Psychrometric Calculations Course

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1.1 Fundamentals of Psychrometrics This tutorial is the first lesson in a series on Psychometrics: Lesson 1.1 review of fundamentals. To begin the study of psychometrics, first we need to have an understanding of some fundamental properties. If you feel comfortable about your understanding of these key concepts, then you can skip this video and go on to lesson 1.2 or 1.3. Some of you, however, might want to take these few minutes for a quick review. By fundamental properties, I mean the following: Mass, Force, Weight, Density, and Specific Volume. In Lesson 1.2 I provide a review of pressure and enthalpy, and finally, the concepts of sensible and latent heat. Then we will proceed to the psychometrics course. First on the list of fundamental properties are mass and force. A common definition frequently given for mass is: a fundamental property of an object-- a measure of the quantity of matter an object contains. Please note that the mass of an object is not affected by any external forces placed upon it. 1.2 Concept of Pressure This is Lesson 1.2. This video is a continuation of fundamentals. 1.2 provides a review of pressure. Pressures of liquids and gasses are of significant importance in HVAC applications. Understanding pressure is fundamental to the study of psychometrics because pressure has a direct effect on the properties of moist air. In order to understand pressure, it is necessary to understand the concept of force. This is because pressure by definition is force per unit of area. Written as an equation, Pressure = Force / Area. Sounds simple, right? Well, actually it is pretty simple. Pressure is a very simple concept. If the force is measured in units of pounds, and the area is measured in units of feet, then the units for pressure would be pounds per square foot, and this would be written as psf. An example in calculating pressure is shown here in example 1.2.1. Assume that we have an open tank and this tank is filled with 3,000 lbs of water. The tank dimensions are 2 feet by 3 feet. What is the pressure exerted on the bottom of the tank? So the solution here is to use the equation for pressure as pressure is equal to the force divided by area. In this example, the force is the weight of the water which is 3,000 lbs. The area is the area on the bottom of the tank, which is 2 feet by 3 feet which equals six square feet. Therefore, the pressure is 3,000 lbs per 6 square feet, which is the same as 500 lbs per square foot written also as 500 psf. Put another way, the water in the tank exerts 500 lbs per every square foot on the bottom of the tank due to the weight of the water. Pressure is a very simple concept but it tends to get confusing due to the variety of applications and units that are used to measure pressure. As in the example, pressure is measure in pounds per square foot. It can also be measure in pounds per square inch, and often we talk about barometric pressure typically given in something called inches of mercury. HVAC systems very often measure pressure in inches per water column, and plumbers like to use a measurement of pressure in feet or feet of of head pressure. Understanding pressure is of great importance in HVAC applications, and different applications measure pressure in different ways and in different units. This slide is to help understand the relationship of measuring pressure. The atmosphere exerts pressure on the earth due to its weight. At sea level, the height of the atmosphere is greater than at higher elevations, for example, Denver. So, the atmospheric pressure is greater in San Francisco than it is in Denver. Pressure gages often measure the difference between the fluid pressure and the atmospheric pressure. So there is a relationship between the pressure measured by the gage and the atmosphere. To understand this relationship, let’s first start with something called zero pressure. The pressure due to the weight of the atmosphere is called atmospheric pressure, which of course is some value above zero pressure, shown as the blue line here in this slide. Pressure measuring devices typically measure the difference between the pressure of the fluid and the atmospheric pressure shown on this slide as the red line. We’ll call that gage pressure. Typically, gages are calibrated to read zero at atmospheric pressure so that the reading you get for the gage pressure is the difference between the measured pressure and atmospheric pressure. Absolute pressure, gage pressure, and atmospheric pressure are related by the equation: P absolute = P atmosphere + P gage. As noted earlier, the pressure exerted by a fluid will depend on the weight of the liquid or the gas. The weight, which is the force exerted, depends on the height of that liquid and the density of the liquid or gas. This makes sense, because if you think about it a liquid that is more dense will exert a greater force than a less dense fluid for the same height of the column of fluid. This relationship is written as an equation P = Density X Height. So there are two columns of liquid of equal height, if they have different densities, they will exert different pressures. If the density is in units of pounds per cubic foot and the height is measured in feet, then the units for pressure are shown in the equation pounds per cubic foot times feet equals pounds per square foot, which equals psf. Let’s consider another example, example 1.2.2. In this example, a 50ft high pipe is filled with water and extends from the condenser on the top floor of a building to cooling tower on the roof. What is the pressure exerted on the condenser due to the water in the pipe? Give the answer in units of psi (pounds per square inch). To solve this, we use the equation for pressure where pressure is equal to the density of the fluid times the height of the fluid. In this case, the density is the density of