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## Chapter 12

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**Chapter 12**Ideal Gas Mixture and Psychrometric Applications**Learning Outcomes**• Describe ideal gas mixture composition in terms of mass fractions and mole fractions. • Use the Dalton model to relate pressure, volume, and temperature and to calculate changes in U, H, and S for ideal gas mixtures. • Apply mass, energy, and entropy balances to systems involving ideal gas mixtures, including mixing processes.**Learning Outcomes, cont.**• Demonstrate understanding of psychrometric terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature. • Use the psychrometric chart to represent common air-conditioning processes and to retrieve data. • Apply mass, energy, and entropy balances to analyze air-conditioning processes and cooling towers.**Gas 1: n1, m1**Gas 2: n2, m2 Gas j: nj, mj Sum: nm … Describing Mixture Composition • Consider a system consisting of a number of gases within a container of volume V. The temperature and pressure of the gas mixture are T and p, respectively. • The composition of the mixture can be described by giving the massmi or the number of molesni for each component present. • The massmi, number of molesni, and molecular weightMi of component i are related by • ni is in kmol when mi is in kg and Mi is in kg/kmol. • ni is in lbmol when mi is in lb and Mi is in lb/lbmol. (Eq. 12.1)**Describing Mixture Composition**• The mass fractionis the relative amount of each component in the mixture. The mass fraction mfi of component i is (Eq. 12.3) wherem is the total mass of mixture. • The sum of the mass fractions of all components in a mixture equals unity.**Describing Mixture Composition**• Alternatively, the mole fractioncan be used to describe the relative amount of each component in the mixture. The mole fraction yi of component i is (Eq. 12.6) wheren is the total moles of mixture. • The sum of the mole fractions of all components in a mixture equals unity. • The apparent (or average) molecular weightM of a mixture is determined as a mole-fraction average of the component molecular weights: (Eq. 12.9)**Describing Mixture Composition**= 34.2 kg/kmol Example: The molar analysis of a gas mixture is 50% N2, 35% CO2, and 15% O2. Determine (a) the apparent molecular weight of the mixture and (b) the analysis in terms of mass fractions. Solution: (a) The apparent molecular weight of the mixture is found using Eq. 12.9and molecular weights (rounded) from Table A-1**Describing Mixture Composition**(b) Although the actual amount of mixture is not known, the calculations can be based on any convenient amount. We use 1 kmol of mixture. • Then, the amount ni of each component, in kmol, is equal to its mole fraction, as shown in column (ii). • Column (iii) lists the respective molecular weights. • Column (iv) gives the massmi of each component, in kg per kmole of mixture, obtained using mi= niMi (Eq. 12.1). • The mass fractions, listed as percentages in column (v), are obtained by dividing the values in column (iv) by the column total and multiplying by 100.**Relating p, V, and T for Ideal Gas Mixtures**• Many systems of practical interest involve mixtures where the overall mixture and each of its components can be modeled as ideal gases. For such mixtures the Dalton mixture modelis commonly used. • The overall mixture is considered an ideal gas (Eq. 12.10) • The Dalton modelalsoassumes each component behaves as an ideal gas as if it were alone at temperatureT and volumeV.**Relating p, V, and T for Ideal Gas Mixtures**• Accordingly, with the Dalton model the individual components do not exert the mixture pressure p but rather a partial pressuredenoted by pi: (Eq. 12.11) • By combining Eqs. 12.10 and 12.11 the partial pressure pi can be determined alternatively from (Eq. 12.12) where the sum of the partial pressures equals the mixture pressure (Eq. 12.13)**Evaluating U, H, and S for Ideal Gas Mixtures**• For an ideal gas mixture, the values of U, H, and S are evaluated by adding the contribution of each component at the condition at which the component exists in the mixture. • Evaluation of the specific internal energy or specific enthalpy of a mixture componentirequires only a single intensive property: the mixture temperature, T. • Evaluation of the specific entropy of a mixture componentirequires two intensive properties. We will use the mixture temperature, T, and the partial pressure, pi.