Thermodynamic Properties of Gas Mixtures

Thermodynamic Properties of Gas Mixtures
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Mixtures of Gases and Vapors 12.1 WPROWADZENIE (INTRODUCTION) In this chapter, we deal with the problem of generating thermodynamic properties for homogenous mixtures of gases and vapors that are not involved in chemical reactions. Properties of chemically reacting mixtures are discussed in detail in Chapter 15. We can define the composition of any mixture based on how we physically create the mixture. For example, we can create a mixture by combining measured masses (or weights) of things or by combining measured volumes of things. The items or things that make up the mixture are called the components of the mixture, and knowing how much of each one is present defines the composition of the mixture. This all seems very simple, so why do we need to dwell on it here? First of all, there are two ways to measure mass, the regular mass (lbm or kg) and the molar mass (lbmole or kgmole), and a composition based on the regular mass is not the same as a composition based on the molar mass. Second, since the conservation of mass law tells us that the total mass of the mixture is simply the sum of the masses of all the components in the mixture,isthetotalvolumeofthemixturethesameasthesumofthevolumesofallthecomponentsinthe mixture? Well, that depends. If the components are immiscible, then the total volume is the sum of the indivi- dual component volumes. But gases and vapors are not immiscible, so how do the component volumes affect the total volume? So, whereas the basic definitions of mixture composition for insoluble solids and liquids seems very easy, the practicality of implementing these definitions for mixtures of soluble gases and vapors is not so easy. However, the mixture composition is not really our primary goal. Our primary goal is to be able to determine the thermodynamic properties of a mixture so that we can apply the first and second laws to an engineering system containing a mixture.Are mixture thermodynamic properties just the sum of the thermodynamic properties of their components? No With the exception of mass, the extensive thermodynamic properties (e.g., mixture total volume V , mixture total m406 CHAPTER 12: Mixtures of Gases and Vapors CRITICAL THINKING Whereas the volumes of insoluble solids and liquids seem easily defined, because their molecules do not intermingle, the same is not true for gases and vapors, where their molecules do intermingle. But, what happens when you mix two soluble liquids? Is the mixture volume in this case just the sum of the volumes of the mixture components before they were mixed? Can you find an example of soluble liquids where the final mixture volume is not equal to the sum of their premixed component volumes? internal energy U , mixture total enthalpy H , and mixture total entropy S ) of a mixture are not generally equal m m m to the sum of the extensive thermodynamic properties of the components. It turns out that the value of any thermodynamic property of a mixture is just the mass weighted sum of the partial specific properties of the mixture’s components. Therefore, to determine the numeric value of a thermodynamic property of a mixture we need to know (a) the exact composition of the mixture and (b) the values of the partial specific properties of all the components in the mixture. This is what the first half of this chapter is all about. The second half of this chapter deals with the application of this material to a very special mixture of gases and vapors, air and water vapor (atmospheric air). This is normally the domain of heating, ventilating, and air condi- tioning (HVAC) engineers, but since the atmosphere affects all of us in our daily life, it provides a good textbook application of gas and vapor mixture theory. The basic elements of HVAC involve applying the first and second laws to systems designed to cool and dehumidify or heat and humidify atmospheric and building air. To carry out this analysis we need the numerical values for specific internal energy, specific enthalpy, and specific entropy for various mixtures of air and water vapor. Since atmospheric air is a fairly complex mixture, it is more convenient to refer to industry prepared tables and charts for accurate thermodynamic property values for this mixture. 12.2 THERMODYNAMIC PROPERTIES OF GAS MIXTURES Unfortunately, there is no single measure of mixture composition. A mixture composition often is given simply in 1 percent, but the percent is calculated on a mass (or weight) basis, a molar basis, a volumebasis,orapressure basis; and the numerical values depend on which basis is used in the calculation. This ambiguity leads us to define four composition percentages or fraction measures for mixtures of gases. Consider a homogeneous mixture made up of N distinct gases, each of which has a unique molecular mass, M. Let i the mass ofeach gas present in the mixturebe m. Thenthe mass balance gives the totalmassofthe mixture m as i m N m = m +m ++m =∑m (12.1) m 1 2 N i i=1 The corresponding number of moles n of gas i with molecular mass M can be determined from Eq. (1.9) as i i n = m/M (1.12) i i i and because the mole unit is just another measure of mass, the total number of moles of mixture n is simply m N n = n +n ++n =∑n (12.2) m 1 2 N i i=1 With these two mass measures, we can define two different mass-based composition measures or fractions as The mass fraction w of chemical species i in the mixture is i (12.3) m i w = i m m and The mole fraction χ of chemical species i in the mixture is i (12.4) n i χ = i n m 1 This is also called a gravimetric basis.12.2 Thermodynamic Properties of Gas Mixtures 407 With the exception of system mass, the extensive properties of a system are not generally conserved in any thermodynamic process, so that their mixture values are not normally equal to the sum of their constituent values. The changes in extensive or intensive properties of a mixture with composition must always be deter- mined experimentally. However, the extensive properties of gases are mathematically homogeneous functions 2 of the first degree in mass. For example, the total volume V of a homogeneous mixture of gases can be m written as a function of the mixture total pressure p , mixture temperature T , and the mass composition of m m the mixture m , m , … , m as 1 2 N V = VðÞ p ,T ,m ,…,m m m m m 1 N and, when the mixture pressure and temperature are held