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- 1. 13-1 Chapter 13 GAS MIXTURESComposition of Gas Mixtures13-1C It is the average or the equivalent gas constant of the gas mixture. No.13-2C No. We can do this only when each gas has the same mole fraction.13-3C It is the average or the equivalent molar mass of the gas mixture. No.13-4C The mass fractions will be identical, but the mole fractions will not.13-5C Yes.13-6C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf),and the ratio of the mole number of a component to the mole number of the mixture is called the molefraction (y).13-7C From the definition of mass fraction, mi N M ⎛M ⎞ mf i = = i i = yi ⎜ i ⎜M ⎟ ⎟ mm N m M m ⎝ m ⎠13-8C Yes, because both CO2 and N2O has the same molar mass, M = 44 kg/kmol.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 13-213-9 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are tobe obtained .Analysis The mass fractions of A and B are expressed as mA N M MA MB mf A = = A A = yA and mf B = y B mm N m M m Mm MmWhere m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparentmolar mass of the mixture is mm N A M A + N B M B Mm = = = y AM A + yB M B Nm NmCombining the two equation above and noting that y A + y B = 1 gives the following convenient relationsfor converting mass fractions to mole fractions, MB yA = and yB = 1 − y A M A (1 / mf A − 1) + M Bwhich are the desired relations.13-10 The definitions for the mass fraction, weight, and the weight fractions are mi (mf) i = m total W = mg Wi ( wf) i = W totalSince the total system consists of one mass unit, the mass of the ith component in this mixture is xi. Theweight of this one component is then Wi = g (mf) iHence, the weight fraction for this one component is g (mf) i ( wf) i = = (mf) i ∑ g (mf) iPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 13-313-11 The moles of components of a gas mixture are given. The mole fractions and the apparent molecularweight are to be determined.Properties The molar masses of He, O2, N2, and H2O are 4.0, 32.0, 28.0 and 18.0 lbm/lbmol, respectively(Table A-1).Analysis The total mole number of the mixture is N m = N He + N O2 + N H2O + N N2 = 1 + 2 + 0.1 + 1.5 = 4.6 lbmoland the mole fractions are N He 1 lbmol y He = = = 0.217 Nm 4.6 lbmol N O2 2 lbmol 1 lbmol He y O2 = = = 0.435 2 lbmol O2 Nm 4.6 lbmol 0.1 lbmol H2O N H2O 0.1 lbmol 1.5 lbmol N2 y H2O = = = 0.0217 Nm 4.6 lbmol N N2 1.5 lbmol y N2 = = = 0.326 Nm 4.6 lbmolThe total mass of the mixture is m m = m He + m O2 + m H2O + + m N2 = N He M He + N O2 M O2 + N H2O M H2O + N N2 M N2 = (1 lbm)(4 lbm/lbmol) + (2 lbm)(32 lbm/lbmol) + (0.1 lbm)(18 lbm/lbmol) + (1.5 lbm)(28 lbm/lbmol) = 111.8 kgThen the apparent molecular weight of the mixture becomes m m 111.8 lbm Mm = = = 24.3 lbm/lbmol N m 4.6 lbmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 13-413-12 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, theaverage molar mass, and gas constant are to be determined.Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1)Analysis (a) The total mass of the mixture is m m = m O 2 + m N 2 + m CO 2 = 5 kg + 8 kg + 10 kg = 23 kgThen the mass fraction of each component becomes mO2 5 kg O2 5 kg mf O 2 = = = 0.217 8 kg N2 mm 23 kg 10 kg CO2 m N2 8 kg mf N 2 = = = 0.348 mm 23 kg m CO 2 10 kg mf CO 2 = = = 0.435 mm 23 kg(b) To find the mole fractions, we need to determine the mole numbers of each component first, mO2 5 kg N O2 = = = 0.156 kmol M O2 32 kg/kmol m N2 8 kg N N2 = = = 0.286 kmol M N2 28 kg/kmol m CO 2 10 kg N CO 2 = = = 0.227 kmol M CO 2 44 kg/kmolThus, N m = N O 2 + N N 2 + N CO 2 = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmoland N O2 0.156 kmol y O2 = = = 0.233 Nm 0.699 kmol N N2 0.286 kmol y N2 = = = 0.428 Nm 0.669 kmol N CO 2 0.227 kmol y CO 2 = = = 0.339 Nm 0.669 kmol(c) The average molar mass and gas constant of the mixture are determined from their definitions: mm 23 kg Mm = = = 34.4 kg/kmol N m 0.669 kmoland Ru 8.314 kJ/kmol ⋅ K Rm = = = 0.242 kJ/kg ⋅ K Mm 34.4 kg/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 13-513-13 The mass fractions of the constituents of a gas mixture are given. The mole fractions of the gas andgas constant are to be determined.Properties The molar masses of CH4, and CO2 are 16.0 and 44.0 kg/kmol, respectively (Table A-1)Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each componentand the total number of moles are m CH 4 75 kg m CH 4 = 75 kg ⎯ ⎯→ N CH 4 = = = 4.688 kmol M CH 4 16 kg/kmol mass m CO 2 25 kg m CO 2 = 25 kg ⎯ ⎯→ N CO 2 = = = 0.568 kmol 75% CH4 M CO 2 44 kg/kmol 25% CO2 N m = N CH 4 + N CO 2 = 4.688 kmol + 0.568 kmol = 5.256 kmolThen the mole fraction of each component becomes