**Evaluating U, H, and S for Ideal Gas Mixtures (Molar Basis)**• Accordingly, when working on a molar basis expressions for U, H, and S of a mixture consisting of several components are:**Evaluating U, H, and S for Ideal Gas Mixtures (Molar Basis)**• The mixture specificheats and are mole-fraction averages of the respective component specific heats. (Eq. 12.23) (Eq. 12.24) • See Sec. 12.4 for applications using these expressions for U, H, S, and the specific heats.**Evaluating U, H, and S for Ideal Gas Mixtures(Mass Basis)**• When working on a mass basis the expressions for U, H, S, and specific heatsof a mixture consisting of two components – a binary mixture– are: Table 12.2**Engineering Applications of Ideal Gas Mixtures**• We encounter ideal gas mixtures in many important areas of application. Two of these are: • Systems involving chemical reactions and, in particular, combustion. For these applications we typically work on a molar basis. Combustion systems are considered in Chapter 13. • Systems for air-conditioning and other applications requiring close control of water vapor in gas mixtures. For these applications we typically work on a mass basis. Systems of this type are considered in the second part of Chapter 12.**Psychrometric Applications**• The remainder of this presentation centers on systems involving moist air. A condensed water phase may also be present in such systems. • The term moist air refers to a mixture of dry air and water vapor in which the dry air is treated as a pure component. • The Dalton model applies to moist air. • By identifying gas 1 with dry air and gas 2 with water vapor, Table 12.2 gives moist air property relations on a mass basis. • The study of systems involving moist air is known as psychrometrics.**Moist Air**• Consider a closed system consisting of moist air occupying a volume V at mixture pressure p and mixture temperature T. • In moist air the amount of water vapor present is much less than the amount of dry air: mv << manv << na. • The Dalton model applies to the mixture of dry air and water vapor:**Moist Air**1. The overall mixture and each component, dry air and water vapor, obey the ideal gas equation of state. 2. Dry air and water vapor within the mixture are considered as if they each exist alone in volume V at the mixture temperature Twhile each exerts part of the mixture pressure. 3. The partial pressurespa and pv of dry air and water vapor are, respectively pa = ya p pv = yv p (Eq. 12.41b) whereya and yvare the mole fractions of the dry air and water vapor, respectively. These moist air expressions conform toEqs. (c)of Table 12.2.**Moist Air**Mixture pressure, p , T 4. The mixture pressure is the sum of the partial pressures of the dry air and the water vapor: p = pa + pv 5. A typical state of water vapor in moist air is fixed using partial pressure pv and the mixture temperature T. The water vapor is superheated at this state. Typical state of the water vapor in moist air**Moist Air**saturated air (pv = pg) 6. When pv corresponds to pg at temperature T, the mixture is said to be saturated. 7. The ratio of pv and pg is called the relative humidity, f: (Eq. 12.44) Relative humidity is usually expressed as a percent and ranges as dry air only (pv = 0) 0 ≤ f ≤ 100%**Humidity Ratio**18.02/28.97 = 0.622 • The humidity ratiow of a moist air sample is the ratio of the mass of the water vapor to the mass of the dry air. (Eq. 12.42) Since mv << ma, the value of w is typically << 1. • Using the ideal gas equation of state and the relationship pa = p–pv (Eq. 12.43)**Mixture Enthalpy**• Values for U, H, and S for moist air can be found by adding contributions of each component. • For example, the enthalpyH is • which conforms to Eq. (d) in Table 12.2. • Dividing by ma and introducing w, the mixture enthalpy per unit mass of dry airis • For moist air, the enthalpy hv is very closely given by the saturated vapor value corresponding to the given temperature. (Eq. 12.45) (Eq. 12.46) (Eq. 12.47)**Heating Moist Air in a Duct**Example: Moist air enters a duct at 10oC, 80% relative humidity, is heated as it flows through the duct, and exits at 30oC. No moisture is added or removed and the mixture pressure remains constant at 1 bar. For steady-state operation and ignoring kinetic and potential energy changes, determine (a) the humidity ratio, w2, and (b) the rate of heat transfer, in kJ per kg of dry air.