constant, this is a homogeneous function of the first degree in the masses m. This means that, if all the remaining variables (the m) are multiplied by an i i arbitrary constant λ, then the mixture volume also is multiplied by λ,or λV = V ðp ,T ,λm ,…,λm Þ m m m m 1 N where λ is an arbitrary variable. Differentiating this equation with respect to λ while holding the pressure and temperature constant and setting λ =1gives   N ∂V ∂V m m V j = m ++ m =∑m v (12.5) m p ,T 1 N i i m m ∂m ∂m 1 N i=1 where  ∂V m v = (12.6) i ∂m i p ,T ,m m m j here v is defined to be the partial specific volume of gas i in the mixture and V = m v = nv is the partial i i i i i i 3 volume of gas i in the mixture. Equation (12.5) leads us to a third common composition measure, the volume fraction: The volume fraction ψ of gas i in a mixture of gases is i V i ψ = (12.7) i V m Even though pressure is not an extensive property, the implicit function theorem from calculus tells us that, if ∂V /∂p ≠0 in Eq. (12.5), we can write the total pressure p of a homogenous mixture of N gases as a function m m m , mixture temperature T , and the mass composition of the mixture m , m ,…, m as of the mixture volume V m m 1 2 N p = p ðV ,T ,m ,m ,:::,m Þ m m m m 1 2 N and, when the total volume and temperature of the mixture are constant, this too is a homogenous function of the first degree in the masses m. Following the development of Eq. (12.5), we can write i  N ∂p m p =∑m =∑mp =∑p = p +p ++p (12.8) m i i i 1 2 N i ∂m i m V ,T ,m i=1 m j where p =ðÞ ∂p /∂p is the partial specific pressure of gas i in the mixture and p = mðÞ ∂p /∂p is the m i V ,T ,m i i m i V ,T ,m i m m j m m j partial pressure of gas i in the mixture. Equation (12.8) provides our fourth common composition measure, the pressure fraction: The pressure fraction π of gas i in a mixture of gases is i p i π = (12.9) i p m 2 See, for example, Kestin, J., 1979. A Course in Thermodynamics. Hemisphere Publishing Corporation, McGraw-Hill Book Company, New York, vol. 1, pp. 326–327. 3 The concept of “partial properties” was introduced by Lewis, G. N., 1907. A new system of thermodynamic chemistry, in Proceedings of the American Academy, 43, p. 273.408 CHAPTER 12: Mixtures of Gases and Vapors Table 12.1 summarizes the four composition measures thus far defined. Dividing both sides of Eq. (12.5) by the mixture mass m produces the specific volume of the mixture as m N V m m i v = =∑ v =∑w v m i i i m m m m i=1 where w is the mass fraction m /m . The concepts of total, specific, and partial specific properties can i i m be extended to other extensive properties, as shown in Table 12.2. On a molar basis, total, molar specific, and partial molar specific properties are defined in Table 12.3. Finally, the constant volume and constant pressure specific heats for the mixture are defined as   ∂u ∂h m m c = and c = v p m m ∂T ∂T m m v v m m These specific heats can also be written on a molar basis and in terms of the partial specific heats of the component gases,asshown in Table 12.4. The mass, mole, volume, and pressure composition fractions have the characteristic that they always sum to unity; that is, N N N N ∑w =∑χ =∑Ψ =∑π = 1:0 (12.10) i i i i i=1 i=1 i=1 i=1 Table 12.1 Four Composition Measures Name of Composition Fraction Defining Equation Mass fraction w w = m/m i i i m Mole fraction χ χ = n/n i i m i Volume fraction ψ ψ = V / V i m i i Partial pressure fraction π π = p/p i i i m Table 12.2 Total, Specific, and Partial Specific Properties of a Mixture of Gases Partial Specific Property Total Property of the Mixture Specific Property of the Mixture of Gas i in the Mixture  N N N ∂V m m v = i i V =∑mv v = V /m =∑ v =∑w v m i i m m m i i i ∂m m i m p ,T ,m m m j i=1 i=1 i=1  N N N ∂U m u = mi i U =∑m u u = U /m =∑ u =∑w u m i i m m m i i i ∂m m i m pm,Tm,mj i=1 i=1 i=1  N N N ∂H m h = m i i H =∑m h h = H /m =∑ h =∑w h m i i m m m i i i ∂m i mm pm,Tm,mj i=1 i=1 i=1  N N N ∂S m m s = i i S =∑ms s = S /m =∑ s =∑ws m i i m m m i i i ∂m i mm p ,T ,m m m j i=1 i=1 i=1 CRITICAL THINKING Is it possible to define a partial temperature of gas i in a mixture of N gases? Since the mixture temperature varies inversely with the system mass in most equations of state for gases, is temperature a homogeneous function of the first degree in the masses m? Could the partial temperature of gas i in the mixture be the temperature exhibited by gas i when it alone occu- i pies the volume of the mixture Vm at the pressure of the mixture p ? (See Problem 67 at the end of this chapter.) m12.2 Thermodynamic Properties of Gas Mixtures 409 Table 12.3 Total, Molar Specific, and Partial Molar Specific Properties of a Mixture of Gases Total Property Molar Specific Property Partial Molar Specific Property of the Mixture of the Mixture of Gas i in the Mixture  N N N ∂V m v = n i i V =∑n v v = V /n =∑ v =∑χ v m i i m m m i i ∂n i i n m p ,T ,n m m j i=1 i=1 i=1  N N N ∂U m n u = i i U =∑n u u = U /n =∑ u =∑χ u m i i m m m i i ∂n i i n m p ,T ,n m m j i=1 i=1 i=1  N N N ∂H m n h = i i H =∑nh h = H /n =∑ h =∑χ h ∂n m i i m m m i i i i n m p ,T ,n m m j i=1 i=1 i=1  N N N ∂S m n s = i i S =∑n s s = S /n =∑ s =∑χ s ∂n m i i m m m i i i n i m p ,T ,n m m j i=1 i=1 i=1 Table 12.4 Specific Heats of a Mixture and the Partial Specific Heats of the Gases in the Mixture Specific Heat of the Mixture Partial Specific Heat of Gas i in the Mixture   N ∂u i ∂u c = m vi c = =∑wc ∂T vm i vi m v ∂T m m v i=1 m   N ∂h i ∂h c = m pi c = =∑w c p i pi ∂T m m p ∂T m m v i=1 m   N ∂u i ∂u c = m vi c = =∑χ c v vi ∂T m i m ∂T vm m v i=1 m  N ∂h ∂h i m c = c = =∑χ c pi p pi m i ∂T ∂T m m v m i=1 v m therefore, when either w, χ, ψ,or π is multiplied by 100, it represents the composition percentage of gas i i i i i present on a mass, molar, volume, or pressure basis, respectively. Note, however, that w and χ do not have i i the same numerical values; therefore, a mass based percentage analysis of the composition depends on which fractional base is used in its determination. If we consider the mixture to be a unique substance, then we can compute its equivalent molecular mass M from m Eqs. (12.1), (1.9), and (12.4) as N N N m nM 1 m i i M = = = ∑m ∑ =∑χ M (12.11) m i i i n n n m m m i=1 i=1 i=1 and using Eqs. (12.2), (1.9), and (12.3) as m m 1 1 m m M = = = = (12.12) m n N N N m ðÞ m/m m i m w i i ∑ ∑ ∑ M M M i i i i=1 i=1 i=1 Using Eqs. (12.11) and (12.12), we can now easily convert back and forth between mass and mole fractions with  m nM M i i i i w = = =χ (12.13) i