water is assumed to be 62.4 lbs per cubic foot. The height is given in the exampl as 50 ft. The pressure, then, is 62.4 pounds per square foot times 50 ft which equals 3120 pounds per square foot. But the question was not in psf, but in psi, pounds per square inch. To do so, we simply apply unit analysis to convert square feet to square inches using the factor: one square foot = 144 inches squared. Doing the math, we see that the answer comes out to 21.7 pounds per square inch. Please note, this is gage pressure. Why? Because if we put a gage at the bottom of the column, the pressure would read 21.7 psi. If we were to move that same gage up to the top, we would notice that the reading would go down until the very top, where it would read zero. This is because the gage is calibrated to read zero at atmospheric pressure, so the gage is reading the results of the pressure due to the water only. Let’s look at one more example. Example 1.2.3. In plumbing applications, the terms head pressure, or feet of head is often used as a metric to measure pressure. We often see a conversion factor that plumbers have to use where 2.31 feet of head is equal to 1 psi. The question here is derive this conversion factor. To solve this, we will use the equation pressure is equal to density times height of the column of fluid, where P, pressure, equals 1 psi and D equals the density of water. First, we need to convert one pounds per square inch to pounds per square foot. Also, we’ll rewrite the equation P = density times height as P divided by density to equal height. Now substitute 144 lbs per square foot, which is the same as 1 psi and insert 62.4 pounds per cubic foot for density. This gives us the following: 144 lbs per square foot divided by 62.4 pounds per cubic foot. Using math and a careful check of the units, we see that this equals 2.31 feet. This concludes a basic review of pressure. The next tutorial provides a review of enthalpy and sensible and latent heat. 1.3 Energy, Enthalpy, Sensible and Latent Heat This is the third lesson in the series on psychrometrics. Lesson 1.3 provides a review of energy, enthalpy, sensible and latent heat, and their importance in psychrometrics. Please note this is intended to be a review only of basic fundamentals. Energy and enthalpy are complex topics. The purpose here is simply to give a brief overview. Energy is a challenging and abstract concept to define. Energy has been described as the ability to do work. In building systems, energy consumption is a primary focus to architects, engineers, and of course, building owners. Energy exists in a number of different forms. A body can have potential energy, kinetic energy and chemical energy as some examples. Energy can also be transferred from one body to another. Energy transfer can occur in a number of different ways. This tutorial will limit the discussion to two forms of energy transfer that are of significance in building environmental systems. The first is work, and the second is heat transfer. Work is the effect created by a force. For work to happen, the force has to result in a displacement of the body. Shown as an equation, work is equal to force times distance. The units for work are quite varied. In the SI, or metric, system, work is expressed in joules. A joule is the force of one newton through a distance of one meter. Also, the kilowatt hour is a unit for work. In the IP, or imperial, system, units used for work can be quite confusing. For instance, there is the foot-pound, the btu and even the horsepower-hour is an interesting unit for work. Example 1.3.1: This is an example on calculating work. A contractor has the job of installing an air conditioning unit on the top of a roof. The unit has a total weight of 10,000 lbs and needs to be hoisted up 30 feet. The question is: How much work is done in lifting the unit to the roof. To solve, we will use the equation for work which is work is equal to the force times the distance. Force, in this example, is the weight of the unit. The distance, of course, is going to be 30 feet. The work, then, is 10,000 lbs times 30 feet. The work is 300,000 foot-pound. Power is also an important concept in building systems. Equipment capacity for chillers, boilers, pumps, fans and cooling towers is based on power. Although in each of these instances, the units to describe the power or capacity is very different. Simply stated, power is work per unit of time and written as the equation power= work divided by time. Please note that this is a highly simplified version of a very complex concept. But it is sufficient enough to illustrate the power in our discussion. Just as with work, power has a number of units commonly used depending on the application. In the metric system, the basic unit for power is the watt and also the kilowatt. Please note the watt is based on the definition for power as the watt by definition is a joule per second. In the imperial system, there are a variety of units used for power. These include horsepower, foot-pounds per minute, where 33,000 foot-pound per minute is equal to 1 horsepower. Btu per hour, and this is sometimes written as Btuh, and ton of refrigeration. Example 1.3.2 For the air conditioning unit, calculate the minimum power required if it takes 5 minutes for the unit to reach the roof. Give the answer in units of ft-lb/min, hp and kW. The solution here is to see the equation for power, where the work is 300,000 ft-lb and the time is 5 min. Then the minimum power would be 300,000 ft-lb divided by 5 minutes, or 60,000 ft-lb per min. A couple of useful conversion factors are: 1 Hp = 33,000 ft-lb per min and 1 Hp = .746 kW. We can convert this using unit analysis. 