SWINE
FACILITY VENTILATION: THEORY AND APPLICATION
North Carolian State University
Biological and Agricultural Engineering
Raleigh, NC 27606
Livestock and Poultry Department
Clinton, NC 28329-0318
Ventilation is the proper movement of air through a building,
replacing the stale air inside with fresh air from outside.
The definition of ventilation has been repeated many times in the classrooms, in research projects and on farms. The definition is simple, but the actual management of the process presents a problem for swine facilities. The definition mentions the movement of air through the building not the management of air in the building. The object of swine facilities ventilation is to replace the air in the building. The problem occurs when it comes to removing the air and replacing it without causing stress on the animals. It is hard to control ventilation, especially on extremely hot or cold days. It is important to focus on the basic purpose of ventilation. It is easy to over look these items on a daily basis and experience problems with facility ventilation. A review of the five basic reasons to ventilate will help focus on the problem.
1. To remove excess heat
2. To remove excess moisture
3. To minimize dust
4. To limit the buildup of harmful gases
5. To provide oxygen for respiration
The picture below shows a graphic image the water holding ability of warm air.
Incoming heated air expands to accept the moisture from the air in a swine facility. The moisture holding capacity of air nearly doubles for every 20-degree rise in temperature. The goal is to use the warm air like a sponge to soak up the excess moisture and transport it outside the building. We would like to maintain the relative humidity level between 50 and 70 percent
Theoretically as cold air moves through the room along the ceiling, it is heated up. The cold air temperature is increased (cut in half) approximately every four feet of entry in the building. Cold dry air enters the building and equal amount of warm moist air exits the building. The temperature and moisture level of the air exiting the building is greater than it was when it entered the building. Heat, excessive moisture, pathogens, and gases are expelled from the building. The example below explains this concept.
Proper static pressure rates will attach the incoming air stream to the ceiling and warm it up as it approaches the center of the room. Every .01 increase in static pressure will move the incoming cold air approximate 2 feet closer to the center of the room. An increase in static pressure increases air velocity. The table below illustrates the relationship between static pressure and air velocity. Setting the static pressure properly will decrease the wind chill on the pigs at floor level. The incoming air will have enough time to warm up and slow down before reaching the floor. The correct static pressure setting directs the air stream across the ceiling to heat up and absorb moisture. The table illustrates the relationship between static pressure and airflow in a 24-foot wide farrowing room.
|
Table: Static Pressure/Air Velocity |
||||
|
.03 |
600 fpm |
= |
6.8 mph |
|
|
.04 |
= |
700 fpm |
= |
7.9 mph |
|
.05 |
= |
800 fpm |
= |
9.1 mph |
|
.06 |
= |
900 fpm |
= |
10.2 mph |
|
.07 |
= |
1000 fpm |
= |
11.4 mph |
· Does this high air speed create a cold draft on pigs?
The following picture shows the cause and effect of proper static pressure. As air moves into the facility along the ceiling, it warms up and slows down. The picture illustrates that a 10 mph intake will slow down to less than 1 mph before it reaches the pigs area when the static pressure is correct.
·
Moisture Removal
and Ventilation
It is important to calculate the proper moisture removal rates for a swine facility. If the proper moisture removal is maintained, many health problems are avoided. The picture below shows the accumulation of moisture in a facility and its removal.
Explanation of the Picture
How moisture
is removed:
1. Cold outside air carries .0003 lb./water in 11 cubic feet.
2. The outside air is –200 F. with 100% RH.
3. Outside air is warmed and expanded after entering the building.
4. The expanded air is now 60 degrees at 75% RH
5. The air is now 13.3 cubic feet and holds .00831 lb./water
Inside air: .00831 Lb./water
Outside air - .00030 lb./water
Difference .00801 lb./water
.00801 ÷ 13.3 cubic ft/air = .0006 lb
.0006 lb. of water removed for every cubic foot air exchanged.
Relation of Moisture
Removal and the Rate of Ventilation
Is moisture removal in your facilities acceptable? If your air exchange rates are not meeting the minimal ventilation rate (MVR), then the answer is, “No”.
A 120-lb. pig produces approximately .19 lb. of water per hour. If .0006 lb. of water were removed per cubic foot air then, each pig would need 320 cubic feet of air per hour to remove the moisture he produces. (.19 lb/hr/pig/. 0006 lb./cubic foot = 320 cubic feet per hour per pig)
This would be equivalent to 5.3 cubic feet per minute per pig. (320 cubic feet ÷ 60 min/hr = 5.3 cubic feet per min per pig) A 1000-pig unit with 120 lb. pigs would need 5,300 cubic feet per minute for moisture removal.