N CH 4 4.688 kmol y CH 4 = = = 0.892 or 89.2% Nm 5.256 kmol N CO 2 0.568 kmol y CO 2 = = = 0.108 or 10.8% Nm 5.256 kmolThe molar mass and the gas constant of the mixture are determined from their definitions, mm 100 kg Mm = = = 19.03 kg/kmol N m 5.256 kmoland Ru 8.314 kJ/kmol ⋅ K Rm = = = 0.437 kJ/kg ⋅ K Mm 19.03 kg/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 13-613-14 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and theapparent gas constant are to be determined.Properties The molar masses of H2, and N2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1)Analysis The mass of each component is determined from ⎯→ m H 2 = N H 2 M H 2 = (8 kmol)(2.0 kg/kmol) = 16 kg N H 2 = 8 kmol ⎯ ⎯→ m N 2 = N N 2 M N 2 = (2 kmol)(28 kg/kmol) = 56 kg N N 2 = 2 kmol ⎯ 8 kmol H2 2 kmol N2The total mass and the total number of moles are m m = m H 2 + m N 2 = 16 kg + 56 kg = 72 kg N m = N H 2 + N N 2 = 8 kmol + 2 kmol = 10 kmolThe molar mass and the gas constant of the mixture are determined from their definitions, mm 72 kg Mm = = = 7.2 kg/kmol N m 10 kmoland Ru 8.314 kJ/kmol ⋅ K Rm = = = 1.155 kJ/kg ⋅ K Mm 7.2 kg/kmol13-15E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and theapparent gas constant are to be determined.Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E).Analysis The mass of each component is determined from ⎯→ m H 2 = N H 2 M H 2 = (5 lbmol)(2.0 lbm/lbmol) = 10 lbm N H 2 = 5 lbmol ⎯ ⎯→ m N 2 = N N 2 M N 2 = (4 lbmol)(28 lbm/lbmol) = 112 lbm 5 lbmol H2 N N 2 = 4 lbmol ⎯ 4 lbmol N2The total mass and the total number of moles are m m = m H 2 + m N 2 = 10 lbm + 112 lbm = 122 lbm N m = N H 2 + N N 2 = 5 lbmol + 4 lbmol = 9 lbmolThe molar mass and the gas constant of the mixture are determined from their definitions, m m 122 lbm Mm = = = 13.56 lbm/lbmol N m 9 lbmoland Ru 1.986 Btu/lbmol ⋅ R Rm = = = 0.1465 Btu/lbm ⋅ R Mm 13.56 lbm/lbmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 13-713-16 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of themixture and the apparent gas constant are to be determined.Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1)Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each componentand the total number of moles are mO2 20 kg m O 2 = 20 kg ⎯ ⎯→ N O 2 = = = 0.625 kmol M O2 32 kg/kmol mass mN2 30 kg m N 2 = 20 kg ⎯ ⎯→ N N 2 = = = 1.071 kmol M N2 28 kg/kmol 20% O2 30% N2 m CO 2 50 kg 50% CO2 m CO 2 = 50 kg ⎯ ⎯→ N CO 2 = = = 1.136 kmol M CO 2 44 kg/kmol N m = N O 2 + N N 2 + N CO 2 = 0.625 + 1.071 + 1.136 = 2.832 kmolNoting that the volume fractions are same as the mole fractions, the volume fraction of each componentbecomes N O2 0.625 kmol y O2 = = = 0.221 or 22.1% Nm 2.832 kmol N N2 1.071 kmol y N2 = = = 0.378 or 37.8% Nm 2.832 kmol N CO 2 1.136 kmol y CO 2 = = = 0.401 or 40.1% Nm 2.832 kmolThe molar mass and the gas constant of the mixture are determined from their definitions, mm 100 kg Mm = = = 35.31 kg/kmol N m 2.832 kmoland Ru 8.314 kJ/kmol ⋅ K Rm = = = 0.235 kJ/kg ⋅ K Mm 35.31 kg/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 13-8P-v-T Behavior of Gas Mixtures13-17C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressuresat which oxygen and nitrogen behave as ideal gases.13-18C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existedalone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but onlyapproximately for real gas mixtures.13-19C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existedalone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but onlyapproximately for real gas mixtures.13-20C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equationof state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-Tbehavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pivi= ZiRiTi, where Zi is the compressibility factor.13-21C Component pressure is the pressure a component would exert if existed alone at the mixturetemperature and volume. Partial pressure is the quantity yiPm, where yi is the mole fraction of component i.These two are identical for ideal gases.13-22C Component volume is the volume a component would occupy if existed alone at the mixturetemperature and pressure. Partial volume is the quantity yiVm, where yi is the mole fraction of component i.These two are identical for ideal gases.13-23C The one with the highest mole number.13-24C The partial pressures will decrease but the pressure fractions will remain the same.13-25C The partial pressures will increase but the pressure fractions will remain the same.13-26C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumeseach gas would occupy if existed alone at the mixture temperature and pressure.”13-27C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of theindividual gas components.”13-28C Yes, it is correct.