Environmental Systems and Facilities Planning
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Transcript of Environmental Systems and Facilities Planning
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Environmental Systemsand Facilities Planning
Doug Overhults University of Kentucky
Biosystems & Agricultural Engineering
University of KentuckyCollege of Agriculture
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Topic OutlinePsychrometrics ReviewEnergy Balances/Loads
Latent heat Sensible heatSolar loads
Insulation Requirements
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Topic OutlineVentilation Systems
Rate requirementsSystem requirements
Moisture Control StandardsAir Quality Standards
HumansAnimalsPlants and Produce
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Psychrometrics
VariablesUsing the Psychrometric ChartPsychrometric 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 oF dry bulb
55 oF dew-point
61 oF wet-bulb
60 % rh
relative humidity
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Psychrometric ProcessesHeating, cooling, humidifying,
dehumidifying air-water vapor mixtures
Five basic processes to knowHeat 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 pointstarting 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 temperatureRH goes up!
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Evaporation: wet bulb increase
Dry Bulb Temperature Scale
“Humidity” Scale
start state
end stateIncrease in vertical scale: humidifiedDecrease 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/ft3/lb da
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ReviewName three temperature variablesName three measures of humidityName the two main axes of the
psychrometric chartName the line between fog and moist
airHeating 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
ProductionMechanical Energy LoadsSolar LoadMoisture Balance
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Heat Gain and LossOccupantsLightingEquipmentVentilationBuilding Envelope
Roof, walls, floor, windowsInfiltration (consider under
ventilation)
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Heat LoadsOccupant (animals, people)
Sensible load (e.g. Btuh/person)Latent load (“)
Lighting, W/m2Appliance W/m2Ventilation air (cfm/person or
animal)Equipment (e.g. Btuh for given
items)
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Building Ventilation RateTemperature control Moisture controlContaminants (CO2, dust, NH3)
controlNeed 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 MP SHL THL500 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:
qs + qso + qm + qh = ΣUA(ti-to) + FP(ti-to) + cpρV (ti-to)
Heat inputs = envelope + floor + ventilation
qs – sensible heat qso – solar heat gainqm – mechanical heat sources qh – supplemental heat
U – building heat transfer coeff.P – floor perimeterF – perimeter heat loss factor cp – specific heat of airV – ventilation rateρ – density of air
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Sensible Energy Balance Leads to Ventilation for
Temperature Control. Rearranging:
V = [ qs - ( Σ UA+ FP)(ti-to)] / 0.24 ρ (ti-to)60
V – cfm (equation for English units)
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Mass Balance
=+mp
Material produced
mvi
Material input rate
mvo
material output rate
Moisture, CO2, and other materials use balance equations.
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Moisture Balance
=mair
Ventilation rate
Mwater
Moisture to be removed
Example balance for moisture control removal rate.
(Wi-Wo)Humidity
ratio difference
/
Q = (V / 60) x [ Wr / (Wi-Wo) ]
Q - cfm V – ft3/lbda Wr – lbm / hr W – lbm / lbda
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Find the minimum winter ventilation rate to maintain60% relative humidity inside a swine nursery that hasa capacity of 800 pigs weighing 10 pounds. Insidetemperature is 85 degrees.
Moisture Balance
ASABE D270.5
Nursery Pigs Bldg. T MP SHL THL4 - 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 maintain60% relative humidity inside a swine nursery that hasa capacity of 800 pigs weighing 10 pounds. Insidetemperature is 85 degrees.
• Find moisture production data• ASABE Standards (EP270.5)• Wr = 0.017 lb/hr/pig
• Get psychrometric data from chart• W0 = 0.0005• Wi = 0.0154• V = 14.1
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Moisture Balance
=mair
Ventilation rate
Mwater
Moisture to be removed
(Wi-Wo)Humidity
ratio difference
/
Q = (V / 60) x [ Wr / (Wi-Wo) ]
• 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 theammonia concentration inside a poultry layer barnfrom rising above 20 ppm. Ammonia production in the barn is estimated to be 21.6 cubic feet per hour. Ammonia concentration in theambient air is 2 ppm.
