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Environmental Systems and Facilities Planning Doug Overhults University of Kentucky Biosystems & Agricultural Engineering University of Kentucky College of Agriculture
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Topic Outline Psychrometrics Review Energy Balances/Loads Latent heat Sensible heat Solar loads Insulation Requirements
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Topic Outline Ventilation Systems Rate requirements System requirements Moisture Control Standards Air Quality Standards Humans Animals Plants and Produce
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Psychrometrics Variables Using the Psychrometric Chart Psychrometric Processes
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Psychrometric Chart Dry Bulb Temperature Scale (axis) “Humidity” Scale or axis State Point
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Psychrometric Chart (temperatures + relative humidity) Dry Bulb Temperature Scale “Humidity” Scale dew-point wet bulb dry bulb Example: 70 o F dry bulb 55 o F dew-point 61 o F wet-bulb 60 % rh relative humidity
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Psychrometric Processes Heating, cooling, humidifying, dehumidifying air-water vapor mixtures Five basic processes to know Heat or Cool (horizontal line) Humidify or De-humidify (vertical line) Evaporative cooling (constant wet-bulb line)
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Heating: dry bulb increase Dry Bulb Temperature Scale “Humidity” Scale ending state point starting state point Horizontal line means no change in dew-point or humidity ratio
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Humidification: dew-point increase Dry Bulb Temperature Scale “Humidity” Scale start state end stateVertical line means no change in dry bulb temperature RH goes up!
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Evaporation: wet bulb increase Dry Bulb Temperature Scale “Humidity” Scale start state end state Increase in vertical scale: humidified Decrease in horizontal scale: cooled Constant wet bulb line Adiabatic process – no heat gained or lost (evaporative cooling)
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Air Density Dry Bulb Temperature Scale “Humidity” Scale Wet bulb line Humid Volume, 1/ ft 3 /lb da
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Review Name three temperature variables Name three measures of humidity Name the two main axes of the psychrometric chart Name the line between fog and moist air Heating or Cooling follow constant line of ? Humidify/Dehumidify follow constant line of ?
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ENERGY AND MASS BALANCES
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Energy and Mass Balances Heat Gain and Loss Latent and Sensible Heat Production Mechanical Energy Loads Solar Load Moisture Balance
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Heat Gain and Loss Occupants Lighting Equipment Ventilation Building Envelope Roof, walls, floor, windows Infiltration (consider under ventilation)
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Heat Loads Occupant (animals, people) Sensible load (e.g. Btuh/person) Latent load (“) Lighting, W/m2 Appliance W/m2 Ventilation air (cfm/person or animal) Equipment (e.g. Btuh for given items)
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Building Ventilation Rate Temperature control Moisture control Contaminants (CO 2, dust, NH 3 ) control Need data for heat, moisture, or contaminant production in building Energy use – VR is a major variable
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Latent and Sensible Heat Production Example from ASAE Standard EP270.5: Table 1. Moisture Production, Sensible Heat Loss, and Total Heat Loss Cattle Bldg. T MPSHL THL 500 kg 21C 1.3 gH2O/kg-h 1.1 W/kg 2.0 W/kg
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Sensible Energy Balance Leads to Ventilation for Temperature Control: q s + q so + q m + q h = Σ UA(t i -t o ) + FP(t i -t o ) + c p ρ V (t i -t o ) Heat inputs = envelope + floor + ventilation q s – sensible heat q so – solar heat gain q m – mechanical heat sources q h – supplemental heat U – building heat transfer coeff. P – floor perimeter F – perimeter heat loss factor c p – specific heat of air V – ventilation rate ρ – density of air
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Sensible Energy Balance Leads to Ventilation for Temperature Control. Rearranging: V = [ q s - ( Σ UA+ FP)(t i -t o )] / 0.24 ρ (t i -t o )60 V – cfm (equation for English units)
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Mass Balance =+ m p Material produced m vi Material input rate m vo material output rate Moisture, CO2, and other materials use balance equations.
