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Remodeling Existing Farm Structures for Commercial Fish CultureRon
Rosati A fish production system using recirculated water may be thought of as analogous to a hog confinement system. In both systems a farmer needs a method to house the animals, control ambient environmental parameters (temperature, noxious gases, etc.), feed the animals, remove waste products, sort animals, control reproduction, harvest animals, treat diseased animals, etc. The primary difference between the two forms of livestock is that the production environment for fish is controlled and manipulated to a much greater degree than is the production environment for hogs. (Other differences between these two forms of livestock are that commercial-scale culture system variables for fish are relatively unknown compared to hogs, indoor food fish production generally involves greater financial risk than confinement hog production, and the market infrastructure for farm raised fish other than catfish is still relatively undeveloped compared to hog marketing systems.) The purpose of this paper is to discuss the criteria used to evaluate the remodeling of an existing farm structure for use as a fish culture environment. This paper was written based on the assumption that commercial fish culture in recirculating systems is a profitable enterprise. However, some fish producers have found that profits were less than anticipated. Each potential producer is encouraged to expend their initial efforts on budgets and profit/loss statements before using capital for fish production. Entrepreneurs who currently own farm structures may be able to use those farm structures to gain the comparative advantage needed to make fish culture profitable. A culture building must contain the equipment required to preform the following function: 1. manipulate the ambient environment, especially
temperature The cost of remodeling verses new construction must be evaluated. Some producers consider remodeling advantageous if the remodeling costs are one-half or less the price of new construction. Turn-key pole building shells can be purchased for approximately $15 per square foot (assuming a 40' x 70' building without a concrete floor, without insulation). Some producers prefer remodeling older farm structures for aesthetic or nostalgic reasons. Other advantages of remodeling older buildings verses new construction is the ability to begin use of the building with minimal investment, and the ability to do most of the remodeling with home labor rather than hired labor. Humstone (1988) presents a 10 point preliminary checklist to begin evaluation of the feasibility of remodeling an existing barn: 1. evaluate framing. Check posts, beams, sills, rafters, and joists to be sure they are solid and free of rot. 2. evaluate the foundation. Check for cracks, settling, and shifting. Look for loose or missing mortar. 3. evaluate the roof. Check the roof covering and flashing. On the inside, look for water stains and rot on sheathing and beams. 4. evaluate the exterior walls. Eyeball the length of the barn at eave level to checks walls for straightness. If the foundation has shifted, walls may have sagged and pulled out. 5. evaluate the building interior. Note the location and existence of drains and gutters. Check the condition of existing floors. Note the amount of free-span space. 6. evaluate the building location. Is the barn conveniently located with good access to and from other buildings and the farmyard, and access to water and electricity? Can delivery and livehaul trucks easily access the fish tanks? Is there room to add on if you want to expand your operation in the future? 7. evaluate the building size. Is the barn big enough for its intended purpose? Is there adequate ceiling clearance and space between interior posts? Is there room to install the equipment you need? 8. evaluate doorways. Will doors need to be enlarged or moved? 9. evaluate the building ambient environment. Is the barn airtight and insulated? Can heating and ventilation systems be installed? 10. evaluate utilities. What is the condition of the plumbing and wiring? Will it need to be updated? Can feeding, water flow and manure handling systems be installed if necessary? If after running through a preliminary checklist the project looks feasible, the following detailed considerations need to be studied while planning for the remodeling: Water SupplyThe most important consideration for fish culture is the supply of water. Even in a recirculating system, substantial quantities of water are required. A well designed recirculating system consumes 10% or less of its total water volume per day. If a culture system contains 50,000 gallons, the producer must plan for a consumption rate of at least 5,000 gallons per day. If a pipe was running 24 hours per day to supply only this use, approximately 3.5 gallon per minute (GPM) must be supplied. In reality, flow rates must be as high as is practically possible in a fish production building. Supply rates for farmsteads which include fish production as a major farm component is at least 20 - 40 GPM. Piping must be sized to accommodate this flow rate. Water well capacity may also be evaluated to assure adequate supply. Well capacity can be increased by pulling the existing pump and installing a larger pump (cost of labor and 1 hp pump are approximately $700). Often, water supply lines throughout the building are made of surface mounted 2" PVC. This piping is inexpensive, easy to install, easy to service and of high capacity. Some consumption devices such as sand filters require large volumes of high pressure water for backflushing. The surge requirements of these devices may dictate the capacity of the building plumbing. For example, a sand filter sized to filter the wastes from a heavily loaded 6,000 gallon system may require 60 GPM at a pressure of 40-50 psi during backwashing. The total consumption of this filter may exceed 700 gallons in a 12 minute time period. Large pumps and large piping is required to handle hydraulic loading at this level. To economize on plumbing, a reservoir tank may be installed. The quality of incoming water needs to be considered in planning the facility. Incoming water should meet these values (concentrations are give in parts per million) (from the US Fish and Wildlife Service, 1982 and Boyd, 1990):