60,000 ft-lb per min times the conversion factor of 1 Hp per 33,000 ft-lb per min gives us 1.82 Hp. Then we can convert 1.82 Hp by multiplying .746 kW per Hp gives us 1.34 kW. As noted earlier, energy can be transferred from one body to another. Work is a form of energy transfer. Another form is heat transfer. For purposes of this tutorial, I’ll use Q to designate heat transfer. In building environmental systems, heat transfer is very often calculated in units of Btu which stands for British Thermal Unit. One Btu is the amount of heat required to raise the temperature of 1 pound of water by 1 degree fahrenheit. Finally, this leads us to enthalpy. Enthalpy is of utmost importance in the study of psychrometrics. In this particular view of a standard psychrometric chart published by ASHRAE, lines of constant enthalpy are shown as diagonal lines from upper left to lower right. So, what is this property called enthalpy? Well, technically speaking enthalpy is a combination of a body’s internal energy plus the product of its pressure times volume. Enthalpy is a measure of the total energy of the system. You might hear the definition of enthalpy as the property of a body that measures its heat content. Scientifically speaking, this is not correct but it is used so often that for the purposes of discussing psychrometrics, we will accept this definition. The units for enthalpy in the IP system is the Btu. Specific enthalpy is Btu per lb. In the metric or SI system of units, the unit of measure for enthalpy is the Joule. Sensible and latent heat: When heat is added to an open container of water, at conditions such as for example an atmospheric pressure at sea level, a noticeable change occurs. Adding heat to the water results in a temperature increase. This is called sensible change. In sensible change, the temperature rises as the enthalpy of the container of water is increased. The increase in enthalpy is called sensible heat. At some point, adding additional heat does not show an observed increase in temperature as the liquid gradually changes into a vapor state. This change is referred to as vaporization, and for water, we call this boiling. Although the temperature does not change the enthalpy is still increasing. This is what is referred to as latent heat. The entire process of sensible change and latent change is the total change and the enthalpy change is the sum of the sensible heat plus the latent heat. Specific Heat: Specific heat of a substance is the amount of heat per unit of mass required to raise the temperature by one degree. In the Imperial System of units, specific heat is in units of Btu per lb per degree fahrenheit. Since psychrometrics is the study of moist air properties, it is helpful to understand the specific heat of the air and water. For water, the specific heat is 1 Btu per pound-degree Fahrenheit. The specific heat of air is .24 Btu per pound-degree Fahrenheit. Please note that this is about one quarter of the value for water. This difference in specific heat between water and air has significant impact on the design of HVAC systems since these systems often involve the removal or addition of heat from air or water. The sensible heat equation is used extensively in HVAC systems. The sensible heat equation quantitatively describes the heat that is added or removed from a substance during sensible change. It’s shown as equation Qs= Mass flow rate times the specific heat times the change in temperature. Shown another way, Qs= m x C x (T2-T1). The dot designates that it is a rate. Units are very important here. When shown to represent a rate, such as Btus per hour, the mass flow rate needs to be in pounds per hour, specific has to be in Btu per pound-degree fahrenheit, and the temperature change is in degrees Fahrenheit. Using the sensible heat equation is very useful, but its limited to processes that involve a sensible change only. The heat added or removed in HVAC processes can also be determined by the enthalpy equation. This is important because it is very useful in understanding complex processes that involve both sensible and latent heat change, such as in dehumidification. Written as an equation, Q is equal to the mass flow rate times the change in enthalpy. To find Q in Btus per hour, the mass flow rate is in pounds per hour and the enthalpy is in Btu per pound. The following example, example 1.3.3 illustrates how sensible heat equation can be used in determining airflow. A 5 kW electric resistance heater is installed in a duct. Measuring the temperature of the air before and after the heater shows the air temperature has an increase of 30 degrees fahrenheit. Use the sensible heat equation to determine the airflow rate in cubic feet per minute or CFM. The first step in solving this problem is to convert the heater capacity from kW to Btu/hr. 5kW times the conversion factor of 3410 Btuh/kW results in 17,050 Btuh. Please note Btuh is the same as Btu per hour. You will see both of these used commonly in describing HVAC applications. The next step is to apply the sensible heat equation. In step 1, the heat transfer from the heater to the air was found to be 17,050 Btuh. This is Q dot in the equation. Since the fluid is air, the specific heat shown here as C is the specific heat of air. We will use 0.24 Btu/lb-degree fahrenheit in this example. The temperature difference is the leaving temperature minus the entering temperature. And now, using some algebraic rearrangement we can solve for the mass flow rate of the air. Solved by entering the values and doing the math, the airflow rate is found to be 20,368 pounds per hour. But pounds per hour is not very practical in HVAC applications, so the final step is to use the average density for air and some unit analysis. So, 20,368 pounds per hour times 1 hour per 60 minutes times 1 cubic foot per 0.075 pound of air, which is the average density of air, will give us 526 cubic feet per minute, also known as CFM. Lesson 2.1: Properties of Moist Air and the Psychrometric Chart Moist air properties: Atmospheric air is the mixture of both dry air and water vapor. It is this mixture that is referred to as moist air. Psychrometrics is the name given to