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Moisture Balance = m air Ventilation rate M water Moisture to be removed Example balance for moisture control removal rate. (W i -W o ) Humidity ratio difference / Q = (V / 60) x [ W r / (W i -W o ) ] Q - cfm V – ft 3 /lb da W r – lb m / hr W – lb m / lb da
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Find the minimum winter ventilation rate to maintain 60% relative humidity inside a swine nursery that has a capacity of 800 pigs weighing 10 pounds. Inside temperature is 85 degrees. Moisture Balance ASABE D270.5 Nursery Pigs Bldg. T MPSHL THL 4 - 6 kg 29C 1.7 gH2O/kg-h 2.2 W/kg 3.3 W/kg
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Find the minimum winter ventilation rate to maintain 60% relative humidity inside a swine nursery that has a capacity of 800 pigs weighing 10 pounds. Inside temperature is 85 degrees. Find moisture production data ASABE Standards (EP270.5) W r = 0.017 lb/hr/pig Get psychrometric data from chart W 0 = 0.0005 Wi = 0.0154 V = 14.1
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Moisture Balance = m air Ventilation rate M water Moisture to be removed (W i -W o ) Humidity ratio difference / Q = (V / 60) x [ W r / (W i -W o ) ] Put data into equation & solve Q = (14.1/60) x [(.017 x 800) / (.0154 -.0005)] Q = 214 cfm
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Find the ventilation rate required to prevent the ammonia concentration inside a poultry layer barn from rising above 20 ppm. Ammonia production in the barn is estimated to be 21.6 cubic feet per hour. Ammonia concentration in the ambient air is 2 ppm. NH 3 Balance
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NH 3 Solution Use volumetric form of mass balance equation V p + V i = V o V p + Q v [NH 3 ] i = Q v [NH 3 ] o Solve for Q v Q v = V p / { [NH 3 ] o - [NH 3 ] i } Volumetric NH 3 production rate per minute V p = (21.6 ft 3 /hr / 60 min/hr) = 0.36 ft 3 /min Plug into equation & solve Q = 0.36 / (.000020 -.000002)] Q = 0.36 /.000018 Q = 20,000 cfm
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What is the ventilation rate for a swine finishing barn that will limit the design temperature rise inside the house to 4 degrees (F) above the ambient temperature? The barn capacity is 1000 pigs at 220 pounds and the inside temperature is approximately 85 F. The overall heat transfer coefficient for the barn is 1200 Btu/hr F. Energy Balance
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What is the ventilation rate for a swine finishing barn that will limit the design temperature rise inside the house to 4 degrees (F) above the ambient temperature? The barn capacity is 1000 pigs at 220 pounds and the inside temperature is approximately 85 F. The overall heat transfer coefficient is 1200 Btu/hr F. Find heat production data ASABE Standards (EP270.5) q = 0.49 W/kg (sensible heat) Convert units & calculate total heat load q = 0.49 W/kg x 100 kg/pig x 1000 pigs = 49,000 W x 3.412 Btu/hr W = 167,188 Btu/hr Density of Air = 0.075 lb/ft 3 Specific heat of air = 0.24 Btu/lb F t i – t o = 4 F
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Continuation... ventilation rate for a swine finishing barn that will limit the design temperature rise inside the house to 4 degrees (F) above the ambient temperature Basic equation Neglect floor heat loss or gain Plug into equation & solve V = [167,188 - (1200 x 4)] / [(0.24 x 0.075) x 4 x 60] V = 37,590 cfm V = [ q s - ( Σ UA+ FP)(t i -t o )] / 0.24 ρ (t i -t o )60
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Building Heat Loss Q b = (A/R) T x ∆t (A/R) T = sum of all (area/resistance) ratios for all components of the building i.e. walls, ceiling, doors, windows, etc. Insulation
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Wall Section - Resitances in Series
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Insulation & Heat Loss Need R-value for each component
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Q b = (A/R) T x ∆t (Btu/hr) Walls - Q w = (A w /R w ) x ∆t Doors - Q d = (A d /R d ) x ∆t Ceiling - Q c = (A c /R c ) x ∆t Proceed through all components Perimeter is special case R-value is per unit of length - essentially assumes a 1 ft width along perimeter Q p = (L p /R p ) x ∆t Insulation & Heat Loss
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Q bldg = (A/R) w x ∆t + (A/R) d x ∆t + (A/R) c x ∆t +..... ∆t is the same for all components Q bldg = (A i /R i ) x ∆t (A/R) Total = (A i /R i ) sum of all (area/resistance) ratios for all components of the building i.e. walls, ceiling, doors, windows, etc. Q bldg = (A/R) Total x ∆t Building Heat Loss
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The wall of a poultry house will be insulated on the inside by adding 2 inches of spray foam insulation. The R-value of the spray foam insulation is 6 per inch of thickness (hr ft 2 F/Btu in). R-values for the top 1/3 and bottom 2/3 of the existing wall are 12 and 6 (hr ft 2 F/Btu), respectively. No other changes are made. What is the heat loss through the wall after the foam insulation is added as a fraction of the heat loss through the existing wall? Insulated Wall Problem
