34.23/(10x500) = 0.201. With Z = 1, E = NTU/(NTU+1) = 0.167, and Tho = 90-0.167(90-50) = 81.6 F.
Nick
Thursday, September 5, 2013
Opaque attic roof heating revisited
With radiation loss, a roof with an absolute Rankine temperature T would have 0.1714E-8(T^4-Ts^4)+(T-(35.5+460)2 = 150 Btu/h, where Ts is the effective sky temperature in Rankine degrees. Duffie and Beckman's 2006 Solar Engineering of Thermal Processes book has equation (3.9.2) for sky temperature Ts = Ta[0.711+0.0056Tdp+0.000073Tdp^2+0.013cos(15t)]^(1/4), where Ts and Ta are in degrees Kelvin and Tdp is the dew point temp in degrees Celsius and t is the number of hours from midnight.
Ta = 35.5+460 = 495.5 degrees R, ie 495.5/1.8 = 275.3 degrees K. NREL says the humidity ratio w = 0.0028 pounds of water per pound of dry air on an average December day in Allentown, which makes the partial pressure of water in air Pa = 29.921/(0.62198/w-1) = 0.135 "Hg, which makes the dew point Tdp = 9621/(17.863-ln(Pa)) = 484 R, ie 484-460 = 24 F, ie -4.2 C. With t = 12 hours (noon), cos(15t) = -1, so Ts = 275.3[0.711-0.0056x4.2+0.000073x4.2^2-0.013)]^(1/4) = 249.6 K, ie 1.8x249.6 = 449.3 R, ie -10.7 F, 46 F degrees less than the air temperature.
And 0.1714E-8(T^4-449.3^4)+(T-495.5)2 = 150 makes T = (1141-(T^4-449.3^4)0.1714E-8)/2. Plugging in T = 530 on the right makes T = 537.8 on the left. Repeating this makes T = 533.7, 535.9, 534.8, 535.3, 535.0, and 535.2 R, ie 75.2 F, which is not much greater than 70 F, so the roof can't provide much space heating in December, even at noon in direct beam sun.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 2.62 kWh/m^2 (2.62x317 = 831 Btu/ft^2) of sun falls on a south roof with an 18.4 degree slope on an average 26.6 F January day with a 34.3 high and a 0.0022 humidity ratio in Allentown. February, March, April, and May bring 3.36, 4.43, 5.32, and 5.65 kWh/m^2 with 29.3, 37.7, 0.0024, 39.4, 48.8, 0.0032, 49.7, 60.4, 0.0046, 60.3, 71.3, and 0.0074 average and high temps and humidity ratios, and so on. If 80% of the sun falls on the roof in all but 3 hours of daylight, opaque attic roof heating does not look at all promising:
10 PI=4*ATN(1)
20 S=1.714E-09'Stephan-Boltzman constant
30 LAT=40.5'north latitude (degrees)
40 DATA 26.6,34.3,2.62,.0022,-20.9
50 DATA 29.3,37.7,3.36,.0024,-13.0
60 DATA 39.4,48.8,4.43,.0032,-2.4
70 DATA 49.7,60.4,5.32,.0046,9.4
80 DATA 60.2,71.3,5.65,.0074,18.8
90 DATA 69.4,80.0,5.80,.0104,23.1
100 DATA 74.1,84.5,6.06,.0122,21.2
110 DATA 72.2,82.3,5.43,.0120,13.5
120 DATA 64.7,75.1,4.70,.0097,2.2
130 DATA 53.2,63.8,3.87,.0064,-9.6
140 DATA 43.1,51.8,2.48,.0044,-18.9
150 DATA 31.8,39.2,2.25,.0028,-23.0
160 FOR MONTH=1 TO 12
170 READ TA,TM,SUN,W,DEC
180 TDF=(TA+TM)/2'average daytime temp (F)
185 TDFP=INT(10*TDF+.5)/10'rounded tdf
190 TDR=TDF+460'daytime temp (R)
200 TD=(TDF+460)/1.8'daytime temp (K)
210 PA=29.921/(.62198/W-1)'vapor pressure ("Hg)
220 TDPF=9621/(17.863-LOG(PA))-460'dew point temp (F)
230 TDP=(TDPF-32)/1.8'dew point temp (C)
240 TSKY=TD*(.698+.0056*TDP+.000073*TDP^2)^.25'sky temp (K)
250 TSKYF=1.8*TSKY-460'sky temp (F)
260 X=-TAN(PI*LAT/180)*TAN(PI*DEC/180)'find day length
270 ICOS=-ATN(X/SQR(-X*X+1))+PI/2'inverse cosine (radians)
280 DAYL=2/15*ICOS*180/PI'day length (hours)
290 SINT=.8*317*SUN/(DAYL-3)'solar intensity (Btu/h-ft^2)
300 TR=530'initial roof temp (R)
305 FOR I=1 TO 10
310 TR=(2*TDR+SINT-S*(TR^4-TSKYR^4))/2'roof temp (R)
315 NEXT I
320 TRF=TR-460'roof temp (F)
330 PRINT 500+MONTH;"'";TDFP,TSKYF,DAYL,SINT,TRF
340 NEXT MONTH
01 30.5 -17.9104 9.462032 102.8209 31.75281
02 33.5 -14.2666 10.48372 113.86 37.80908
03 44.1 -1.409027 11.72648 128.7402 50.34619
04 55.1 13.38834 13.08379 133.7942 59.52539
05 65.8 30.77084 14.25374 127.3213 64.54016
06 74.7 45.41767 14.84857 124.1399 69.43659
07 79.3 52.98224 14.5795 132.7187 75.28638
08 77.3 50.6933 13.57763 130.1849 73.11206
09 69.9 39.54871 12.2507 128.8465 67.81403
10 58.5 21.59882 10.89256 124.3491 58.66553
11 47.4 5.783173 9.73297 93.41019 40.42139
12 35.5 -10.78049 9.165865 92.54176 31.68912
The average roof shingle temps are less than 70 F except for the months of July and August, when the outdoor temperature is warm enough that the house doesn't need heat.
This could be refined with a simulation using hourly TMY2 weather data, but so far it looks doomed...
Nick
Ta = 35.5+460 = 495.5 degrees R, ie 495.5/1.8 = 275.3 degrees K. NREL says the humidity ratio w = 0.0028 pounds of water per pound of dry air on an average December day in Allentown, which makes the partial pressure of water in air Pa = 29.921/(0.62198/w-1) = 0.135 "Hg, which makes the dew point Tdp = 9621/(17.863-ln(Pa)) = 484 R, ie 484-460 = 24 F, ie -4.2 C. With t = 12 hours (noon), cos(15t) = -1, so Ts = 275.3[0.711-0.0056x4.2+0.000073x4.2^2-0.013)]^(1/4) = 249.6 K, ie 1.8x249.6 = 449.3 R, ie -10.7 F, 46 F degrees less than the air temperature.
And 0.1714E-8(T^4-449.3^4)+(T-495.5)2 = 150 makes T = (1141-(T^4-449.3^4)0.1714E-8)/2. Plugging in T = 530 on the right makes T = 537.8 on the left. Repeating this makes T = 533.7, 535.9, 534.8, 535.3, 535.0, and 535.2 R, ie 75.2 F, which is not much greater than 70 F, so the roof can't provide much space heating in December, even at noon in direct beam sun.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 2.62 kWh/m^2 (2.62x317 = 831 Btu/ft^2) of sun falls on a south roof with an 18.4 degree slope on an average 26.6 F January day with a 34.3 high and a 0.0022 humidity ratio in Allentown. February, March, April, and May bring 3.36, 4.43, 5.32, and 5.65 kWh/m^2 with 29.3, 37.7, 0.0024, 39.4, 48.8, 0.0032, 49.7, 60.4, 0.0046, 60.3, 71.3, and 0.0074 average and high temps and humidity ratios, and so on. If 80% of the sun falls on the roof in all but 3 hours of daylight, opaque attic roof heating does not look at all promising:
10 PI=4*ATN(1)
20 S=1.714E-09'Stephan-Boltzman constant
30 LAT=40.5'north latitude (degrees)
40 DATA 26.6,34.3,2.62,.0022,-20.9
50 DATA 29.3,37.7,3.36,.0024,-13.0
60 DATA 39.4,48.8,4.43,.0032,-2.4
70 DATA 49.7,60.4,5.32,.0046,9.4
80 DATA 60.2,71.3,5.65,.0074,18.8
90 DATA 69.4,80.0,5.80,.0104,23.1
100 DATA 74.1,84.5,6.06,.0122,21.2
110 DATA 72.2,82.3,5.43,.0120,13.5
120 DATA 64.7,75.1,4.70,.0097,2.2
130 DATA 53.2,63.8,3.87,.0064,-9.6
140 DATA 43.1,51.8,2.48,.0044,-18.9
150 DATA 31.8,39.2,2.25,.0028,-23.0
160 FOR MONTH=1 TO 12
170 READ TA,TM,SUN,W,DEC
180 TDF=(TA+TM)/2'average daytime temp (F)
185 TDFP=INT(10*TDF+.5)/10'rounded tdf
190 TDR=TDF+460'daytime temp (R)
200 TD=(TDF+460)/1.8'daytime temp (K)
210 PA=29.921/(.62198/W-1)'vapor pressure ("Hg)
220 TDPF=9621/(17.863-LOG(PA))-460'dew point temp (F)
230 TDP=(TDPF-32)/1.8'dew point temp (C)
240 TSKY=TD*(.698+.0056*TDP+.000073*TDP^2)^.25'sky temp (K)
250 TSKYF=1.8*TSKY-460'sky temp (F)
260 X=-TAN(PI*LAT/180)*TAN(PI*DEC/180)'find day length
270 ICOS=-ATN(X/SQR(-X*X+1))+PI/2'inverse cosine (radians)
280 DAYL=2/15*ICOS*180/PI'day length (hours)
290 SINT=.8*317*SUN/(DAYL-3)'solar intensity (Btu/h-ft^2)
300 TR=530'initial roof temp (R)
305 FOR I=1 TO 10
310 TR=(2*TDR+SINT-S*(TR^4-TSKYR^4))/2'roof temp (R)
315 NEXT I
320 TRF=TR-460'roof temp (F)
330 PRINT 500+MONTH;"'";TDFP,TSKYF,DAYL,SINT,TRF
340 NEXT MONTH
01 30.5 -17.9104 9.462032 102.8209 31.75281
02 33.5 -14.2666 10.48372 113.86 37.80908
03 44.1 -1.409027 11.72648 128.7402 50.34619
04 55.1 13.38834 13.08379 133.7942 59.52539
05 65.8 30.77084 14.25374 127.3213 64.54016
06 74.7 45.41767 14.84857 124.1399 69.43659
07 79.3 52.98224 14.5795 132.7187 75.28638
08 77.3 50.6933 13.57763 130.1849 73.11206
09 69.9 39.54871 12.2507 128.8465 67.81403
10 58.5 21.59882 10.89256 124.3491 58.66553
11 47.4 5.783173 9.73297 93.41019 40.42139
12 35.5 -10.78049 9.165865 92.54176 31.68912
The average roof shingle temps are less than 70 F except for the months of July and August, when the outdoor temperature is warm enough that the house doesn't need heat.
This could be refined with a simulation using hourly TMY2 weather data, but so far it looks doomed...
Nick
Tuesday, September 3, 2013
Solar attic heating
Richard asks a $20 question :-)
>I live in Kutztown, PA in a 3,000 SQ FT detached 2 story "colonial" - I've had the idea that there must be some way to recover the attic's heat in winter for assistance in living space heat...
You can recover some heat from the attic with a blower in series with a heating thermostat in the house and a cooling thermostat in the attic, eg 2 of these $20 line voltage thermostats: http://www.zorotools.com/g/Line%20Voltage%20Thermostats/00053175/
Kutztown has a 40.5 degree north latitude, so the max sun elevation at noon on 12/26 is 90-40.5-23.5 = 26 degrees. A 4/12 south roof with an 18.4 degree elevation would receive about 250cos(90-18.4-26) = 150 Btu/h/ft^2 in full beam sun at noon, which could raise a dark shingle temp by 150/2 = 75 F on a calm day with a 2 Btu/h-F-ft^2 still airfilm conductance and no radiation loss.
Solar Attic, Inc of Minneapolis claims a 30% space energy savings using their system, which seems optimistic. IIRC, their system sold for $1400 until the 40% tax credit disappeared in 1984, when the price dropped about 40% to $900. These days, they mostly focus on swimming pool heating with air-to-water heat exchangers (like car radiators) in attics.