the study of the air and water vapor mixture. Seven properties of atmospheric air are shown on the psychrometric chart. These properties are: dry bulb temperature, wet bulb temperature, dew point, humidity ratio, relative humidity, specific volume, and specific enthalpy. Dry Bulb and Wet Bulb Temperature: So, let’s take a look at a basic thermometer. If the thermometer is measuring the air temperature and the bulb of the thermometer is dry and open to the air, the temperature rating is called the dry bulb temperature. If the bulb is surrounded by a water soaked wick and moving through the air, then the temperature will be different than the dry bulb temperature and that is what is called wet bulb temperature. The difference between the dry bulb reading and the wet bulb readings of the same air is referred to as wet bulb depression. A tool called a sling psychrometer or sling thermometer is used to measure both dry bulb and wet bulb temperature of an air sample. It consists of two thermometers side by side, where one bulb is covered by a wet cloth and the other is dry. The psychrometer is rotated through the air to get the readings. Shown here is an example of a simple and very easy to use sling psychrometer that is manufactured by forestry suppliers. You can see other examples of sling psychrometers online. The Dew Point: If air is cooled at a constant pressure, there is a point at which the water vapor in the air will condense. That is, change from a vapor to a liquid state. Evidence of this change is seen by moisture forming on surfaces. This temperature is referred to as the dew point. Humidity Ratio: Humidity ratio is also known as moisture content. The formula for humidity ratio is shown here and looks pretty complicated. Basically, humidity ratio is simply the weight of the water vapor per pound of dry air. Humidity ratio gives us an indication of the actual weight of the water in the air. Presented another way, if we could take a pound sample of air and squeeze out all of the water, the humidity ratio would be the amount of water in pounds relative to the amount of dry air in pounds. Since air has typically a relatively small amount of water vapor, in terms of pounds, another expression for humidity ratio is commonly used called grains of moisture. There are 7,000 grains of water in a pound. Relative Humidity: Relative humidity is discussed more often in everyday life than humidity ratio, and is a little easier for most people to understand. Relative humidity is the ratio of the actual water vapor pressure in the air relative to the water vapor present if the air were saturated. Saturated air means that the condition of the air when it contains the maximum amount of water vapor that in can hold. The maximum amount of water vapor that the air can hold is dependent on the air temperature at a given atmospheric pressure. In other words, relative humidity is the amount of moisture the air contains relative to the amount of moisture it could contain, expressed as a percentage. To illustrate relative humidity, consider a six ounce container. If the container contains six ounces of water, the container is full and therefore we could say it is saturated. If only three ounces of water were in the six ounce container, the container would be half full, or fifty percent. Specific Volume and Specific Enthalpy: Specific volume was presented earlier in lesson 1.1:review of fundamentals. Specific Volume is the inverse of density, or the volume of air per unit weight of dry air. In the imperial system of units, shown on the psychrometric chart earlier, it is in units of cubic feet per pound of dry air. Specific enthalpy on the psychrometric chart is often referred to as heat content of the air per unit weight of air. The enthalpy of the air is actually the sum of the individual enthalpies of the water vapor plus the enthalpy of the dry air. This includes both the sensible heat of the dry air and the latent and sensible heat of the water vapor. The properties of moist air that were just presented can be illustrated in tables or maybe more conveniently in graphical form. The graphical representation of these properties is the psychrometric chart. At first glance, the psychrometric chart can be confusing and intimidating. We will go through the chart and explain what each of the lines and curves shown represent. The chart shown here is the ASHRAE chart. Notice this chart is for normal temperatures and a barometric pressure of 29.92 inches of mercury. This is what is considered to be sea level. Recall that air density changes at different elevations. Because of this, the air properties will be different depending on the atmospheric pressure. If you are working on a project at higher elevations, you will need to use a chart based on lower pressure. These charts are available from ASHRAE and other various sources. Another item to note is that the chart shown here is in the IP system of units. Of course, charts based on metric or SI system are also available. Any condition of the air is shown as a point on the psychrometric chart. So, for instance, 75 degrees dry bulb and 50% relative humidity is shown here. Each property is represented by a line or a curve as with relative humidity. Once a condition is found using any two properties, the other properties can be read directly off the chart. So, at 50% relative humidity, 75 degrees dry bulb, we can find the humidity ratio, specific volume, wet bulb, specific enthalpy and dew point. On the psychrometric chart, the dry bulb temperature is shown on the bottom axis. Notice that this chart has dry bulb temperature range from 32 degress fahrenheit at the farthest left and goes up to 100 degrees fahrenheit for the highest value. Lines of constant dry bulb are shown as vertical lines. The lines of dry bulb temperature on this chart are in 5 degree increments. Humidity ratio values are shown on the far right of the chart. Since this chart