Outline of Today’s Talk
Ventilation Systems Basics
Fan Selection Using BESS Lab Performance Test Results
Ventilation Requirements for Pigs
◦ Cold, Mild, Hot Weather
◦ Different Production Stages
Air Distribution in the Building
…Using Homework Examples
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Why Ventilate?
Pig well-being ◦ Pig comfort
Delivery of major inputs ◦ Air
◦ Water
◦ Feed
Removal of products of metabolism ◦ Gases
◦ Particulates
◦ Manure
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Ventilation Design What do we do to consistently maintain an optimum environment? ◦ Engineering design
How do we know if our design is successful? ◦ Facility commissioning
Will a properly designed and commissioned system always work? ◦ Facility instrumentation, observation, and maintenance
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Fan Performance
Ventilation rate (cfm) depends on air pressure (inch water) against which the fan works
Variation in airflow rates among 0.9-m (36 inch) and 1.2-m (48 inch) fans, as measured with the FANS
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18Static Pressure, inch WC
Fan
airf
low
rate
, cfm
48" fans
36" fans
1 cfm = 1.7 m3/hr
1 inch WC = 248.9 pa
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Ventilation Rate
How much air are these fans actually moving?
All fans are different – even “certified” fan models
But let’s look at performance measurement
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Handout 1: Fan Selection Method from BESS Lab Database
Given: A desired ventilation rate for a specific task. For example, 2,000 cubic feet per minute (cfm) at 0.10 inch static pressure (SP) is needed for 1,000 newly weaned pigs (12 lb body weight).
Required: Find a 14” diameter propeller fan from the BESS database (bess.illinois.edu), which delivers no less than the desired airflow at the specified building static pressure. The fan must not have any sort of steel blades or housing because of concern for corrosion.
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Solution Method
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1. Go to the BESS lab website bess.illinois.edu 2. Select “Performance Tests” from the tabs at the top of the page 3. Select “Current Performance Tests” from the tab to get recent data 4. Select “60 Hz (North America)” for a U.S. application 5. Use the drop-down tabs to select: All Manufacturers, and choose a 14”
diameter, then press Submit 6. Review the fans. For example, look at the Exacon 07479e (Model
AF14LP) and note its Air Flow at 0.10 in SP is 2480 cfm. Click on the Test # for that model fan to open a fan test summary sheet. (see next page for the fan test summary format from BESS).
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Note that it has a plastic housing and plastic shutters (not metal). It has a wire guard and a discharge cone. It has a 6-blade plastic propeller. It is providing 2,480 cfm at 0.10 in. H2O (static pressure), running 1554 rpm, using 230.4 volts, 1.35 Amps, and 306 Watts. The fan efficiency (VER, ventilation efficiency ratio) is 8.1 cfm/Watt.
Fan Performance Curves: 48” fans in one poultry barn*
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8,000
10,000
12,000
14,000
16,000
18,000
20,000
22,000
0.00 0.05 0.10 0.15 0.20 0.25
Air
flow
(CFM
)
Static Pressure (inches of water)
Source: D.G. Overhults University of Kentucky
On-site Determination of Fan Performance Curve
Upstream (inside)
Downstream (outside)
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Comparison to BESS Lab Test Data*
bess.Illinois.edu – free fan performance data
EMILI 2015
13
0
20
40
60
80
100
120
Brand A DD Brand C BD Brand B DD Brand D BD
% of L
ab Tes
t Airflow
36” fans DD = direct drive BD = belt drive
Source: D.G. Overhults University of Kentucky
University of Illinois
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Dirty Fans and Shutters
1/8 inch of dirt/dust can cause up to a 40% reduction in fan and shutter air flow.
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Handout 2: Ventilation Rate Requirements for Swine
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Given: We need to determine the proper amount of ventilation for different size and type of pigs. We will use this information to determine building minimum and maximum ventilation rates for each type of production facility.
For example, how many cubic feet per minute (cfm) for 1,200 newly weaned pigs (12 lb body weight)?
Or how many cfm are required for 1,200 finishing pigs?
Required: Use the recommended rates (next slide) as a basis. Note, these are in cfm/pig.
