MECE-251 Thermodynamics, Lesson 2: Psychrometrics: Properties of Dry Air and Water Vapor Mixtures We have studied water in various phases ranging from solid to liquid to gas using data tables in the appendix. We studied “dry air” as an ideal gas. Now, let’s look at “moist air” as a mixture of water and dry air In our previous lesson we learned about the properties of pure water as it changes phase, particularly focusing on the transition between liquid and gas phases. We used tabulated data to observe very accurate relationships between the temperature, pressure and specific volume of water. We also studied dry air using the ideal gas equation of state, and the conditions under which a gas may or may not be treated as an ideal gas. We learned about the compressibility factor as a function of the reduced temperature and reduced pressure. Next, we will mix the dry air and water vapor together, to create a mixture that we will call “moist air.” This is a crucial element of pneumatic control systems. Failure to properly account for the moisture content of air as a function of temperature, system pressure and ambient pressure can cause numerous difficulties in a pneumatically operated mechatronic system. R·I·T MECE-251 1

Mixtures – A Simple View Psychrometrics: Dry air and water vapor – a special case of mixtures. Volume Fraction (Molar) View N = Nair + Nvapor yair = Nair / N yvapor = Nvapor / N Consider the isolated control volume that consists of a cylinder with a gaseous mixture of dry air and water vapor. The dry air and the water vapor are uniformly distributed throughout the control volume. The control volume contains N_air moles of air and N_vapor moles of water, for a total of N moles of mixture. The mole fraction of each constituent is the ratio of the moles of the constituent to the moles of mixture. The molar fraction is sometimes referred to as the volume fraction. Now, imagine that all of the water molecules (still in the gas phase) are collected in the bottom of the cylinder and all of the air content is in the upper portion of the cylinder. While obviously not an equilibrium condition, this view helps us to understand the mass fraction interpretation of mixtures. The total mass in the control volume is the sum of the mass of dry air and the mass of water vapor. The mass fractions of each are the ratio of the mass of each constituent to the mass of the mixture. The mass fractions of each component are different from the volume fractions of each constituent. This is due to the fact that the density (and specific volume) of each constituent is different. Mass Fraction View m = mair + mvapor mfair = mair / m mfvapor = mvapor / m Reference: Engineering Thermodynamics - A Graphical Approach by Israel Urieli http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter10b.html Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 2

Convert Mole Fractions to Mass Fractions Psychrometrics: Dry air and water vapor – a special case of mixtures. For Dry Air TC = 133 K = 239 R PC = 37.7 Bar ≈ 547 psia M = 28.97 kg/kmol = 28.97 lbm/lbmol R = 0.287 kJ/Kg·K = 53.3 ft·lbf / lbm·R For Pure Water TC = 647.3K = 1165. R PC = 220.9 Bar ≈ 218 atm M = 18.02 kg/kmol R = 0.4615 kJ/Kg·K = 85.76 ft·lbf / lbm·R Component y (kmole/kmole) M (kg/kmole) m = y M (kg) mf (%) Dry Air 0.95 28.97 27.52 96.8 Water Vapor 0.05 18.02 0.90 3.2 Sum 1.0 28.42 100.0 We regularly need to convert between volume (or molar) fractions and mass fractions. For our mechatronics applications, we will first look at the composition of moist air that is used as inlet air at the intake of an air compressor. While dry air is not a pure substance, we are going to treat it as one for the purposes of modeling ambient air. The effective molecular weight of dry air is taken to be 28.97 kg per kmole. Water is a pure substance, but we know from our previous work that water often exists in solid, liquid, and gas phases over the temperature and pressure range of interest for mechatronics applications. For the moment, let’s also consider water vapor to be an ideal gas, with a molecular weight of 18.02 kg/kmol. Now, let’s make a simple state table, to understand the composition of moist air. The first column of the state table contains the names of the constituents. The second column contains the mole fractions of each constituent, with units of kmole of the constituent to kmole of the mixture. The third column contains the effective molecular weight of each constituent. The mass of each constituent present in the control volume, in column 4, can be computed as the product of the mole fraction and the molecular weight of each constituent. The total mass of the mixture is found by summing the mass of all constituents. Finally, the mass fraction of each constituent is computed as the ratio of the mass of each constituent over the mass of the mixture. Reference: Engineering Thermodynamics - A Graphical Approach by Israel Urieli http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter10b.html Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 3