provides humidity ratio in pounds per moisture per pound of dry air, the values are quite small, ranging from zero and increasing in increments of 0.0002 pound of moisture per pound of dry air. Relative humidity is shown on the chart as these curved lines. The farthest to the left is the line of saturation, or 100% relative humidity. This chart shows lines in increments of 10% with the lowest at 10% relative humidity. Lines of constant specific volume shown here in cubic feet per pound of dry air are these lines that extend from the upper left to the lower right. The values begin at 12.5 cubic feet per pound of dry air at the left and continue up to 15.3 cubic feet per pound at the far upper right. Enthalpy is shown here in Btus per pound of dry air. Because the enthalpy and wet bulb lines are very close, the enthalpy values are given in both the left sloped line shown here and the bottom axis. This allows the user of the table to better read the enthalpy values. Wet bulb temperature lines are shown almost parallel to the lines of contant enthalpy. These extend from the saturation curve down to the left. This chart shows wet bulbs in degrees fahrenheit ranging from 32 degrees in the far left to 94 degrees in the far upper right and are in increments of 5 degrees fahrenheit. Moisture on a surface is condensed from the air surrounding it if the surface temperature is less than the dew point. Dew point is the temperature which the air is saturated with water vapor. Cooling the air past this temperature causes the moisture to condense. Condensation and dew point are important to understand within the built environment. Condensation on window surfaces, pipes, and ductwork must be avoided since this water can result in damage to the building elements. Please note at saturation, or 100% relative humidity, the dry bulb wet bulb, and dew point temperatures are the same. Psychrometrics Lesson 2.2: Air Conditioning Processes and the Psychrometric Chart The purpose of air conditioning equipment is to change the condition of the air from some entering condition to a leaving condition. We refer to this change of condition in the air sample as a process. In this sectional view of an air-handling unit, there are a number of air conditioning processes that can occur. The outside air enters the unit as an outside air condition and is mixed with return air from the spaces. Both the outside air and the return air will undergo a process change due to the mixing process itself. The mixed air may experience further changes depending on the processes that occur next. It could experience a change in dry bulb temperature due to heating or cooling as it passes over the heating and cooling coils, it may experience a change in moisture content if moisture condenses out, and of course, a change in enthalpy depending on the application. In this tutorial, we will discuss these processes and how they are represented on a psychrometric chart. On a psychrometric chart, a process, or change in properties of air, can be represented by drawing a line from the entering condition to the leaving condition. The first process that we will look at is sensible heat change. In sensible heating, or cooling, heat is added or removed resulting in a change in the dry bulb temperature. Since moisture is not added or removed, this process follows a line of constant humidity ratio. Sensible heating results in an increase in the dry bulb temperature. Notice also some other changes in properties that will occur. The relative humidity is reduced, and the specific volume is increased. In addition, we see the specific enthalpy also increases. Sensible cooling is the process in which the dry bulb temperature is reduced but without a change in humidity ratio. During the sensible cooling process, the relative humidity increases, the specific enthalpy is reduced, and the specific volume is also reduced. Latent heat change: When water vapor is added or removed from the air, it is referred to as humidification or dehumidification, which results in a latent heat change. In this process, the dry bulb temperature remains constant. The humidity ratio is either increased as in humidification, or decreased or dehumidification. Because the dry bulb temperature is not affected, there is not a sensible change in the air temperature. Note that in the humidification process, the specific enthalpy of the air is increased and in dehumidification the enthalpy decreases. Air can undergo a combination of sensible and latent change during an air conditioning process. The four scenarios that are possible are sensible heating and humidification, sensible heating and dehumidification, sensible cooling and humidification, and sensible cooling and dehumidification. Please note that, in general, the dry bulb, wet bulb and specific enthalpy also change during these processes. For example, in the cooling and dehumidification process, the dry bulb temperature is decreased, the humidity ratio along with the specific enthalpy also decreases. This is due to both sensible and latent heat removal. This is one of the most common and often used processes in air-cooling applications. Sensible change is shown on the psychrometric chart as a process lying along the constant humidity ratio as illustrated earlier. The sensible heat equation can be applied to the moist air in the following equation, where Q sensible is equal to the mass of air times 0.24, which is the specific heat of air, and the change in temperature plus the mass of the water vapor times 0.45, the specific heat of the water vapor, and the change in temperature of the air sample. For air conditioning applications, it is common to see the second term for water vapor neglected since it is typically too small to be of any significance and therefore the sensible heat equation for air is sometimes simply shown as Q is equal to 0.24 times the mass of air times the temperature change. Since air conditioning