Ventilation Rate

Ventilation is the movement of air through a building. The movement may be caused by natural convection, or wind, or by forcing air movement with fans. Air is used to remove moisture, reduce the temperature if the outside air is cooler than the inside, and replace gases inside the building with outside air. The minimum ventilation rate is determined by which purpose of ventilation has the greatest requirement. For livestock buildings, if the ventilation rate is sufficient to remove excess moisture, it usually will be more than enough for the other needs.

In this section, a procedure involving the use of a psychrometric chart is used to make estimates of the amount of ventilation required to remove the excess moisture from a livestock building. For buildings used for other purposes, you must determine which aspect of the environment is critical, and base your calculations on that factor. Appendix contains values for moisture and heat released by different animals and the heat produced by mechanical equipment.

When outside air is introduced into a building, it must be uniformly distributed within the building and mixed with the building air to prevent drafts. One method commonly used is to introduce the air through small openings distributed around or throughout the building. Another method is to use heat exchangers. A heat exchanger uses the heat of the exhaust air to preheat incoming air, thus reducing the chilling effects of drafts and reducing the amount of heat that must be added. Actual airflow rates required for animal buildings will vary with the type and size of animal, management system, type of floor and floor drainage, and the number of animals in the building.

As moisture evaporates into the air it changes the physical properties of the air. The amount of change can be determined by locating the state points for both conditions. The numerical differences in the moisture content and the total heat of both conditions provide the information used to determine the amount of air and the heat required to evaporate the moisture.

Table 23.1. Psychrometric data for ventilation problem.

Properties

Inside

Outside

Difference

Dry bulb (°F)

55

40

Wet Bulb (°F)

48

35

Moisture (lb H2O/lb of air)

0.0054

0.0028

0.0026

Total heat (BTU/lb)

19.00

13.0

6.0

Specific volume (ft3/lb of air)

13.1

12.65

Problem: How much air and heat will be needed to evaporate 2.0 lb of water per hour from within a building when the inside air temperature is 55°F dry bulb (db) and 48°F wet bulb (wb) and the outside air conditions are 40°Fdb and 35°Fwb?

Solution: In this problem we will use values for five of the properties of air for two different air conditions. Constructing a table, Table 23.1 is very helpful and is recommended for solving problems of this type.

The first step is to use the psychrometric chart to find the values for moisture content, total heat, and specific volume for both the inside and outside air. These values are recorded in the columns "Inside" and "Outside." Note: the values used are from the chart in Figure 23.4. Numbers from an ASHRAE chart will be slightly different. An ASHRAE or similar chart should be used when solving a real problem.

The next step is to determine the differences in moisture content and in total heat between the inside and the outside air. These values are recorded in the column labeled "Difference." Values for the differences in temperature and in specific volume are not needed in this problem.

The third step is to determine the amount of air (lb) required to remove the moisture. The amount of air required to evaporate the moisture is determined from the difference in the moisture content of the inside and the outside air. A difference of 0.0026 lb of water per pound of air means that every pound of outside air brought in and raised to inside conditions is capable of absorbing 0.0026 lb of moisture. If each pound of air absorbs the maximum amount of water, 0.0026 lb, and the amount of water that must be removed is 2.0 lb per hour, then the required amount of air (lb) is:

hr 0.0026 lb water 1 hr hr

For these inside and outside conditions, 770 lb of air per hour moving through the building will be needed to remove 2.0 lb of water per hour, assuming that the air absorbs the maximum amount of water.

Before water in liquid form can be absorbed by air, it must be vaporized, and this requires heat. In addition, if the inside of the building is warmer than the outside, ventilation causes a loss of heat from the building. The total amount of heat lost is determined by using the difference in the heat of the incoming and the outgoing air. The amount of heat in the inside air is 19.0 BTU/lb, and that of the outside air is 13.0 BTU/lb—a difference of 6.0 BTU/lb. This amount of heat will be lost with each pound of air ventilated. We have already determined that 770 lb of air per hour will be needed to evaporate the water produced in the building; so the heat loss due to ventilation is:

BTU lb air 6.0 BTU BTU

hr hr lb air hr

For these conditions, 4,600 BTU per hour of heat will be transported outside the building with the ventilation air.

Ventilation fans usually are sized in units of cubic feet per minute. This will require a units conversion from weight to volume. This conversion is made using the specific volume values from the psychrometric chart. When the ventilation fan is exhausting building air (the most common situation), then the inside specific volume is used. If the fan is blowing in outside air, the outside specific volume is used. Because it takes 770 lb of air per hour to evaporate the water produced in the building, the volume of air (ft3/hr) that is required is determined by multiplying the pounds of air per hour by the specific volume. The size of fan required is:

ft3 lb air ft3 ft3

hr hr lb air hr

A fan, or several fans with a combined capacity of 10,000 cubic feet per hour is required to provide enough ventilation to remove the excess moisture from the building.

The values calculated in the previous problem are accurate only as long as the inside and the outside environments do not change. This only occurs for short periods of time because the outside temperature is constantly changing, the respiration rate of the animals changes with activity, and heat exchanges also occur because of conduction, convection, and radiation. Heating, ventilating, and cooling systems must be able to react to change, and if they are required to maintain a constant inside environment, they must have the capacity for the most extreme inside and outside situations.

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