If the house is cooler than (say) 70 F and the attic is warmer than 80 F, the thermostats would allow the blower to circulate attic air through the house through supply and return air paths. One of the paths could have a one-way plastic film damper to avoid warm house air flowing up into the attic at night. The blower would draw down warm air from the highest point in the attic with the ridge vent (if any) blocked and gable vent doors closed. A vertical east-west plastic film partition could confine warm air to the space under the south roof. This can work a lot better with a steep transparent south roof, as described at: http://www.ece.vill.edu/~nick/Soldier...On.pdf
You could also add winter heat to the house with one of Gary Reysa's simple thermosyphoning air heaters over an insulated south house wall, described at: http://www.builditsolar.com/Projects/SpaceHeating/solar_barn_project.htm
NREL says 800 Btu/ft^2 of sun falls on a south wall on an average 31.8 F December day with a 39.2 high and an average 35.5 F daytime temp in Allentown, PA, so an average 0.9x8ft^2x800/6h = 960 Btu/h of heat would enter a 1 ft wide x 8' tall strip of R1 polycarbonate air heater south glazing with 90% solar transmission.
One empirical chimney formula says C = 16.6Asqrt(HdT) cubic feet per minute of air will naturally flow up through a warm vertical H' tall duct with A ft^2 of cross-sectional area and a dT (F) temp diff between the top and the bottom, with CdT Btu/h of heatflow. If the solar heat that flows into the 8' strip equals the heat that flows into the room plus the heat that flows from the warm glazing to the outdoors, with a 3% vent area and A = 0.03x8 = 0.24 ft^2, 16.6x0.24sqrt(8)dT^1.5+(70+dT/2-35.5)8ft^2/R1 = 960 Btu/h makes dT = (60-0.355dT)^(2/3). Plugging in dT = 20 on the right makes dT = 14.1 on the left. Repeating makes dT = 14.4, then 14.4, with a T = 84.4 F air heater outlet temp and C = 43 cfm and a 43x14.4 = 616 Btu/h useful heat output for the 8' strip and a 100x616/800 = 77% solar collection efficiency and a 616x6h/8ft^2 = 462 Btu/ft^2 air heater gain on an average December day. A 4'x8' air heater would collect the heat equivalent of 4.6 therms of natural gas during the month of December.
Thanks for asking a question. Please let me know if you'd like further clarification.
Nick
>I live in Kutztown, PA in a 3,000 SQ FT detached 2 story "colonial" - I've had the idea that there must be some way to recover the attic's heat in winter for assistance in living space heat...
You can recover some heat from the attic with a blower in series with a heating thermostat in the house and a cooling thermostat in the attic, eg 2 of these $20 line voltage thermostats: http://www.zorotools.com/g/Line%20Voltage%20Thermostats/00053175/
Kutztown has a 40.5 degree north latitude, so the max sun elevation at noon on 12/26 is 90-40.5-23.5 = 26 degrees. A 4/12 south roof with an 18.4 degree elevation would receive about 250cos(90-18.4-26) = 150 Btu/h/ft^2 in full beam sun at noon, which could raise a dark shingle temp by 150/2 = 75 F on a calm day with a 2 Btu/h-F-ft^2 still airfilm conductance and no radiation loss.
Solar Attic, Inc of Minneapolis claims a 30% space energy savings using their system, which seems optimistic. IIRC, their system sold for $1400 until the 40% tax credit disappeared in 1984, when the price dropped about 40% to $900. These days, they mostly focus on swimming pool heating with air-to-water heat exchangers (like car radiators) in attics.
If the house is cooler than (say) 70 F and the attic is warmer than 80 F, the thermostats would allow the blower to circulate attic air through the house through supply and return air paths. One of the paths could have a one-way plastic film damper to avoid warm house air flowing up into the attic at night. The blower would draw down warm air from the highest point in the attic with the ridge vent (if any) blocked and gable vent doors closed. A vertical east-west plastic film partition could confine warm air to the space under the south roof. This can work a lot better with a steep transparent south roof, as described at: http://www.ece.vill.edu/~nick/Soldier...On.pdf
You could also add winter heat to the house with one of Gary Reysa's simple thermosyphoning air heaters over an insulated south house wall, described at: http://www.builditsolar.com/Projects/SpaceHeating/solar_barn_project.htm
NREL says 800 Btu/ft^2 of sun falls on a south wall on an average 31.8 F December day with a 39.2 high and an average 35.5 F daytime temp in Allentown, PA, so an average 0.9x8ft^2x800/6h = 960 Btu/h of heat would enter a 1 ft wide x 8' tall strip of R1 polycarbonate air heater south glazing with 90% solar transmission.
One empirical chimney formula says C = 16.6Asqrt(HdT) cubic feet per minute of air will naturally flow up through a warm vertical H' tall duct with A ft^2 of cross-sectional area and a dT (F) temp diff between the top and the bottom, with CdT Btu/h of heatflow. If the solar heat that flows into the 8' strip equals the heat that flows into the room plus the heat that flows from the warm glazing to the outdoors, with a 3% vent area and A = 0.03x8 = 0.24 ft^2, 16.6x0.24sqrt(8)dT^1.5+(70+dT/2-35.5)8ft^2/R1 = 960 Btu/h makes dT = (60-0.355dT)^(2/3). Plugging in dT = 20 on the right makes dT = 14.1 on the left. Repeating makes dT = 14.4, then 14.4, with a T = 84.4 F air heater outlet temp and C = 43 cfm and a 43x14.4 = 616 Btu/h useful heat output for the 8' strip and a 100x616/800 = 77% solar collection efficiency and a 616x6h/8ft^2 = 462 Btu/ft^2 air heater gain on an average December day. A 4'x8' air heater would collect the heat equivalent of 4.6 therms of natural gas during the month of December.
Thanks for asking a question. Please let me know if you'd like further clarification.
Nick
Wednesday, August 7, 2013
Solar pool heating
"Rushforth Solar" <ar@...> wrote:
> I would think your quickest cheapest pool heater would be covering the pool
use with black plastic.
Clear plastic works better, since it's hard to warm water from above.
For one thing, warm water rises.
For another, http://www.rsifibre.com/technical-papers/thermal-diffusivity.php
says
Tr = (1/a)(2s/Pi)^2ln[8(Ts-Ta)/(Pi^2(Ta-Tb)],
(dividing by (Ta-Tb) on the right), where
Tr = time to obtain average temp Tb
a = thermal diffusivity
s = slab thickness
Ts = constant surface slab temp
Ta = initial slab temp
Tb = average slab temp at time Tr
In their example, warming a 2" air slab would take 25.3 seconds.
http://en.wikipedia.org/wiki/Thermal_diffusivity says water has a 0.143x10^-6
m^2/s thermal diffusivity,
So warming a 2" water slab would take 3920 seconds (65 minutes.)
And warming a 60" water slab would take 980 hours.
Nick
> I would think your quickest cheapest pool heater would be covering the pool
use with black plastic.
Clear plastic works better, since it's hard to warm water from above.
For one thing, warm water rises.
For another, http://www.rsifibre.com/technical-papers/thermal-diffusivity.php
says
Tr = (1/a)(2s/Pi)^2ln[8(Ts-Ta)/(Pi^2(Ta-Tb)],
(dividing by (Ta-Tb) on the right), where
Tr = time to obtain average temp Tb
a = thermal diffusivity
s = slab thickness
Ts = constant surface slab temp
Ta = initial slab temp
Tb = average slab temp at time Tr
In their example, warming a 2" air slab would take 25.3 seconds.
http://en.wikipedia.org/wiki/Thermal_diffusivity says water has a 0.143x10^-6
m^2/s thermal diffusivity,
So warming a 2" water slab would take 3920 seconds (65 minutes.)
And warming a 60" water slab would take 980 hours.
Nick
Sunday, July 28, 2013
Revisiting Dan's PA greenhouse
Daniel Baker <baker_daniel@...> wrote:
>... the calculations don't support what I was hoping to accomplish.
> Is that about right?
I think so. For starters, it seems to me that the greenhouse needs more insulation at night and on cloudy days, and the heat store needs to move more heat into greenhouse air at night.
Filling the space between 2 layers of polyethylene film with R20 soap bubble foam at night would reduce a 500 ft^2 cover conductance to 25 Btu/h-F, so the greenhouse would only need (64-26.6)25 = 935 vs 18.7K Btu/h to stay 64 F on a 26.6 F night, or 5x24x935 = 112.2K Btu for 5 cloudy days in a row.
This could come from N 55 gallon drums with 25N ft^2 of surface and a 1.5x25N Btu/h-F conductance to slow-moving air and a Tm = 64+935/(1.5x25N) F minimum water temp.
Drums in an insulated box (a solar closet (tm)) with an R1 transparent south wall over an insulated south wall (a passive thermosyphoning air heater) under a bench that gain and lose 0.9x0.8x830 = 598 Btu/ft^2 = 6h(Ta-82)1ft^2/R1 in a greenhouse that's 82 F for 6 hours on an average December day would have temperature Ta = 182 F.
A 55 gallon drum has 55x8.33 = 458 Btu/F of thermal capacitance. Storing 112.2K Btu = (182-Tm)458N = (182-64-24.93/N)458N makes N = 2.28. N = 4 makes Tm = 70.2 F, storing 205K Btu, enough heat for 219 cloudy hours, ie 9.1 days, with an approximate 1-2^-9.1 = 0.998 solar heating fraction.
If the drum tops and bottoms are not exposed to air, that lowers the heat transfer surface to about 75 ft^2, with a 1.5x75 = 112 Btu/h-F airfilm conductance, which raises Tm to 64+935/112 = 72.3 F. And the drum box needs a 62 F thermostat and a fan, eg a http://www.grainger.com/Grainger/DAYTON-Axial-Fan-3VU71 665 cfm fan, which further raises Tm to 64+935/(1/112+1/665) = 73.7 F. And with finite insulation, the box will lose some heat at night, making the water cooler than the surrounding air during the day, which lowers Ta.
And a 9'x10' tomato greenhouse needs about 0.4x9'x10' = 36 lb/day of winter dehumidification, which normally comes from daytime ventilation with dry outdoor air, which precludes beneficial CO2 enrichment with compost, which can also generate heat. Air at 82 F and 70% RH has vapor pressure Pa = 0.7e^(17.863-9621/(460+82)) = 0.783 "Hg and humidity ratio wa = 0.62198/(29.921/Pa-1) = 0.01671 pounds of water per pound of dry air.
NREL's Blue Book http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/state.html says an average 33.6 F December day in Harrisburg has a 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The average December humidity ratio wo = 0.0028, and air weighs about 0.075 lb/ft^2, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wo) = 36lb, with a C = 96 cfm fan. With an average (33.6+40.6)/2 = 37.1 F daytime temp, this costs about 6h(82-37.1)96 = 25.9K Btu/day of heat.
The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F. If we cool it to Tg < 71 (F) with condensation, it will have vapor pressure Pg = e^(17.863-9621/(460+Tg)) "Hg at 100% RH and humidity ratio wg = 0.62198/(29.921/Pg-1). For example, cooling it to 43 F makes Pg = 0.282 "Hg and wg = 0.0059, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wg) = 36lb, with a C = 124 cfm fan, at a cost of about 6h(82-43)124 = 29K Btu of sensible heat + 36K Btu for the latent heat of condensation, totaling 65K Btu/day.
If the condensation happens on the inside of 500 ft 2 of outer glazing film with a 3 Btu/h-F-ft^2 conductance to 37.1 F outdoor air, the glazing temp Tg will be 37.1+65K/6h/500/3 = 44.3 F, which is warmer than 43 F, so let's try cooling the air to (44.3+43)/2 = 43.7. Then Pg = 0.290 "Hg and wg = 0.0061 and C = 1.33/(0.01671-0.0061) = 125.4 cfm and the glazing gains 6h(82-43.7)125.4+6K = 64.8K Btu and Tg = 37.1+64.8K/6h/500/3 = 44.3 F. That's closer.
With tall indeterminate tomato plants and an 8' height and 80% solar transmission, the greenhouse could gain 0.8x10'(9'x520+8'x830') = 90.6K Btu/day. Dehumidifcation with 64.9K Btu/day leaves 90.6K-64.9K = 25.7K Btu for 18 hours of nighttime heat and more efficient dehumidification by night ventilation. With 18h(64-33.6)(500ft^2/R20+C) = 25.7K, C = 22 cfm, which can remove 18hx60x22x0.075(0.01671-0.0028) = 25 lb/day of water vapor. We could reduce the dew point of
greenhouse air to 64 F by condensation before night ventilation to avoid dripping water on plants.
Meanwhile, page 1 of http://www.imok.ufl.edu/docs/pdf/vegetable_hort/trans_pp2.pdf says "on average, yields of crops should increase by 33% with a doubling of C02 concentration in the earth's atmosphere," ie 700 vs 350 ppm, and page 549 of Joe Hanon's Greenhouses book (CRC, 1998) says a 1 kg dry weight compost loss can provide 1.5 kg of CO2 and suggests 7-14 kg/m^2 of wet compost to provide sufficient CO2 for 20 days, eg 88 kg (193 lb) for a 9'x10' 8.4 m^2 greenhouse. Some growers spread straw between crops in windrows and keep it moist and turn it every 2-3 days, but it seems more efficient to keep compost in a closed and insulated container with an oxygen sensor and a blower to maintain at least 1% O2 and a humidity sensor and a solenoid valve to maintain a 50% moisture content, or less, if less CO2 is required.
http://www.omafra.gov.on.ca/english/crops/facts/00-077.htm says:
>... about 0.50–0.60 kg of CO2/hr/100 m2 must be added in a ‘standard’ glass greenhouse to maintain 1,300 ppm. For double-polyethylene houses supplementation is 0.25–0.35 kg of CO2/hr/100 m2. For glass houses, supplementation is primarily used to offset the dilution due to air infiltration, while for double-poly houses the amount of CO2 required is about equal for the natural air exchange and photosynthesis.