i m n M M m m m m410 CHAPTER 12: Mixtures of Gases and Vapors and  n M i m χ = = w (12.14) i i n m M i Last, if the mixture is a gas or a vapor, we can determine its equivalent gas constant R from the universal gas m . . . constantℜ = 8:313kJ/ðÞ kgmole K = 1545:35ft lbf/ðÞ lbm R and Eq. (12.11) or (12.12) as ℜ R = (12.15) m M m At this point, we have developed general formulae for determining the values of thermodynamic properties and other important characteristics of mixtures of substances in their solid, liquid, vapor, or gaseous states. Before we can continue further, we need to know how these mixture thermodynamic properties are related to each other. Since the number of possible mixture compositions is infinite, the construction of thermodynamic tables or charts for all possible mixtures is impractical except for very common mixtures, such as pure air and air–water vapor mixtures. EXAMPLE 12.1 A new gas furnace requires a mixture of 50.0% propane and 50.0% air on a mass basis. Determine (a) the equivalent mole- cular mass of the mixture, (b) the mixture composition on a molar basis, and (c) the equivalent gas constant of the mixture. Solution a. The mixture composition on a mass basis is w = 0.500, and w = 0.500. The molecular masses of the components propane air are found in Table C.13 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics as M = 44:09kg/kgmole and M = 28:97kg/kgmole propane air Then, the equivalent molecular mass of the mixture is given by Eq. (12.12) as kg 1 1 = = M = 35:0 m N 0:500 0:500 kgmole + w i ∑ 44:09 28:97 M i i=1 b. The equivalent molar values are given by Eq. (12.14) as  n M i m χ = = w i i n M m i and this is used as follows:  35:0 kg/kgmole M m χ = w = 0:500 = 0:397 propane propane M 44:1 kg/kgmole propane and the remaining component is  35:0 kg/kgmole M m χ = w = 0:500 = 0:603 air air M 28:97 kg/kgmole air so that the composition is 50.0% propane and 50.0% air on a mass basis, but it is 39.7% propane and 60.3% air on a molar basis. (Note that, once we knew the propane molar concentration, we could have determined the air molar concentration through Eq. (12.10), since w +w = 1.0 andχ +χ = 1.0, soχ = 1.0− χ = propane air propane air air propane 1.0− 0.397 = 0.603.) c. The equivalent gas constant for this mixture is given by Eq. (12.15) as . 8:3143 kJ/kgmole K ℜ kJ R = = = 0:238 m . M 34:97 kg/kgmole kg K m The following example is slightly more complicated than Example 12.1, because it deals with a mixture contain- ing four components. The solution method, however, is the same as in Example 12.1.12.2 Thermodynamic Properties of Gas Mixtures 411 EXAMPLE 12.2 The molar composition of air that is normally used to determine the thermodynamic properties of air at standard tem- perature and pressure is Component Molar % Nitrogen 78.09 Oxygen 20.95 Argon 0.930 CO and trace gases 0.0300 2 Total 100.00% Determine (a) the equivalent molecular mass, (b) the gas constant for this mixture, and (c) the composition of air on a mass (or weight) basis. Solution a. Since we are given the molar composition for air, we can find its equivalent molecular weight from Eq. (12.11). Assuming that argon and carbon dioxide are the only minor components present, Table C.13 provides the necessary molecular masses as M = 28:02kg=kgmole nitrogen M = 32:00kg=kgmole oxygen M = 39:94kg=kgmole argon M = 44:01kg=kgmole carbondioxide Then, Eq. (12.11) gives 4 M =∑χ M =χ M +χ M +χ M +χ M air i N O Ar CO i N 2 O 2 Ar CO 2 2 2 2 i=1 = 0:7809ð28:02kg=kgmoleÞ+0:2095ð32:00kg=kgmoleÞ +0:00930ð39:94kg=kgmoleÞ+0:000300ð44:10kg=kgmoleÞ = 28:97kg=kgmole b. Equation (12.15) gives the gas constant of this mixture as . 8:3143kJ=ðkgmole KÞ kJ ℜ R = = = 0:287 air . M 28:97kg=kgmole kg K air whichagreeswiththe values givenin Table C.13bin Thermodynamic Tables to accompany Modern Engineering Thermodynamics. c. Equation (12.13) can be used to determine the corresponding mass or weight fraction composition as   M 28:02kg=kgmole N 2 w =χ = 0:7809 = 0:7553 = 75:53%bymass N2 N 2 M 28:97kg=kgmole air   M 32:00kg=kgmole O 2 w =χ = 0:2095 = 0:2314 = 23:14%bymass O 2 O 2 M 28:97kg=kgmole air   39:94kg=kgmole M Ar w =χ = 0:00930 = 0:0128 = 1:28%bymass Ar Ar M 28:97kg=kgmole air   M 44:01kg=kgmole CO 2 w =χ = 0:000300 = 0:00046 = 0:0456%bymass CO 2 CO 2 M 28:97kg=kgmole air Note the difference between the mass and mole fraction composition values. Exercises 1. Research has suddenly discovered that the composition given in Example 12.1 was wrong. It should have been 30.0% propane and 70.0% air on a mass basis. Determine the proper molar basis composition for this new mixture. Answer: χ = 0.220 and χ = 0.780. propane air (Continued)412 CHAPTER 12: Mixtures of Gases and Vapors EXAMPLE 12.2 (Continued) 2. Oops, Research discovered still another error; it turns out that the composition given in Example 12.1 was correct except that it is 50.0% propane and 50.0% air on a molar basis. Now , Research wants you to determine the corresponding mass based composition. Answer: w = 0.603 and w = 0.397. propane air 3. A more detailed composition for air than that given in Example 12.2 is Component Molar % (Answer: Mass %) Nitrogen (N ) 78.084 75.519 2 Oxygen (O ) 20.948 23.143 2 Argon (Ar) 0.934 1.288 Carbon dioxide (CO ) 0.0314 0.0477 2 Neon (Ne) 0.00182 0.00127 Helium (He) 0.000520 0.0000720 Methane (CH ) 0.000200 0.000110 4 Krypton (Kr) 0.000110 0.000320 Hydrogen (H ) 0.0000500 0.00000350 2 Dinitrogen monoxide (N O) 0.0000500 0.0000620 2 Xenon (Xe) 0.00000800 0.0000360 Total = 100.000% 100.000% Determine the equivalent molecular mass and the corresponding composition on a mass (or weight) basis. Answer: = 28.97 kg/kgmole = 28.97 lbm/lbmole, and see the table. M m 12.3 MIXTURES OF IDEAL GASES A mixture of ideal gases behaves as a unique ideal gas with an equivalent molecular mass M and gas m constant R given by Eqs. (12.11) or (12.12) and (12.15). Ideal gas mixtures obey all of the ideal gas m equations of state: p V = m R T m m m m m Z T m2 u −u = c dT m2 m1 vm m Tm1 Z T m2 h −h = c dT m2 m1 pm m T m1 and Z T m2 s −s = ðÞ c /T dT +R lnðÞ v /v m2 m1 vm m m m m2 m1 Tm1 Z T m2  = c /T dT −R lnðÞ p /p pm m m m m2 m1 T m1 where p and T are the mixture pressure and temperature, respectively. If the mixture can be considered to m m have constant specific heats, then these equations reduce to u −u = cðÞ T −T m2 m1 vm m2 m1 h −h = cðÞ T −T m2 m1 pm m2 m1 s −s = c lnðÞ T /T +R lnðÞ v /v m2 m1 vm m2 m1 m m2 m1 lnðÞ T /T −R lnðÞ p /p = c pm m2 m1 m m2 m1 From p V = m R T and Eqs. (12.3), (12.12), and (12.15), we find that, for a mixture of ideal gases, m m m m m  m R T T m m m ℜ m V = = m m m p M p m m m  N N w T ℜT m i m m i = m ℜ ∑ = ∑ m M p p M i m m i i=1 i=112.3 Mixtures of Ideal Gases 413 Then, Eq. (12.6) can be used to find   ∂V T RT m ℜ m i m v = = = = v (12.16) i i ∂m M p p i i m m p ,T ,m m m j where v is the specific volume of gas i at the pressure and temperature of the mixture. Similarly, using the i appropriate equations for the partial specific internal energy u, enthalpy h,andentropy s, we can show that, i i i for a mixture of ideal gases with constant specific heats, the partial specific properties of gas i in the mixture are the same as the corresponding specific properties; that is, u = c ðT −T Þ i vi m 0 h = c ðT −T Þ i pi m 0 and s = s = c lnðT /T Þ−R lnðp /p Þ i i pi m 0 i m 0 where T and p are arbitrary reference state values. 