VARIABLE WARMWATER FISH
--------- -----------------
dissolved oxygen 6-saturation
carbon dioxide 0-10
total alkalinity (as CaCO3) 50-400
pH 6.5-9.0
total hardness (as CaCO3) 20-400
calcium 10-160
iron 0-0.5
hydrogen sulfide <0.005 un-ionized ammonia
<0.05
total dissolved gases <105% total gas pressure
salinity (value for catfish) 100-8,000
temperature appropriate for species
Culture water may need to be treated to meet these requirements. Planning must allow for the treatment and storage of incoming water to allow the oxidation of iron, gas exchange, settling, warm-up, etc. DrainageA 50,000 gallon system reusing 90% of its culture water must plan on disposing of 5,000 gallons per day. A building must have a drainage system to allow this amount of water to flow into a disposal system. Floor drains must be of a size to accommodate large flows. Floor ConstructionFloors in a culture room are often concrete to allow easy accomplishment of husbandry activities. Floors should be sloped towards drain. Level floors will actually have some slope and it may be away from floor drains. Reinforcement must be used in the floor to allow the floor to withstand the heavy loading resulting from culture tanks. When constructing floors, give consideration to casting concrete tanks directly into the floor. LightLaboratory and hatchery areas should be equipped with lighting at a level of 2.75 watts per cubic foot of floor space. Culture areas may receive less lighting. A hatchery manager should have the ability to manipulate photoperiod to correspond with culture activities. UV light at high intensities may be harmful to larval fish. Algal growth on biofilters may be stimulated by high light intensities. Since algae out competes nitrosomonas and nitrobacter bacteria, biofilters should not be exposed to high light intensities. Vapor-proof lamps should be used near culture tanks. DoorsWhen evaluating the remodeling of a building, consider the placement and size of doors. A building suitable for aquaculture must have doors allowing access by trucks, forklifts, etc. In addition, tanks and filters will need to be replaced - a large door may be required to install a 5000 gallon rectangular tank. The flow of materials throughout the structure should also be evaluated during remodeling. Plan for conveyors, alleyways, augers, etc as required to move fish, feed, supplies or equipment. Avoid the temptation to "make-do" with inefficient, high-labor techniques. Once a system grows to a commercial scale, a producer will have difficulty performing repetitive, heavy tasks such as harvesting fish, loading and unloading livehaul trucks, and unloading and moving feed without mechanized equipment. Bulk bins and auger conveyors may be used for feed handling. Some feed equipment companies such as Choretime currently market tested and proven automatic feed delivery systems for fish farmers. Electrical ServiceA recirculating aquaculture system will be a consumer of large amounts of electrical power. The facility may be supplied with its own electrical drop. A 200 amp power supply should be installed even if current needs are less than 200 amps. Three phase power is required for pumps larger than approximately 2 HP. Both 110 and 220 volt service should be installed. All wiring should be installed according to the National Electrical Code by a competent electrician. Wiring throughout the facility should be exterior quality. Each circuit should be protected by a ground fault circuit interrupter. Underground feeder (UF) cable should be surface mounted throughout the facility. Wire size must be based upon current and anticipated loads. Planning is required to determine the location of large pump motors so individual circuits can be installed for those motors. HeatingExtensive heating is required for the culture of all warmwater species such as tilapia and limited heating is required even for coldwater species to allow for optimum fish growth and for operator comfort. The ideal heating plant would be inexpensive, efficient, easily installed into a retrofitted building, and easily maintained. Infrared heaters fit these requirements best. LP or natural gas is a less expensive fuel than electricity so its use is preferred. See Appendix B for a comparison of heat sources. Avoid the use of unvented gas heaters in northern climates. Building ventilation rates must be increased to compensate for the increased indoor pollution created by these heaters and the economic efficiency of the heater is quickly lost due to these increased ventilation rates. The most applicable heater for a retrofitted building is a tube style infrared gas heater. Tube style infrared gas heaters are easily vented through a 6" hole cut into the side of the building. Their efficiency is higher than 90%. A tube-style heater is easily installed in a remodeled barn by ceiling hangers. A major advantage to tube-style infrared heaters is they can be used to heat culture water directly. Infrared heat heats the surface it strikes so an infrared heater suspended over a tank will heat that tank rather than heating air between the heater and the tank. Submerged electrical heaters are short-lived, potentially dangerous and consume expensive fuel. Convective air furnaces (hot air furnaces) have the advantage of being readily available, easily repaired, and easy to install. Modern, high-efficiency gas furnaces can be vented directly through building sidewalls without the use of an extensive chimney system