**Heating Moist Air in a Duct**Since the mass flow rates of the dry air and water vapor do not change from inlet to exit, they are denoted for simplicity as ma and mv. Moreover, since no moisture is added or removed, the humidity ratio does not change from inlet to exit: w1 = w2. The common humidity ratio is denoted by w. ∙ ∙ Solution: (a) At steady state, mass rate balances for the dry air and water vapor read:**Heating Moist Air in a Duct**The humidity ratio is evaluated using data at the inlet: • The partial pressure of the water vapor at the inlet, pv1, can be evaluated from the given inlet relative humidity f1 and the saturated pressure pg1 at 10oC from Table A-2: pv1 = f1pg1 = 0.8(0.01228 bar) = 0.0098 bar • The humidity ratio can be found from:**Heating Moist Air in a Duct**0 ∙ • Solving for Qcv (b) The steady-state form of the energy rate balance reduces to: ∙ ∙ • Noting that mv = wma, we get**Heating Moist Air in a Duct**For the dry air, ha1 and ha2 are obtained from ideal gas table Table A-22 at 10oC and 30oC, respectively. For the water vapor, hv1 and hv2 are obtained from steam table Table A-2 at 10oC and 30oC, respectively, using hv≈ hg The contribution of the water vapor to the heat transfer magnitude is relatively minor.**Dew Point Temperature**• When moist air is cooled, partial condensation of the water vapor initially present can occur. This is observed in condensation of vapor on window panes, pipes carrying cold water, and formation of dew on grass. • An important special case is cooling of moist air at constant mixture pressure, p. • The figure shows a sample of moist air, initially at State 1, where the water vapor is superheated. The accompanying T-v diagram locates states of water. • Let’s study this system as it is cooled in stages from its initial temperature.**Dew Point Temperature**• In the first part of the cooling process, the mixture pressure and water vapor mole fraction remain constant. • Since pv = yv p, the partial pressure of the water vapor remains constant. • Accordingly, the water vapor cools at constant pv from state 1 to state d, called the dew point. • The temperature at state d is called the dew point temperature. • As the system cools below the dew point temperature, some of the water vapor initially present condenses. The rest remains a vapor.**Dew Point Temperature**• At the final temperature, the system consists of the dry air initially present plus saturated water vapor and saturated liquid. • Since some of the water vapor initially present has condensed, the partial pressure of the water vapor at the final state, pg2, is less than the partial pressure initially, pv1. • The amount of water that condenses, mw, equals the difference in the initial and final amounts of water vapor: mw = mv1 – mv2**Dew Point Temperature**• Using mv = wma and the fact that the amount of dry air remains constant, the amount of water condensed per unit mass of dry air is where andp denotes the mixture pressure, which remains constant while cooling occurs.**Dry-bulb Temperature and Wet-bulb Temperature**• In engineering applications involving moist air, two readily-measured temperatures are commonly used: the dry-bulb and wet-bulb temperatures. • The dry-bulb temperature, Tdb, is simply the temperature measured by an ordinary thermometer placed in contact with the moist air. • The wet-bulb temperature, Twb, is the temperature measured by a thermometer whose bulb is enclosed by a wick moistened with water.**Dry-bulb Temperature and Wet-bulb Temperature**Moist Air in • The figure shows wet-bulb and dry-bulb thermometers mounted on an instrument called a psychrometer. Flow of moist air over the two thermometers is induced by a battery-operated fan. • Owing to evaporation from the wet wick to the moist air, the wet-bulb temperature reading is less than the dry-bulb temperature: Twb < Tdb. • Each temperature is easily read from its respective thermometer.