0 0 These relations were also discovered experimentally in the 19th century and are now known as the Gibbs-Dalton and Amagat laws. In 1801, John Dalton (1766–1844) carried out a series of experiments that led him to con- clude that the total pressure p of a mixture of ideal gases was equal to the sum of the partial pressures of the m individual component gases in the mixture, where the partial pressure p of gas i in a mixture of ideal gases is i the pressure gas i would exert if it alone occupied the volume of the mixture at the temperature of the mixture. This is known as Dalton’s law of partial pressures, and it can be written as N p =∑p = p +p +p ++p (12.17) m i 1 2 3 N i=1 where mRT i i m p = (12.18) i V m Later, Emile Amagat (1841–1915) discovered experimentally that the total volume V of a mixture of ideal m gases was equal to the sum of the partial volumes V of the individual component gases in the mixture, where i the partial volume V of gas i in a mixture of ideal gases is the volume gas i would occupy if it alone was at i thepressureandtemperatureofthemixture.Thisisknownas Amagat’s law of partial volumes,anditcanbe written as N V =∑V = V +V +V ++V (12.19) m i 1 2 3 N i=1 where  mRT RT i i m i m V = or v = (12.20) i i p p m m WHAT IS DALTON’S LAW OF PARTIAL PRESSURES? Dalton’s law of partial pressures states that the partial pressure p of ideal gas i in a mixture of ideal gases is equal to the i pressure gas i would exert if it alone occupied the volume of the mixture at the temperature of the mixture. WHAT IS AMAGAT’S LAW OF PARTIAL VOLUMES? Amagat’s law of partial volumes states that the partial volume V of ideal gas i in a mixture of ideal gases is equal to the i volume gas i would occupy if it alone was at pressure and temperature of the mixture.414 CHAPTER 12: Mixtures of Gases and Vapors Finally, the thermodynamic description of a mixture of ideal gases was completed through the work of Josiah Willard Gibbs (1838–1903), who generalized Dalton’s law to define all the partial properties (except volume) of the components in the mixture to be equal to the values that those properties would have if each component gas alone occupied the volume of the mixture at the temperature of the mixture and at the partial pressure of that component. The Gibbs-Dalton ideal gas mixture lawpresumesthatnomolecularinteractionstakeplace between the components of the mixture, because each component is presumed to behave as though the other components were not present. Undertheseconditions,weconcludethatalltheextensiveproperties of a mixture of ideal gases are conserved, and the mixture value of any extensive property can be determined by summing the contributions made by each gas present in the mixture. Therefore, for ideal gases only, N N V =∑V =∑mv (12.21) m i i i i=1 i=1 and the specific volume of this mixture can be calculated from N N V m m i v = =∑ v =∑wv (12.22) m i i i m m m m i=1 i=1 where w is the mass fraction of ideal gas i in the mixture and v is the specific volume of that gas determined at i i thepressureandtemperatureofthemixture(i.e., v = RT /p ). Table 12.5 lists the mass and molar equations i i m m for all the total and specific properties of a mixture of ideal gases. The mass fraction w,themolefraction χ,thevolumefraction ψ and the pressure fraction π make four i i i, i composition measures that can be used to describe a mixture. However, for ideal gases, there is a simple relation between these four quantities. From Eqs. (12.16) and (12.18), we can write the pressure and volume fractions as p mR T /V i i i m m wR wM n i i i m i π = = = = = =χ i i p m R T /V R M n m m m m m m i m and mRT /p V wR wM n i i i m m i i i m i ψ = = = = = =χ i i V m R T /p R M n m m m m m m i m Table 12.5 Mass and Molar Total and Specific Thermodynamic Properties of Ideal Gas Mixtures Total Property Mass Specific Property Molar Specific Property N N N N V =∑V =∑m v v =∑wv v =∑χ v m i i m i i m i i i i=1 i=1 i=1 i=1 N N N N U =∑U =∑mu u =∑w u u =∑χ u m i i i m i i m i i i=1 i=1 i=1 i=1 N N N N H =∑H =∑mh h =∑w h h =∑χ h m i i i m i i m i i i=1 i=1 i=1 i=1 N N N N S =∑S =∑ms s =∑w s s =∑χ s m i i i m i i m i i i=1 i=1 i=1 i=1 N N c =∑χ c vm i vi c =∑w c vm i vi i=1 i=1 N N c =∑χ c pm pi c =∑w c i pm i pi i=1 i=112.3 Mixtures of Ideal Gases 415 Consequently, for a mixture of ideal gases, we have  p V n M i i i m =π = =ψ = =χ = w i i i i p V n M m m m i or  M m π =ψ =χ = w (12.23) i i i i M i which relates all four composition measures. Thus, the pressure fraction, volume fraction, and mole fraction are all equal and differ from the mass fraction by only a molecular mass ratio. EXAMPLE 12.3 An analysis of the exhaust gas from an experimental engine produces the following results on a molar basis: Carbon dioxide = 9.51% Water = 19.01% Nitrogen = 71.48% Assuming ideal gas behavior, (a) determine the volume fraction, pressure fraction, and mass fraction composition of the mixture, and (b) if the total pressure of the mixture is 14.7 psia, determine the partial pressure of the water vapor in the exhaust gas mixture. Solution a. For ideal gas behavior, Eq. (12.23) tells us that the mole fractions, volume fractions, and the pressure fractions are all the same, or χ =ψ =π = 9:51% CO CO CO 2 2 2 χ =ψ =π = 19:01% H O H O H2O 2 2 χ =ψ =π = 71:48% N N N2 2 2 The equivalent molecular mass of this ideal gas mixture is given by Eq. (12.11) as N M =∑χ M =χ M + χ M +χ M m i CO H O N i CO 2 H O 2 N 2 2 2 2 i=1 = 0:0951ð44:01kg/kgmoleÞ+0:1901ð18:02kg/kgmoleÞ+0:7148ð28:02kg/kgmoleÞ = 27:64kg/kgmole and the corresponding mass fractions are given by Eq. (12.23) as   