but older furnaces require an extensive exhaust gas venting system which can be cost prohibitive. Supplemental water heating devices are often required when warmwater fish are cultured in buildings heated by convective air furnaces. Radiant floor heaters can be cast into the concrete when floors and tanks are being cast. These heaters are safe, do not become biofouled and do not build-up scale but their inaccessibility is an obvious disadvantage. See Appendix A to determine the size of heater required. Add 25% to the heater size to allow for the heating of water. For a general approximation, a 100,000 BTU infrared gas heater at the Illinois State University aquaculture facility is more than adequate to supply air and water heat for a 1000 square foot building with 10,000 gallons of culture water. InsulationInsulation must be installed in most barns when they are retrofitted for fish culture. Insulation is important because it allows the control of both temperature and condensation. The form of insulation installed is determined by cost, ease of installation and health consideration. Fiberglass insulation is the most common and likely the most cost-effective per unit of R-value. Batts may be installed in walls and over ceilings and blown or loose fill fiberglass may be used over ceilings. Unfortunately, fiberglass batt insulation is often difficult to install in pole type buildings. Rigid board insulation is made from cellulose fiber, fiberglass, polystyrene, polyurethane, and polyisocyanurate. Sheets are 1/2 to 2" thick by 2 or 4' wide. Sheets are usually rated at high R-value per inch of thickness but they are also more expensive per unit of R-value when compared to fiberglass batt insulation. Despite its relatively high cost, board insulation is a good choice for remodeling a barn into a fish culture building because the insulation is easily installed into most existing barns styles, board insulation can be moisture resistant, and board insulation has a high R-value per inch. Some board insulations are flammable are release toxic gases when burned. Foamed-in-place insulation are composed of polystyrene or polyurethane. If "open-celled", foam insulation is not moisture resistant. Closed cell foam is moisture resistant. Foam insulations are easily installed into existing buildings by spraying. Some foam insulation have been found to release formaldehyde gases as they age. Safety and cost problems need to be compared against the ease on installation and high R-values of this type on insulation. Concrete is an excellent construction material but it has a very low R-value. Its primary function is to provide structural strength but from a thermal perspective it is little better than an excellent wind break. See Appendix B to determine wall R-values. Air Moisture ControlIs excessive humidity a problem in a fish culture building? Not from the perspective of the fish - the primary inhabitants of the building. In other confinement livestock buildings, managers are concerned about air quality because it impacts on the health of the building inhabitants but in fish confinement buildings air quality must deteriorate to severely debilitated condition before fish health is affected. Excessive humidity can be a problem, however, because it causes deterioration of building structural components, electronic monitoring equipment, stored feed and other corrosion sensitive materials. Humidity is best handled through a multi-tiered approach: 1. Keep moisture sensitive materials out of a fish culture room. Store feed in a separate room. Store electronic monitors in an office or "clean" room. Use lights and other structural components which can tolerate high moisture conditions. Plywood, for example, should be exterior quality at minimum. CDX plywood is suitable for use in a fish culture building if the "C" grade veneer faces inside the room. "D" grade veneers cannot tolerate high moisture conditions even through the plywood contains exterior grade glue. A better choice for interior sheathing is to use "white board" - plywood coated with fiberglass or plastic. This material is commonly used as sheathing in hog farrowing houses and is readily available from farm supply construction centers. Avoid use of water-borne preservatives in the construction materials used inside a fish production building since preservatives may leach out of the wood and cause fish health problems. Oil-borne preservative such as pentachlorophenol will not leach into water as easily. 2. Construct the room so that moisture does not move into the walls. If moist air moves into a wall the air will eventually cool to its dewpoint and condense inside the wall leading to matted and ineffective fiberglass insulation, rotting of wall structural members, shorting of electrical components, and paint blistering. A vapor barrier is of obvious premier importance in any fish culture building when temperatures and ventilation rates will be controlled. Vapor barriers placed on the warm side of walls prevent moisture movement into the wall. Warm air holds greater amounts of water vapor than does cold air. Four to six-mil polyethylene is the most common vapor barrier. If whiteboard is used as wall sheathing, the seams of the sheathing should be sealed with silicone to provide a vapor barrier. Ceilings must also be moisture-proofed. 3. Install air heat exchangers. A heat exchanger will allow the venting of high-humidity air and replace it low-humidity, outside air. The purpose of the heat exchanger is to transfer heat from the exhaust air to the incoming air. Heat exchangers are sized to keep interior moisture levels at 50-80% relative humidity. Higher humidities may lead to condensation problems and lower levels lead to dust problems. Some studies have demonstrated that air held at 50 - 80% R.H. is detrimental to air-borne bacteria. A heat exchanger transfers approximately half of the exhaust heat to incoming air. A small disadvantage to the use of heat exchangers in a fish production building is the slight loss of heat energy associated with the moisture content of the vented air. While the temperature of the vented air may be lower, the latent heat content of the exhaust air is still high due to the vapor still present in the air. Dry air in a fish culture building can easily absorb heat energy without an accompanying increase in temperature. As the relative humidity of air increase, the air is less likely to absorb heat without also changing temperature. 