Years ago, I visited a prizewinning orchid grower in Pennsylvania. He was a retired chemist. His greenhouse was rather dark and completely closed in, with a huge air conditioner and a natural gas heater with a thermostat and mister nozzles with a humidistat and a solenoid valve to maintain 90 F at 90% RH for 24 hours per day and alcohol lamps for CO2 enrichment and powerful fans to make plants move for better gas diffusion absorption by stomata...
Nick
>... the calculations don't support what I was hoping to accomplish.
> Is that about right?
I think so. For starters, it seems to me that the greenhouse needs more insulation at night and on cloudy days, and the heat store needs to move more heat into greenhouse air at night.
Filling the space between 2 layers of polyethylene film with R20 soap bubble foam at night would reduce a 500 ft^2 cover conductance to 25 Btu/h-F, so the greenhouse would only need (64-26.6)25 = 935 vs 18.7K Btu/h to stay 64 F on a 26.6 F night, or 5x24x935 = 112.2K Btu for 5 cloudy days in a row.
This could come from N 55 gallon drums with 25N ft^2 of surface and a 1.5x25N Btu/h-F conductance to slow-moving air and a Tm = 64+935/(1.5x25N) F minimum water temp.
Drums in an insulated box (a solar closet (tm)) with an R1 transparent south wall over an insulated south wall (a passive thermosyphoning air heater) under a bench that gain and lose 0.9x0.8x830 = 598 Btu/ft^2 = 6h(Ta-82)1ft^2/R1 in a greenhouse that's 82 F for 6 hours on an average December day would have temperature Ta = 182 F.
A 55 gallon drum has 55x8.33 = 458 Btu/F of thermal capacitance. Storing 112.2K Btu = (182-Tm)458N = (182-64-24.93/N)458N makes N = 2.28. N = 4 makes Tm = 70.2 F, storing 205K Btu, enough heat for 219 cloudy hours, ie 9.1 days, with an approximate 1-2^-9.1 = 0.998 solar heating fraction.
If the drum tops and bottoms are not exposed to air, that lowers the heat transfer surface to about 75 ft^2, with a 1.5x75 = 112 Btu/h-F airfilm conductance, which raises Tm to 64+935/112 = 72.3 F. And the drum box needs a 62 F thermostat and a fan, eg a http://www.grainger.com/Grainger/DAYTON-Axial-Fan-3VU71 665 cfm fan, which further raises Tm to 64+935/(1/112+1/665) = 73.7 F. And with finite insulation, the box will lose some heat at night, making the water cooler than the surrounding air during the day, which lowers Ta.
And a 9'x10' tomato greenhouse needs about 0.4x9'x10' = 36 lb/day of winter dehumidification, which normally comes from daytime ventilation with dry outdoor air, which precludes beneficial CO2 enrichment with compost, which can also generate heat. Air at 82 F and 70% RH has vapor pressure Pa = 0.7e^(17.863-9621/(460+82)) = 0.783 "Hg and humidity ratio wa = 0.62198/(29.921/Pa-1) = 0.01671 pounds of water per pound of dry air.
NREL's Blue Book http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/state.html says an average 33.6 F December day in Harrisburg has a 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The average December humidity ratio wo = 0.0028, and air weighs about 0.075 lb/ft^2, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wo) = 36lb, with a C = 96 cfm fan. With an average (33.6+40.6)/2 = 37.1 F daytime temp, this costs about 6h(82-37.1)96 = 25.9K Btu/day of heat.
The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F. If we cool it to Tg < 71 (F) with condensation, it will have vapor pressure Pg = e^(17.863-9621/(460+Tg)) "Hg at 100% RH and humidity ratio wg = 0.62198/(29.921/Pg-1). For example, cooling it to 43 F makes Pg = 0.282 "Hg and wg = 0.0059, so we can remove 36 pounds of water vapor in 6 hours if 6hx60m/hxCx0.075lb/ft^3(wa-wg) = 36lb, with a C = 124 cfm fan, at a cost of about 6h(82-43)124 = 29K Btu of sensible heat + 36K Btu for the latent heat of condensation, totaling 65K Btu/day.
If the condensation happens on the inside of 500 ft 2 of outer glazing film with a 3 Btu/h-F-ft^2 conductance to 37.1 F outdoor air, the glazing temp Tg will be 37.1+65K/6h/500/3 = 44.3 F, which is warmer than 43 F, so let's try cooling the air to (44.3+43)/2 = 43.7. Then Pg = 0.290 "Hg and wg = 0.0061 and C = 1.33/(0.01671-0.0061) = 125.4 cfm and the glazing gains 6h(82-43.7)125.4+6K = 64.8K Btu and Tg = 37.1+64.8K/6h/500/3 = 44.3 F. That's closer.
With tall indeterminate tomato plants and an 8' height and 80% solar transmission, the greenhouse could gain 0.8x10'(9'x520+8'x830') = 90.6K Btu/day. Dehumidifcation with 64.9K Btu/day leaves 90.6K-64.9K = 25.7K Btu for 18 hours of nighttime heat and more efficient dehumidification by night ventilation. With 18h(64-33.6)(500ft^2/R20+C) = 25.7K, C = 22 cfm, which can remove 18hx60x22x0.075(0.01671-0.0028) = 25 lb/day of water vapor. We could reduce the dew point of
greenhouse air to 64 F by condensation before night ventilation to avoid dripping water on plants.
Meanwhile, page 1 of http://www.imok.ufl.edu/docs/pdf/vegetable_hort/trans_pp2.pdf says "on average, yields of crops should increase by 33% with a doubling of C02 concentration in the earth's atmosphere," ie 700 vs 350 ppm, and page 549 of Joe Hanon's Greenhouses book (CRC, 1998) says a 1 kg dry weight compost loss can provide 1.5 kg of CO2 and suggests 7-14 kg/m^2 of wet compost to provide sufficient CO2 for 20 days, eg 88 kg (193 lb) for a 9'x10' 8.4 m^2 greenhouse. Some growers spread straw between crops in windrows and keep it moist and turn it every 2-3 days, but it seems more efficient to keep compost in a closed and insulated container with an oxygen sensor and a blower to maintain at least 1% O2 and a humidity sensor and a solenoid valve to maintain a 50% moisture content, or less, if less CO2 is required.
http://www.omafra.gov.on.ca/english/crops/facts/00-077.htm says:
>... about 0.50–0.60 kg of CO2/hr/100 m2 must be added in a ‘standard’ glass greenhouse to maintain 1,300 ppm. For double-polyethylene houses supplementation is 0.25–0.35 kg of CO2/hr/100 m2. For glass houses, supplementation is primarily used to offset the dilution due to air infiltration, while for double-poly houses the amount of CO2 required is about equal for the natural air exchange and photosynthesis.
Years ago, I visited a prizewinning orchid grower in Pennsylvania. He was a retired chemist. His greenhouse was rather dark and completely closed in, with a huge air conditioner and a natural gas heater with a thermostat and mister nozzles with a humidistat and a solenoid valve to maintain 90 F at 90% RH for 24 hours per day and alcohol lamps for CO2 enrichment and powerful fans to make plants move for better gas diffusion absorption by stomata...
Nick
Friday, July 26, 2013
A PA greenhouse solar heating project
Dan Baker writes:
>I am building a backyard greenhouse about 10 x 9 and am thinking of trying to extend my season as long as possible. I live in S. Central PA.
NREL's Blue Book http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/state.html says an average 33.6 F December day in Harrisburg has 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The 30-year record December low is -8 F. The deep ground temp is 52.9.
The Harrisburg TMY2 EPW weather data stat file at http://apps1.eere.energy.gov/buildings/energyplus/cfm/weather_data3.cfm/region=4_north_and_central_america_wmo_region_4/country=1_usa/cname=USA says the soil is 5.0 C (41.0 F) at a 0.5 meter depth in December.
> My plan after viewing many YouTube videos is to dig out a 4 x 4 x 8 trench in the greenhouse, line it with 1" reflective coated rigid board insulation, place 2 55 gallon (plastic) barrels connected with PVC and cover it all with sand.
Here's an electrical equivalent circuit on an average night, with 128 ft^2 of R5 board insulation and an average 2' sand depth at R0.44 per inch (an R21 vertical bed resistance) and a 1.5x4x8 = 48 Btu/h-F bed surface conductance and 500 ft^2 of R1 greenhouse glazing:
Ts Tg
5/128 10.5/32 | 10.5/32 1/48 | 1/500
41.0---www-----www---*-----www----www-------www---26.6
--> |
I --- Cs = 916+113ft^3x18Btu/ft^3-F = 2956 Btu/F
---
|
-
With no collector, I = (41.0-26.6)/0.718 = 20.1 Btu/h, which makes
the Thevenin equivalent temp Tt = 41.0-20.1(5/128+10.5/32) = 33.6 F.
To find the Thevenin resistance, ground the 41.0 and 26.6 sources and calculate the resistance from Ts to ground: 0.367 "ohms" to the left in parallel with 0.351 on the right, ie 0.179 "ohms."
So the circuit above is equivalent to:
0.179
Ts-------www---
| |
--- 2956 --- 33.6 F
--- -
| |
- -
with an RC = 0.179F-h/Btux2956Btu/F = 529 hour time constant.
> The collector would be a 4x4 insulated panel using corrugated metal and running copper piping from bottom to top and sealing it with glass or something that would help hold the heat. Then water would be circulated by a swimming pool pump and go into the storage in the ground. It would circulate all day and then at night I would turn off the pump and let the heat dissipate into the sand which would radiate into the greenhouse through the night.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 2.75 kWh/m^2 (872 Btu/ft^2) of sun falls on a south wall with a 60 degree tilt on an average December day, so a 4'x4' panel with R1 glazing with 90% solar transmission and perfect back and side insulation and lots of water flow for 6 hours per day would have this equivalent circuit:
0.9x16ft^2x872/6h = 2093 Btu/h
---
|-------|-->|--------->
--- |
|
R1/16ft^2 |
33.6 -----www-----
which is equivalent to:
1/16
------www----->
|
--- 33.6+2093/16 = 164.4 F.
-
|
-
Adding the first equivalent circuit and a switch...
1/16 Ts 0.179
------www---c d---www---
| --> \ --> |
--- 164.4 F Ic | Ts Id --- 33.6 F
- --- -
| --- |
- | -
-
to charge the sand bed for 6 hours and discharge it
for 18 hours on an average day...
6Ic = 18Id makes 6(164.4-Ts)16 = 18(Ts-33.6)/0.179,
ie the sand bed temp Ts = 97.5 F.
And (97.5-26.6)/(10.5/32+1/48+1/500) = 202 Btu/h flows from the top of the sand bed into the greenhouse at night, which raises the greenhouse air temp to 26.6+202/500 = 27.0 on an average night.
Nick
>I am building a backyard greenhouse about 10 x 9 and am thinking of trying to extend my season as long as possible. I live in S. Central PA.
NREL's Blue Book http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/bluebook/state.html says an average 33.6 F December day in Harrisburg has 26.6 low and a 40.6 high, with 520 Btu/ft^2 of sun on the ground and 830 on a south wall. The 30-year record December low is -8 F. The deep ground temp is 52.9.
The Harrisburg TMY2 EPW weather data stat file at http://apps1.eere.energy.gov/buildings/energyplus/cfm/weather_data3.cfm/region=4_north_and_central_america_wmo_region_4/country=1_usa/cname=USA says the soil is 5.0 C (41.0 F) at a 0.5 meter depth in December.
> My plan after viewing many YouTube videos is to dig out a 4 x 4 x 8 trench in the greenhouse, line it with 1" reflective coated rigid board insulation, place 2 55 gallon (plastic) barrels connected with PVC and cover it all with sand.
Here's an electrical equivalent circuit on an average night, with 128 ft^2 of R5 board insulation and an average 2' sand depth at R0.44 per inch (an R21 vertical bed resistance) and a 1.5x4x8 = 48 Btu/h-F bed surface conductance and 500 ft^2 of R1 greenhouse glazing:
Ts Tg
5/128 10.5/32 | 10.5/32 1/48 | 1/500
41.0---www-----www---*-----www----www-------www---26.6
--> |
I --- Cs = 916+113ft^3x18Btu/ft^3-F = 2956 Btu/F
---
|
-
With no collector, I = (41.0-26.6)/0.718 = 20.1 Btu/h, which makes
the Thevenin equivalent temp Tt = 41.0-20.1(5/128+10.5/32) = 33.6 F.
To find the Thevenin resistance, ground the 41.0 and 26.6 sources and calculate the resistance from Ts to ground: 0.367 "ohms" to the left in parallel with 0.351 on the right, ie 0.179 "ohms."