M 44:01kg/kgmole CO 2 w =ψ =ð9:51%Þ = 15:14% CO 2 CO 2 M 27:64kg/kgmole m   M 18:02kg/kgmole H2O w =ψ =ð19:01%Þ = 12:39% H O 2 H O 2 M 27:64kg/kgmole m   M 28:02kg/kgmole N 2 w =ψ =ð71:48%Þ = 72:46% N N 2 2 M 27:64kg/kgmole m b. If the total pressure is 14.7 psia, then the partial pressure of the water vapor in the exhaust gas mixture is given by Eq. (12.23) as p = p χ =ð14:7psiaÞð0:1901Þ = 2:79psia H O m 2 H O 2 Exercises 4. Find the partial volume of the nitrogen in the exhaust gas in Example 12.3 if the total volume of the gas mixture is 3 3 8.00 ft . Answer: V = 2:14ft : N 2 5. If there are 20.0 moles of the exhaust gas mixture in Example 12.3, how many moles of carbon dioxide would be in the mixture? Answer: n = 14:3moles: CO 2 6. If there is 0.650 lbm of the exhaust gas mixture in Example 12.3, how many lbm of water vapor would be in the mixture? Answer: m = 0.081 lbm. water vapor416 CHAPTER 12: Mixtures of Gases and Vapors EXAMPLE 12.4 Though oxygen is necessary to sustain life, breathing oxygen at elevated pressure has toxic effects. It causes changes in lung tissue and affects the liver and brain. Acute oxygen poisoning at high pressures can cause convulsions that can lead to death (even at atmospheric pressure, pure oxygen can be breathed safely for only two hours). Oxygen poisoning at elevated envir- onmental pressure can be avoided by maintaining the oxygen partial pressure equal to that of atmospheric air at standard temperature and pressure (STP). Also, since atmospheric nitrogen is very soluble in blood and body tissue, rapid depressuri- zation causes nitrogen bubbles to form in the blood and tissue (nitrogen embolisms) producing a condition commonly called the bends. Divers going to great depths in the sea are able to circumvent this problem somewhat by breathing a compressed helium- oxygen mixture in which the oxygen partial pressure is adjusted so that it is always equal to its value in atmospheric air at STP. Helium, being much less soluble in body tissue than nitrogen, decreases the time required for depressurization when the diver returns to the surface. The engineering problem that we must solve is stated as follows: a. For a deep water diver, determine the proper helium-oxygen breathing mixture composition for a dive to 100. m below 2 the surface of the water, where the pressure is 1.08 MN/m . Give your answer in mole, volume, and mass fractions. b. Determine the effective gas constant, the specific heats, and the specific heat ratio for this mixture. Assume helium and oxygen behave as ideal gases with constant specific heats. Solution a. From Example 12.3 and Eq. (12.23), we find that the partial pressure of oxygen in air at STP is 2 p =χ p = 0:2095ðÞ 0:1013MN/m O m 2 O 2 2 = 0:0212MN/m 2 Therefore, at a total pressure in 100.m of water of 1.08 MN/m , this same partial pressure requires a mole and volume fraction of oxygen of only χ =ψ =π = p /p = 0:0212/1:08 = 0:0196 O O m O O 2 2 2 2 and, from Eq. (12.10), the helium mole and volume fractions are χ =ψ = 1− χ = 1−0:0196 = 0:980 He He O 2 The equivalent mass fractions are given by Eq. (12.13), where the mixture equivalent molecular mass can be com- puted from Eq. (12.11), as M =χ M +χ M = 0:0196ð32:00Þ+0:980ð4:003Þ = 4:55kg=kgmole m O He O 2 He 2 then, w =χðÞ M /M = 0:0196ð32:00/4:55Þ = 0:138 O O O m 2 2 2 and w = 1−w = 0:862 He O 2 Therefore, the required oxygen concentration is only 1.96% on a volume or molar basis, but it is 13.8% on a mass or weight basis. b. The mixture equivalent gas constant can be computed from Eq. (12.15) as . 8:3143kJ/ðÞ kgmole K ℜ . R = = = 1:86kJ/ðkg KÞ m M 4:55kg/kgmole m and the mixture specific heats can be determined from the equations in Table 12.5 and Table C.13b as . c = w c +w c = 0:138ð0:657Þ+0:862ð3:123Þ = 2:78kJ=ðkg KÞ vm O vO He vHe 2 2 and . c = w c +w c = 0:138ð0:917Þ+0:862ð5:200Þ = 4:61kJ=ðkg KÞ pm O2 pO2 He pHe12.4 Psychrometrics 417 Finally, the specific heat ratio of the mixture is c pm 4:61 k = = = 1:66 m c 2:78 vm (Note that k ≠∑wk because of its definition as a ratio.) m i i Exercises 2 7. Rework Example 12.4 for a dive to 200. m below the surface of the water, where the pressure is 2.063 MN/m . Answer: χ = 0:0103, χ = 0.990, R = 1.94 kJ/kg·K, c = 2.93 kJ/kg·K, c = 4.87 kJ/kg·K, and k = 1.66. He m vm pm O 2 2 8. Rework Example 12.4 for an argon-oxygen mixture subject to the same condition of p = 0:0212MN/m . Answer: O 2 χ = 0:0196, χ = 0.980, R = 0.209 kJ/kg·K, c = 3.08 kJ/kg·K, c = 5.13 kJ/kg·K, and k = 1.66. He m vm pm O 2 Ideal gas mixture equations are used to produce property values in thermodynamic problems just as though the mixture is a single unique gas. This is illustrated in the next example. EXAMPLE 12.5 Determine the power per unit mass flow rate required to isentropically compress the helium-oxygen mixture described in 2 2 Example 12.2 from atmospheric pressure (0.101 MN/m ) and 20.0°Cto1.08MN/m in a steady flow, steady state process. Assume the mixture has constant specific heats. Solution _ The unknown here is the power per unit mass flow rate W/m _ . Since this is an open system, the energy rate balance for a m steady state, steady flow, single-inlet, single-outlet system is (neglecting kinetic and potential energy effects) ERB(SS,SF,SI,SO) _ _ Q−W +m _ðÞ h −h = 0 m m1 m2 and, since an isentropic process is normally also adiabatic, _ W/m _ = h −h = cðÞ T −T m m1 m2 pm m1 m2 For an ideal gas in an isentropic process, Eq. (7.38) gives ðk −1Þ/k m m T = T ðp /p Þ m2 m1 m2 m1 and, using the results of Example 12.4, this gives 0:66/1:66 T =ðÞ 20:0+273:15KðÞ 1:08/0:101 = 803K = 530°C m2 Then, _ . W/m _ = cðÞ T −T =½ 4:61kJ/ðkg KÞðÞ 293−803K = −2350kJ/kg m pm m1 m2 Exercises 9. Determine the isentropic power per unit mass flow rate required in Example 12.5 if the inlet temperature is 10.0°C _ rather than 20.0°C. Answer: W/m _ =−2030 kJ/kg. m 2 10. If the exit pressure in Example 12.5 is increased to 2.06 MN/m , determine the required isentropic power input per unit _ mass flow rate. Answer: W/m _ =−3110 kJ/kg. m 11. Determine the aergonic (zero work) heat transfer required to cool the compressed mixture in Example 12.5 from 530°C _ _ back to 20.0°C again. Answer: Q/m =−2350 kJ/kg. m 12.4 PSYCHROMETRICS Psychrometrics is the study of atmospheric air, which is a mixture of pure air and water vapor at atmospheric 4 pressure. The pure air portion of an air–water vapor mixture is commonly called dry air; consequently, atmospheric air is said to consist of a mixture of dry air