4. Vent the cold side of the wall. Regardless of the quality of the vapor barrier, a small amount of water vapor will find its way into a wall in a fish culture building. All building cavities should be vented to allow the escape of this moisture. Roof ridge vents may be installed to vent attic spaces. In some cases, it may be necessary to vent walls by drilling 5/8" holes through the top plate. Three holes are recommended for every stud in the wall. Vent holes may be drilled into the top and bottoms of exterior sheathing. Small louvers should be placed into these vent holes to prevent the entrance of Monitor, Provide Automatic ControlAn alarm and monitor system is strongly recommended for commercial- scale fish culture systems. Monitor and control devices should be planned into the building design during remodeling. The system should monitor oxygen level, high and low water level, temperature, water flow, and electrical flow. Systems currently on the market will measure these parameters and call 4 telephone numbers with an alarm message if the parameters are outside of specified ranges. In addition, a monitor system can automatically adjust electrical component to control water quality. According to manufacturers data, temperature, oxygen levels, water flow rates and water level can all be manipulated automatically by a system costing approximately $6,000. Historic PreservationMany older barns are remodeled for aesthetic reasons. If an agricultural building is more than 50 years old and has notoriety for its architecture or association with important persons or events, the building may be registered on the National Register of Historic Places. The registration of a building does not limit the use or even destruction of the structure but it does give its owner eligibility for remodeling tax credits and perhaps property tax freezes. To qualify for a 20% tax credit, a building must meet these criteria: 1. the building is listed on the National Register of Historic Places. (A 10% tax credit is available for nonregistered buildings built before 1936.) 2. the building is used for income-producing purposes, 3. rehabilitation costs are greater than $5,000, and 4. at least 75% of the existing internal structural framework is retained. To discuss the potential for tax credits as you remodel your old barn into a fish culture facility, contact the National Trust for Historic Preservation, 53 West Jackson Blvd., Suite 1135, Chicago, Illinois 60604, (312-939-5547). BibliographyBoyd, Claude E. Water Quality in Ponds for Aquaculture. Auburn University. Alabama Agricultural Experiment Station. 1990. Degroot, Rodney C. Your Wood Can Last Forever. USDA Forest Service Humstone, Mary. Barn Again! A Guide to Rehabilitation of Older Farm Buildings. Meredith Corporation. 1988. (1-800-228-4630). Johnson, Dexter. Using Old Farm Buildings. North Dakota State University Agricultural Engineering Department, Fargo, ND 58105. 1988. Publication AERR 88-1. Midwest Plan Service. Structures and Environment Handbook. Eleventh edition. Iowa State University, Ames, Iowa. 1983. Piper, Robert G., et al. Fish Hatchery Management. U.S. Department of the Interior Fish and Wildlife Service. 1982. Vara, Jon. Giving Old Barns New Life. Country Journal. June, 1985. pp 48-60. APPENDIX A(from Midwest Plan Service Structures and Environment Handbook 1) Calculating Heat LossThe rate of heat loss through each building component is proportional to its area and the difference between the inside and outside temperatures. The rate of heat flow is also determined by the RT value of the building component; the higher the RT value, the lower the rate of heat flow. The rate of heat loss from each building component, q, is given by:
Eq 631-1.
q = (A/RT) x (ti - to)
q = rate of heat loss from the building com-
ponent, Btu/hr
A = area of the building component, ft2
RT = total resistance to heat flow of the com-
ponent, F-ft2-hr/Btu
ti = inside temperature, F
to = outside temperature, F
The floor perimeter is a special case and the RT value in Table 631-3 is given per foot of length. In this case the area, A, is replaced by the length of the exterior wall. To obtain the total heat loss from a building, the losses through each building component are simply added together. The following sample problem illustrates the procedure. Sample ProblemFind the amount of heat loss that will occur in a 24' x 36' building constructed as illustrated in Fig 631-18. The inside design temperature is 60 F and the outside design temperature is -10 F. The building has two 3' x 7' doors, insulated with 1", 1 pcf of molded polystyrene. Step I List the length, width, wall height, and foundation height. Calculate the perimeter, frame wall area (excluding windows and doors), concrete wall area, ceiling area, window area, and door area. The resistance of the frame wall is 12.47 from Table 631-2. The RT for a 6" concrete wall with 2" polystyrene insulation is 11.58 from Table 631-2. The RTfor the ceiling is 13.47 and for the doors is 7.99 from Table 631-3. Step II The heat losses, q, for the ceiling, walls, and perimeter are found by placing the appropriate values from Step I into the heat loss equation. The RT value of 2.22 in the perimeter equation assumes 2" x 24" polystyrene perimeter insulation, Table 631-3. The total heat loss, qb, from the building is the sum of the ceiling, wall, perimeter window, and door losses.