So the circuit above is equivalent to:
0.179
Ts-------www---
| |
--- 2956 --- 33.6 F
--- -
| |
- -
with an RC = 0.179F-h/Btux2956Btu/F = 529 hour time constant.
> The collector would be a 4x4 insulated panel using corrugated metal and running copper piping from bottom to top and sealing it with glass or something that would help hold the heat. Then water would be circulated by a swimming pool pump and go into the storage in the ground. It would circulate all day and then at night I would turn off the pump and let the heat dissipate into the sand which would radiate into the greenhouse through the night.
http://rredc.nrel.gov/solar/calculators/PVWATTS/version1/US/code/pvwattsv1.cgi says 2.75 kWh/m^2 (872 Btu/ft^2) of sun falls on a south wall with a 60 degree tilt on an average December day, so a 4'x4' panel with R1 glazing with 90% solar transmission and perfect back and side insulation and lots of water flow for 6 hours per day would have this equivalent circuit:
0.9x16ft^2x872/6h = 2093 Btu/h
---
|-------|-->|--------->
--- |
|
R1/16ft^2 |
33.6 -----www-----
which is equivalent to:
1/16
------www----->
|
--- 33.6+2093/16 = 164.4 F.
-
|
-
Adding the first equivalent circuit and a switch...
1/16 Ts 0.179
------www---c d---www---
| --> \ --> |
--- 164.4 F Ic | Ts Id --- 33.6 F
- --- -
| --- |
- | -
-
to charge the sand bed for 6 hours and discharge it
for 18 hours on an average day...
6Ic = 18Id makes 6(164.4-Ts)16 = 18(Ts-33.6)/0.179,
ie the sand bed temp Ts = 97.5 F.
And (97.5-26.6)/(10.5/32+1/48+1/500) = 202 Btu/h flows from the top of the sand bed into the greenhouse at night, which raises the greenhouse air temp to 26.6+202/500 = 27.0 on an average night.
Nick
Wednesday, July 24, 2013
Should we insulate hot water pipes?
Malcolm Shealy discusses this at http://www.leaningpinesoftware.com/hot_water_pipes_pipe_cooling.shtml
The SRCC OG-300 solar water heater test procedure uses 6 hourly 3 gpm 10.7 gallon 135 F bursts in a 67.5 F environment...
With no insulation, 1' of 135 F 1/2" horizontal bare copper pipe with a 0.625" OD in 67.5 F still air would have a 0.27((135-67.5)/0.625"/12))^0.25 = 1.62 Btu/h-F-ft^2 convective thermal conductance, using equation 2.20US for laminar free convection from a horizontal pipe in Kreider and Rabl's 1994 Heating and Cooling of Buildings book. With pipe conductance Gb = Pi0.625"/12x1'x1.62 = 0.265 Btu/h-F-ft, it would lose (135-67.5)0.265 = 17.9 Btu/h per linear foot. A 2 gpm (1000 Btu/h-F) water stream would lose 2 F with a 2000 Btu/h pipe loss after flowing through 2000/17.9 = 112 feet of bare pipe.
How much will a $1.18 piece of R2.3 6' pipe insulation save in 1 year, if we heat water with electricity at $0.10/kWh and use it in 6 10.7 gallon 3 gpm hourly bursts each day? http://www.homedepot.com/p/Armacell-Tubolit-3-4-in-x-6-ft-Polyethylene-Pipe-Wrap-Insulation-OEP07838/100539941#specifications
With "R2.3 insulation" (ie a system R-value after installation, including the outer air film), Gi = Pi0.625"/12x1'/R2.3 = 0.071 Btu/h-F-ft, so the pipe would lose (135-67.5)0.071 = 4.8 Btu/h per linear foot. A 2 gpm insulated water stream would be 2 F warmer than a bare pipe stream after flowing through 2000/(17.9-4.8) = 153 feet of insulated pipe.
Type M 1/2" copper pipe http://www.homedepot.com/h_d1/N-5yc1v/R-203540899/h_d2/ProductDisplay?catalogId=10053&langId=-1&keyword=1%2F2%22+copper+pipe&storeId=10051 weighs 0.933lb/5' = 0.1866 lb/foot, with a 0.0915x0.1866 = 0.0171 Btu/F thermal capacitance. Its 0.5" ID holds 62.33Pi(0.25/12)^2x1'= 0.085 lb of water, making the total (tiny) capacitance C = 6x0.1021 = 0.6126 Btu/F.
If hot water uses are 3.6 minutes long, with (60-3.6)/60 = 0.94 hours between uses, bare pipe with a C/Gb = 0.39 hour time constant would cool from 135 to 67.5+(135-67.5)e^(-0.94/0.39) = 73.4 F before the next use during the day, and 67.5 F overnight. Reheating 6' of water to 135 F would take (5((135-73.4)+(135-67.5)0.6126 = 169 Btu/day, or 61644 Btu/year, worth about $0.10x61644/3412 = $1.81/year.
Insulated pipe with a C/Gi = 1.44 hour time constant would cool to 67.5+(135-67.5)e^(-0.94/1.44) = 102.6 F during the day, and 67.5 F overnight. Reheating it takes (5((135-102.6)+(135-67.5)0.6126 = 141 Btu/day, or 51341 Btu/year, worth about $0.10x51341/3412 = $1.50/year.The reheating difference is 31 cents per year.
So this $1.18 investment has a 100x$0.31/$1.18 = 26% tax free return and a $1.18/$0.31 = 3.8 year simple payback, if the pipe is in an unheated space and labor is free, with hourly hot water uses.
Pipe insulation within a few feet of a water heater makes even more sense, but plumbing heat traps don't seem to make sense, given pipe insulation, since they only slow convection inside a single pipe, unless there's a hot-cold connection above. And a 4 watt mechanical timer that turns off a well-insulated water heater at night would use more electrical power than it saves.
How about a simple greywater heat exchanger? Home Depot sells 300' of 1" HDPE 100 psi NSF plastic pipe for $96.64... http://www.homedepot.com/p/Advanced-Drainage-Systems-1-in-x-300-ft-IPS-100-PSI-NSF-Poly-Pipe-2-1100300/202282480#.Ue-zaDbD_IU
With a 1.049" ID and a 1.119" OD, and a 0.035" wall thickness and still clean water on the inside and some crud on the outside, 1' of pipe could have a 1'xPi1.049/12x20 = 5.50 Btu/h-F conductance and a Pi(1.049/24)^2x62.33 = 0.374 Btu/F conductance and an RC = C/G = 0.374/5.50 = 0.068 hour (4.1 minute) time constant, so a 300' pressurized pipe coil near the top of a tall unpressurized stratified tank with a much larger volume and non-conductive walls and a drain from the bottom (eg a 4' diameter x 8' tall culvert or ferrocement or thin bent plywood tank in a basement with an insulated lining and a greywater dip tube to the bottom) could heat Pi(1.049/24)^2x7.48x200 = 13.5 gallons of 60 F water to 105-(105-60)e^(-0.94/0.068) = 104.9999554 F between hourly uses, with a 105 F shower drain temp.
Every piece of NSF pipe is tested to 5 times the nominal pressure rating at 73 F, and it loses about 10% of its strength for each 10 F temperature increase above 73 F.
"Scott L" <shiva@...> wrote from Fairbanks:
>... we insulate our cold and hot water pipes is to keep them from freezing as fast if the heat quits.
NREL says the average temp is -10.1 F in January in Fairbanks, with a -18.5 min and a -61.0 30-year record low and a 0.0006 humidity ratio. The deep ground temp is 26.9 F.
Insulation doesn't help much with small pipes. At -10.1, a 1/2" pipe with R2.3 insulation and a 1.43 hour time constant would cool from 70 to 32 F in -1.43ln((32-(-10))/(70-(-10)) = 0.92 hours and freeze solid after another 0.085x144/((32-(-10))0.071) = 4.1 hours. A battery or generator could run (32-(-10))0.071/3.412 = 0.87 watts/foot of heat tape under pipe insulation when the heat quits. A 12V 200 Ah battery could keep 200' of R2.3 pipe from freezing for 13.7 hours.
>The cold water pipes sweat and drip if left uninsulated and exposed to a heated area with our cold water.
The air would need significant moisture for that to happen. Air with a w = 0.0006 humidity ratio has vapor pressure P = 29.921/(0.62198/w+1) = 0.0288 "Hg and dew point Tdp = 9621/(17.863-ln(P))-460 = -10.6 F.
> Lots of us haul or have our water delivered. $80 for 800 gallons in the town area and easily twice that if you live outside the city area by much. You can not afford to run the water until it gets hot over a long distance.
We could insulate both hot and cold pipes and use a motion detector and pump to circulate water back through the heater before hot water uses. That could also help avoid frozen pipes when the heat quits. Cooling a 50 gallon 135 F tank to 32 F with 200' of R2.3 pipe in -10 F air would take (135-32)50x8.33Btu/(200x(32-(-10))0.071) = 72 hours.
Nick
The SRCC OG-300 solar water heater test procedure uses 6 hourly 3 gpm 10.7 gallon 135 F bursts in a 67.5 F environment...
With no insulation, 1' of 135 F 1/2" horizontal bare copper pipe with a 0.625" OD in 67.5 F still air would have a 0.27((135-67.5)/0.625"/12))^0.25 = 1.62 Btu/h-F-ft^2 convective thermal conductance, using equation 2.20US for laminar free convection from a horizontal pipe in Kreider and Rabl's 1994 Heating and Cooling of Buildings book. With pipe conductance Gb = Pi0.625"/12x1'x1.62 = 0.265 Btu/h-F-ft, it would lose (135-67.5)0.265 = 17.9 Btu/h per linear foot. A 2 gpm (1000 Btu/h-F) water stream would lose 2 F with a 2000 Btu/h pipe loss after flowing through 2000/17.9 = 112 feet of bare pipe.
How much will a $1.18 piece of R2.3 6' pipe insulation save in 1 year, if we heat water with electricity at $0.10/kWh and use it in 6 10.7 gallon 3 gpm hourly bursts each day? http://www.homedepot.com/p/Armacell-Tubolit-3-4-in-x-6-ft-Polyethylene-Pipe-Wrap-Insulation-OEP07838/100539941#specifications
With "R2.3 insulation" (ie a system R-value after installation, including the outer air film), Gi = Pi0.625"/12x1'/R2.3 = 0.071 Btu/h-F-ft, so the pipe would lose (135-67.5)0.071 = 4.8 Btu/h per linear foot. A 2 gpm insulated water stream would be 2 F warmer than a bare pipe stream after flowing through 2000/(17.9-4.8) = 153 feet of insulated pipe.
Type M 1/2" copper pipe http://www.homedepot.com/h_d1/N-5yc1v/R-203540899/h_d2/ProductDisplay?catalogId=10053&langId=-1&keyword=1%2F2%22+copper+pipe&storeId=10051 weighs 0.933lb/5' = 0.1866 lb/foot, with a 0.0915x0.1866 = 0.0171 Btu/F thermal capacitance. Its 0.5" ID holds 62.33Pi(0.25/12)^2x1'= 0.085 lb of water, making the total (tiny) capacitance C = 6x0.1021 = 0.6126 Btu/F.
If hot water uses are 3.6 minutes long, with (60-3.6)/60 = 0.94 hours between uses, bare pipe with a C/Gb = 0.39 hour time constant would cool from 135 to 67.5+(135-67.5)e^(-0.94/0.39) = 73.4 F before the next use during the day, and 67.5 F overnight. Reheating 6' of water to 135 F would take (5((135-73.4)+(135-67.5)0.6126 = 169 Btu/day, or 61644 Btu/year, worth about $0.10x61644/3412 = $1.81/year.
Insulated pipe with a C/Gi = 1.44 hour time constant would cool to 67.5+(135-67.5)e^(-0.94/1.44) = 102.6 F during the day, and 67.5 F overnight. Reheating it takes (5((135-102.6)+(135-67.5)0.6126 = 141 Btu/day, or 51341 Btu/year, worth about $0.10x51341/3412 = $1.50/year.The reheating difference is 31 cents per year.
So this $1.18 investment has a 100x$0.31/$1.18 = 26% tax free return and a $1.18/$0.31 = 3.8 year simple payback, if the pipe is in an unheated space and labor is free, with hourly hot water uses.
Pipe insulation within a few feet of a water heater makes even more sense, but plumbing heat traps don't seem to make sense, given pipe insulation, since they only slow convection inside a single pipe, unless there's a hot-cold connection above. And a 4 watt mechanical timer that turns off a well-insulated water heater at night would use more electrical power than it saves.