and water vapor. Both the air and the water vapor in 4 The term psychrometer is from the Greek psychros (cold) and meter (measure).418 CHAPTER 12: Mixtures of Gases and Vapors CRITICAL THINKING On page 77 of the July/August 1993 issue of Family Handyman magazine, a helpful hint is given on how to determine the amount of propane remaining a cook stove tank. According to this magazine you just “Pour a cup of hot water over the outside of the tank. A condensation line will appear on the tank surface at the level of the remaining propane.” Since the formation of a “condensation line” requires reducing the liquid-vapor interface inside the tank to a temperature below the dew point temperature, can you explain how this test works? Are there any conditions under which this test would not work? (Hint: Look at the thermodynamics of the propane’s evaporation and condensation processes that result from the heating by the hot water and the subsequent cooling by the local atmosphere.) this mixture are treated as ideal gases (even though we say water vapor and not water gas). This particular mixture of ideal gases is important because of its meteorological and environmental comfort implications. To begin this discussion, we define two new composition measures for the amount of water vapor present in the 5 mixture. Both measures are a type of humidity, as is shown. 1. The relative humidity ϕ is the ratio of the actual partial pressure of the water vapor present in the mixture to the saturation pressure of the water vapor at the temperature of the mixture, or p w Relativehumidity =ϕ = (12.24) p sat The value of p can be found in Table C.1 in Thermodynamic Tables to accompany Modern Engineering Thermo- sat dynamics at the temperature of the mixture. Since 0≤ ϕ≤ 1, the relative humidity is normally reported as a percentage. This is the common meteorological humidity measure. 2. The humidity ratioω is the ratio of the mass of water vapor present in the mixture divided by the mass of dry air present in the mixture, or m w Humidityratioω = (12.25) m a where m = m + m , and p = p + p = atmospheric pressure. Assuming ideal gas behavior for both the m a w m a w air and water vapor, we can write m = p V /ðÞ R T and m = p V /ðÞ R T , then w w m w m a a m a m  p R p M 18:016p p p w a w w w w w ω = (12.26a) = = = 0:622 = 0:622 p R p M 28:97p p p −p a w a a a a m w From Eq. (12.24), we find that p =ϕp , and substituting this into Eq. (12.26a) w sat providesaformulathatrelatesthetwohumiditymeasures: Dew point temperature  p ϕp isotherm sat sat ω = 0:622ϕ = 0:622 (12.26b) p p −p a m w p Atmospheric T w A colorful term from the meteorological profession is the dew point tempera- temperature ture T , which is the temperature at which liquid water (dew) condenses isotherm DP out of the atmosphere at constant atmospheric pressure (and consequently at constant water vapor partial pressure): T = TðÞ evaluatedatp (12.27) DP sat w ω FIGURE 12.1 If the partial pressure of the water vapor (p ) is known, then the dew point w The partial pressure and dew point temperature of temperature can be found in Table C.2. Figure 12.1 illustrates these concepts a mixture of water vapor and dry air. on a pressure-specific volume schematic. 5 Since neither of these two humidity measures corresponds to any of the four composition measures previously discussed, this brings the number of composition measures used in this chapter to six.12.4 Psychrometrics 419 EXAMPLE 12.6 On a particular day, the weather forecast states that the relative humidity is 56.8% when the atmospheric temperature and pressure are 25.0°C and 0.101 MPa, respectively. Determine: a. The partial pressure of the water vapor in the atmosphere. b. The humidity ratio of the atmosphere. c. The dew point temperature of the atmosphere. Solution a. From Table C.1b, we find that pðÞ 25:0°C = 0:003169MPa sat and, from Eq. (12.24), we can calculate the partial pressure of the water vapor present in the mixture as p =ϕp = 0:568ðÞ 0:003169MPa = 0:00180MPa = 1:80kPa w sat b. From Dalton’s law for partial pressure, we can find the partial pressure of the dry air in the mixture as p = p −p = 101−1:8 = 99:2kPa a m w then, Eq. (12.26a) gives the humidity ratio ω as ω = 0:622ðÞ p /p = 0:622ðÞ 1:80/99:2 = 0:0113kgH Operkgof dryair w a 2 Note that, since the value of ω is not constrained to lie between 0 and 1, it is not reported as a percentage. c. Using Eq. (12.27) and Table C.2b, we find the dew point temperature to be T = TðÞ 0:00180MPa = 15:8°C DP sat Exercises 12. If the relative humidity in Example 12.6 is 45.0% rather than 56.8% and all the remaining variables are the same, determine the new dew point temperature. Answer: T = 12.1°C. DP 13. Suppose the atmospheric temperature in Example 12.6 is 20.0°C rather than 25.0°C and all other variables remain the same. Determine the humidity ratio of this mixture. Answer: ω = 0.00830 kg H O per kg of dry air. 2 14. Rework Example 12.6 for a relative humidity of 35.0%, an atmospheric temperature of 20.0°C, and an atmospheric −3 pressure of 0.101 MPa. Answer: (a) p = 0.820 kPa, (b)ω = 5.00 × 10 kg H O per kg of dry air, (c) T = 4.00°C. w 2 DP The steady state, steady flow, isothermal boundary energy and entropy rate balances for a mixture of dry air and water vapor with negligible flow stream kinetic and potential energies can be written either on an unmixed component basis as _ _ _ _ Q−W +mðÞ h −h +mðÞ h −h = 0 (12.28) a 1 2 w 1 2 a w and _ _ Q/T +m _ðÞ s −s +m _ðÞ s −s +S = 0 (12.29) b a 1 2 w 1 2 p a w or on a premixed mixture basis as _ _ Q−W +m _ðÞ h −h = 0 m 1 2 m and _ _ _ Q/T +mðÞ s −s +S = 0 b m 1 2 p m where the mixture enthalpy and entropy changes are given by ðh −h Þ = w ðh −h Þ +w ðh −h Þ 1 2 a 1 2 w 1 2 m a w and ðs −s Þ = w ðs −s Þ +w ðs −s Þ 1 2 a 1 2 w 1 2 m a w420 CHAPTER 12: Mixtures of Gases and Vapors In these formulae, h is found in the gas tables (Table C.16), h is found in the superheated steam tables, and w a w a and w are the mass fractions of the dry air and water vapor. However, since psychrometrics involves only a two- w component mixture, there is no particular advantage to using the complicated premixed mixture formula. There- fore, we confine our attention to the simpler unmixed component form illustrated in Eqs. (12.28) and (12.29). 