Lesson plans a. Introduction
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Figure 21. Just as everything else on the Earth, air feels the force of gravity. It is the force of gravity which allows Earth to have an atmosphere. We’ve stated before, air has a mass. Since gravity is acting on the air, the air also has a weight. By definition, weight is the gravitational force which acts upon a mass: W = mg Where:
W = weight m = mass
g = gravity = 9.8 m/s2 on Earth Said in another way, gravity is pulling all of the air in the Earth’s atmosphere towards the Earth. This pulling action is what keeps the air close to the earth and creates our atmosphere. Now let’s reverse how we think of the air in the Earth’s atmosphere. Think of the atmosphere as slices or layers of air from the surface of the Earth to the top of the atmosphere. Imagine each of the slices having a mass of 1kg and thus from the equation would have a weight of 1kg x 9.8 m/s2, or 9.8N Since weight is a force, the weight of the slice at the very top of the atmosphere will push down on the second slice. The weight of the first and second slice will push down on the third slice. This scenario continues all the way to the bottom of the atmosphere or at the Earth’s surface. This pushing, the force created by the weight of the air, is what causes air pressure, illustrated in Figure 22. So as you can see, air pressure varies according to how high above the Earth’s surface you are. The air pressure at the Earth’s Surface is greater than the air pressure on top of the Rocky Mountains which is greater than the air pressure at the altitude at which air planes fly. 
Figure 22. To know the air pressure at various attitudes is of importance in the study of aeronautics so a method to measure the air pressure was developed. Scientists created a standard constant atmospheric pressure to relate other pressures to. This pressure is called standard atmospheric pressure, and it is the average pressure of the atmosphere measured at sea level. Standard atmospheric pressure is a constant and is expressed by various means: 1 atm 14.7 PSI (Pounds per Square Inch) 1.013 x 105 N/m2 760 Torr
Figure 23. The density of mercury is then used to convert the inches or millimeters of mercury measured to another unit of measure for pressure, such as PSI or N/m2. For example, suppose a barometer’s column of mercury raises 30.5 in. The density of mercury is 0.491 lb/in3 (Densities are known for most common substances and can be looked up in tables). Then to find the air pressure: Pair = r x inches of mercury Pair = 0.491 lbs x 30.5 in in3 Pair = 14.98 lbs in2 Similarly, if the barometer was calibrated in millimeters of mercury, it would read, 774.7 millimeters of mercury. The density of mercury in metric units is1.33 x105 N/m3, therefore: Pair = r x millimeters of mercury Pair = 1.33 x 105 N x 774.7 mm x 1m m3 1000mm
m2 Go back and notice the standard atmospheric pressure expressed in Torr. 1 Torr is equal to 1 millimeter of mercury. Therefore at standard atmospheric pressure, a column of mercury will rise 760 millimeters. In aeronautics we often use inches of water to measure pressure. The density of water is 0.036 lb/in3. The pressure of 14.7 PSI will raise a column of water 406.8 in. Patm = rwater x inches of water
in2 in3 The reason the column of water rises much higher than the mercury column is because of the difference in the densities of mercury and water. Since the density of mercury is 13.6 x’s heavier than water, the column of mercury will raise13.6 x’s less than the water column will. The water column is lighter and thus can be raised higher with the same amount of pressure pushing down on it. Air pressure is often referenced to the standard atmospheric pressure. It is the difference between the air pressure read on a gauge or test equipment and the standard atmospheric pressure, often referred to as the DP (Delta Pressure) reading, which is generally of importance in the study of aeronautics. Equation of State Exercises:
V Where: p = pressure m = mass
R = specific gas constant, 1716 ft lb/(slug)(R) or 287 J/(kg)(K) for air T = temperature V = volume
a) To find the mass of the air use the density equation. r = m V
m = mass V = Volume solve in terms of m m = r x V The density of air is 1.20 kg/m3. The volume of air from the problem is 2.5m3. Substitute these values into the equation: m = 1.20 kg x 2.5m3 m3 m = 3.0 kg b) Use the equation of state to find the new volume of air. p = mRT V