processes typically involve the flow of air, and because most instrument read the airflow in cubic feet per minute, or CFM, another convenient form of the sensible heat equation is often used. The relationship between mass flow-rate in pounds per hour and CFM is shown here. Pounds per hour times 60 minutes per hour times 1 pound per 13.3 cubic feet of air. This is the specific volume of air at certain conditions. Typically, we use this condition as standard air which is at 68 degrees fahrenheit and 29.92 inches of mercury. We will often here this referred to as standard air condition. Now, substituting into the sensible heat equation, we arrive at the familiar and very useful equation for sensible change of air as Qsis equal to 1.08 times CFM times change in temperature. We will put a dot over the Q to represent that this is an airflow rate and not a mass volume. A latent change is the process of adding or removing moisture without a change in the dry bulb temperature. As noted earlier, a latent change results in the change of the specific enthalpy but not a change in the dry bulb temperature. The rate of water that is added or removed during humidification or dehumidification is shown in the equation Mass flow rate of water is equal to the mass of the air sample times the change in humidity ratio. On the psychrometric chart, latent change is shown as a vertical line along some dry bulb temperature. As discussed earlier, latent change for moist air results in a change of the enthalpy of the air. In the dehumidification process, heat is removed from the air in order to condense out the water vapor. This can be shown on the psychrometric chart by observing the change in enthalpy during the dehumidification process. The rate of heat removal can be expressed using the enthalpy equation and is shown QLis equal to the mass flow rate of the air times the change in enthalpy which is represented as h2-h1. Please note that in these equation, Maoften with a dot over it to represent flow rate is the mass flow rate of air in pounds per hour. The change in latent heat will be in Btuhs. Enthalpy is represented in Btus per pound. It is very common in air conditioning systems for the air to undergo both sensible and latent change. These processes may occur separately or simultaneously but in either case the analysis uses both the sensible and latent heat equations. The total heat removed or added is the sum of the latent change and the sensible change. This can be demonstrated on the psychrometric chart. Dehumidification and sensible cooling occur frequently in air conditioning systems. Air that pasess over cooling coil will often cool to the point that water vapor condenses out so that the air is both cooled and dehumidified. The sensible change is the horizontal line as shown here. The latent change is represented by a vertical line along the dry bulb temperature. The total change is the sum of the change in enthalpy of both the sensible and the latent changes. Evaporative cooling is a process that is very special cooling and dehumidification process that deserves a detailed discussion in itself. Evaporative cooling occurs frequently in air conditioning systems. Two noted examples are cooling tower applications and evaporative cooler. In evaporative cooling, a noticeable change occurs in the air stream. The dry bulb temperature of the air decreases even though no external cooling source is used. Evaporative cooling is an adiabatic process, meaning it is a process in which there is no change in the total heat content. In addition, evaporative cooling is a constant enthalpy process. So if this is true, then how does the cooling effect occur? The cooling effect comes from the evaporation of the water into the airstream. Evaporation is a latent process that requires heat, which comes from the airstream. It is an adiabatic process because in this process there is an exchange of heat within the air mixture. The sensible heat decrease is the same as the increase in latent heat and the net effect is that there is no heat added or removed from the air stream. In addition, it is observed that the wet bulb temperature of the air stream does not change during this process. Evaporative cooling is only practical for air conditioning in very dry climates. In looking at the psychrometric chart for a typical summer outdoor design conditions in a humid climate of say, 90 degrees dry bulb and 74 degrees wet bulb for example, evaporative cooling can only result in air cooled to 74 degrees dry bulb. Please note that evaporative cooling follows the line of constant wet bulb as shown here. However, if we started with air at say, 94 degrees dry bulb and 10% relative humidity, which we might see occur in a warm and dry climate, evaporative cooling could produce supply air at 62 degrees fahrenheit which would be sufficient for cooling applications for comfort. Evaporative cooling also occurs in a cooling tower. But in this application, the process is to cool the water. Water is sprayed into an air stream and some of this water evaporates. The heat required to evaporate the water comes from the air stream and also from the water that does not evaporate. The cooled water is then returned to the system where it can be reused. This is the end of this tutorial, lesson 2.2. In lesson 2.3, there are example calculations on using the equations and processes discussed here. Lesson 2.3: Example Problems in Psychrometrics Example 2.3.1: Sensible Heat Change In this example, an electric resistance heater is installed in an air duct. The heater is needed to heat 650 pounds per hour of air from 75 degrees fahrenheit to 95 degrees fahrenheit. The problem is to find the capacity of the heater in kW. The process line is shown here on the psychrometric chart. The electric resistance of the heater is converted into heat. This problem is an example of a sensible change in the air stream as it is heated. On the