Fig 631-18. Sample problem building wall.WORKSHEET-HEAT LOSSStep I
Building dimensions (ft) Surface area (ft2) RT values
--------------------------- ---------------------------- ------------------------------
Length (L) 36 Ceiling area 864 ceiling 13.47
Width (W) 24 Window area 0 Window
Frame wall height (H) 6 Door area 42 Door 7.99
Concrete wall height (F) 2 Frame wall area less Frame wall 12.47
Perimeter 120 window & door area 678 Concrete wall 11.58
Concrete wall area 240 Perimeter 2.22
Design temperatures (F)
to (outside temp) = -10 ti (inside temp) = 60
Deltat = 70
Step II
Heat loss from building, qb
Deltat x ceiling area 70 x 864
Ceiling qc = ---------------------- qc = ---------- = 4490 Btu/hr
ceiling RT 13.47
Deltat x window area
Windows qwi = --------------------- qwi = 0
window RT
Deltat x door area 70 x 42
Doors qd = -------------------- qd = -------- = 368 Btu/hr
door RT 7.99
Deltat x frame wall area 70 x 678
Frame walls qw = ------------------------- qw = -------- = = 3806 Btu/hr
frame wall RT 12.47
concrete Deltat x concrete wall area 70 x 240
walls qf = ----------------------------- qf = ----------- = 1451 Btu/hr
concrete wall RT 11.58
Deltat x perimeter 70 x 120
Perimeter qp = ------------------- qp = ---------- = 3784 Btu/hr
perimeter RT 2.22
qb = qc + qwi + qd + qw + qf + qp = 13,899 Btu/hr
Or: Building heat loss, qb can be expressed in terms of the inside-outside temperature difference, At:
Eq 631-2.
qb = A/R x Deltat
A/R =sum of all (area/resistance) ratios of
the building
Using the above sample problem:
Building (A/R) = ceiling (A/R) + frame wall (A/R)
+ concrete (A/R) + perimeter (A/R) + window
(A/R) + door (A/R)
= 864/13.47 + 678/12.47 + 240/11.58 + 120/2.22
+ 0 + 42/7.99
= 64.14 + 54.37 + 20.73 + 54.05 + 5.26
= 198.55 Btu/hr-F
Therefore, the heat loss from the building is:
qb =198.55 x Deltat
= 198.55 x 70
= 13,899 Btu/hr
Effects of Changing Insulation ValuesNote three important items in the example that can affect the quantity of heat loss: 1.Use enough insulation. Suppose the insulation in the frame walls is decreased from R =11 batt insulation to 25/32" insulating sheathing (see Table 631-2). This is a decrease in RT value from 12.47 to 4.43. Recalculating the heat loss from the frame walls and building yields:
Frame wall:
qb = Deltat x frame wall area e frame wall RT
= 70 x 678 + 4.43
= 10,713 Btu/hr
Therefore, the new building heat loss is:
qb =20,806 Btu/hr
This represents an increase in heat loss through the frame wall of 180%. The total building heat loss increases 50%. 2.Insulate all areas. The use of insulation on concrete walls and around the building perimeter often went unnoticed until recent years. The illustrate the effectiveness of insulating the concrete wall and perimeter, assume the 2" x 48" insulation covering the concrete wall and perimeter in Fig 631-18 is neglected. Recalculation of the heat loss from the concrete wall, perimeter, and building yields:
Concrete wall:
RT = 1.33 (Table 631-1)
Perimeter:
RT = 1.23 (Table 631-3)
Concrete wall:
qf = Deltat x concrete wall area
÷ concrete wall RT
= 70 x 240 ÷ 1.33
= 12,632 Btu/hr
Perimeter:
= Deltat x perimeter ÷ perimeter RT
= 70 x 120 ÷ 1.23
= 6,829 Btu/hr
Therefore, the new building heat loss is:
= 28,125 Btu/hr
The heat loss through the concrete wall increases 771%; and the perimeter, 80%. Not insulating the concrete wall and perimeter causes the total heat loss of the building to almost double. 3.Install windows only when necessary. Single pane glass is poor insulation. If windows are required, use thermopane or windows with storms to cut heat loss. Th illustrate heat loss through windows, assume there are eight in this building. The windows are single thickness having an area of 8.75 ft2 (2'-6" x 3'-6"). Recalculating the building heat loss yields:
Windows:
qwi = Deltat x window area ÷ window RT
= 70 x (8.75 x 8) ÷ 0.91
= 5385 Btu/hr
Frame walls:
= Deltat x (frame wall area - window area)
÷ frame wall RT
= 70 x (678 - 70) ÷ 12.47
= 3413 Btu/hr
Therefore, the new building heat loss is:
qb = 18,891 Btu/hr