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 13-913-29C With Kays rule, a real-gas mixture is treated as a pure substance whose critical pressure andtemperature are defined in terms of the critical pressures and temperatures of the mixture components as ′ Pcr , m = ∑y P i cr ,i ′ and Tcr , m = ∑yT i cr ,iThe compressibility factor of the mixture (Zm) is then easily determined using these pseudo-critical pointvalues.13-30 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of themixture are to be determined.Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases, and the mixture asan ideal gas mixture.Analysis The total number of moles is N m = N Ar + N N 2 = 0.5 kmol + 2 kmol = 2.5 kmoland 0.5 kmol Ar N m Ru Tm (2.5 kmol)(8.314 kPa ⋅ m /kmol ⋅ K)(280 K) 3 2 kmol N2 Vm = = = 23.3 m 3 Pm 250 kPa 280 KAlso, 250 kPa Q P2V 2 P1V1 T 400 K = ⎯ ⎯→ P2 = 2 P1 = (250 kPa ) = 357.1 kPa T2 T1 T1 280 KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 13-1013-31 The masses of the constituents of a gas mixture at a specified pressure and temperature are given.The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixtureas an ideal gas mixture.Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1)Analysis The mole numbers of the constituents are mCO 2 1 kg mCO 2 = 1 kg ⎯→ ⎯ N CO 2 = = = 0.0227 kmol MCO 2 44 kg / kmol 1 kg CO2 3 kg CH4 mCH 4 3 kg mCH 4 = 3 kg ⎯→ ⎯ N CH 4 = = = 0.1875 kmol MCH 4 16 kg / kmol 300 K 200 kPa N m = N CO 2 + N CH 4 = 0.0227 kmol + 0.1875 kmol = 0.2102 kmol N CO 2 0.0227 kmol yCO 2 = = = 0108 . Nm 0.2102 kmol N CH 4 0.1875 kmol yCH 4 = = = 0.892 Nm 0.2102 kmolThen the partial pressures become PCO 2 = y CO 2 Pm = (0.108)(200 kPa ) = 21.6 kPa PCH 4 = y CH 4 Pm = (0.892)(200 kPa ) = 178.4 kPaThe apparent molar mass of the mixture is mm 4 kg Mm = = = 19.03 kg / kmol N m 0.2102 kmol13-32 The masses of the constituents of a gas mixture at a specified temperature are given. The partialpressure of each gas and the total pressure of the mixture are to be determined.Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture asan ideal gas mixture.Analysis The partial pressures of constituent gases are ⎛ mRT ⎞ (0.6 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(300 K) 0.3 m3 PN 2 =⎜ ⎟ = = 178.1 kPa ⎝ V ⎠ N2 0.3 m 3 0.6 kg N2 0.4 kg O2 ⎛ mRT ⎞ (0.4 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(300 K) PO 2 = ⎜ ⎟ = = 103.9 kPa 300 K ⎝ V ⎠ O2 0.3 m 3and Pm = PN 2 + PO 2 = 178.1 kPa + 103.9 kPa = 282.0 kPaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 13-1113-33 The masses, temperatures, and pressures of two gases contained in two tanks connected to each otherare given. The valve connecting the tanks is opened and the final temperature is measured. The volume ofeach tank and the final pressure are to be determined.Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture asan ideal gas mixtureProperties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1)Analysis The volumes of the tanks are ⎛ mRT ⎞ (1 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(298 K)V N2 = ⎜ ⎟ = = 0.295 m 3 1 kg N2 3 kg O2 ⎝ P ⎠ N2 300 kPa 25°C 25°C ⎛ mRT ⎞ (3 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(298 K)V O2 = ⎜ ⎟ = = 0.465 m 3 300 kPa 500 kPa ⎝ P ⎠ O2 500 kPa V total = V N 2 +V O 2 = 0.295 m 3 + 0.465 m 3 = 0.76 m 3Also, mN2 1 kg m N 2 = 1 kg ⎯ ⎯→ N N 2 = = = 0.03571 kmol M N2 28 kg/kmol mO2 3 kg m O 2 = 3 kg ⎯ ⎯→ N O 2 = = = 0.09375 kmol M O2 32 kg/kmol N m = N N 2 + N O 2 = 0.03571 kmol + 0.09375 kmol = 0.1295 kmolThus, ⎛ NRu T ⎞ (0.1295 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K) Pm = ⎜ ⎜ V ⎟ = ⎟ = 422.2 kPa ⎝ ⎠m 0.76 m 313-34 A container contains a mixture of two fluids whose volumes are given. The density of the mixture isto be determined.Assumptions The volume of the mixture is the sum of the volumes of the two constituents.Properties The specific volumes of the two fluids are given to be 0.0003 m3/kg and 0.00023 m3/kg.Analysis The mass of the two fluids are VA 0.001 m 3 mA = = = 3.333 kg v A 0.0003 m 3 /kg 1 L fluid A V 0.002 m 3 2 L fluid B mB = B = = 8.696 kg v B 0.00023 m 3 /kgThe density of the mixture is then m A + m B (3.333 + 8.696) lbf ρ= = = 4010 kg/m 3 V A +V B (0.001 + 0.002) ft 3PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 13-1213-35E A mixture is obtained by mixing two gases at constant pressure and temperature. The volume andspecific volume of the mixture are to be determined.Properties The densities of two gases are given in the problem statement.Analysis The volume of constituent gas A is mA 1 lbm VA = = = 1000 ft 3 ρA 0.001 lbm/ft 