**Psychrometric Chart**• Graphical representations of moist-air data are provided by psychrometric charts. • Psychrometric charts in SI and English units are given in Figs. A-9 and A-9E, respectively. These charts are constructed for a moist air mixture pressure of 1 atm. • Several important features of the psychrometric chart are discussed in Sec. 12.7, including**Psychrometric Chart**• Dry-bulb temperature, Tdb. Moist air state Tdb**Psychrometric Chart**• Humidity ratio, w. Moist air state w**Psychrometric Chart**• Dew point temperature, Tdp. • Since the dew point is the state where moist air becomes saturated when cooled at constant pressure, the dew point for a given state is determined from the chart by following a line of constant w (constant pv) to the saturation line where f= 100%. Moist air state Tdp**Psychrometric Chart**• Relative humidity, f. Moist air state**Psychrometric Chart**The value of (ha + whv) is calculated using ha = cpaT Fig. 12.9: TinoC,cpa= 1.005 kJ/kg-K Fig. 12.9E: TinoF,cpa = 0.24 Btu/lb-R • Mixture enthalpy per unit mass of dry air, (ha + whv). (ha + whv) Moist air state**Psychrometric Chart**• Wet-bulb temperature, Twb. • Lines of constant wet-bulb temperature are approximately lines of constant mixture enthalpy. Twb Moist air state**Psychrometric Chart**• Volume per unit mass of dry air, V/ma. • Lines giving V/ma can be interpreted as the volume of dry air or of water vapor (each per unit mass of dry air) because in keeping with the Dalton model each component is considered to fill the entire volume. Moist air state V/ma**Psychrometric Chart**Example: Using Fig. A-9, determine relative humidity, humidity ratio, and mixture enthalpy, in kJ/kg (dry air) corresponding to dry-bulb and wet-bulb temperatures of 30oC and 25oC, respectively.**Psychrometric Chart**(ha + whv) = 76 kJ/kg dry air f = 67% 25oC w = 0.0181 kg water/kg dry air Solution:**Analyzing Air-Conditioning Systems**• The next series of slides demonstrates the application of mass and energy rate balances together with property data to typical air-conditioning systems using the psychrometric principles introduced thus far. • Featured applications include • Dehumidification • Humidification • Mixing of two moist air streams • An application of psychrometric principles to a cooling tower is also considered.**Dehumidification**• The aim of a dehumidifier is to remove some of the water vapor in the moist air passing through the unit. • This is achieved by allowing the moist air to flow across a cooling coil carrying a refrigerant at a temperature low enough that some water vapor condenses.**Dehumidification**• The figure shows a control volume enclosing a dehumidifier operating at steady state. • Moist air enters at state 1. • As the moist air flows over the cooling coil, some water vapor condenses. • Saturated moist air exits at state 2 (T2 < T1). • Condensate exits as saturated liquid at state 3. Here, we take T3 = T2. 2 1 f2 = 100%, T2 < T1, w2 < w1 ∙ ma, T1, w1 3 ∙ mw T3 = T2**Dehumidification**• For the control volume, let us evaluate • The amount of condensate exiting per unit mass of dry air: mw/ma and • The rate of heat transfer between the moist air and cooling coil, per unit mass of dry air: Qcv/ma. ∙ ∙ 1 2 f2 = 100%, T2 < T1, w2 < w1 ∙ ma, T1, w1 ∙ ∙ 3 ∙ mw T3 = T2**Dehumidification**∙ ∙ ∙ ∙ Then, with mv1 = w1ma and mv2 = w2ma, where ma denotes the common mass flow rate of the dry air, we get the following expression for the amount of water condensed per unit mass of dry air ∙ (1) • Mass rate balances. At steady state, mass rate balances for the dry air and water are, respectively (dry air) (water) 1 2 Solving for the mass flow rate of the condensate f2 = 100%, T2 < T1, w2 < w1 ∙ ma, T1, w1 3 ∙ mw T3 = T2**Dehumidification**(2) ∙ ∙ ∙ ∙ With mv1 = w1ma, mv2 = w2ma, and Eq. (1), Eq.(2) becomes (3) Since heat transfer occurs from the moist air to the cooling coil, Qcv/ma will be negative in value. ∙ ∙ ∙ • Energy rate balance. With Wcv = 0 and no significant kinetic and potential energy changes, the energy rate balance for the control volume reduces at steady state to**Dehumidification**(3) • For the condensate, hw = hf (T2), where hf is obtained from Table A-2. • Options for evaluating the underlined terms of Eq. (3) include • w1 and w2 are known. Since T1 and T2 are also known, ha1 and ha2 can be obtained from ideal gas table Table A-22, while hv1 and hv2 can be obtained from steam table Table A-2 using hv = hg. (ha + whv)1 (ha + whv)2 • Alternatively, using the respective temperature and humidity ratio values to fix the states, (ha + whv) at states 1 and 2 can be read from a psychrometric chart. T2 T1