Convert Mass Fractions to Mole Fractions Psychrometrics: Dry air and water vapor – a special case of mixtures. For Dry Air TC = 133 K = 239 R PC = 37.7 Bar ≈ 547 psia M = 28.97 kg/kmol = 28.97 lbm/lbmol R = 0.287 kJ/Kg·K = 53.3 ft·lbf / lbm·R For Pure Water TC = 647.3K = 1165. R PC = 220.9 Bar ≈ 218 atm M = 18.02 kg/kmol R = 0.4615 kJ/Kg·K = 85.76 ft·lbf / lbm·R Component mf (%) M (kg/kmole) N = mf / M (kg) y (kmole/kmole) Dry Air 0.97 28.97 0.035 95.4 Water Vapor 0.03 18.02 0.0017 4.6 Sum 1.0 0.0367 100.0 We also need to convert from mass fractions to mole fractions. As before, we make a state table, and the first column of the state table contains the names of the constituents. This time, the second column contains the mass fractions of each constituent, with units of kg of the constituent to kg of the mixture. The third column contains the effective molecular weight of each constituent. The number of moles of each constituent present in the control volume, in column 4, can be computed as the quotient of the mass fraction and the molecular weight of each constituent. The total moles of the mixture are found by summing the moles of all constituents. Finally, the mole fraction of each constituent is computed as the ratio of the moles of each constituent over the moles of the mixture. Reference: Engineering Thermodynamics - A Graphical Approach by Israel Urieli http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter10b.html Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 4

Ideal Gas EOS for Mixtures (Dalton Model) Consider both dry air and water vapor as ideal gases. Dalton Mixture Model V = Vair = Vvapor T = Tair = Tvapor Partial Pressure Model P = Pair + Pvapor Pair = yair P Pvapor = yvapor P Ideal Gas, dry air Pair vair = Rair T Pair V = mair Rair T Pair = ρair RairT Pair V = Nair Runiv T Ideal Gas, water vapor Pvapor vair = Rvapor T Pvapor V = mvapor Rvapor T Pvapor = ρvapor Rvapor T Pvapor V = Nvapor Runiv T Now that we have a good understanding of mass fraction and mole fraction, we can introduce the Dalton Model for mixtures of gases. We consider both the dry air and the water vapor to be ideal gases. In the Dalton Model, let’s imagine that we have an isolated control volume containing a mixture of dry air and water vapor. The water vapor and dry air collectively occupy the entire control volume, as is the case with all gases. Also, since we assume that the mixture is at thermodynamic equilibrium, we can state that the temperature of the mixture equals the temperature of the dry air which equals the temperature of the water vapor. However, we have to handle the contribution of each component of the mixture to the pressure in a different manner. Both gases contain molecules that behave with certain motion characteristics that reflect their molecular weight. The static pressure in the control volume consists is the sum of the partial pressure of the dry air and the partial pressure of the water vapor. The partial pressure is computed from the product of the mole fraction of each component and the static pressure in the control volume. You will learn more about static pressure in the fluids module. Using the partial pressure of the dry air, we can use the ideal gas equation of state to compute the specific volume of the dry air component at a given temperature, and using the effective ideal gas constant for the dry air (which is itself a mixture of components.) These four forms of the ideal gas equation of state are equivalent… they are simply algebraic manipulations of the same principle. Next, using the partial pressure of the water vapor, we can use the ideal gas equation of state to compute the specific volume of the water vapor component at a given temperature and partial pressure. As an exercise, you should check that the reduced partial pressure and the absolute temperature of the water vapor support the assumption of an ideal gas using the compressibility chart from the previous lecture. Reference: Engineering Thermodynamics - A Graphical Approach by Israel Urieli http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter10b.html Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 5

Humidity Ratio Psychrometrics: Dry air and water vapor – a special case of mixtures. Define Humidity Ratio (aka Specific Humidity) ω ≡ mvapor / mair Since mvapor << mair , we get w << 1. Using the ideal gas equation of state and Pair = P – Pvapor The humidity ratio of moist air is defined as the ratio of the mass of the water vapor to the mass of the dry air. The mass of water vapor is much less than the mass of dry air. Using the ideal gas equation of state for both the vapor and dry air components, we can develop the simplified model for humidity ratio. The ratio of the effective molecular weight of water vapor and dry air is 0.622. The humidity ratio is a function of the partial pressure of the vapor and the static pressure of the mixture. 18.02/28.97 = 0.622 Reference: Moran, Shapiro, Boettner, Bailey: Fundamentals of Engineering Thermodynamics, 7th Edition, Dec 2010, © 2011, Wiley. R·I·T MECE-251 6