3and the volume of constituent gas B is 1 lbm gas A mB 2 lbm 2 lbm gas B VB = = = 1000 ft 3 ρB 0.002 lbm/ft 3Hence, the volume of the mixture is V = V A + V B = 1000 + 1000 = 2000 ft 3The specific volume of the mixture will then be V 2000 ft 3 v= = = 666.7 ft 3 /lbm m (1 + 2) lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 13-1313-36 The masses of components of a gas mixture are given. The apparent molecular weight of thismixture, the volume it occupies, the partial volume of the oxygen, and the partial pressure of the helium areto be determined.Properties The molar masses of O2, CO2, and He are 32.0, 44.0, and 4.0 kg/kmol, respectively (Table A-1).Analysis The total mass of the mixture is m m = m O2 + m CO2 + m He = 0.1 + 1 + 0.5 = 1.6 kgThe mole numbers of each component are m O2 0.1 kg 0.1 kg O2 N O2 = = = 0.003125 kmol 1 kg CO2 M O2 32 kg/kmol 0.5 kg He m CO2 1 kg N CO2 = = = 0.02273 kmol M CO2 44 kg/kmol m He 0.5 kg N He = = = 0.125 kmol M He 4 kg/kmolThe mole number of the mixture is N m = N O2 + N CO2 + N He = 0.003125 + 0.02273 + 0.125 = 0.15086 kmolThen the apparent molecular weight of the mixture becomes mm 1.6 kg Mm = = = 10.61 kg/kmol N m 0.15086 kmolThe volume of this ideal gas mixture is N m Ru T (0.1509 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K) Vm = = = 3.764 m 3 P 100 kPaThe partial volume of oxygen in the mixture is N O2 0.003125 kmol V O2 = y O2V m = Vm = (3.764 m 3 ) = 0.07795 m 3 Nm 0.1509 kmolThe partial pressure of helium in the mixture is N He 0.125 kmol PHe = y He Pm = Pm = (100 kPa) = 82.84 kPa Nm 0.1509 kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 13-1413-37 The mass fractions of components of a gas mixture are given. The mole fractions of eachconstituent, the mixture’s apparent molecular weight, the partial pressure of each constituent, and theapparent specific heats of the mixture are to be determined.Properties The molar masses of N2, He, CH4, and C2H6 are 28.0, 4.0, 16.0, and 30.0 kg/kmol, respectively(Table A-1). The constant-pressure specific heats of these gases at room temperature are 1.039, 5.1926,2.2537, and 1.7662 kJ/kg⋅K, respectively (Table A-2a).Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 15 kg N N2 = N2 = = 0.5357 kmol M N2 28 kg/kmol m He 5 kg N He = = = 1.25 kmol 15% N2 M He 4 kg/kmol 5% He m CH4 60 kg 60% CH4 N CH4 = = = 3.75 kmol 20% C2H6 M CH4 16 kg/kmol (by mass) m C2H6 20 kg N C2H6 = = = 0.6667 kmol M C2H6 30 kg/kmolThe mole number of the mixture is N m = N N2 + N He + N CH4 + N C2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 kmoland the mole fractions are N 0.5357 kmol y N2 = N2 = = 0.08637 Nm 6.2024 kmol N He 1.25 kmol y He = = = 0.2015 Nm 6.2024 kmol N CH4 3.75 kmol y CH4 = = = 0.6046 Nm 6.2024 kmol N C2H6 0.6667 kmol y C2H6 = = = 0.1075 Nm 6.2024 kmolThe apparent molecular weight of the mixture is m 100 kg Mm = m = = 16.12 kg/kmol N m 6.2024 kmolThe partial pressure of each constituent for a mixture pressure of 1200 kPa are PN2 = y N2 Pm = (0.08637)(1200 kPa) = 103.6 kPa PHe = y He Pm = (0.2015)(1200 kPa) = 241.8 kPa PCH4 = y CH4 Pm = (0.6046)(1200 kPa) = 725.5 kPa PC2H6 = y C2H6 Pm = (0.1075)(1200 kPa) = 129.0 kPaThe constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.15 ×1.039 + 0.05 × 5.1926 + 0.60 × 2.2537 + 0.20 × 1.7662 = 2.121 kJ/kg ⋅ KThe apparent gas constant of the mixture is R 8.314 kJ/kmol ⋅ K R= u = = 0.5158 kJ/kg ⋅ K Mm 16.12 kg/kmolThen the constant-volume specific heat is cv = c p − R = 2.121 − 0.5158 = 1.605 kJ/kg ⋅ KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 13-1513-38 The volume fractions of components of a gas mixture are given. The mixture’s apparent molecularweight and the apparent specific heats of the mixture are to be determined.Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 kg/kmol, respectively(Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 1.039,0.846, and 2.2537 kJ/kg⋅K, respectively (Table A-2).Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the molefractions, mass of each component are m O2 = N O2 M O2 = (30 kmol)(32 kg/kmol) = 960 kg m N2 = N N2 M N2 = (40 kmol)(28 kg/kmol) = 1120 kg m CO2 = N CO2 M CO2 = (10 kmol)(44 kg/kmol) = 440 kg m CH4 = N CH4 M CH4 = (20 kmol)(16 kg/kmol) = 320 kg 30% O2The total mass is 40% N2 10% CO2 m m = m O2 + m N2 + m CO2 + m CH4 20% CH4 = 960 + 1120 + 440 + 320 = 2840 kg (by volume)Then the mass fractions are m O2 960 kg mf O2 = = = 0.3380 mm 2840 kg m N2 1120 kg mf N2 = = = 0.3944 mm 2840 kg m CO2 440 kg mf CO2 = = = 0.1549 mm 2840 kg m CH4 320 kg mf CH4 = = = 0.1127 mm 2840 kgThe apparent molecular weight of the mixture is mm 2840 kg Mm = = = 28.40 kg/kmol N m 100 kmolThe constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf N2 c p , N2 + mf CO2 c p ,CO2 + mf CH4 c p ,CH4 = 0.3380 × 0.918 + 0.3944 × 1.039 + 0.1549 × 0.846 + 0.1127 × 2.2537 = 1.1051 kJ/kg ⋅ KThe apparent gas constant of the mixture is Ru 8.314 kJ/kmol ⋅ K R= = = 0.2927 kJ/kg ⋅ K Mm 28.40 kg/kmolThen the constant-volume specific heat is cv = c p − R = 1.1051 − 0.2927 = 0.8124 kJ/kg ⋅ KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 13-1613-39 The mass fractions of components of a gas mixture are given. The volume occupied by 100 kg ofthis mixture is to be determined.Properties The molar masses of CH4, C3H8, and C4H10 are 16.0, 44.0, and 58.0 kg/kmol, respectively(Table A-1).Analysis The mole numbers of each component are m CH4 60 kg 60% CH4 N CH4 = = = 3.75 kmol M CH4 16 kg/kmol 25% C3H8 m C3H8 25 kg 15% C4H10 N C3H8 = = = 0.5682 kmol (by mass) M C3H8 44 kg/kmol m C4H10 15 kg N C4H10 = = = 0.2586 kmol M C4H10 58 kg/kmolThe mole number of the mixture is N m = N CH4 + N C3H8 + N C4H10 = 3.75 + 0.5682 + 0.2586 = 4.5768 kmolThe apparent molecular weight of the mixture is mm 100 kg Mm = = = 21.85 kg/kmol N m 4.5768 kmolThen the volume of this ideal gas mixture is N m Ru T (4.5768 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(310 K) Vm = = = 3.93 m 3 P 3000 kPaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 13-1713-40E The mass fractions of components of a gas mixture are given. The mass of 5 ft3 of this mixture andthe partial volumes of the components are to be determined.Properties The molar masses of N2, O2, and He are 28.0, 32.0, and 4.0 lbm/lbmol, respectively (Table A-1E).Analysis We consider 100 lbm of this mixture for calculating the molar mass of the mixture. The molenumbers of each component are m N2 60 lbm N N2 = = = 2.1429 lbmol M N2 28 lbm/lbmol 5 ft3 m 30 lbm N O2 = O2 = = 0.9375 lbmol 60% N2 M O2 32 lbm/lbmol 30% O2 m He 10 lbm 10% He N He = = = 2.5 lbmol (by mass) M He 4 lbm/lbmolThe mole number of the mixture is N m = N N2 + N O2 + N He = 2.1429 + 0.9375 + 2.5 = 5.5804 lbmolThe apparent molecular weight of the mixture is mm 100 lbm Mm = = = 17.92 lbm/lbmol N m 5.5804 lbmolThen the mass of this ideal gas mixture is PVM m (300 psia)(5 ft 3 )(17.92 lbm/lbmol) m= = = 4.727 lbm Ru T (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(530 R)The mole fractions are N N2 2.1429 lbmol y N2 = = = 0.3840 Nm 5.5804 lbmol N O2 0.9375 lbmol y O2 = = = 0.1680 Nm 5.5804 lbmol N He 2.5 lbmol y He = = = 0.4480 Nm 5.5804 lbmolNoting that volume fractions are equal to mole fractions, the partial volumes are determined from V N2 = y N2V m = (0.3840)(5 ft 3 ) = 1.92 ft 3 V O2 = y O2V m = (0.1680)(5 ft 3 ) = 0.84 ft 3 V He = y HeV m = (0.4480)(5 ft 3 ) = 2.24 ft 3PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 13-1813-41 The mass fractions of components of a gas mixture are given. The partial pressure of ethane is to bedetermined.Properties The molar masses of CH4 and C2H6 are 16.0 and 30.0 kg/kmol, respectively (Table A-1).Analysis We consider 100 kg of this mixture. The mole numbers of each component are m CH4 70 kg N CH4 = = = 4.375 kmol M CH4 16 kg/kmol m C2H6 30 kg N C2H6 = = = 1.0 kmol 70% CH4 M C2H6 30 kg/kmol 30% C2H6The mole number of the mixture is (by mass) 100 m3 N m = N CH4 + N C2H6 = 4.375 + 1.0 = 5.375 kmol 130 kPa, 25°CThe mole fractions are N CH4 4.375 kmol y CH4 = = = 0.8139 Nm 5.375 kmol N C2H6 1.0 kmol y C2H6 = = = 0.1861 Nm 5.375 kmolThe final pressure of ethane in the final mixture is PC2H6 = y C2H6 Pm = (0.1861)(130 kPa) = 24.19 kPa13-42E The Orsat analysis (molar fractions) of components of a gas mixture are given. The mass flow rateof the mixture is to be determined.Properties The molar masses of CO2, O2, N2, and CO are 44.0, 32.0, 28.0, and 28.0 lbm/lbmol,respectively (Table A-1E).Analysis The molar fraction of N2 is y N2 = 1 − y CO2 − y O2 − y CO = 1 − 0.15 − 0.15 − 0.01 = 0.69 15% CO2 15% O2The molar mass of the mixture is determined from 1% CO 69% N2 M m = y CO2 M CO2 + y O2 M O2 + y CO M CO + y N2 M N2 (by mole) = 0.15 × 44 + 0.15 × 32 + 0.01× 28 + 0.69 × 28 = 31.00 lbm/lbmolThe specific volume of the mixture is Ru T (10.73 psia ⋅ ft 3 /lbmol ⋅ R)(660 R)v= = = 15.54 ft 3 /lbm Mixture MmP (31.00 lbm/lbmol)(14.7 psia) 20 ft/s, 1 atm 200°FThe mass flow rate of these gases is then AV (10 ft 2 )(20 ft/s) m= & = = 12.87 lbm/s v 15.54 ft 3 /lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 13-1913-43 The volumetric fractions of components of a gas mixture before and after a separation unit are given.The changes in partial pressures of the components in the mixture before and after the separation unit are tobe determined.Analysis The partial pressures before the