12.5 THE ADIABATIC SATURATOR Evaporative humidification processes normally occur without external heat transfer and are therefore adiabatic. If the outlet of an evaporative humidifier is saturated with water vapor (ϕ = 100%), then the device is known as an adiabatic saturator. A simple adiabatic saturator is shown in Figure 12.2. It consists of an inlet air–water vapor flow stream at temperature T , a liquid makeup water flow stream at temperature T , and an outlet air–water 1 2 vapor flow stream. If the unit is insulated and made long enough, the outlet flow stream will be saturated with water vapor, and the temperature T of the outlet flow stream is then called the adiabatic saturation temperature. 3 Since this device is adiabatic and aergonic, its steady state, steady flow energy rate balance (ERB) reduces to (neglecting changes in flow stream kinetic and potential energy) _ _ _ _ _ m h +m h +m h −m h −m h = 0 a1 a1 w1 w1 w2 w2 a3 a3 w3 w3 wherewehavechosentoseparatethecontributions from the air and water components according to Eq. (12.28). From the conservation of mass, m _ = m _ = m _ and m _ = m _ −m _ . a1 a3 a w2 w3 w1 Then, the ERB becomes _ _ _ _ _ m ðh −h Þ+ðm −m Þh +m h −m h = 0 a a1 a3 w3 w1 w2 w1 w1 w3 w3 or _ _ _ m ðh −h Þ+m ðh −h Þ+m ðh −h Þ = 0 a a1 a3 w1 w1 w2 w3 w2 w3 _ _ _ _ _ Dividing by m and introducing the humidity ratios ω = m /m and ω = m /m , and solving for ω gives a 1 w1 a 3 w3 a 1 ðh −h Þ+ω ðh −h Þ a3 a1 3 w3 w2 ω = (12.30) 1 h −h w1 w2 Since we can treat the air here as an ideal gas and assuming T = T , 3 2 h −h = c ðT −T Þ = c ðT −T Þ a3 a1 pa 3 1 pa 2 1 and since the liquid makeup water is only a slightly compressed liquid, we can write h ≈h ðT Þ = h w2 f 2 f2 Finally, since the outlet state contains saturated water vapor at the adiabatic saturation temperature, T = T , 3 2 h = h ðT Þ = h ðT Þ = h w3 g 3 g 2 g2 The water vapor in the inlet region is superheated. A quick check of the Mollier diagram (Figure 7.15) reveals that the isotherms in the low-pressure superheated region are very nearly horizontal. Therefore, the enthalpy of water vapor in this region depends only on temperature, so we can take h = h ðT Þ = h w1 g 1 g1 Insulation Evaporation 1 3 Air−water Saturated vapor air−water mixture vapor mixture Water Liquid makeup water 2 FIGURE 12.2 An adiabatic saturator.T = constant WB 12.6 The Sling Psychrometer 421 Then, Eq. (12.30) becomes Adiabatic saturator inlet humidity ratio c ðT −T Þ+ω ðh Þ pa 2 1 3 fg2 ω = (12.31) 1 h −h g1 f2 Thus, by simply measuring the inlet temperature T and the outlet adiabatic saturation temperature T = T,we 1 2 3 can calculate the inlet humidity ratio of the air–water vapor mixture, ω,directlyfromEq.(12.31).However, 1 an adiabatic saturator must be extremely long to obtain 100% relative humidity at the outlet. This difficulty is overcome by the sling psychrometer discussed next. 12.6 THE SLING PSYCHROMETER Figure 12.3 illustrates a simple device for determining air humidity, called a sling psychrometer.Itcontainstwo thermometers, one of which is covered with a wick saturated with ambient temperature liquid water. These two thermometers are called dry bulb and wet bulb. When the sling psychrometer is spun rapidly in the air, the eva- poration of the water from the wick causes the wet bulb thermometer to read lower than the dry bulb thermo- meter. After the psychrometer has been spun long enough for the thermometers to reach equilibrium temperatures, the unit is stopped and the two thermometers are quickly read. A psychrometric chart (or table) is then used to convert the dry bulb temperature T and the wet bulb temperature T into humidity informa- DB WB tion. The wet bulb temperature is approximately equal to the adiabatic saturation temperature, so T ≈ T = T WB 2 3 in Eq. (12.31). Figure 12.4 illustrates the major characteristics of a psychrometric chart. Larger charts of professional engineering quality can be found in Charts D.5 and D.6 of Thermodynamic Tables to accompany Modern Engineering Wet bulb thermometer 010 20 30 40 50 60 70 80 90 100 010 20 30 40 50 60 70 80 90 100 Swivel Dry bulb Water soaked thermometer (wet) wick Handle FIGURE 12.3 A sling psychrometer. T WB State point T = T WB DP p ω w T DB Dry bulb temperature FIGURE 12.4 The elements of a psychrometric chart. The intersection of the dry bulb and wet bulb constant temperature lines determine the state of the water vapor in the system, from which T , p , ϕ, andω can then be found. DP w Saturation curve φ = constant φ = 100% Water vapor partial pressure p w Humidity ratio ω422 CHAPTER 12: Mixtures of Gases and Vapors CRITICAL THINKING A sling psychrometer can be used to determine the humidity of the surrounding air because its wet bulb temperature is nearly the same as the adiabatic saturation temperature, and Eq. (12.31) can then be used to find ω = ω in Figure 12.2. 1 air The combined heat and mass transfer rate analysis of a wet bulb thermometer yields the following result:   2/3 ω −ω ω −ω adiabatic air WB air Pr = T −T T −T Sc air adiabatic air WB where ω = ω in Figure 12.2, ω is the humidity ratio in the vicinity of the wet bulb, T = T = T in Figure 12.2, adiabatic 3 WB air DB 1 Pr = c μ/k is the Prandtl number, and Sc =μ/(ρD) is the Schmidt number. Pr and Sc are traditional dimensionless numbers p composed of viscosity μ, constant pressure specific heat c , specific heat ratio k, density ρ, and mass diffusivity D. From the p 2/3 equation, we see that T = T only if (Pr/Sc) = 1. It turns out that, for water vapor in air, the Prandtl to Schmidt WB adiabatic number ratio is about 1.0, so the wet bulb and adiabatic saturation temperatures are about the same. However, for any 2/3 other chemical vapor–air mixture, (Pr/Sc) generally is not equal to 1, and the wet bulb and adiabatic saturation tempera- tures are not equal. Suppose you had a serious leak or spill of a dangerous liquid chemical, like a refrigerant that subsequently evaporated into the air of a closed room, and you wanted to use a sling psychrometer to estimate the resulting concentration of the chemi- cal in the air. Is it possible to use this equation to correct the wet bulb temperature (with the bulb wetted with the spilled liquid) so that it could be used with a psychrometric chart (or Eq. (12.31)) to give an estimate of the concentration of the spilled chemical in the air? Thermodynamics. Note that the dry bulb temperature is just the temperature registered on any ordinary thermo- meter and the psychrometric chart is just part of the p-T diagram for saturated and superheated water vapor in the low-pressure region. When the mixture is saturated with water vapor (ϕ = 100%), no water can evaporate from the wet bulb wick and T = T = T . Note also that a psychrometric chart is drawn for a fixed total WB DB DP pressure, thus Charts D.5 and D.6 are valid only for mixtures at 1 atm total pressure. EXAMPLE 12.7 Wet and dry bulb temperature measurements made outside on a cold day reveal that T = 5.0°Cand T = 4.0°C. Using DB WB the psychrometric chart, determine a. ϕ, ω, T , and p for the outside air DP w b. The values ofϕ,ω, T , and p if this mixture is heated at constant pressure to 25.0°C. WB w Solution a. From Chart D.6 at T = 5.0°C and T = 4.0°C, we readϕ = 80.