How about a simple greywater heat exchanger? Home Depot sells 300' of 1" HDPE 100 psi NSF plastic pipe for $96.64... http://www.homedepot.com/p/Advanced-Drainage-Systems-1-in-x-300-ft-IPS-100-PSI-NSF-Poly-Pipe-2-1100300/202282480#.Ue-zaDbD_IU
With a 1.049" ID and a 1.119" OD, and a 0.035" wall thickness and still clean water on the inside and some crud on the outside, 1' of pipe could have a 1'xPi1.049/12x20 = 5.50 Btu/h-F conductance and a Pi(1.049/24)^2x62.33 = 0.374 Btu/F conductance and an RC = C/G = 0.374/5.50 = 0.068 hour (4.1 minute) time constant, so a 300' pressurized pipe coil near the top of a tall unpressurized stratified tank with a much larger volume and non-conductive walls and a drain from the bottom (eg a 4' diameter x 8' tall culvert or ferrocement or thin bent plywood tank in a basement with an insulated lining and a greywater dip tube to the bottom) could heat Pi(1.049/24)^2x7.48x200 = 13.5 gallons of 60 F water to 105-(105-60)e^(-0.94/0.068) = 104.9999554 F between hourly uses, with a 105 F shower drain temp.
Every piece of NSF pipe is tested to 5 times the nominal pressure rating at 73 F, and it loses about 10% of its strength for each 10 F temperature increase above 73 F.
"Scott L" <shiva@...> wrote from Fairbanks:
>... we insulate our cold and hot water pipes is to keep them from freezing as fast if the heat quits.
NREL says the average temp is -10.1 F in January in Fairbanks, with a -18.5 min and a -61.0 30-year record low and a 0.0006 humidity ratio. The deep ground temp is 26.9 F.
Insulation doesn't help much with small pipes. At -10.1, a 1/2" pipe with R2.3 insulation and a 1.43 hour time constant would cool from 70 to 32 F in -1.43ln((32-(-10))/(70-(-10)) = 0.92 hours and freeze solid after another 0.085x144/((32-(-10))0.071) = 4.1 hours. A battery or generator could run (32-(-10))0.071/3.412 = 0.87 watts/foot of heat tape under pipe insulation when the heat quits. A 12V 200 Ah battery could keep 200' of R2.3 pipe from freezing for 13.7 hours.
>The cold water pipes sweat and drip if left uninsulated and exposed to a heated area with our cold water.
The air would need significant moisture for that to happen. Air with a w = 0.0006 humidity ratio has vapor pressure P = 29.921/(0.62198/w+1) = 0.0288 "Hg and dew point Tdp = 9621/(17.863-ln(P))-460 = -10.6 F.
> Lots of us haul or have our water delivered. $80 for 800 gallons in the town area and easily twice that if you live outside the city area by much. You can not afford to run the water until it gets hot over a long distance.
We could insulate both hot and cold pipes and use a motion detector and pump to circulate water back through the heater before hot water uses. That could also help avoid frozen pipes when the heat quits. Cooling a 50 gallon 135 F tank to 32 F with 200' of R2.3 pipe in -10 F air would take (135-32)50x8.33Btu/(200x(32-(-10))0.071) = 72 hours.
Nick
Saturday, July 20, 2013
Personal cooling
This $184 Polar system http://www.polarproducts.com/polarshop/pc/-Active-Ice-Therapy-System-26p20.htm#details has an icewater cooler and a pump and a U-shaped bladder and an optional lumbar compression wrap for the bladder.
We might circulate antifreeze through an optional 12.5"x22.5" rectangular bladder under ice-cube trays in a small freezer http://www.walmart.com/ip/Igloo-3.2-cu.-ft.-2-Door-Refrigerator-and-Freezer/15162473 in series with the U-bladder, in order to cool the cooler without adding ice.
This system has adjustable flow rate temperature control, ie we can vary the pump flow rate manually to vary the cooling. A more advanced system could have an adjustable bladder thermostat to run the pump as needed and measure the dew point of the surrounding air and limit the lower bladder temp to avoid condensation. Air at T (F) (absolute temp 460+T R) and RH% has an approximate (T+460)/(1-ln(RH/100)(T+460)/9621))-460 F dewpoint. For example, 70 F air at 50% RH has a 530/(1-ln(0.5)530/9621))-460 = 50.5 F dewpoint.
If it's humid, air movement helps. The Berkeley online comfort calculator http://www.cbe.berkeley.edu/comforttool/ says 25.4 C (77.7 F) air with a 25 C mean radiant (wall) temp and a 0.15 m/s air speed and a 50% RH and a 1.2 metabolic rate and 0.5 clo clothing is comfortable (PMV = 0.01 on a scale of -3 (very cold) to +3 (very hot)). Increasing the RH to 60% makes the PMV = 0.08 (slightly warm.) Increasing the air speed to 0.17 m/s makes things comfortable again, with PMV = -0.01. So does removing some clothing, with PMV = -0.02 at 0.45 clo. See also http://sustainabilityworkshop.autodesk.com/buildings/human-thermal-comfort and http://wikihelp.autodesk.com/Simulation_CFD/enu/2014/Help/0083-Tutorial83/0334-Tutorial334 and http://www.labeee.ufsc.br/sites/default/files/disciplinas/Thermal%20Booklet.pdf
Losing weight also helps, allometrically-speaking, ie increasing our heat-losing-surface-to-heat-generation-volume ratio. The ASHRAE Handbook of Fundamentals says a W-pound animal generates P = 6.6W^0.75 Btu/h of heat, and a W pound x H inch tall person has A = 0.108W^0.425H^0.725 ft^2 of DuBois surface area, eg P = 289 Btu/h and A = 19.6 ft^2 for a 154 lb 68" tall ASHRAE-standard human, who might be dT = P/(1.5A) = 9.8 F warmer than surrounding still air if naked, with no sweating or shivering or other adaptations. Raising W to 200 lb makes P = 351 Btu/h and A = 21.9 ft^2 and dT = 10.7 F, with H = 68". A proportional height gain to H = 12.7x200^(1/3) = 74.2" raises A to 23.3 ft^2 and lowers dT to 10.0 F.
Nick
We might circulate antifreeze through an optional 12.5"x22.5" rectangular bladder under ice-cube trays in a small freezer http://www.walmart.com/ip/Igloo-3.2-cu.-ft.-2-Door-Refrigerator-and-Freezer/15162473 in series with the U-bladder, in order to cool the cooler without adding ice.
This system has adjustable flow rate temperature control, ie we can vary the pump flow rate manually to vary the cooling. A more advanced system could have an adjustable bladder thermostat to run the pump as needed and measure the dew point of the surrounding air and limit the lower bladder temp to avoid condensation. Air at T (F) (absolute temp 460+T R) and RH% has an approximate (T+460)/(1-ln(RH/100)(T+460)/9621))-460 F dewpoint. For example, 70 F air at 50% RH has a 530/(1-ln(0.5)530/9621))-460 = 50.5 F dewpoint.
If it's humid, air movement helps. The Berkeley online comfort calculator http://www.cbe.berkeley.edu/comforttool/ says 25.4 C (77.7 F) air with a 25 C mean radiant (wall) temp and a 0.15 m/s air speed and a 50% RH and a 1.2 metabolic rate and 0.5 clo clothing is comfortable (PMV = 0.01 on a scale of -3 (very cold) to +3 (very hot)). Increasing the RH to 60% makes the PMV = 0.08 (slightly warm.) Increasing the air speed to 0.17 m/s makes things comfortable again, with PMV = -0.01. So does removing some clothing, with PMV = -0.02 at 0.45 clo. See also http://sustainabilityworkshop.autodesk.com/buildings/human-thermal-comfort and http://wikihelp.autodesk.com/Simulation_CFD/enu/2014/Help/0083-Tutorial83/0334-Tutorial334 and http://www.labeee.ufsc.br/sites/default/files/disciplinas/Thermal%20Booklet.pdf
Losing weight also helps, allometrically-speaking, ie increasing our heat-losing-surface-to-heat-generation-volume ratio. The ASHRAE Handbook of Fundamentals says a W-pound animal generates P = 6.6W^0.75 Btu/h of heat, and a W pound x H inch tall person has A = 0.108W^0.425H^0.725 ft^2 of DuBois surface area, eg P = 289 Btu/h and A = 19.6 ft^2 for a 154 lb 68" tall ASHRAE-standard human, who might be dT = P/(1.5A) = 9.8 F warmer than surrounding still air if naked, with no sweating or shivering or other adaptations. Raising W to 200 lb makes P = 351 Btu/h and A = 21.9 ft^2 and dT = 10.7 F, with H = 68". A proportional height gain to H = 12.7x200^(1/3) = 74.2" raises A to 23.3 ft^2 and lowers dT to 10.0 F.
See also: http://theoatmeal.com/comics/running
Nick
Thursday, July 18, 2013
Recycle a freezer into a heat storage tank
This 24.9 ft^3 chest freezer could hold 24.9x7.48 = 186 gallons of water in a 10'x12' folded EPDM liner... http://www.homedepot.com/p/GE-24-9-cu-ft-Chest-Freezer-in-White-CM25SBWW/202986204#.UdagDzbD_IU
It uses 568 kWh/year, eg 3x568x3412Btu/kWh/365d/24h = 664 Btu/h with a COP of 3 at 0 F in a 70 F room, with a 664Btu/h/70F = 9.5 Btu/h-F thermal conductance and about 80 ft^2 of surface with an R-value of 80ft^2/9.5Btu/h-F = R8.4 ft^2-F-h/Btu, which could be raised with fiberglass or foamboard. R10 foamboard over the top and sides would reduce the conductance to 65ft^2/R18.4+15ft^2/R8.4 = 5.3 Btu/h-F.
It could live above a 4'x16' bare fin 2.56 gpm DIY collector made from $67 worth of 0.018" brown- or black-painted aluminum flashing http://www.lowes.com/pd_474412-205-69124_0__?productId=4650669&Ntt=amerimax+trim+coil&pl=1¤tURL=%3FNtt%3Damerimax%2Btrim%2Bcoil&facetInfo= as described at http://www.builditsolar.com/Experimental/PEXCollector/PEXCollector.htm
The collector could have 128 ft of 1/2" PEX-Al-PEX tubing on 6" centers, with 2 Ts to make 2 64' 1.28 gpm flow paths. http://www.calctool.org/CALC/eng/civil/hazen-williams_g says 64' of 1/2" PEX tubing with a 0.475" ID would have a 1.28 gpm flow with a 3.7' pressure drop, and an $85 Grundfos 15-58C pump http://www.pexsupply.com/Grundfos-59896341-UPS15-58FC-3-Speed-Circulator-Pump-1-25-HP-115-volt-4701000-p can deliver 2.56 gpm with a 8' pressure drop on speed 1, using 60 watts.
A $100 sun-pump.com differential thermostat controller http://www.sun-pump.com/ or an Arduino controller http://www.nateful.com/differduino/differduino.html could run the pump when the collector is less than 40 F to avoid freezing or when the tank is warmer than 160 F to avoid overheating the EPDM liner. The tank could could contain a 1"x300' pressurized PEX coil as a heat exchanger, or it could be a gravity-pressurized hot water supply with a float valve. (Rich Komp has a gravity hot water supply with a 3' head (a 55 gallon drum) for a shower with a rainwater supply in his off-grid house in Maine. The flow is adequate, but it won't knock you back against the wall like a shower in a hotel.)
Where I live near Philadelphia (not an easy climate for solar heating), 1000 Btu/ft^2 of sun falls on a south wall on a 30 F average January day with a 34 F daytime temp. If 0.8x8'x16'x1000 = 102.4K Btu enters 8'x16' of R2 twinwall glazing with 80% solar transmission over 6 hours and 51.2K falls on a 4'x16' 160 F bare collector which also gains 6h(Ta-160)4'x16'x3 Btu from both sides from Ta (F) air inside the glazing and
102.4K = 51.2K + 6h(Ta-160)4'x16'x3 + 6h(Ta-34)8'x16'/R2,
then Ta = 161.8 F, with 51.2K + 6h(Ta-160)4'x16'x3 = 53.3K Btu of water heating on an average January day, ie 30% more than the 41K Btu/day SRCC 0G-300 spec, in colder, cloudier weather, eg 53.3K/(110-60)/8.33 = 128 gallons of 110 F water heated from 60 F.
On the 1st cloudy day after a long string of average days, the tank can supply 53.3K Btu and cool to 160-53.3K/(186x8.33) = 125.6 F, with no backup heat requirement.
On the 2nd cloudy day, as it cools from 125.6 to 110 F, the tank can supply (125.6-110)186x8.33 = 24.2K Btu, enough to heat 24.2K/(110-60)/8.33 = 58 gallons of water from 60 to 110 F. Adding 128-58 = 70 gallons at 60 F will reduce the tank temp to (60x70+110x186)/(70+186) = 96.3 F, with an average (110+96.3)/2 = 103.2 exit temp, and heating that 70 gallons to 110 requires (110-103.2)70x8.33 = 3986 Btu, making the backup fraction 3986/53.3K = 0.075 after the 2nd cloudy day.
On the 3rd cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+96.3x186)/(128+186) = 81.5, with an average (96.3+81.5)/2 = 88.9 exit temp, and heating that 128 gallons to 110 requires (110-88.9)128x8.33 = 22.5K Btu, making the backup fraction 22.5K/53.3K = 0.422 after the 3rd day.