separation unit are PCH4 = y CH4 Pm = (0.60)(100 kPa) = 60 kPa PC2H6 = y C2H6 Pm = (0.20)(100 kPa) = 20 kPa PC3H8 = y C3H8 Pm = (0.10)(100 kPa) = 10 kPaThe mole fraction of propane is 0.10 after the separation unit. The corresponding mole fractions ofmethane and ethane are determined as follows: x = 0.01 ⎯⎯→ x = 0.00808 0.6 + 0.2 + x 0.60 60% CH4 y CH4 = = 0.7425 20% C2H6 0.6 + 0.2 + 0.00808 10% C3H8 0.20 y C2H6 = = 0.2475 0.6 + 0.2 + 0.00808 (by volume) 0.00808 y C3H8 = = 0.01 0.6 + 0.2 + 0.00808The partial pressures after the separation unit are PCH4 = y CH4 Pm = (0.7425)(100 kPa) = 74.25 kPa PC2H6 = y C2H6 Pm = (0.2475)(100 kPa) = 24.75 kPa PC3H8 = y C3H8 Pm = (0.01)(100 kPa) = 1kPaThe changes in partial pressures are then ΔPCH4 = 74.25 − 60 = 14.25 kPa ΔPC2H6 = 24.75 − 20 = 4.75 kPa ΔPC3H8 = 1 − 10 = −9 kPa13-44 The partial pressure of R-134a in atmospheric air to form a 100-ppm contaminant is to bedetermined.Analysis Noting that volume fractions and mole fractions are equal, the molar fraction of R-134a in air is 100 y R134a = = 0.0001 10 6The partial pressure of R-134a in air is then PR134a = y R134a Pm = (0.0001)(100 kPa) = 0.01 kPaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 13-2013-45E The volumetric analysis of a mixture of gases is given. The volumetric and mass flow rates are tobe determined using three methods.Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 lbm/lbmol,respectively (Table A-1E).Analysis (a) We consider 100 lbmol of this mixture. Noting that volume fractions are equal to the molefractions, mass of each component are m O2 = N O2 M O2 = (30 lbmol)(32 lbm/lbmol) = 960 lbm m N2 = N N2 M N2 = (40 lbmol)(28 lbm/lbmol) = 1120 lbm 30% O2 m CO2 = N CO2 M CO2 = (10 lbmol)(44 lbm/lbmol) = 440 lbm 40% N2 10% CO2 m CH4 = N CH4 M CH4 = (20 lbmol)(16 lbm/lbmol) = 320 lbm 20% CH4The total mass is (by volume) m m = m O2 + m N2 + m CO2 + m CH4 = 960 + 1120 + 440 + 320 = 2840 lbmThe apparent molecular weight of the mixture is mm 2840 lbm Mixture Mm = = = 28.40 lbm/lbmol N m 100 lbmol 1500 psia 70°FThe apparent gas constant of the mixture is Ru 10.73 psia ⋅ ft 3 /lbmol ⋅ R R= = = 0.3778 psia ⋅ ft 3 /lbm ⋅ R Mm 28.40 lbm/lbmolThe specific volume of the mixture is RT (0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) v= = = 0.1335 ft 3 /lbm P 1500 psiaThe volume flow rate is πD 2 π (1/12 ft) 2 V& = AV = V= (10 ft/s) = 0.05454 ft 3 /s 4 4and the mass flow rate is V& 0.05454 ft 3 /s m= & = = 0.4085 lbm/s v 0.1335 ft 3 /lbm(b) To use the Amagat’s law for this real gas mixture, we first need the mole fractions and the Z of eachcomponent at the mixture temperature and pressure. The compressibility factors are obtained using Fig. A-15 to be Tm 530 R ⎫ 530 R ⎫ T R ,O2 = = = 1.902 ⎪ T R , N2 = = 2.334 Tcr,O2 278.6 R ⎪ 227.1 R ⎪ ⎪ ⎬ Z O2 = 0.94 1500 psia ⎬ Z N2 = 0.99 Pm = 2.038 ⎪ ⎪ 1500 psia PR ,O2 = = PR ,CN = = 3.049 Pcr,O2 736 psia ⎪ 492 psia ⎪ ⎭ ⎭PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 13-21 530 R ⎫ 530 R ⎫ T R ,CO2 = = 0.968 ⎪ T R ,CH4 = = 1.541 ⎪ 547.5 R ⎪ 343.9 R ⎪ 1500 psia ⎬ Z CO2 = 0.21 1500 psia ⎬ Z CO2 = 0.85 PR ,CO2 = = 1.401 ⎪ PR ,CH4 = = 2.229 ⎪ 1071 psia ⎪ ⎭ 673 psia ⎪ ⎭and Zm = ∑y Z i i = y O2 Z O2 + y O2 Z O2 + y CO2 Z CO2 + y CH4 Z CH4 = (0.30)(0.94) + (0.40)(0.99) + (0.10)(0.21) + (0.20)(0.85) = 0.869Then, Z m RT (0.869)(0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) v= = = 0.1160 ft 3 /lbm P 1500 psia V& = 0.05454 ft 3 /s V& 0.05454 ft 3 /s m= & = = 0.4702 lbm/s v 0.1160 ft 3 /lbm(c) To use Kays rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure ofthe mixture using the critical point properties of mixture gases. ′ Tcr , m = ∑yT i cr ,i = y O2 Tcr ,O2 + y N2 Tcr , N2 + y CO2 Tcr ,CO2 + y CH4 Tcr ,CH4 = (0.30)(278.6 R) + (0.40)(227.1 R) + (0.10)(547.5 R) + (0.20)(343.9 R) = 298.0 R ′ Pcr , m = ∑y P i cr ,i = y O2 Pcr ,O2 + + y N2 Pcr , N2 + y CO2 Pcr ,CO2 + y CH4 Pcr ,CH4 = (0.30)(736 psia) + (0.40)(492 psia) + (0.10)(1071 psia) + (0.20)(673 psia) = 659.3 psiaand Tm 530 R ⎫ TR = = = 1.779 ⎪ Tcr,m 298.0 R ⎪ ⎬ Z m = 0.915 (Fig. A-15) Pm = 2.275 ⎪ 1500 psia PR = = Pcr,m 659.3 psia ⎪ ⎭Then, Z m RT (0.915)(0.3778 psia ⋅ ft 3 /lbm ⋅ R)(530 R) v= = = 0.1221 ft 3 /lbm P 1500 psia V& = 0.05454 ft 3 /s V& 0.05454 ft 3 /s m= & = = 0.4467 lbm/s v 0.1221 ft 3 /lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 13-2213-46 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume ofthe mixture is to be determined using three methods.Analysis (a) Under specified conditions both O2 and N2 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas, ⎛ PV ⎞ (8000 kPa)(0.3 m3 ) ⎜NO2 = ⎜ ⎟ = = 1.443 kmol 0.3 m3 O2 ⎟ ⎠O 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K) 