psychrometric chart, it looks like a straight line along some line of constant humidity ratio. To solve, we will use the sensible heat equation where Qsis equal to the mass flow rate of the air times the specific heat of the air times the change in dry bulb temperature. In this application Qsis the heat required in Btuhs. The specific heat of the air is 0.24 Btus per pound-degree fahrenheit. Please note that the mass flow rate of the air has to be in pounds per hour for the units to work. So, applying the math we find that Q sensible is equal to 0.24 times 650 times 95 minus 75 degrees fahrenheit and this works out to 3120 Btuhs. To find the required capacity of the heater, we simply need to convert Btus per hour to kW using the conversion factor 1kW is equal to 3410 Btuh and we find that the required capacity is 0.91 kW. Our recommendation would be to provide a 1 kW heater. This is example 2.3.2 In the previous example we used the sensible heat equation to find the capacity needed to heat up the air by 20 degrees fahrenheit and at an airflow rate of 650 pounds per hour. Recall from lesson 2.2 that in HVAC applications, the sensible heat for the change in enthalpy of water vapor is typically considered negligible and is ignored. For this example, we will determine how much air results in neglecting the enthalpy change of the water vapor. To do this, we need to know something about the water vapor content of our air sample. For this example, we will assume the humidity ratio W is 0.008 pounds of water per pound of dry air. To find the sensible change of the moist air sample, taking into consideration both the dry air and the water vapor, our equation would be Qsis equal to the mass flow rate of the air times the specific heat of the air times the dry bulb temperature change of the air sample plus the mass flow rate of the water vapor times the specific heat of the water vapor times the change in dry bulb temperature of the water vapor. The mass flow rate of the air sample is 650 pounds per hour. So what is the mass flow rate of the water vapor? The mass flow rate of the water vapor is the humidity ratio, which is pounds per water per pounds of dry air, times the pounds per air hour, which is in this example, 650 pounds per hour. So the mass flow rate of the water would be 0.008 times 650, which is 5.2 pounds of water vapor per hour. And now we can apply this to our equation, where we would have Qsis equal to 0.24 times 650 times 20, which is the sensible heat change of the dry air, plus 0.45 times 0.52 times 20, which is going to be the sensible heat of the water vapor. This comes up to 3169 Btuhs. Qsin kW using the same conversion factor as in our previous example would be 0.93 kW. In the previous example, by neglecting the water vapor in our sample, we had 0.91 kW, so the amount of air is approximately 2%. In this application, we would probably specify 1kW heater regardless. Example 2.3.3 In the previous examples, we worked with the airflow rate in terms of pounds per hour, which is the mass flow rate. But common practice is to specify airflow rate in terms of CFM. What is the airflow rate of the air entering the heater in our previous example in CFM? Well, to solve this, we actually need to look at the psychrometric chart because we see if we take 650 pounds per hour and multiply that by 1 hour per 60 minutes, we need to multiply that also by the specific volume of the air in cubic feet per pound of air to convert to CFM. The problem is, what is the specific volume of the air sample at what conditions? Looking at the process line on the psychrometric chart, we see that the specific volume of the air entering condition looks to be about 13.6 cubic feet per pound of air. If we put this into our equation, we would have 650 pounds per hour times 1 hour per 60 minutes times 13.6 cubic feet per pound and this would come up to approximately 148 CFM. But looking at the process line on the psychrometric chart, we see that the specific volume of the air leaving the heater is greater than the specific volume of the air entering the heater. Therefore, the CFM is going to be different and in fact the CFM does change and is dependant upon the conditions. We can determine the CFM at the leaving conditions by doing the same equation but using the specific volume of the air at the leaving conditions, and this would be 650 pounds per hour times 1 hour per 60 minutes, and in this case, we would use 14.16 cubic feet per pound of air, and we find that the CFM of the air leaving the heater is about 153 CFM. This can actually be confusing, so it is important to understand the conditions at which CFM is specified. Manufactures of mechanical equipment know this, so in rating equipment you might see reference to standard air conditions. Standard air may be expressed as having the following conditions: specific volume of 13.3 cubic feet per pound, the density of 0.075 pounds per cubic foot, dry bulb temperature of 60 degrees, and pressure at 29.92 inches of mercury. It is important to always check what the manufacture means when they specify standard air because it actually may be different than these conditions. The specific volume and density of the air depends on the condition it exists, so the CFM can actually change. If we were to apply these conditions to our previous example, we would see that the 650 pounds per hour times 1 hour per 60 minutes times the specific volume of 13.3 would yield an airflow of 144 CFM. Example 2.3.4 This is an example of latent heat addition in humidification. As noted in lesson 2.2, humidification can be shown on the psychrometric chart as a process line of constant dry bulb temperature with increasing humidity ratio as shown here. The psychrometric chart shows that the conditions of winter air may have very low humidity, which is probably too low for comfort conditions. Therefore, humidification is a common air conditioning process useful for colder climates