On the 4th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+81.5x186)/(128+186) = 72.7, with an average (81.5+72.7)/2 = 77.1 exit temp, and heating that 128 gallons to 110 requires (110-77.1)128x8.33 = 35.1K Btu, making the backup fraction 35.1K/53.3K = 0.634 after the 4th day.
On the 5th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+72.7x186)/(128+186) = 67.5, with an average (72.7+67.5)/2 = 70.1 exit temp, and heating that 128 gallons to 110 requires (110-70.1)128x8.33 = 42.5K Btu, making the backup fraction 42.5K/53.3K = 0.769 after the 5th cloudy day.
(If I did that right. I should learn how to use spreadsheets :-)
Assuming cloudy days are like coin flips, the expected backup heating fraction is approximately
0x2^-1+0x2^-2+0.075x2^3+0.422x2^-4+0.634x2^-5+0.769x2^-6 =
0 +0 +9.4E-3 +26.4E-3 +19.8E-3 +12.0E-3 = 0.0676,
so, ignoring the tank heat loss, the expected solar heating fraction is 1-0.0676 = 93% in January.
Nick
Now that all of us are plagued with the pollution resulting from the overabundance of devices we have purchased, perhaps government or church groups should sponsor a series of "you don't need it" commercials. Instead of the bright uniformed "service personnel" of the Ace Air Conditioning Company briskly delivering and installing the latest gadgets, the commercials would show the expensive equipment misused: a bored housewife growing geraniums in her new dishwashing machine; a small child casually dismantling a TV-stereo combo with a claw hammer...
from the book "Sunspots," by Steve Baer, 1979.
It uses 568 kWh/year, eg 3x568x3412Btu/kWh/365d/24h = 664 Btu/h with a COP of 3 at 0 F in a 70 F room, with a 664Btu/h/70F = 9.5 Btu/h-F thermal conductance and about 80 ft^2 of surface with an R-value of 80ft^2/9.5Btu/h-F = R8.4 ft^2-F-h/Btu, which could be raised with fiberglass or foamboard. R10 foamboard over the top and sides would reduce the conductance to 65ft^2/R18.4+15ft^2/R8.4 = 5.3 Btu/h-F.
It could live above a 4'x16' bare fin 2.56 gpm DIY collector made from $67 worth of 0.018" brown- or black-painted aluminum flashing http://www.lowes.com/pd_474412-205-69124_0__?productId=4650669&Ntt=amerimax+trim+coil&pl=1¤tURL=%3FNtt%3Damerimax%2Btrim%2Bcoil&facetInfo= as described at http://www.builditsolar.com/Experimental/PEXCollector/PEXCollector.htm
The collector could have 128 ft of 1/2" PEX-Al-PEX tubing on 6" centers, with 2 Ts to make 2 64' 1.28 gpm flow paths. http://www.calctool.org/CALC/eng/civil/hazen-williams_g says 64' of 1/2" PEX tubing with a 0.475" ID would have a 1.28 gpm flow with a 3.7' pressure drop, and an $85 Grundfos 15-58C pump http://www.pexsupply.com/Grundfos-59896341-UPS15-58FC-3-Speed-Circulator-Pump-1-25-HP-115-volt-4701000-p can deliver 2.56 gpm with a 8' pressure drop on speed 1, using 60 watts.
A $100 sun-pump.com differential thermostat controller http://www.sun-pump.com/ or an Arduino controller http://www.nateful.com/differduino/differduino.html could run the pump when the collector is less than 40 F to avoid freezing or when the tank is warmer than 160 F to avoid overheating the EPDM liner. The tank could could contain a 1"x300' pressurized PEX coil as a heat exchanger, or it could be a gravity-pressurized hot water supply with a float valve. (Rich Komp has a gravity hot water supply with a 3' head (a 55 gallon drum) for a shower with a rainwater supply in his off-grid house in Maine. The flow is adequate, but it won't knock you back against the wall like a shower in a hotel.)
Where I live near Philadelphia (not an easy climate for solar heating), 1000 Btu/ft^2 of sun falls on a south wall on a 30 F average January day with a 34 F daytime temp. If 0.8x8'x16'x1000 = 102.4K Btu enters 8'x16' of R2 twinwall glazing with 80% solar transmission over 6 hours and 51.2K falls on a 4'x16' 160 F bare collector which also gains 6h(Ta-160)4'x16'x3 Btu from both sides from Ta (F) air inside the glazing and
102.4K = 51.2K + 6h(Ta-160)4'x16'x3 + 6h(Ta-34)8'x16'/R2,
then Ta = 161.8 F, with 51.2K + 6h(Ta-160)4'x16'x3 = 53.3K Btu of water heating on an average January day, ie 30% more than the 41K Btu/day SRCC 0G-300 spec, in colder, cloudier weather, eg 53.3K/(110-60)/8.33 = 128 gallons of 110 F water heated from 60 F.
On the 1st cloudy day after a long string of average days, the tank can supply 53.3K Btu and cool to 160-53.3K/(186x8.33) = 125.6 F, with no backup heat requirement.
On the 2nd cloudy day, as it cools from 125.6 to 110 F, the tank can supply (125.6-110)186x8.33 = 24.2K Btu, enough to heat 24.2K/(110-60)/8.33 = 58 gallons of water from 60 to 110 F. Adding 128-58 = 70 gallons at 60 F will reduce the tank temp to (60x70+110x186)/(70+186) = 96.3 F, with an average (110+96.3)/2 = 103.2 exit temp, and heating that 70 gallons to 110 requires (110-103.2)70x8.33 = 3986 Btu, making the backup fraction 3986/53.3K = 0.075 after the 2nd cloudy day.
On the 3rd cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+96.3x186)/(128+186) = 81.5, with an average (96.3+81.5)/2 = 88.9 exit temp, and heating that 128 gallons to 110 requires (110-88.9)128x8.33 = 22.5K Btu, making the backup fraction 22.5K/53.3K = 0.422 after the 3rd day.
On the 4th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+81.5x186)/(128+186) = 72.7, with an average (81.5+72.7)/2 = 77.1 exit temp, and heating that 128 gallons to 110 requires (110-77.1)128x8.33 = 35.1K Btu, making the backup fraction 35.1K/53.3K = 0.634 after the 4th day.
On the 5th cloudy day, adding 128 gallons of 60 F water lowers the tank temp to (60x128+72.7x186)/(128+186) = 67.5, with an average (72.7+67.5)/2 = 70.1 exit temp, and heating that 128 gallons to 110 requires (110-70.1)128x8.33 = 42.5K Btu, making the backup fraction 42.5K/53.3K = 0.769 after the 5th cloudy day.
(If I did that right. I should learn how to use spreadsheets :-)
Assuming cloudy days are like coin flips, the expected backup heating fraction is approximately
0x2^-1+0x2^-2+0.075x2^3+0.422x2^-4+0.634x2^-5+0.769x2^-6 =
0 +0 +9.4E-3 +26.4E-3 +19.8E-3 +12.0E-3 = 0.0676,
so, ignoring the tank heat loss, the expected solar heating fraction is 1-0.0676 = 93% in January.
Nick
Now that all of us are plagued with the pollution resulting from the overabundance of devices we have purchased, perhaps government or church groups should sponsor a series of "you don't need it" commercials. Instead of the bright uniformed "service personnel" of the Ace Air Conditioning Company briskly delivering and installing the latest gadgets, the commercials would show the expensive equipment misused: a bored housewife growing geraniums in her new dishwashing machine; a small child casually dismantling a TV-stereo combo with a claw hammer...
from the book "Sunspots," by Steve Baer, 1979.
Sunday, July 14, 2013
DIY air-air heat exchangers
"yreysa" <gary@...> mentions Shurcliff's example:
http://www.builditsolar.com/Experimental/HRV/ShurcliffHRV.htm
It has about 36x65"x19"/12^2 = 309 ft^2 of heat exchange surface with an approximate U = 0.75 Btu/h-F-ft^2 2-sided slow airfilm conductance, so AU = 309x0.75 = 232 Btu/h-F, with NTU = 232/100 = 2.32 and effectiveness E = NTU/(NTU+1) = 70% at 100 cfm and 82% at 50 cfm.
Room air at 70 F and 50% RH has 0.00787 pounds of water per pound of dry air and weighs 0.075 lb/ft^3. Freezing all the water out of a 30 cfm airstream would make 24hx60m/hx30cfmx0.075lb/ft^3x0.00787lb/ft^3 = 25.5 pounds of ice per day. Melting ice takes 144 Btu/lb. If the exhaust fan runs without the intake fan for T hours per day to melt ice and T(70-32)30 = (24-T)60x30x0.075x0.00787x144, T = 2.84 hours per day, with a 12% duty cycle. Condensation and ice could form in the outgoing airstream between Coroplast flat plates, with incoming cold air inside the plate corrugations.
But (IMO) this is so rarely needed in the US that it makes more sense to just run an exhaust fan with a humidistat if the RH reaches say, 60% in an "airtight house." Swedish houses have 0.025 ACH. A SIP house might have 0.1 or 0.2. An average US house has about 1 ACH. On an average day, a 4,000 ft^2 house can provide ASHRAE's standard 15 cfm for a full-time occupant with a natural air leakage of only 15x60/(4Kx8') = 0.028 ACH, ie about 0.56 ACH on a 50 Pa blower door test.
Bill Shurcliff also suggested a "lung" with an external bellows that could turn all the cracks and crevices in a house envelope into very efficient bidirectional latent and sensible air-air heat exchangers. ORNL's preliminary experiments used a bellows that was too small to exhaust all the air from a 2x6 stud cavity.
To avoid that restriction, a 90 watt 2470 cfm Lasko 2155a reversible window fan could be in an interior wall that divides the house in two partitions with a humidistat in series with a repeat cycle timer like Grainger's $83.90 2A179 (with its $4.26 5X582 socket) which can reverse the fan airflow with adjustable off and cycle times from 1.2 seconds to 300 hours.
If every 10'x10' envelope section contains a 4x10'x6"x1/32" crack and 30 cfm flows through 4,096 ft^2 of envelope and each section has a "duct" with 40/32/12 = 0.104 ft^2 of cross-sectional area, the entire envelope has 4096/100x0.104 = 4.27 ft^2, and the average air velocity is 2x30/4.27 = 14 fpm. Each wall section has 40x0.5x2 = 40 ft^2 of heat exchange surface, so Cmin = 30 and NTU = AU/Cmin = 4096/100x40x1.5/30 = 82 and E = 1-exp(-82), ie 100%. Not bad :-)
The cycle time should be long enough that stale house air clears the path to the outdoors before the fan reverses. If it isn't, we'll know, because the humidity and CO2 concentration in the house won't decrease much as the fan runs.
This could also dehumidify a house... http://www.consumerreports.org/cro/dehumidifiers/buying-guide.htm says Energy Star dehumifidifiers typically remove 4 pounds of water per kWh (and add heat to a house if not part of an air-conditioner.) Philadelphia has an average 76.7 F temp and a 67.2 daily min and 0.0133 humidity ratio in July. ASHRAE says 80 F with a 0.0120 humidity ratio is comfortable.
A Lasko fan with a smart ventilation controller could remove 60x2470x0.075(0.0120-0.0100) = 22.2 pints per hour of water vapor from house air with a 0.0120 humidity ratio when the outdoor humidity ratio drops to 0.0100 using 90Wh, at 22.3/0.090 = 247 pints/kWh, using 988 times less energy than an Energy Star dehumidifier.
And this could happen at night in summertime, when outdoor air is cooler, since house materials can store dryness as well as coolth. Concrete weighs about 150 lb/ft^3 and stores about 25 Btu/F-ft^3 and 1% of its weight by volume as the RH of the surrounding air rises from 40 to 60%.
Nick
http://www.builditsolar.com/Experimental/HRV/ShurcliffHRV.htm
It has about 36x65"x19"/12^2 = 309 ft^2 of heat exchange surface with an approximate U = 0.75 Btu/h-F-ft^2 2-sided slow airfilm conductance, so AU = 309x0.75 = 232 Btu/h-F, with NTU = 232/100 = 2.32 and effectiveness E = NTU/(NTU+1) = 70% at 100 cfm and 82% at 50 cfm.
Room air at 70 F and 50% RH has 0.00787 pounds of water per pound of dry air and weighs 0.075 lb/ft^3. Freezing all the water out of a 30 cfm airstream would make 24hx60m/hx30cfmx0.075lb/ft^3x0.00787lb/ft^3 = 25.5 pounds of ice per day. Melting ice takes 144 Btu/lb. If the exhaust fan runs without the intake fan for T hours per day to melt ice and T(70-32)30 = (24-T)60x30x0.075x0.00787x144, T = 2.84 hours per day, with a 12% duty cycle. Condensation and ice could form in the outgoing airstream between Coroplast flat plates, with incoming cold air inside the plate corrugations.