3 ⎝ RuT 200 K 8 MPa N 2 + O2 ⎛ PV ⎞ (8000 kPa)(0.5 m 3 )NN2 = ⎜ ⎜RT ⎟ = ⎟ = 2.406 kmol ⎠ N 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K) 3 200 K ⎝ u 0.5 m3 N2 8 MPaN m = N O 2 + N N 2 = 1.443 kmol + 2.406 kmol 200 K = 3.849 kmol 8 MPa N m RuTm (3.849 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(200 K)Vm = = = 0.8 m 3 Pm 8000 kPa(b) To use Kays rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure ofthe mixture using the critical point properties of O2 and N2 from Table A-1. But we first need to determinethe Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15), Tm 200 K ⎫ T R ,O 2 = = = 1.292 ⎪ Tcr,O 2 154.8 K ⎪ O 2: ⎬ Z O 2 = 0.77 Pm 8 MPa PR ,O 2 = = = 1.575 ⎪ Pcr,O 2 5.08 MPa ⎪ ⎭ Tm 200K ⎫ TR, N 2 = = = 1.585 ⎪ Tcr, N 2 126.2K ⎪ N 2: ⎬ Z N 2 = 0.863 Pm 8 MPa PR , N 2 = = = 2.360 ⎪ Pcr, N 2 3.39 MPa ⎪ ⎭ ⎛ PV ⎞ (8000 kPa)(0.3 m 3 ) N O2 = ⎜ ⎜ ZR T ⎟ = ⎟ = 1.874 kmol ⎠ O 2 (0.77)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K) 3 ⎝ u ⎛ PV ⎞ (8000 kPa)(0.5 m 3 ) N N2 = ⎜ ⎜ ZR T ⎟ = ⎟ = 2.787 kmol ⎠ N 2 (0.863)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K) 3 ⎝ u N m = N O 2 + N N 2 = 1.874 kmol + 2.787 kmol = 4.661 kmolThe mole fractions are N O2 1.874kmol y O2 = = = 0.402 Nm 4.661kmol N N2 2.787kmol y N2 = = = 0.598 Nm 4.661kmol ′ Tcr , m = ∑yT i cr ,i = y O 2 Tcr ,O 2 + y N 2 Tcr , N 2 = (0.402)(154.8K) + (0.598)(126.2K) = 137.7K ′ Pcr , m = ∑y P i cr ,i = y O 2 Pcr ,O 2 + y N 2 Pcr , N 2 = (0.402)(5.08MPa) + (0.598)(3.39MPa) = 4.07 MPaThen,PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 13-23 Tm 200 K ⎫ TR = = = 1.452⎪ Tcr,O 2 137.7 K ⎪ ⎬ Z m = 0.82 (Fig. A-15) Pm 8 MPa PR = = = 1.966 ⎪ Pcr,O 2 4.07 MPa ⎪ ⎭Thus, Z m N m Ru Tm (0.82)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) Vm = = = 0.79 m 3 Pm 8000 kPa(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixturetemperature and pressure, which are determined in part (b). Then, Zm = ∑y Z i i = y O 2 Z O 2 + y N 2 Z N 2 = (0.402 )(0.77 ) + (0.598)(0.863) = 0.83Thus, Z m N m Ru Tm (0.83)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K) Vm = = = 0.80 m 3 Pm 8000 kPaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 13-2413-47 [Also solved by EES on enclosed CD] The mole numbers, temperatures, and pressures of two gasesforming a mixture are given. The final temperature is also given. The pressure of the mixture is to bedetermined using two methods.Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gasbehavior. Treating the mixture as an ideal gas,Initial state : PV1 = N1RuT1 ⎫ N 2T2 (4)(200 K) ⎬ P2 = P = (5 MPa ) = 18.2 MPa 1Final state : P2V 2 = N 2 RuT2 ⎭ 1 N1T1 (1)(220 K)(b) Initially, T1 220 K ⎫ 1 kmol ArTR = = = 1.457 ⎪ Tcr,Ar 151.0 K ⎪ 220 K ⎬ Z Ar = 0.90 (Fig. A-15) 3 kmol N2 P1 5 MPa 5 MPaPR = = = 1.0278 ⎪ 190 K Pcr,Ar 4.86 MPa ⎪ ⎭ 8 MPaThen the volume of the tank is ZN Ar Ru T (0.90)(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K) V = = = 0.33 m 3 P 5000 kPaAfter mixing, Tm 200K ⎫ T R ,Ar = = = 1.325 ⎪ Tcr,Ar 151.0K ⎪ ⎪ v Ar V m / N Ar ⎪ Ar: V R , Ar = = ⎬ PR = 0.90 (Fig. A-15) Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar ⎪ ⎪ (0.33 m 3 )/(1 kmol) = = 1.278 ⎪ (8.314 kPa ⋅ m 3 /kmol ⋅ K)(151.0 K)/(4860 kPa) ⎪ ⎭ Tm 200K ⎫ TR, N 2 = = = 1.585 ⎪ Tcr, N 2 126.2K ⎪ v N2 V m / N N2 ⎪ ⎪ N 2: V R, N 2 = = ⎬ PR = 3.75 (Fig. A-15) Ru Tcr, N 2 / Pcr, N 2 Ru Tcr, N 2 / Pcr, N 2 ⎪ 3 ⎪ (0.33 m )/(3 kmol) = = 0.355 ⎪ (8.314 kPa ⋅ m 3 /kmol ⋅ K)(126.2 K)/(3390 kPa) ⎪ ⎭Thus, PAr = ( PR Pcr ) Ar = (0.90)(4.86 MPa) = 4.37 MPa PN 2 = ( PR Pcr ) N 2 = (3.75)(3.39 MPa) = 12.7 MPaand Pm = PAr + PN 2 = 4.37 MPa + 12.7 MPa = 17.1 MPaPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 13-2513-48 EES Problem 13-47 is reconsidered. The effect of the moles of nitrogen supplied to the tank on thefinal pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibilitychart with Daltons law.Analysis The problem is solved using EES, and the solution is given below."Input Data"R_u = 8.314 [kJ/kmol-K] "universal Gas Constant"T_Ar = 220 [K]P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially."N_Ar = 1 [kmol]{N_N2 = 3 [kmol]}T_mix = 200 [K]T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text"P_cr_Ar=4860 [kPa]T_cr_N2=126.2 [K]P_cr_N2=3390 [kPa]"Ideal-gas Solution:"P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank."P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure"N_mix=N_Ar + N_N2 "Moles of mixture""Real Gas