V p = mRT V = mRT p
1 Pa = 1N/m2 thus our pressure = p = 3.0 x 105 N/m2 Since the specific gas constant, R, is in terms of Kelvin, we must covert our temperature to Kelvin T (K) = 273 + C T (K) = 273 + 30C = 303 (K) Substitute in values V = 3.0kg 287 J 303 K ___________ kg K _______ 3.0 x 105 N/m2
Substituting this into the equation and canceling units: V = 3.0kg 287 N m 303 K m2 3.0 x 105 N kg K V = 0.87 m3
2. An automobile tire has a volume of 1000 in3 and contains air at a gauge pressure of 24 lb/in2 when the temperature is 0C. What is the gauge pressure of the air in the tire when its temperature rises to 27C and its volume increases to 1020in3? Assume the tire cannot leak air. Since the tire is sealed and no air can escape, so the mass of the air will stay the same at both pressures, volumes and temperatures. Additionally, R is a constant for both situations. For the first situation let us designate: p1 = 24lb/in2 T1 = 0C + 273 = 273K V1 = 1000 in3
If we solve the equation of state for mR we get mR = pV T
1. mR = p1V1 T1 Similarly for the second situation:
T2 = 27C + 273 = 300K V2 = 1020 in3
T2 As we said above, m and R stay constant in both situations, so the two equations can be set equal to each other: p1V1 = p2V2 T1 T2 Since we need to know the new gauge pressure of the tire, p2, we will solve the equation for p2: p2V2 = T2 p1V1 T1 p2 = T2 p1V1 V2T1 Substituting in the values: p2 = 300K24lb 1000 in3 in2 1020 in3 273K p2 = 25.86 lb/in2 C. THE PROPERTIES OF AIR IN MOTION Objective: The study of aerodynamics is a sub set of the study of physics. The term aerodynamics means, using the most simplistic definition, the movement of air. Wind is air that is moving. A fan generates air movement. A car driving down the highway is moving through air, causing air movement. The physical principles used in the study of physics to define movement of solids, which may be more familiar to us, also can be applied to the movement of liquids and gases. Moving air has all of the same properties explained in the lesson on static air: temperature, pressure, volume and mass, plus one more. This additional property is velocity. Air in Motion Lesson: In the lesson on static air we began the lesson by explaining the relationships between the various properties of static air. We will begin this lesson the same way by explaining the relationships between the various properties of moving or flowing air. However, before we do that, we need to make a few assumptions about the air we will be studying. The assumptions we make will make the air we study an ideal gas. Once we understand how an ideal gas behaves, we will have a sound background in the physics of the flow of gas to build upon as we talk about gas that is not ideal. Our first assumption is that the air is incompressible, meaning that the density of the air will have a constant value. Also at low velocities, which we will be concentrating on in this lesson, the temperature of the moving gas is not of importance. In other words, a change in the temperature of the flowing gas will not affect any of the other properties of the flowing gas. Having made our assumptions, let us go back and look at the properties of flowing air again. Since we are neglecting density and temperature, we are only left with pressure and velocity. (Remember that density is the mass per volume of a substance, therefore if the density does not change, the mass and the volume will not change either.) At this point it will be beneficial to visualize flowing air. Take ten pieces of string about four inches long and tie them to the guard of a household fan. Turn the fan on. What happens? The strings all protrude out 90 degrees from the fan, as demonstrated in Figure 24. 
Figure 24. The strings can be defined as following the flowline or streamline of the air flowing from the fan. Notice also that the strings all follow their own flowline, they do not cross one and other. Now visualize instead, a stream of beads following the flowlines instead of the string, and think of each bead as a particle within the flow path. Now imagine the bead getting smaller and smaller, until it is the size of the individual molecules of air. The individual molecules of air are following the flowlines as well. The velocity of the molecules of air or the beads flowing in the flowline may speed up or slow down, but they will always follow the flowline. This is visualized in Figure 25. 
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