The heat loss through the eight windows is over 1 1/2 times the heat loss of the four frame walls. The total building heat loss increases 36%. WORKSHEET-HEAT LOSS
Step I Building dimensions (ft) Length (L) ________ Width (W) ________ Frame wall height (H) ________ concrete wall height (F) ________ Perimeter ________ RT values Ceiling ________ Window ________ Door ________ Frame wall ________ concrete wall ________ Perimeter ________ Surface area (ft2) Ceiling area ________ Window area ________ Door area ________ Frame wall area less window & door area ________ Concrete wall area ________ Design temperatures, F to (outside temp) = ________ , ti (inside temp) = ________ Deltat = ______ Step II
Building heat loss, qb
Deltat x ceiling area ( ) x ( )
Ceiling qc = ----------------------- qc = ----------------------- = ______ Btu/hr
ceiling RT ( )
Deltat x window area ( ) x ( )
Windows qwi = ---------------------- qwi = ----------------------- = ______ Btu/hr
window RT ( )
Deltat x door area ( ) x ( )
Doors qd = --------------------- qd = ----------------------- = ______ Btu/hr
door T ( )
Deltat x door area ( ) x ( )
Frame walls qw = --------------------- qw = ---------------------- = _______ Btu/hr
frame wall T ( )
Deltat x door area ( ) x ( )
Concrete walls qf = --------------------- qf = ---------------------- = _______ Btu/hr
concrete wall T ( )
Deltat x door area ( ) x ( )
Perimeter qp = --------------------- qp = ---------------------- = _______ Btu/hr
perimeter T ( )
qb + qc + qwi + qd + qw + qf + qp = _______ Btu/hr
APPENDIX BHeating Fuel Cost ComparisonChoosing a FuelYou can run electricity through your heat pump, burn corn cobs in your biomass furnace, stoke up the wood stove--there are numerous ways to keep your home warm in the winter. What you pay for that warmth depends largely on what fuel you use and the efficiency of your furnace. Unfortunately, finding out which fuel is cheapest for you is about as easy as comparing corn cobs to kilowatts. That's because fuels are sold in a variety of units, ranging from the ton, to the gallon, to the cord. The worksheet on the back of this page helps you compare costs of different types of fuels. It takes into account both fuel costs in your area and the efficiency of your particular furnace or stove. After doing the computations, you'll know approximately what it will cost you to put one million Btu's (British thermal units) into your house, using a particular type of fuel and furnace. During the heating season, it takes about 100 to 150 million Btu's to heat the average Iowa home. Using the WorksheetTo use the worksheet, do the following computations for each fuel you want to price: Get local fuel prices. Write them in column 2. Make sure the figures correspond to the units listed in column 2. For example, LP gas is priced by the gallon, electricity by the kilowatt-hour, wood by the cord. Find out the annual efficiency of your heating system (or the heating system you're considering). If you have an older heating system, you may not know its efficiency In that case use the typical efficiency figure already listed in column 3. Efficiencies are available for most new furnaces and stoves. If you know the system's efficiency, cross out the given figure in column 3 and put the actual efficiency In the blank. (The actual efficiency of heating systems varies. Your heating system efficiency might be much higher or lower than the typical efficiencies given.) Multiply column 1 by column 2. Divide the result by column 3. Put this figure in column 4. Column 4 shows the cost of putting 1 million Btu's of heat into your house, using a specific fuel. Other ConsiderationsFuel cost isn't the only factor to consider when choosing a heating system. Other factors to consider include the cost of buying the system, the convenience and availability of the fuel, storage space needed, reliability of the system, maintenance expenses, and safety and environmental concerns. Prepared by Tom Greiner, extension agricultural engineer, and Diana Pounds, communications specialist at Iowa State University.
