Environmental Systemsand Facilities Planning

Doug Overhults University of Kentucky

Biosystems & Agricultural Engineering

University of KentuckyCollege of Agriculture

Topic OutlinePsychrometrics ReviewEnergy Balances/Loads

Latent heat Sensible heatSolar loads

Insulation Requirements

Topic OutlineVentilation Systems

Rate requirementsSystem requirements

Moisture Control StandardsAir Quality Standards

HumansAnimalsPlants and Produce

Psychrometrics

VariablesUsing the Psychrometric ChartPsychrometric Processes

Psychrometric Chart

Dry Bulb Temperature Scale (axis)

“Humidity” Scale

or axis

State Point

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

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)

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

Humidification: dew-point increase

Dry Bulb Temperature Scale

“Humidity” Scale

start state

end stateVertical line means no change in dry bulb temperatureRH goes up!

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)

Air Density

Dry Bulb Temperature Scale

“Humidity” Scale

Wet bulb line

Humid Volume, 1/ft3/lb da

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 ?

ENERGY AND MASS BALANCES

Energy and Mass Balances Heat Gain and Loss Latent and Sensible Heat

ProductionMechanical Energy LoadsSolar LoadMoisture Balance

Heat Gain and LossOccupantsLightingEquipmentVentilationBuilding Envelope

Roof, walls, floor, windowsInfiltration (consider under

ventilation)

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)

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

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

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

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)

Mass Balance

=+mp

Material produced

mvi

Material input rate

mvo

material output rate

Moisture, CO2, and other materials use balance equations.

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

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

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

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

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

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

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

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

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

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

Insulation

Wall Section -Resitances in Series

Insulation & Heat Loss

Need R-value for each component

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

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

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

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

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

Fan Operating Cost

= ÷W

Power input, Watts

VVentilation volumetric flow rate

cfm / WattFan Test Efficiency

Electrical Power Cost

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

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

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 = $

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.

ReferenceMWPS - 32

Contains ASABE heat & moistureproduction data & example problems

Midwest Plan ServiceIowa State UniversityAmes, IA

ReferenceMWPS - 1

Broad reference to cover agriculturalfacilities, structures, & environmental control

Midwest Plan ServiceIowa State UniversityAmes, IAwww.mwps.org

STRUCTURES andENVIRONMENT

HANDBOOK

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

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

FACILITIES Manure Management Example

Manure Management FacilitiesAnimal Manure ProductionNutrient ProductionDesign Storage VolumesLagoon – Minimum Design VolumeReferences

ASAE – EP 384.2, 393.3, 403.3, 470NRCS – Ag. Waste Field Handbook

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

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

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.

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)

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

Thank You

and

Best Wishes for Success on Your PE Exam ! !

University of KentuckyCollege of Agriculture Biosystems & Agricultural Engineering