Relative Humidity Psychrometrics: Dry air and water vapor – a special case of mixtures. The mixture pressure is the sum of the partial pressures of the dry air and the water vapor: P = Pa + P v The water vapor is typically superheated at partial pressure Pv and mixture temperature T. When Pv corresponds to Pg at temperature T, the mixture is said to be saturated. Mixture pressure, p , T The mixture pressure is the sum of the partial pressures of the dry air and the water vapor. The water vapor is typically superheated at partial pressure Pv mixture temperature T, shown here far to the right of the vapor line on the steam dome. When Pv corresponds to Pg at temperature T, the mixture is said to be saturated. The saturation pressure corresponds to the saturation temperature of the mixture. The ratio of the partial pressure of the vapor to the saturation pressure of the mixture at temperature T is called the relative humidity, phi. When the relative humidity is 0% there is no water in the air at all. When the relative humidity is 100% any decrease in temperature will result in condensation (also known as rain or snow!). Typical state of the water vapor in moist air The ratio of Pv and Pg is called the relative humidity, f: dry air only (pv = 0) 0 ≤ f ≤ 100% saturated air (pv = pg) Reference: Moran, Shapiro, Boettner, Bailey: Fundamentals of Engineering Thermodynamics, 7th Edition, Dec 2010, © 2011, Wiley. R·I·T MECE-251 7

A Simplified Psychrometric Chart for a Mixture of Dry Air and Water Vapor at 1 atmosphere The vertical lines of dry bulb temperature represent the temperature of the mixture. The horizontal lines represent constant humidity ratio. The slanted lines represent constant specific volume. A psychrometric chart is a convenient method for determining the state of moist air, and for plotting processes involving moist air. A psychrometric chart is unique to the ambient air pressure – the barometric pressure. The chart shown here is for sea level. ASHRAE, the American Society for Heating Refrigeration, and Air conditioning Engineers, publishes high quality charts for use at various barometric pressures. The psychrometric chart is a combination of the steam tables at one pressure and the Ideal Gas Equation of State for dry air. The equations presented previously may be used instead of the psychrometric chart. The vertical (black) lines on the chart represent the temperature of the mixture, called the dry bulb temperature. The horizontal (purple) lines represent the humidity ratio, or specific humidity, of the mixture. The slanted (green) lines represent lines of constant specific volume for the mixture. The curved (blue) lines represent lines of constant relative humidity for the mixture. As an example, if we have a moist air mixture at P = 1 atmosphere, with a dry bulb temperature of 30 C and humidity ratio of omega = 10 gr of water vapor per kg of dry air, then then intersection of the temperature and humidity ratio lines permits us to determine that the specific volume of the mixture is approximately 0.87 m3/kg and the relative humidity is about 40%. The curved lines represent constant relative humidity. At P = 1 atm (the entire chart) T = 30C and ω = 10 gr/kg we can find v = 0.87 m3/kg and Φ = 40% Reference: Engineering Thermodynamics - A Graphical Approach by Israel Urieli http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter10b.html Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 8

Next Steps L2 Task 4B: Please scan through sections 12.1, 12.2 and 12.4, 12.5 and 12.6 in the textbook. Please review the example problems on line. L2 Task 4C: Then, solve the review problem. L2 Task 5: Solve and TURN IN the case study problem. L2 Task 6: Take the Lesson 2 quiz. The material that we covered in lecture today corresponds to three sections in Chapter 12. We have not covered the content from section 12.3, and scan over the examples about enthalpy (h). We will come back to enthalpy later. After you scan chapter 12, please solve the review problem, and then move on to the case study and the quiz for lesson 2. As before, please use the lesson 2 discussion board to ask questions and discuss the material with your peers in the class. I will join in as necessary if issues remain unresolved. Reference: Schaum’s Outline of Thermodynamics for Engineers, Second Edition, M.C. Potter and C.W. Somerton, McGraw Hill R·I·T MECE-251 9