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R-value of added insulation (2 inches) R foam = 2 x 6 = 12 New R-values R upper = 12 + 12 = 24 R lower = 12 + 6 = 18 No area given – solve for a unit area (1/3 upper & 2/3 lower) Insulated Wall Solution
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What is Q after /Q before No ∆t given but no change between before & after The end result is a ratio of heat losses, so ∆t will be the same in numerator & denominator. All that remains is a ratio of the new & old A/R values. Existing – Wall A/R= 0.33/12 + 0.67/6 = 0.139 New – Wall A/R= 0.33/24 + 0.67/18 = 0.051 Ratio New/Old= 0.051/0.139 = 0.367 Insulated Wall Solution
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Fan Operating Cost =÷ W Power input, Watts V Ventilation volumetric flow rate cfm / Watt Fan Test Efficiency Electrical Power Cost
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Calculate Operating Costs Design Ventilation Rate – 169,700 cfm Fan Choices Brand A – 21,300 cfm @ 19.8 cfm/watt Brand B – 22, 100 cfm @ 16.2 cfm/watt Fans operate 4000 hours per year Electricity cost - $0.10 per kWh Calculate annual operating cost difference
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Calculate Operating Costs Determine number of fans required Brand A - 169,700/21,300 = 7.97 Brand B - 169,700/22,100 = 7.68 8 fans required for brand A or B
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Calculate Operating Costs Use EP 566, Section 6.2 Annual cost is for all 8 fans Watts * hrs * $/kWh * kWh/Wh = $
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References – Env. Systems Albright, L.D. 1990. Environment Control for Animals and Plants. ASAE Hellickson, M.A. and J.N. Walker. 1983. Ventilation of Agricultural Structures. ASAE ASHRAE Handbook of Fundamentals. 2009.
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Reference MWPS - 32 Contains ASABE heat & moisture production data & example problems Midwest Plan Service Iowa State University Ames, IA
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Reference MWPS - 1 Broad reference to cover agricultural facilities, structures, & environmental control Midwest Plan Service Iowa State University Ames, IA www.mwps.org STRUCTURES and ENVIRONMENT HANDBOOK
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Useful References – Env Sys MidWest Plan Service. 1990. MWPS-32, Mechanical Ventilation Systems for Livestock Housing. Greenhouse Engineering (NRAES – 33) ISBN 0-935817-57- 3http://palspublishing.cals.cornell.edu/nra _order.taf
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References – ASAE Standards EP270.5 – Ventilation systems for poultry and livestock EP282.2 – Emergency ventilation and care of animals EP406.4 – Heating, ventilating cooling greenhouses EP460 – Commercial Greenhouse Design and Layout EP475.1 – Storages for bulk, fall-crop, irish potatoes EP566 – Selection of energy efficient ventilation fans
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FACILITIES Manure Management Example
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Manure Management Facilities Animal Manure Production Nutrient Production Design Storage Volumes Lagoon – Minimum Design Volume References ASAE – EP 384.2, 393.3, 403.3, 470 NRCS – Ag. Waste Field Handbook
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Size a Manure Storage 1 year storage Above ground 90’ dia. tank – uncovered 2500 hd capacity – grow/finish pigs Building turns over 2.7 times/yr Manure production 20 ft 3 /finished animal Net annual rainfall = 14 inches 25 yr. – 24 hr storm = 5.8 inches
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Size a Manure Storage Use EP 393, sections 5.1 & 5.3 Total volume has 5 components Manure, Net rainfall, 25 yr-24 hr storm Incomplete removal, Freeboard for agitation
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Size a Manure Storage Manure Depth = 21.22 ft. Net rain = 1.17 ft 25 yr-24 hr storm = 0.48 ft Incomplete removal = 0.67 ft Freeboard = 1 ft Total Tank Depth = 24.54 ft.
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References - Facilties Agricultural Wiring Handbook, 15th edition, Rural Electricity Resource Council Farm Buildings Wiring Handbook, MWPS-28 (now updated to 2005 code) Equipotential Plane in Livestock Containment Areas ASAE, EP473.2 Designing Facilities for Pesticide and Fertilizer Containment, MWPS-37 On-Farm Agrichemical Handling Facilities, NRAES- 78 Farm and Home Concrete Handbook, MWPS-35 Farmstead Planning Handbook, MWPS-2 (download only)
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References – ASAE Standards D384.2 – Manure Production and Characteristics EP393.3 – Manure Storages EP403.4 – Design of Anaerobic Lagoons for Animal Waste Management EP470.1 – Manure Storage Safety S607 – Ventilating Manure Storages to Reduce Entry Risks
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Thank You and Best Wishes for Success on Your PE Exam ! ! University of Kentucky College of Agriculture Biosystems & Agricultural Engineering
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