But (IMO) this is so rarely needed in the US that it makes more sense to just run an exhaust fan with a humidistat if the RH reaches say, 60% in an "airtight house." Swedish houses have 0.025 ACH. A SIP house might have 0.1 or 0.2. An average US house has about 1 ACH. On an average day, a 4,000 ft^2 house can provide ASHRAE's standard 15 cfm for a full-time occupant with a natural air leakage of only 15x60/(4Kx8') = 0.028 ACH, ie about 0.56 ACH on a 50 Pa blower door test.
Bill Shurcliff also suggested a "lung" with an external bellows that could turn all the cracks and crevices in a house envelope into very efficient bidirectional latent and sensible air-air heat exchangers. ORNL's preliminary experiments used a bellows that was too small to exhaust all the air from a 2x6 stud cavity.
To avoid that restriction, a 90 watt 2470 cfm Lasko 2155a reversible window fan could be in an interior wall that divides the house in two partitions with a humidistat in series with a repeat cycle timer like Grainger's $83.90 2A179 (with its $4.26 5X582 socket) which can reverse the fan airflow with adjustable off and cycle times from 1.2 seconds to 300 hours.
If every 10'x10' envelope section contains a 4x10'x6"x1/32" crack and 30 cfm flows through 4,096 ft^2 of envelope and each section has a "duct" with 40/32/12 = 0.104 ft^2 of cross-sectional area, the entire envelope has 4096/100x0.104 = 4.27 ft^2, and the average air velocity is 2x30/4.27 = 14 fpm. Each wall section has 40x0.5x2 = 40 ft^2 of heat exchange surface, so Cmin = 30 and NTU = AU/Cmin = 4096/100x40x1.5/30 = 82 and E = 1-exp(-82), ie 100%. Not bad :-)
The cycle time should be long enough that stale house air clears the path to the outdoors before the fan reverses. If it isn't, we'll know, because the humidity and CO2 concentration in the house won't decrease much as the fan runs.
This could also dehumidify a house... http://www.consumerreports.org/cro/dehumidifiers/buying-guide.htm says Energy Star dehumifidifiers typically remove 4 pounds of water per kWh (and add heat to a house if not part of an air-conditioner.) Philadelphia has an average 76.7 F temp and a 67.2 daily min and 0.0133 humidity ratio in July. ASHRAE says 80 F with a 0.0120 humidity ratio is comfortable.
A Lasko fan with a smart ventilation controller could remove 60x2470x0.075(0.0120-0.0100) = 22.2 pints per hour of water vapor from house air with a 0.0120 humidity ratio when the outdoor humidity ratio drops to 0.0100 using 90Wh, at 22.3/0.090 = 247 pints/kWh, using 988 times less energy than an Energy Star dehumidifier.
And this could happen at night in summertime, when outdoor air is cooler, since house materials can store dryness as well as coolth. Concrete weighs about 150 lb/ft^3 and stores about 25 Btu/F-ft^3 and 1% of its weight by volume as the RH of the surrounding air rises from 40 to 60%.
Nick
Low-flow showers
The March/April 2013 issue of Sierra (the club magazine) says wind is
now
cheaper than any other form of electrical energy generation, at $48-95/MWH, vs
$61-89 for a natural gas combined cycle plant with 60% efficiency or coal at
$62-141 or $200-231 for an industrial gas turbine (eg Capstone?), IF we remove
all subsidies for all forms of generation. But conservation and energy
efficiency are still cheaper, at $0-50/MWH.
I use a fully-enclosed shower with a continuous 1/4 gpm brass nozzle head and
swivel fitting from McMaster-Carr in parallel with occasional 1 gpm sprays from a hose
with a hand valve...
http://www.homedepot.com/p/RINSE-ACE-Snap-n-Spray-6-ft-Quick-Connect-and-Detacha\
ble-Hose-with-On-and-Off-Sprayer-4170/202519744#.UeJLZTbD_IU Drilling a hole
through the plastic disk shutoff valve in the sprayer T will keep the 1/4 gpm
shower head going when you are using the sprayer to wash off soap.
http://www.sunfrost.com/efficient_shower.html describes the benefits of a fully-enclosed shower.
This $9.17 nozzle uses 0.2 gpm at 40 psi with a 60 degree spray cone...
http://www.mcmaster.com/#32885k11/=nm45ot
Here's a $15.60 swivel fitting... http://www.mcmaster.com/#34615k71/=nm48wo
This uses about 5 times less water than a 1.5 gpm "low-flow" shower head...
http://www.homedepot.com/p/American-Standard-FloWise-Single-Function-1-7-8-in-Wa\
ter-Saving-Showerhead-in-Polished-Chrome-1660-710-002/100649323#.UeJN0TbD_IU
McGill U's "Water Conservation and the Mist Experience" pamphlet suggests using a garden sprayer, eg this $8.57 1-gallon sprayer... http://www.walmart.com/ip/RL-Flo-Master-1-Gallon-Sprayer/13376324
Bucky Fuller's Dymaxion pressure mist shower used 1 pint per hour, ie 0.002 gpm, ie 2 milligpm, 750 times less than a "low-flow showerhead."
Nick
cheaper than any other form of electrical energy generation, at $48-95/MWH, vs
$61-89 for a natural gas combined cycle plant with 60% efficiency or coal at
$62-141 or $200-231 for an industrial gas turbine (eg Capstone?), IF we remove
all subsidies for all forms of generation. But conservation and energy
efficiency are still cheaper, at $0-50/MWH.
I use a fully-enclosed shower with a continuous 1/4 gpm brass nozzle head and
swivel fitting from McMaster-Carr in parallel with occasional 1 gpm sprays from a hose
with a hand valve...
http://www.homedepot.com/p/RINSE-ACE-Snap-n-Spray-6-ft-Quick-Connect-and-Detacha\
ble-Hose-with-On-and-Off-Sprayer-4170/202519744#.UeJLZTbD_IU Drilling a hole
through the plastic disk shutoff valve in the sprayer T will keep the 1/4 gpm
shower head going when you are using the sprayer to wash off soap.
http://www.sunfrost.com/efficient_shower.html describes the benefits of a fully-enclosed shower.
This $9.17 nozzle uses 0.2 gpm at 40 psi with a 60 degree spray cone...
http://www.mcmaster.com/#32885k11/=nm45ot
Here's a $15.60 swivel fitting... http://www.mcmaster.com/#34615k71/=nm48wo
This uses about 5 times less water than a 1.5 gpm "low-flow" shower head...
http://www.homedepot.com/p/American-Standard-FloWise-Single-Function-1-7-8-in-Wa\
ter-Saving-Showerhead-in-Polished-Chrome-1660-710-002/100649323#.UeJN0TbD_IU
McGill U's "Water Conservation and the Mist Experience" pamphlet suggests using a garden sprayer, eg this $8.57 1-gallon sprayer... http://www.walmart.com/ip/RL-Flo-Master-1-Gallon-Sprayer/13376324
Bucky Fuller's Dymaxion pressure mist shower used 1 pint per hour, ie 0.002 gpm, ie 2 milligpm, 750 times less than a "low-flow showerhead."
Nick
Wednesday, July 10, 2013
A tomato greenhouse in Nashville
"brad.howard308" <bradhoward@...> wrote:
>I am trying to design a solar heating system for a new 32 ft x 96 ft greenhouse on my farm.
NREL says 730 Btu/ft^2 of sun falls on the ground and 1030 falls on a south wall on an average 36.2 F January day in Nashville, with a 26.5 and 45.9 low and high and a (36.2+45.9)/2 = 41 average daytime temp and a wa = 0.0035 humidity ratio (0.0035 pounds of water per pound of dry air.) The deep ground temp is 59.1. The 30-year record low is -17 F. The average July temp is 79.3, with a 68.9 and 89.5 low and high and 1980 Btu/ft^2 of sun on the ground and 770 on a south wall and a 0.0152 humidity ratio.
>I need to try and keep temps in the greenhouse between 50F - 90F with the optimum range 60F - 80F.
http://msucares.com/pubs/publications/p1828.pdf recommends a 70 to 82 F day temperature at 60-70% RH and a 64 F minimum night temperature, with nutritional deficiencies below 60 F and fruit damage below 57 F.
http://edepot.wur.nl/12805 says a tomato greenhouse can evaporate 0.4 lb/ft^2 per day of water vapor, about 0.4x32'x96' = 1229 lb for a 32'x96' greenhouse.
Air at 82 F and 70% RH has vapor pressure Pg = 0.7e^(17.863-9621/(82+460)) = 0.783 "Hg and humidity ratio wg = 0.62198/(29.921/Pg-1) = 0.0167, and it weighs about 0.075 lb/ft^3, so it looks like maintaining a 70% RH requires 1229/(6hx60m/h/0.075lb/ft^3(wg-wa)) = 3448 cfm of ventilation for 6 hours per day. The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F.
>Idea 1... a south facing 4 ft x 96 ft screen collector tilted at about 50 degrees, laying on its side down the length of the greenhouse blowing into 4 inch corrugated plastic non-perforated drain pipe buried 3 ft under the green house with 2 ft perimeter insulation.
Why not omit the screen and its plant shading and just blow warm air down from the peak of the greenhouse through perforated pipes, to allow any condensation to drain from the pipes (altho there won't be any condensation if the ground is warmer than 71 F)... http://www.sunnyjohn.com/indexpages/shcs.htm suggests something like this, but I'd still worry about mold and mildew in the pipes infecting the plants.
The Kissock equation says A ft^2 of soil with Rf floor insulation and Rp perimeter insulation would lose Qf = UA(Tg-Ts) Btu/h, with U = 0.11/(4+Rf+Rp)+0.89/(16+Rf), where Tg is the greenhouse air temp and Ts is the deep ground temp. With A = 32'x96' and Rf = 0 and Rp = 10 and Tg = 59.1, Qf = 205(Tg-59.1) Btu/h.
With 2 R1 polyethylene film covers, the greenhouse thermal conductance would be about 32'x96'/R2 = 1536 Btu/h-F.
With tall indeterminate tomato plants and a 12' height and 80% solar transmission, it could gain about 0.8x96'(32'x730+1030x12') = 2.74 million Btu/day.
With lots of thermal mass and airflow and a constant Tg (F) temp 24 hours per day,
2.64M = 6(Tg-41)3448+24(Tg-36.2)1536+24(Tg-59.1)205
= 20688Tg-848208+36864Tg-1334477+4920Tg-290722
makes Tg = 83.5, but the greenhouse could be 70 F during the day and 64 at night, so there's plenty of solar heat on an average day.
Keeping the greenhouse 64 F on an average night requires (64-26.5)1536 = 57.6K Btu/h. Soil with a 32x96x1.5 = 4608 slow-moving airfilm conductance could provide this at a 64+57.6K/4608 = 76 F surface temp.
Keeping the greenhouse 64 F on a 24-hour average 36.2 F day requires (64-36.2)1536 = 42.7K Btu/h... 16 96' x 4" pipes would have 1608 ft^2 of surface with a 1608x1.5 = 2413 Btu/h-F airfilm conductance... http://www.engineeringtoolbox.com/duct-friction-pressure-loss-d_444.html says 55 cfm flowing through 48' of 4" duct would have a 0.1 H20 pressure drop, so 16 pipes fed from the center would have about 0.1 "H20 with a 16x2x55 = 1760 cfm flow. With a combined surface and pipe conductance of 4608+1/(1/2413+1/1760) = 5626 Btu/h-F, the soil can provide 42.7K Btu/h at a 64+42.7K/5626 = 72 F temp.
The greenhouse needs 5dx24hx57.6K = 6.9 million Btu to stay 64 F for 5 cloudy 36.2 F days in a row. This could come from 6.9M/(82-72) = 690K Btu/F of mass cooling from 82 to 72 F, eg 690K/30Btu/F-ft^3 = 23K ft^3 of soil with a 30 Btu/ft^3-F specific heat by volume, eg a 23K/32'x96' = 7.5' deep x 32'x96' greenhouse floor. That's very deep, because the temperature swing of this mass is only 10 F, and the floor probably needs more than 1 level of pipes, given the R1 per foot soil thermal resistance.
>Idea 2... a south facing 4 ft x 96 ft PEX or CPVC type collector in the same configuration but use a "tank", basically an insulated hole lined with pond liner as a heat storage tank. Use pex buried in the ground to heat the ground up and inside the greenhouse.
This seems more promising. Water has twice the thermal mass by volume of soil, and it could have a higher temp on an average day and a larger temperature swing. At 140 F, a 4'x96' water tray with an R1 cover under the ceiling would collect 4'x96'(0.8x730-6h(140-82)/R1) = 90.6K Btu/day. With 4'x96'x2x1.5 = 1152 Btu/h-F of airfilm conductance and a 10K cfm ceiling fan and a 1/(1/1152+1/10K) = 1033 Btu/h-F total conductance, tray water at 64+57.6K/1033 = 120 F could keep the greenhouse 64 F on a 26.5 F night.
That's a high minimum usable tray water temp, so the greenhouse needs a large water volume to store enough heat for 5 cloudy days in a row: 6.9M/(140-120)/62.33 = 5469 ft^3 of water, eg an 18' cube. Adding 8 1000 Btu/h-F car radiators and fans would raise the total conductance to 9033 Btu/h-F with a minimum tray water temp of 64+57.6K/9033 = 70 F and a 6.9M/(140-70)/62.33 = 1581 ft^3 tank, eg a 12' cube, but it seems more practical to fill the space between 2 poly film covers with soap bubble foam on cloudy days.
