Air Conditioning Processes

 HEATING PROCESS

    The heating process adds sensible heat to the system and follows a constant, horizontal moisture line. When air is heated by a steam or hot water coil, electric heat, or furnace, no moisture is added. Image A illustrates a fan system with a heating coil. Image B illustrates a psychrometric chart for this heating coil.

Image A

Image B

Air is heated from 13°C dry-bulb to 30°C dry-bulb represented by Line A-B. This is the process line for heating. The relative humidity drops from 40 percent to 12 percent and the moisture content remains 3.8 grams of moisture per kilogram of air. Determine the total heat added as follows:

1. Draw diagonal lines parallel to the constant enthalpy lines from Points A and B to the enthalpy scale.

2. Read the enthalpy on the enthalpy scale.

3. Calculate the enthalpy added as follows:

            Total heat at Point B - total heat at Point A = total heat added.

            40.0 - 22.8 = 17.2 kilojoules per kilogram of dry air.

Since there is no change in moisture content, the total heat added is all sensible. Whenever the process moves along a constant moisture line, only sensible heat is changed.

COOLING PROCESS:

        The cooling process removes sensible heat and, often, latent heat from the air. Consider a condition where only sensible heat is removed. Image C illustrates a cooling process where air is cooled from 32°C to 21°C but no moisture is removed.

Image C

Line A-B represents the process line for cooling. The relative humidity in this example increases from 50% (Point A) to 95% (Point B), because air at 21°C cannot hold as much moisture as air at 32°C. Consequently, the same amount of moisture results in a higher percentage relative humidity at 21°C than at 32°C. Calculate the total heat removed as follows:

        Total heat at Point A - total heat at Point B = total heat removed.

        70.5 - 59.0 = 11.5 kilojoules per kilogram of dry air

This is all sensible heat since there is no change in moisture content.

HUMIDIFYING PROCESS

BASIC PROCESS:

        The humidifying process adds moisture to the air and crosses constant moisture lines. If the dry bulb remains constant, the process involves the addition of latent heat only.

        Relative humidity is the ratio of the amount of moisture in the air to the maximum amount of moisture the air can hold at the same temperature and pressure. If the dry-bulb temperature increases without adding moisture, the relative humidity decreases. The psychrometric charts in Image A and B illustrates what happens.

Image A

Image B

Referring to Chart No. 2 (Image A), outdoor air at -18°C dry-bulb and 75% RH (Point A) contains about 0.55 grams of moisture per kilogram of dry air. The 0.55 grams of moisture per kilogram of dry air is carried over to Chart 1 (Image B) and a horizontal line ( constant moisture line ) is drawn.

    The outdoor air (-18°C at 75% RH) must be heated to a comfortable indoor air level. If the air is heated to 21°C, for example, draw a vertical line at that dry-bulb temperature. The intersection of the dry-bulb line and the moisture line determines the new condition. The moisture content is still 0.55 grams of moisture per kilogram of dry air, but the relative humidity drops to about 4.5%  (Point A, Figure B). This indicated a need to add moisture to the air. Two examples of humidifying process are discussed below for your better understanding.

EXAMPLE 1:

    Determine the amount of moisture required to raise the relative humidity from 4.5 percent to 35 percent when the air temperature is raised from -18°C to 21°C and then maintained at a constant 21°C.

Image C

Image C provides an example of raising the relative humidity by adding moisture to the air. Assume this example represents a room that is 9 by 12 meters with a 2.5 meter ceiling and two air changes per hour. Determine how much moisture must be added to raise the relative humidity to 35 percent (Point B).

        To raise the relative humidity from 4.5 percent (Point A) to 35 percent (Point B) at 21°C, the moisture to be added can be determined as follows:

1. The moisture content required for 21°C air at 35 percent rh is 5.5 grams of moisture per kilogram of dry air.

2. The moisture content of the heated air at 21°C and 4.5 percent rh is 0.55 grams of moisture per kilogram of dry air.

3. The moisture required is :

                        5.5g/kg - 0.55g/kg = 4.95 grams of moisture per kilogram of dry air.

Line A-B, Image C , represents this humidifying process on the psychrometric chart.    

The space contains the following volume :

                        9m x 12m x 2.5m = 270 cubic meters

Two air changes per hour is as follows:

                        2 x 270 cubic meters = 540 cubic meters per hour

                        or

                        540 / (60 x 60) = 150 liters per second

        This amount of air brought into the room, heated to 21°C, and humidified. Chart no 2 (Image A) illustrates that outdoor air at - 18°C has a volume of 0.712 cubic meters per kilogram. The reciprocal of this provides the density or 1.404 kilograms per cubic meter. Converting the cubic meters per hour of air to kilograms per hour provides :

                        540 cubic meters per hour x 1.404 kg/ cubic meter = 758.2 kg of air per hour.

For the space in the example, the moisture that must be added is : 

                        758.2 kg/hr x 4.95 g/kg = 3753 grams

                                                               = 3.75 kilograms of water per hour.

EXAMPLE 2:

        Determine the moisture required to provide 24°C air at 50 percent rh using 10°C air at 52 percent rh.

In this example, assume that 4700 liters of air per second must be humidified. First, plot the supply air Point A, Image D, at 10°C and 52 percent rh.

Image D

Then, establish the condition after the air is heated to 24°C dry bulb. Since the moisture content has not changed, this is found at the intersection of the horizontal, constant moisture line (from point A) and then vertical 24°C dry-bulb temperature line (Point B).

The air at Points A and B has 4.0 grams of moisture per kilogram of air. While the moisture content remains the same after the air is heated to 24°C (Point B), the relative humidity drops from 52 percent to 21 percent. To raise the relative humidity to 50 percent at 24°C, find the new point on the chart at the intersection of the  dry-bulb line and the 50 percent rh curve or Point C. The moisture content at this point is 9.3 grams of moisture per kilogram of dry air. Calculate the moisture to be added as follows:

                9.3 g/kg - 4.0 g/kg = 5.3 grams of moisture per kilogram of dry air.

    Line B-C in Image D represents this humidifying process on the psychrometric chart.

    At 24°C and 21 percent relative humidity, the psychrometric chart shows that the volume of one kilogram of air is about 0.847 cubic meters. There are two ways to find the weight of the air. One way is to use the volume to find the weight. Assuming 4700 liters (4.7 cubic meter) per second of air :

                        4.7 cubic meter per second/ 0.847 cubic meter per kg = 5.55 kg of air per second.

    The other way is to use the density to find the weight. The reciprocal of the volume provides the density as follows:

                        1/ 0.847 cubic meter per kg = 1.18 kg per cubic meter.

    The weight is then :

                        4.7 cubic meter per kg x 1.18 kg per cubic meter = 5.55 kg of air per second.

    If each kilogram of dry air requires 5.3 grams of moisture, then the following moisture must be added:

                        5.55 kg/s x 5.3 g/kg = 29.4 grams of moisture per second.

    Thus, a humidifier must provide 105.8 kilograms of water per hour to raise the space humidity to 50 percent at 24°C.

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