NH3 Balance
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NH3 Solution
• Use volumetric form of mass balance equation• Vp + Vi = Vo• Vp + Qv[NH3]i = Qv[NH3]o• Solve for Qv• Qv = Vp / { [NH3]o - [NH3]i }
• Volumetric NH3 production rate per minute• Vp = (21.6 ft3/hr / 60 min/hr) = 0.36 ft3/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/ft3
• Specific heat of air = 0.24 Btu/lb F• ti – to = 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 = [ qs - ( Σ UA+ FP)(ti-to)] / 0.24 ρ (ti-to)60
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Building Heat Loss
Qb = (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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Insulation
Wall Section -Resitances in Series
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Insulation & Heat Loss
Need R-value for each component
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Qb = (A/R)T x ∆t (Btu/hr)Walls - Qw = (Aw/Rw) x ∆tDoors - Qd = (Ad/Rd) x ∆tCeiling - Qc = (Ac/Rc) x ∆tProceed through all componentsPerimeter is special case
R-value is per unit of length - essentially assumes a 1 ft width along perimeter
Qp = (Lp/Rp) x ∆t
Insulation & Heat Loss
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Qbldg = (A/R)w x ∆t + (A/R)d x ∆t + (A/R)c x ∆t + . . . . .
∆t is the same for all components Qbldg = (Ai/Ri) x ∆t (A/R)Total = (Ai/Ri) sum of all
(area/resistance) ratios for all components of the building i.e. walls, ceiling, doors, windows, etc.
Qbldg = (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 ft2 F/Btu in). R-values for the top 1/3 and bottom 2/3 of the existing wall are 12 and 6 (hr ft2 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) Rfoam = 2 x 6 = 12
New R-values Rupper = 12 + 12 = 24 Rlower = 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 Qafter/Qbefore
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
= 0.139New –
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
VVentilation volumetric flow rate
cfm / WattFan Test Efficiency
Electrical Power Cost
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Calculate Operating CostsDesign Ventilation Rate – 169,700
cfmFan Choices
Brand A – 21,300 cfm @ 19.8 cfm/wattBrand B – 22, 100 cfm @ 16.2
cfm/wattFans operate 4000 hours per yearElectricity cost - $0.10 per kWhCalculate annual operating cost
difference
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Calculate Operating CostsDetermine number of fans required
Brand A - 169,700/21,300 = 7.97Brand B - 169,700/22,100 = 7.68
8 fans required for brand A or B
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Calculate Operating CostsUse EP 566, Section 6.2
Annual cost is for all 8 fans
$923.010.001*$0.10*4,000*19.8
170,40016.2
176,800
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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ReferenceMWPS - 32
Contains ASABE heat & moistureproduction data & example problems
Midwest Plan ServiceIowa State UniversityAmes, IA
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ReferenceMWPS - 1
Broad reference to cover agriculturalfacilities, structures, & environmental control
Midwest Plan ServiceIowa State UniversityAmes, IAwww.mwps.org
STRUCTURES andENVIRONMENT
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 FacilitiesAnimal Manure ProductionNutrient ProductionDesign Storage VolumesLagoon – Minimum Design VolumeReferences
ASAE – EP 384.2, 393.3, 403.3, 470NRCS – Ag. Waste Field Handbook
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Size a Manure Storage1 year storageAbove ground 90’ dia. tank –
uncovered 2500 hd capacity – grow/finish pigsBuilding turns over 2.7 times/yrManure production 20 ft3/finished
animalNet annual rainfall = 14 inches25 yr. – 24 hr storm = 5.8 inches
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Size a Manure StorageUse EP 393, sections 5.1 & 5.3Total volume has 5 components
Manure, Net rainfall, 25 yr-24 hr storm
Incomplete removal, Freeboard for agitation 3000,1357.2*2500*20 ftManureVol
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Size a Manure StorageManure Depth = 21.22 ft.Net rain = 1.17 ft25 yr-24 hr storm = 0.48 ftIncomplete removal = 0.67 ftFreeboard = 1 ftTotal 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 KentuckyCollege of Agriculture Biosystems & Agricultural Engineering