%,ω = 0.004 kg of water vapor per kg of dry air, DB1 WB1 1 1 2 T = 20°C, and p = 700. N/m . DP1 w1 b. Now the mixture is heated at constant pressure until its dry bulb temperature increases to 25.0°C. Note that, when the temperature is stated without a modifier (i.e., “wet” or “dry”), we presume it is the ordinary, or dry bulb, temperature. Then, Chart D.6 givesϕ ≈ 20.%,ω =ω , T = 13°C, T = T , and p = p . 2 2 1 WB2 DP2 DP1 w2 w1 This is shown in Figure 12.5. Notice that, under these conditions, the relative humidity and the wet and dry bulb temperatures change, but none of the other characteristics change. This is because the amount of water vapor and the amount of air present do not change. Exercises 15. Determine the values of ϕ, ω, T , and p when the mixture in Example 12.7 is reheated to 20.0°C rather than 25.0°C WB w and all the other variables remain the same. Answer: ϕ = 28%,ω = 0.004 kg H O per kg of dry air, T = 11°C, and 2 WB 2 p = 700. N/m . w 16. If the dry bulb temperature in Example 12.7 is 8.0°C rather than 5.0°C and all the other variables remain the same, 2 determineϕ,ω, and p for the outside air. Answer: ϕ = 50%,ω = 0.0035 kg H O per kg of dry air, and p = 550 N/m . w 2 w 17. Rework Example 12.7 for a dry bulb temperature of 10.0°C and a wet bulb temperature of 8.0°C. Answer:(a) ϕ = 2 75%, ω = 0.006 kg H O per kg of dry air, T = 6.0°C, and p = 900 N/m;and(b) ϕ = 30%, ω = 0.006 kg HOper 2 DP w 2 2 kg of dry air, T = 14°C, and p = 900 N/m . WB w12.6 The Sling Psychrometer 423 T = 4°C WB1 p w T = 2°C ω DP1 2 p = 700. N/m ω = 0.004 kg/kg w1 1 5°C T DB T = 13°C WB2 T = 4°C WB1 p ω w T =T = 2°C DP2 DP1 p = p w2 w1 ω = ω 2 1 2 = 700. N/m State 2 = 0.004 kg/kg State 1 5°C 25°C T DB FIGURE 12.5 Example 12.7. If a sling psychrometer is spun such that the air velocity over the wick is greater than 3.0 m/s, then the wet bulb temperature is essentially equal to the adiabatic saturation temperature T in Eq. (12.31). The following example 2 illustrates this point. EXAMPLE 12.8 The wet and dry bulb temperatures measured in a dormitory room are 60.0°F and 70.0°F, respectively, when the barometric pressure is 14.7 psia. Assuming that the wet bulb temperature is equal to the adiabatic saturation temperature, use Eq. (12.31) to find the humidity ratio (ω) in the room and compare your answer with that obtained from the psychrometric chart, Chart D.5. Solution Here, we have T = 60.0°F and T = 70.0°F. Then, from Table C.1a in Thermodynamic Tables to accompany Modern Engineer- WB DB ing Thermodynamics, we find h = h ð70:0°FÞ= 1092:0Btu/lbm g1 g h = h ð60:0°FÞ= 1059:6Btu/lbm fg2 fg h = h ð60:0°FÞ= 28:1Btu/lbm f2 f and p = p ð60:0°FÞ¼0:2563psia w3 sat Then, Eq. (12.26a) gives ω = 0:622ðÞ 0:2563/1ðÞ 4:7−0:2563 = 0:0110lbmwaterperlbmof dryair 3 6 and, from Eq. (12.31), we get 0:240ð60−70Þ+0:0110ð1059:6Þ ω = = 0:00874lbmwaterperlbmof dryair 1 1092:0−28:1 = 0:00874ð7000Þ = 61:2grainsof waterperlbmof dryair (Continued) φ = 100% φ = 80% 1 φ = 100% φ = 80% 1 φ = 20% 2424 CHAPTER 12: Mixtures of Gases and Vapors EXAMPLE 12.8 (Continued) which, with T = 60.0°F and T = 70.0°F, the psychrometric chart, Chart D.5, gives approximately WB DB ω = 61 grainsof waterperlbmof dryair 1 which is essentially the same as that calculated from Eq. (12.31). Exercises 18. If the dry bulb temperature in the room discussed in Example 12.8 is 80.0°F rather than 70.0°F, calculate the humidity = 45.2 grains ratio in the room and compare your answer with that obtained from the psychrometric chart. Answer:ω 1 of water per lbm of dry air. 19. Rework Example 12.8 for wet and dry bulb temperatures of 65.0°F and 85.0°F, respectively. Answer: ω = 60.2 grains of 1 water per lbm of dry air. 20. Using the method of your choice, determine the relative humidity ratio in the room discussed in Example 12.8 when the wet and dry bulb temperatures are 18.0°C and 22.0°C, respectively. Answer:ω = 0.0115 kg of water per kg of dry air. 6 The grain is the smallest of the ancient Egyptian measures of weight (see Chapter 1) and originally represented the average weight of a grain of barley corn. Today, it is still used in some engineering fields (e.g., in the heating, ventilating, and air conditioning field) as a mass unit, with 7000 grains = 1 lbm. Equation (12.31) gives essentially the same values as obtained from the psychrometric chart, but the chart is much easier and quicker to use. 12.7 AIR CONDITIONING Complete air conditioning involves producing an environment with desired pressure, temperature, humidity, purity, and circulation characteristics. In this section, we are concerned only with altering the temperature and the humidity in typical air conditioning applications. In Example 12.8, we see how winter air is severely dehumidified if it is simply heated up to room temperature. Water vapor must be added to bring its humidity up into the 40 to 50% relative humidity range. This can easily be done by blowing the heated air across a moist surface, as shown in Figure 12.6. This is the technique used in a common room humidifier. The humidification process 2–3 shown in Figure 12.6 is also an example of evaporative cooling. When unsatu- rated air is brought into contact with liquid water at the same (dry bulb) temperature, some of the water evapo- rates (thus cooling the mixture) and the resulting air–water vapor mixture moves upward along the T = WB constant line, as shown in Figure 12.6b. The minimum dry bulb temperature that can be produced by evapora- tive cooling occurs when the outlet air becomes saturated with water vapor (ϕ = 100%), then T = T .This DB WB concept is illustrated in the following example. Electrical work Liquid water input input φ 1 T WB Humidification Cold Warm φ φ 1 2 3 and cooling outside 1 2 conditioned ω 3 p w 2–3 air air 2 1 φ Heating 2 1–2 T Moist DB Heater Fan screen (a) (b) FIGURE 12.6 Temperature and humidity conditioning of cold winter air. φ = 100%

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