and heating applications. The following example demonstrates the latent heat process. A water humidifier in a warm air duct system increases the moisture content from 0.004 to 0.016 pounds of water per pound of air. If the air flow rate is 5,000 CFM and the dry bulb temperature is at 95 degrees dry bulb, then use the psychrometric chart to determine the heat required in this latent process. From the psychrometric chart, we can determine the enthalpy of the air entering and leaving the humidifier. The enthalpy of the air entering the humidifier is approximately 26.8 Btus per pound. The enthalpy leaving the humidifier is at 38.8 Btus per pound. By applying the enthalpy equation, which is Q is equal to the mass flow rate of the air multiplied by the change in enthalpy we can determine the heat required for this process. Please note that this equation to get our Q in Btuhs, the flow rate of the air is in pounds per hour which is the mass flow rate. So just as in our previous examples, we have to determine what the mass flow rate is for 5,000 CFM. Note that I have decided to use the conditions for standard air to determine the mass flow rate based on this 5,000 CFM as follows: 5,000 CFM multiplied by 60 minutes per hour times the specific volume of air as one pound per 13.3 cubic feet times the change in enthalpy. In doing the math, this comes out to 270,677 Btu/hr. This is example 2.3.5: Combined Sensible and Latent Heat Process Air conditioning for occupant comfort often requires that the air be both cooled sensibly and dehumidified. To determine the total heat removed for this process, both the sensible and latent components must be considered. Q total is going to be the sensible component plus the latent component. If our airflow rate is given in CFM, then our sensible component Qsis going to be some constant multiplied by the CFM times the change in temperature. The latent component is going to be a different constant times CFM times the change in humidity ratio. As you can see, this gets pretty complicated right off the bat, so there is actually an easier way to do this using the psychrometric chart and our enthalpy equation. The enthalpy equation can be used for all processes, whether they are latent, sensible or a combination of the two. Let’s consider the following example: and air conditioning unit has a cooling coil that cools and dehumidifies 400,000 CFM of air from 78 degrees fahrenheit and at 50% relative humidity to 62 degrees fahrenheit dry bulb and 56 degrees wet bulb. In this process, the air is going to be both cooled and dehumidified. This can be shown on the psychrometric chart using the enthalpy equation. Find the sensible, latent and total capacity of the cooling coil in units of Btuhs and also tons of refrigeration. Let’s consider the total change first, from the entering and leaving conditions. The Q total is going to be the mass flow rate of the air times the enthalpy of the air change as it goes through this process. The mass flow rate of the air is 40,000 CFM times 60 minutes per hour times 1 over 13.3 cubic feet, which is the specific volume of the air sample. Please note that for this example, I decided to use the standard air conditions to convert to the mass flow rate. So the total change is going to be the enthalpy change as it goes through this process. The entering enthalpy is h1 as shown on the psychrometric chart to be about 30 Btus per pound. At the leaving conditions, the enthalpy is 23.8. This gives us a total change of approximately 6.2 Btus per pound and applying the enthalpy equation, we get 1,118,797 Btuhs. Using the conversion factor for tons of refrigeration as 1 ton equals to 12,000 Btuhs, we see that the total cooling capacity of this cooling coil is approximately 93 tons. Let’s look at what the latent change is for this example. Using the enthalpy equation, we can determine the latent change by determining the change in enthalpy for the latent process. The entering condition is the change as 30 Btus per pound but the enthalpy change for the latent process is shown here as 26.8 Btus per pound in the leaving conditions. So this gives a change in enthalpy of about 3.2 Btus per pound. Applying this into our enthalpy equation, we get 577,444 Btuhs, or about 48 tons of refrigeration due to the latent change. Of course, the same application can be done to determine the sensible change of the air sample. In this example, the enthalpy change due to the sensible change can be found where h1 is 26.8 Btus per pound off of the psychrometric chart, and h2 is 23.8 Btus per pound. So applying the enthalpy equation, the sensible change is approximately 541,353 Btus per hour, or about 45 tons of refrigeration. You will find that the sensible change plus the latent change will equal the total change and the total cooling capacity. Example 2.3.6 For the previous example, how much moisture will condense out of the cooling coil? Well, to determine this we need to determine the change in moisture content or humidity ratio as it goes through this process. The humidity ratio entering the coil at the entering conditions can be found on the psychrometric chart as about 0.0102 pounds of water per pound of air. The humidity ratio leaving the coil is about 0.0082. To find the rate of condensation in pounds per water, we multiply the mass flow rate of the air by the change in humidity ratio. Here we see that the mass flow rate of the air is equal to the 40,000 CFM times 60 minutes per hour times 1 pound per 13.3 cubic feet, and in doing the math, we will find that we will get approximately 360 pounds of water per hour that condenses out of this cooling coil. So how much is this in gpm? Well, I’m not going to solve that but I will let you figure that out. I will give you a hint, and that is to use 8.33 pounds per gallon and of course, 60 minutes per hour, and apply that to the 360 pounds per hour and see what you get in terms of gallons per minute.