(1) (2) (3) (4)
Fuel Quantity for Fuel Annual Fuel cost per
one million price efficiency million Btu's
Btu's
-----------------------------------------------------------------------------------------
Natural $_______
gas 10.0 Ccf X per Ccf ÷: (0.67 or ______) = $ ______
LP $_______
gas 11.11 gal. X per gal. ÷: (0.67 or ______) = $ ______
Fuel $_______
oil 7.14 gal. X per gal. ÷: (0.55 or ______) = $ ______
Electricity $_______
(resistance) 293 kWh X per kWh ÷: (1.00 or ______) = $ ______
Electricity $_______
(heat pump) 293 kWh X per kWh ÷: (1.50 or ______) = $ ______
Hard $_______
coal 0.0417 ton X per ton ÷: (0.50 or ______) = $ ______
Hard $_______
wood 0.0357 cord X per cord ÷: (0.50 or ______) = $ ______
Medium $_______
wood 0.0476 cord X per cord ÷: (0.50 or ______) = $ ______
Soft $_______
wood 0.0714 cord X per cord ÷: (0.50 or ______) = $ ______
Kerosene - $_______
7.41 gal. X per gal. ÷: (0.99 or ______) = $ ______
Biomass
(garbage, $_______
pulp, etc.) 143 lbs. X per lb. ÷: (0.50 or ______) = $ ______
-----------------------------------------------------------------------------------------
Three examplesFollowing are three examples. In the first, heating fuel cost is computed, using a 67 percent-efficient furnace and natural gas price of 51 cents. The second example shows the fuel cost when natural gas remains at 51 cents per Ccf, but the furnace efficiency is 92 percent. In the third example, heating fuel cost is figured for a 100 percent-efficient resistance electric furnace and an electricity cost of 7.04 cents per kWh.
------------------------------------------------------------------------
$.51
Natural --------
gas 10.0 Ccf X per Ccf ÷ (0.67 or X ) = $ 7.61
$.51
Natural --------
gas 10.0 Ccf X per Ccf ÷ ( X or .92 ) = $ 5.54
$.0704
Electricity ---------
(resistance) 293 kWh X per kWh ÷ (1.00 or X ) = $20.63
--------------------------------------------------------------------------
Solar energy To compare solar costs, the cost of installing solar equipment must be considered. See extension publication Active Solar Collectors-Are They a Good Investment? (Pm-1034). *This figure represents the efficiency of the heat pump It's the heat output of the unit divided by the electricity used, usually called the coefficient of performance (COP). Units Ccf = 100 cubic feet of as therm = about 1 Ccf kWh = 1 kilowatt-hour cord = a 4 x 4 x 8 foot stack of wood Cooperative Extension service, Iowa State University of Science and Technology and the United States Department of Agriculture cooperating. Robert L. Crom, director, Ames, Iowa. Distributed in furtherance of the Acts of Congress of May 8 and June 30, 1914. and justice for all The Iowa Cooperative Extension Service's programs and policies are consistent with pertinent federal and state laws and regulations on non-discrimination regarding race, color, national origin, religion, sex, age, and handicap. APPENDIX C - Calculating R-Values (from MWPS Structures and Environment Handbook 1)Table 631-1. Insulation values. From 1981 ASHRAE Handbook of Fundamentals. Values do not include surface conditions unless noted otherwise. All values are approximate.
R-value
Per inch for thickness
Material (approximate) listed
1/k 1/C
--------------------------------------------------------------
Batt and blanket insulation
Glass or mineral wool, fiberglass 3.00-3.80*
Fill-type insulation
Cellulose 3.13-3.70
Glass or mineral wool 2.50-3.00
Vermiculite 2.20
Shavings or sawdust 2.22
Hay or straw, 20" 30+
Rigid insulation
Exp. polystyrene,
extruded, plain 5.00
molded beads, 1 pcf 5.00
molded beads, over 1 pcf 4.20
Expanded rubber 4.55
Expanded polyurethane, aged 6.25
Glass fiber 4.00
Wood or cane fiberboard 2.50
Polyisocyanurate 7.04
Foamed-in-place insulation
Polyurethane 6.00
Building materials
Concrete, solid 0.08
Concrete block, 3 hole, 8" 1.11
lightweight aggregate, 8" 2.00
lightweight, cores insulated 5.03
Brick, common 0.20
Metal siding 0.00
hollow-backed 0.61
insulated-backed, 3/8" 1.82
Softwoods, fir and pine 1.25
Hardwoods, maple and oak 0.91
Plywood, 3/8" 1.25 0.47
Plywood, 1/2" 1.25 0.62
Particleboard, medium density 1.06
Hardboard, tempered, 1/4" 1.00 0.25
Insulating sheathing, 25/32" 2.06
Gypsum or plasterboard, 1/2" 0.45
Wood siding, lapped, 1/2" x 8" 0.81
Asphalt shingles 0.44
Wood shingles 0.94
Windows (includes surface conditions)
Single glazed 0.91
with storm windows 2.00
Insulating glass, 1/4" air space
double pane 1.69
triple pane 2.56
Doors (exterior, includes surface conditions)
Wood, solid core, 1 3/4" 3.03
Metal, urethane core, 1 3/4" 2.50
Metal, polystyrene core, 1 3/4" 2.13
Air space (3/4" to 4") 0.90
Surface conditions
Inside surface 0.68
Outside surface 0.17
*The insulation value of fiberglass varies with ban thickness.