Air Flow Requirement For Each Pig, cubic feet of air per minute (cfm)
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Cold Weather Rates (for moisture control)
Fully slo)ed
Partly slo)ed Solid floor Rate for odor
control Mild
Weather Hot
Weather Sow and Li9er 10 17 20 35 80 325 Pre-‐nursery pig
(12-‐30lb) 1.0 1.6 2 3.5 10 25
Nursery pig (30-‐75lb) 1.5 2.5 3 5 15 35
Growing pig (75-‐150 lb) 3.5 5.5 7 10 24 75
Finishing pig (150-‐220+ lb) 5 8 10 18 35 120
GestaJng sow (325 lb) 6 10 12 20 40 150
Boar (400 lb) 7 12 14 24 50 180
Examples:
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1. How many cfm are needed for 1,200 newly weaned pigs at the cold weather rate? Answer: depends on floor type. For fully slotted floor we need 1 cfm/pig thus 1,200 cfm; for a solid floor we need 2 cfm/pig, or 2,400 cfm total.
2. How many cfm are needed for 1,200 finishing pigs at the cold weather rate? Answer: also depends on floor type. For a fully slotted floor: 5 cfm/pig or 6,000 cfm. For a solid floor: 10 cfm/pig or 12,000 cfm.
3. How many cfm are needed for 1,200 finishing pigs at the hot weather rate? Answer: 120 cfm/pig, or 144,000 cfm
Conclusion: Ventilation rate requirements vary tremendously with the size of the animal and to a lesser extent with the type of housing. We need to determine these values in order to properly design a ventilation system, whether it is natural ventilation, mechanical ventilation, or a hybrid combination of both methods.
We calculate the fresh air requirement by season:
In winter, just enough air to remove the moisture produced by the pigs
In summer, enough air to remove the heat produced by the pigs
Between times, a smooth transition of air exchange
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Bring in the Fresh Air Fresh air must mix
Inlets need to be properly sized
Inlet air velocity is the key
The role of static pressure is critical
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Air speeds at the pig level are very important
Small pigs – less than 0.15 m/s (30 feet/minute)
Larger pigs – less than 0.25 m/s in winter (50 feet/minute)
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Fan Performance and Building Static Pressure
Ventilation rate depends on the pressure against which the fan works
Fan creates static pressure (if building is tight
Static pressure determines air velocity through inlets
We want 800-1000 feet/min inlet velocity
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Measuring Static Pressure Outside the building
Inside the building
Very Important! Measuring Static Pressure: An Inclined Manometer, or Digital Manometer. Scale: mm H2O or Pa
Pd
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The Air Inlets Determine:
Fresh air placement in the room
Proper mixing of fresh air with stale air
Air speeds at pig level
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For good air mixing, fresh air should enter the room in a “jet” – 800 to 1,000 feet per minute
Outside the building
Inside the building
LOW SP: more air from fan, lower inlet air velocity
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High SP difference: less air from fan, higher inlet air velocity
Outside the building
Inside the building
HIGH SP
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Manage the numer of inlets based on size of pigs (i.e. ventilation rate needed
Not all openings needed when pigs are small
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Handout 3: Fresh Air Inlets Area Determination
Given: Determine how much inlet area is needed for a required ventilation rate.
The formula (in English units) is:
A = 144*Flow/Velocity, where A is square inches, Flow is in cfm, and Velocity is in feet/min. Design values for Velocity are about 800-1,000 fpm so this simplifies to
A=0.144*Flow, or 1 ft2 /1000 cfm
(The 144 converts square feet to square inches). You can also remember it is 1 square foot of opening needed per 1000 cfm.
Required: You must figure the area based on the ventilation rate provided. Calculate the total opening area needed. Then, you should be able to figure out how big an opening is needed for different inlet lengths.
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Examples of Calculations 1. Given a 1,200 head finisher building with a winter time ventilation rate of 2,400 cfm, how many square inches of opening do we need? Answer: Area = 0.144 x 2,400 = 345.6 in2 (2.4 ft2)
2. Given 12 ceiling inlets, each with 4 ft (48 inches) of opening, approximately how wide should be the opening in this example? Answer: Width of Opening = Area/Length = 345.6 in2 /(48 in./inlet x 12 inlets) = 0.6 in. Very small! (Shut down some openings!)
3. For the same facility at summer conditions the required ventilation is 144,000 cfm. How much opening do we need now? Answer: Area = 0.144 x 144,000 = 144 ft2 = 20,736 in2
4. Will the 12 ceiling inlets have enough opening area, if they have 8 ft per inlet? Answer: Width = 20,736 in2 /(96 in./inlet x 12 inlets) = 18 inches….Very large! We need more inlets or different design.
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Conclusion
Since the building’s ventilation rate requirements vary tremendously with the size of the animal, so will the required inlet area.
This is a key problem in mechanically ventilated systems – to provide the proper inlet area to promote good air exchange and good mixing of fresh air with indoor air, without drafts and without dead spots.
Understanding the role of static pressure difference is helpful for proper inlet design and operation
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