R20 foam would reduce the heat requirement to (64-26.5)32'x96'/R20 = 5.76K Btu/h on an average night, and the tray and ceiling fan could provide that with a minimum tray water temp of 64+5.76K/1033 = 70 F. The greenhouse would only need 5dx24hx5.76K = 690K Btu to stay 64 F for 5 cloudy 36.2 F days in a row, eg a 690K/(140-70)/62.33 = 158 ft^3 tank, eg a 6' cube or a 4' wide x 3' tall x 14' long plywood box with a 10'x20' folded EPDM liner under a bench.
Zelon’s Swedish patent (US No. 3672184, June, 1972) described insulating shop windows at night with soap bubble foam. Professor John Groh at U. Arizona measured US R3 per inch for soap bubble greenhouse insulation in 1968. Professor Otho Wells at U. New Hampshire did later greenhouse experiments. In 1995, Bill Sturm built a 12,000 ft^2 tomato greenhouse with a soap bubble foam roof in Calgary, Alberta and measured an 84% propane energy savings with and without foam on alternate nights at 20 F below zero.
Bill showed me a simple foam generator with a shop vac blower connected to a horizontal 2" PVC pipe full of holes in a shallow trough near the ground containing a 10% detergent solution which expands by a factor of 300 when foamed. He suggests some window screen to push on a microswitch in an air return at the top of the glazing cavity to turn off the shop vac for all but a few seconds every half-hour at night, along with a standard greenhouse inflation blower above the trough water line to fill the space between the 2 films with air during the day, with a check valve to reduce blower power... http://www.sunnyjohn.com/photos/devices_tools/dt_inflate_blower/index.htm
How about summer cooling?
Nick
>I am trying to design a solar heating system for a new 32 ft x 96 ft greenhouse on my farm.
NREL says 730 Btu/ft^2 of sun falls on the ground and 1030 falls on a south wall on an average 36.2 F January day in Nashville, with a 26.5 and 45.9 low and high and a (36.2+45.9)/2 = 41 average daytime temp and a wa = 0.0035 humidity ratio (0.0035 pounds of water per pound of dry air.) The deep ground temp is 59.1. The 30-year record low is -17 F. The average July temp is 79.3, with a 68.9 and 89.5 low and high and 1980 Btu/ft^2 of sun on the ground and 770 on a south wall and a 0.0152 humidity ratio.
>I need to try and keep temps in the greenhouse between 50F - 90F with the optimum range 60F - 80F.
http://msucares.com/pubs/publications/p1828.pdf recommends a 70 to 82 F day temperature at 60-70% RH and a 64 F minimum night temperature, with nutritional deficiencies below 60 F and fruit damage below 57 F.
http://edepot.wur.nl/12805 says a tomato greenhouse can evaporate 0.4 lb/ft^2 per day of water vapor, about 0.4x32'x96' = 1229 lb for a 32'x96' greenhouse.
Air at 82 F and 70% RH has vapor pressure Pg = 0.7e^(17.863-9621/(82+460)) = 0.783 "Hg and humidity ratio wg = 0.62198/(29.921/Pg-1) = 0.0167, and it weighs about 0.075 lb/ft^3, so it looks like maintaining a 70% RH requires 1229/(6hx60m/h/0.075lb/ft^3(wg-wa)) = 3448 cfm of ventilation for 6 hours per day. The dew point of 82 F (542 R) air at 70% RH is approximately 542/(1-542ln(0.7)/9621)-460 = 71 F.
>Idea 1... a south facing 4 ft x 96 ft screen collector tilted at about 50 degrees, laying on its side down the length of the greenhouse blowing into 4 inch corrugated plastic non-perforated drain pipe buried 3 ft under the green house with 2 ft perimeter insulation.
Why not omit the screen and its plant shading and just blow warm air down from the peak of the greenhouse through perforated pipes, to allow any condensation to drain from the pipes (altho there won't be any condensation if the ground is warmer than 71 F)... http://www.sunnyjohn.com/indexpages/shcs.htm suggests something like this, but I'd still worry about mold and mildew in the pipes infecting the plants.
The Kissock equation says A ft^2 of soil with Rf floor insulation and Rp perimeter insulation would lose Qf = UA(Tg-Ts) Btu/h, with U = 0.11/(4+Rf+Rp)+0.89/(16+Rf), where Tg is the greenhouse air temp and Ts is the deep ground temp. With A = 32'x96' and Rf = 0 and Rp = 10 and Tg = 59.1, Qf = 205(Tg-59.1) Btu/h.
With 2 R1 polyethylene film covers, the greenhouse thermal conductance would be about 32'x96'/R2 = 1536 Btu/h-F.
With tall indeterminate tomato plants and a 12' height and 80% solar transmission, it could gain about 0.8x96'(32'x730+1030x12') = 2.74 million Btu/day.
With lots of thermal mass and airflow and a constant Tg (F) temp 24 hours per day,
2.64M = 6(Tg-41)3448+24(Tg-36.2)1536+24(Tg-59.1)205
= 20688Tg-848208+36864Tg-1334477+4920Tg-290722
makes Tg = 83.5, but the greenhouse could be 70 F during the day and 64 at night, so there's plenty of solar heat on an average day.
Keeping the greenhouse 64 F on an average night requires (64-26.5)1536 = 57.6K Btu/h. Soil with a 32x96x1.5 = 4608 slow-moving airfilm conductance could provide this at a 64+57.6K/4608 = 76 F surface temp.
Keeping the greenhouse 64 F on a 24-hour average 36.2 F day requires (64-36.2)1536 = 42.7K Btu/h... 16 96' x 4" pipes would have 1608 ft^2 of surface with a 1608x1.5 = 2413 Btu/h-F airfilm conductance... http://www.engineeringtoolbox.com/duct-friction-pressure-loss-d_444.html says 55 cfm flowing through 48' of 4" duct would have a 0.1 H20 pressure drop, so 16 pipes fed from the center would have about 0.1 "H20 with a 16x2x55 = 1760 cfm flow. With a combined surface and pipe conductance of 4608+1/(1/2413+1/1760) = 5626 Btu/h-F, the soil can provide 42.7K Btu/h at a 64+42.7K/5626 = 72 F temp.
The greenhouse needs 5dx24hx57.6K = 6.9 million Btu to stay 64 F for 5 cloudy 36.2 F days in a row. This could come from 6.9M/(82-72) = 690K Btu/F of mass cooling from 82 to 72 F, eg 690K/30Btu/F-ft^3 = 23K ft^3 of soil with a 30 Btu/ft^3-F specific heat by volume, eg a 23K/32'x96' = 7.5' deep x 32'x96' greenhouse floor. That's very deep, because the temperature swing of this mass is only 10 F, and the floor probably needs more than 1 level of pipes, given the R1 per foot soil thermal resistance.
>Idea 2... a south facing 4 ft x 96 ft PEX or CPVC type collector in the same configuration but use a "tank", basically an insulated hole lined with pond liner as a heat storage tank. Use pex buried in the ground to heat the ground up and inside the greenhouse.
This seems more promising. Water has twice the thermal mass by volume of soil, and it could have a higher temp on an average day and a larger temperature swing. At 140 F, a 4'x96' water tray with an R1 cover under the ceiling would collect 4'x96'(0.8x730-6h(140-82)/R1) = 90.6K Btu/day. With 4'x96'x2x1.5 = 1152 Btu/h-F of airfilm conductance and a 10K cfm ceiling fan and a 1/(1/1152+1/10K) = 1033 Btu/h-F total conductance, tray water at 64+57.6K/1033 = 120 F could keep the greenhouse 64 F on a 26.5 F night.
That's a high minimum usable tray water temp, so the greenhouse needs a large water volume to store enough heat for 5 cloudy days in a row: 6.9M/(140-120)/62.33 = 5469 ft^3 of water, eg an 18' cube. Adding 8 1000 Btu/h-F car radiators and fans would raise the total conductance to 9033 Btu/h-F with a minimum tray water temp of 64+57.6K/9033 = 70 F and a 6.9M/(140-70)/62.33 = 1581 ft^3 tank, eg a 12' cube, but it seems more practical to fill the space between 2 poly film covers with soap bubble foam on cloudy days.
R20 foam would reduce the heat requirement to (64-26.5)32'x96'/R20 = 5.76K Btu/h on an average night, and the tray and ceiling fan could provide that with a minimum tray water temp of 64+5.76K/1033 = 70 F. The greenhouse would only need 5dx24hx5.76K = 690K Btu to stay 64 F for 5 cloudy 36.2 F days in a row, eg a 690K/(140-70)/62.33 = 158 ft^3 tank, eg a 6' cube or a 4' wide x 3' tall x 14' long plywood box with a 10'x20' folded EPDM liner under a bench.
Zelon’s Swedish patent (US No. 3672184, June, 1972) described insulating shop windows at night with soap bubble foam. Professor John Groh at U. Arizona measured US R3 per inch for soap bubble greenhouse insulation in 1968. Professor Otho Wells at U. New Hampshire did later greenhouse experiments. In 1995, Bill Sturm built a 12,000 ft^2 tomato greenhouse with a soap bubble foam roof in Calgary, Alberta and measured an 84% propane energy savings with and without foam on alternate nights at 20 F below zero.
Bill showed me a simple foam generator with a shop vac blower connected to a horizontal 2" PVC pipe full of holes in a shallow trough near the ground containing a 10% detergent solution which expands by a factor of 300 when foamed. He suggests some window screen to push on a microswitch in an air return at the top of the glazing cavity to turn off the shop vac for all but a few seconds every half-hour at night, along with a standard greenhouse inflation blower above the trough water line to fill the space between the 2 films with air during the day, with a check valve to reduce blower power... http://www.sunnyjohn.com/photos/devices_tools/dt_inflate_blower/index.htm
How about summer cooling?
Nick
Monday, July 1, 2013
Improving indoor evaporative cooling
ASHRAE says 80 F air with a 0.0120 max humidity ratio (pounds of water per pound
of dry air) is comfortable. Vapor pressure Pi = 29.921/(0.62198/0.0120+1) =
0.5663 "Hg, and RH = 0.5663/e^(17.863-9621/(460+80)) = 0.54, ie 54% max.
In a dry climate, a portable indoor evaporative cooler (eg
http://www.amazon.com/SPT-SF-608R-Portable-Evaporative-Cooler/dp/B000OKHY2S/ref=\
sr_1_2?s=appliances&ie=UTF8&qid=1372588973&sr=1-2&keywords=evaporative+cooler )
can be efficient if a line-voltage thermostat (
http://www.zorotools.com/g/00057840/k-G2595704?&gcsct=0ChMIsJ6kztGLuAIVBK7mCh01F\
QAAEAE ) runs the cooler when the room air temp rises to 80 F and a humidistat (
http://www.zorotools.com/g/Humidity%20Controls/00053793/ ) runs an exhaust fan (
http://www.amazon.com/Lasko-2155A-Electrically-Reversible-Window/dp/B00002N5Z9/r\
ef=sr_1_1?ie=UTF8&qid=1372590555&sr=8-1&keywords=lasko+2155a ) when or if the
room RH rises to 54%.
If the outdoor air temp is Ta = 86.7 F with a wa = 0.008 humidity ratio (an
average July afternoon in Billings MT) and the cooler evaporates 10 pounds of
water per hour and the exhaust fan moves C cfm of 0.075lb/ft^3 air and 10 =
60C0.075(wi-wa), C = 555 cfm, and the net cooling is 10x1000 - (86.7-80)555 =
6278 Btu/h, like a typical window AC with about 10X less electrical power.
Cool exhaust air could desirably fill a house attic ("up ducts") with a one-way
plastic film damper to prevent warm outdoor air from flowing down through the
house when the fan isn't running and allow cooler outdoor air to flow up through
the house when it is available. If the house has enough natural air leakage, the
fan may never run.
A cooler near an outdoor air inlet can evaporate more water than one handling
cooler and moister house air. A supply plenum could enclose the exhaust fan and
the cooler inlet, with a plastic film damper that allows house air to enter the
plenum when the exhaust fan is not running. The exhaust fan could become a
2-watt motorized damper, raising the COP.
A solenoid valve scrounged from an old washing machine and some misters that
spray water on a rock pile in a tub could replace the indoor evaporative cooler
and lower the cost and raise the SEER to about 6278Btu/h/(90Wx555/2470) = 310 vs
10 for a window AC, using the 90 watt 2470 cfm Lasko fan...
http://www.amazon.com/Orbit-10057W-Arizona-Outdoor-Misting/dp/B000JJT9MM/ref=sr_\
1_3?s=lawn-garden&rps=1&ie=UTF8&qid=1372600688&sr=1-3&keywords=mister
Fancier controls could turn off evaporation and run conventional AC when the
weather is unsuitable.
Nick
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