The table below illustrates the moisture production tables for various size pigs.
Controlling Ventilation Rate
Exhaust fans move air as a capacity of cubic feet of air moved per minute (cfm). Exhaust fans create a negative static pressure in a building. Static pressure and the amount of air inlet area work in conjunction. A decrease in the area of opening at an inlet will increase static pressure and inlet jet velocity. On the other hand, an increase in open inlet area will decrease the static pressure and the inlet air speed. In either case, the ventilation rate is constant as long as the function of the fan is not affected.
However, a fan works at optimum when it is fed air at specific level of static pressure. An increase in fan speed above optimum increases static pressure when inlet area is constant. The air inlet is too small and the fan motor is loaded harder to create a higher static pressure to move in enough air to feed the fan. This is considered operating past the point of diminishing ventilation returns. The graph below illustrates the cfm/static pressure relationship. In order for ventilation system to hold static pressure in the proper range, inlet area must be planned and controlled. Unplanned inlet openings like leaky doorways and holes into adjoining rooms should be sealed. Some unplanned openings create airflow that is not fresh. Other unplanned openings may provide fresh air, but they reduce the ability of fans to maintain proper static pressure at low speeds. Fans have little capacity to maintain proper static pressure in a state of poor facility maintenance. It is also not good when air velocity is increased to maintain a workable ventilation rate. The extra air movement results in more stress on the pigs. It is better to manage the unplanned openings and reduce pig stress.
The main objective in warm/hot weather ventilation is to remove excess heat from the animals. Proper removal of heat also reduces excess moisture in the building. Two types of heat that affect housed animals are sensible heat and latent heat. Sensible heat is driven by a temperature difference. For example, when heat is added with a gas heater, a temperature difference is created. Latent heat exchange affects the physical properties of the animals without changing the surrounding room temperature (dry bulb temperature). For example, when a sow is “dripped”, her physical temperature changes without changing the air temperature around the sow.
The methods of heat transfer are important concepts to controlling heat in a swine facility.
1. Conduction: Transfer of heat from one object to another by direct contact. (Example: warm pigs, cold floor)
2. Convection: Transfer of heat from an object to the air around it. (Example: warm pigs, cool air)
3. Radiation: Transfer of heat from an object to the animal without direct contact. (Example: hot tin warmed by the sun, pig inside the building)
Cooling swine in the summer in eastern North Carolina is a hard job. The best method cools by moving large volumes of air across pigs with tunnel ventilation or circulation fans. When the pigs’ temperature and the circulating air approach the same temperature, the effective cooling potential is greatly reduced. When this happens, the air velocity must be supplemented with evaporative cooling (example, cool cells) to lower the air temperature. The lower air temperature can absorb excess body heat and expels it outside.
The chart below illustrates the effect of air velocity to cool pigs. As air blows across pigs, heat is given off to the cooler air. The heat from metabolism is reduced and pig feels cooler. The effective environment the pigs experience seems temperature reduced.
Using evaporative cooling a ventilated swine facility reduces air temperature, but the relative humidity is increased. Evaporation will decrease temperature by 1 degree F. for every 2.3 to 2.5% increase in relative humidity. Evaporative cooling uses the latent heat to convert liquid water to vapor. The process of evaporation removes the sensible heat from the air. The potential for cooling the air with evaporative cooling is the difference between the dry bulb and the wet bulb. The potential for cooling a facility is around 15-20 degrees F. (Formula: dry bulb – wet bulb)
The following chart illustrates the relationship of relative humidity to the dry and wet bulb temperature. The relationship explains the difficulty involved in regulating temperature in a swine facility on a hot summer day. It is difficult as the environment is constantly changing.
The next chart graphically represents the relationship between the air temperature and relative humidity. Warm air expands and can hold more moisture. As temperature increases, the relative humidity decreases.
Adding evaporative cooling reduces the air temperature. The following chart illustrates the amount of cooling that can be expected from evaporative cooling in a finishing building. As the incoming air becomes more humid there is less potential to cool the animals with evaporative cooling.
Consider the amount of heat produced in a finishing barn by the metabolic processes of every pig.