Check package label.
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Ways of expressing the value of insulation are: R = thermal resistance, hr-ft2-F/Btu. It is the resistance to heat flow of 1 ft2 of material when the temperature difference between the two sides is 1 F. R is an additive quantity; 2" of a material has twice the R-value of 1". Also the individual R- values for all materials in a given section of a structure can be added together to obtain a total R-value. RT = total thermal resistance. It is the total resistance of an entire wall, ceiling, etc. section, including the air film coefficients. U = overall coefficient of heat transmission, Btu/hr-ft2-F = 1/RT It is the heat in Btu/hr that passes through an entire wall, ceiling, etc. section of 1 ft2, in one hour per 1 F temperature difference between the air on the warm side and the air on the cold side. k = thermal conductivity, Btu-in./ft2-F-hr. It is the heat in Btu/hr that passes through a piece of material 1" thick and 1 ft2, when the temperature difference between the two sides is 1F. C = thermal conductance, Btu/ft2-F-hr. C is like k, except it is given for the total thickness: k for glass wool = 0.29; C for 3" glass wool = 0.10. By convention, C does not usually include the effects of boundary layer resistances. In the following discussion, R is used because the insulation value of a wall is easier to calculate, and many insulations are marked with their R-value. Example 1: Given the wall in Fig 631-7, find the total R-value. From Table 631-1 find the R-values for each material.
Fig 631-7. R-value of a wall section.By adding the individual R-values, we find the wall has a total R-value of 13.28. Note that the blanket insulation provides more than 80% of the total R. Example calculations for other wall constructions are shown in Table 631-2.
Outside surface (15 mph wind) 0.17
Concrete (6") 0.48
Inside surface (still air) 0.68
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Total resistance, RT 1.33
Outside surface (15 mph wind) 0.17
Plywood (1/2") 0.62
Inside surface (still air) 0.68
------
Total resistance, RT 1.47
Outside surface (15 mph wind) 0.17
Sheet metal 0.00
Inside surface (still air) 0.68
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Total resistance, RT 0.85
Outside surface (15 mph wind) 0.17
8' lightweight concrete block 2.00
Inside surface (still air) 0.68
-----
Total resistance, RT 2.85
Outside surface (15 mph wind) 0.17
Plywood (1/2") 0.62
Air space 0.90
Supplemental vapor barrier 0.00
Plywood (1/2") 0.62
Inside surface (still air) 0.68
-----
Total resistance, RT 2.99
Outside surface (15 mph wind) 0.17
Sheet metal 0.00
Fiber board insulating sheathing,
(25/22") 2.06
Air space 0.90
Supplemental vapor barrier 0.00
Plywood (1/2") 0.62
Inside surface (still air) 0.68
------
Total resistance, RT 4.43
Outside surface (15 mph wind) 0.17
8" lightweight concrete block,
cores filled with vermiculite 5.03
Inside surface (still air) 0.68
------
Total resistance, RT 5.88
Outside surface (15 mph wind) 0.17
Plywood (1/2") 0.62
Air space 0.90
Supplemental vapor barrier 0.00
25/32" insulating sheathing 2.06
Inside surface (still air) 0.68
------
Total resistance, RT 4.43
Outside surface (15 mph wind) 0.17
Sheet metal 0.00
Expanded polyurethane (aged) 6.25
Inside surface (still air) 0.68
------
Total resistance, RT 7.10
Outside surface (15 mph wind) 0.17
Concrete (6") 0.48
Extruded polystyrene (2") 10.00
Hardboard, tempered 0.25
Inside surface (still air) 0.68
------
Total resistance, RT 11.58
Outside surface (15 mph wind) 0.17
Plywood (1/2") 0.62
Batt insulation, R=11 11.00
Supplemental vapor barrier 0.00
Plywood (1/2") 0.62
Inside surface (still air) 0.68
------
Total resistance, RT 13.09
Outside surface (15 mph wind) 0.17
Sheet metal 0.00
Ban insulation 11.00
Supplemental vapor barrier 0.00
Plywood (1/2") 0.62
inside surface (still air) 0.68
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Total resistance, RT 12.47
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