Solution:"P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank"T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar"P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar"Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar"P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture"T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture"P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture"Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture"P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture"T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture"P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture"Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture"P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Daltons law. 23800" 225000 NN2 Pmix Pmix,IG [kmol] [kPa] [kPa] 180000 1 9009 9091 2 13276 13636 S olu tio n M ethod 3 17793 18182 135000 P mix [kPa] 4 23254 22727 C h art 5 30565 27273 Ideal G as 6 41067 31818 90000 7 56970 36364 8 82372 40909 45000 9 126040 45455 10 211047 50000 0 1 2 3 4 5 6 7 8 9 10 N N 2 [km ol]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 13-26Properties of Gas Mixtures13-49C Yes. Yes (extensive property).13-50C No (intensive property).13-51C The answers are the same for entropy.13-52C Yes. Yes (conservation of energy).13-53C We have to use the partial pressure.13-54C No, this is an approximate approach. It assumes a component behaves as if it existed alone at themixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 13-2713-55 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabaticcompression process. The makeup of the mixture on a mass basis and the internal energy change per unitmass of mixture are to be determined.Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and themixture as an ideal gas mixture.Properties 1 The molar masses of CO2, CO, O2, and N2 are 44.0, 28.0, 32.0, and 28.0 kg/kmol, respectively(Table A-1). 2 The process is reversible.Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass ofthe mixture is determined to be M m = y CO 2 M CO 2 + y CO M CO + y O 2 M O 2 + y N 2 M N 2 = (0.15)(44) + (0.05)(28) + (0.10)(32) + (0.70)(28) = 30.80 kg/kmolThe mass fractions are M CO 2 44 kg/kmol mf CO 2 = y CO 2 = (0.15) = 0.2143 Mm 30.80 kg/kmol M CO 28 kg/kmol mf CO = y CO = (0.05) = 0.0454 Mm 30.80 kg/kmol M O2 32 kg/kmol mf O 2 = y O 2 = (0.10) = 0.1039 15% CO2 Mm 30.80 kg/kmol 5% CO M N2 28 kg/kmol 10% O2 mf N 2 = y N 2 = (0.70) = 0.6364 Mm 30.80 kg/kmol 70% N2 300 K, 1 barThe final pressure of mixture is expressed from ideal gas relation to be T2 T P2 = P1 r = (100 kPa )(8) 2 = 2.667T2 (Eq. 1) T1 300 Ksince the final temperature is not known. We assume that the process is reversible as well being adiabatic(i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determinedfrom EES to be: T = 300 K, P = (0.2143 × 100) = 21.43 kPa ⎯ ⎯→ s CO 2 ,1 = 5.2190 kJ/kg.K T = 300 K, P = (0.04545 × 100) = 4.55 kPa ⎯ ⎯→ s CO,1 = 79483 kJ/kg.K T = 300 K, P = (0.1039 × 100) = 10.39 kPa ⎯ ⎯→ s N 2 ,1 = 6.9485 kJ/kg.K T = 300 K, P = (0.6364 × 100) = 63.64 kPa ⎯ ⎯→ s O 2 ,1 = 7.0115 kJ/kg.KThe final state entropies cannot be determined at this point since the final pressure and temperature are notknown. However, for an isentropic process, the entropy change is zero and the final temperature and thefinal pressure may be determined from Δs total = mf CO 2 Δs CO 2 + mf CO Δs CO + mf O 2 Δs O 2 + mf N 2 Δs N 2 = 0and using Eq. (1). The solution may be obtained using EES to be T2 = 631.4 K, P2 = 1684 kPaThe initial and final internal energies are (from EES)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 13-28 u CO 2 ,1 = −8997 kJ/kg u CC,1 = −4033 kJ/kg T1 = 300 K ⎯ ⎯→ u O 2 ,1 = −76.24 kJ/kg u N 2 ,1 = −87.11 kJ/kg, u CO 2 , 2 = −8734 kJ/kg u CO, 2 = −3780 kJ/kg T2 = 631.4 K ⎯ ⎯→ u O 2 ,2 = 156.8 kJ/kg u N 2 ,2 = 163.9 kJ/kgThe internal energy change per unit mass of mixture is determined from Δu mixture = mf CO 2 (u CO 2 , 2 − u CO 2 ,1 ) + mf CO (u CO, 2 − u CO,1 ) + mf O 2 (u O 2 , 2 − u O 2 ,1 ) + mf N 2 (u N 2 , 2 − u N 2 ,1 ) = 0.2143[(−8734) − (−8997)] + 0.0454[(−3780) − (−4033)] + 0.1039[156.8 − (−76.24)]6 + 0.6364[163.9 − (−87.11)] = 251.8 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 13-2913-56 Propane and air mixture is compressed isentropically in an internal combustion engine. The workinput is to be determined.Assumptions Under specified conditions propane and air can be treated as ideal gases, and the mixture asan ideal gas mixture.Properties The molar masses of C3H8 and air are 44.0 and 28.97 kg/kmol, respectively (TableA-1).Analysis Given the air-fuel ratio, the mass fractions are determined to be AF 16 mf air = = = 0.9412 AF + 1 17 1 1 mf C3H8 = = = 0.05882 AF + 1 17The molar mass of the mixture is determined to be 1 1 Propane Mm = = = 29.56 kg/kmol mf air mf C3H8 0.9412 0.05882 Air + + 95 kPa M air M C3H8 28.97 kg/kmol 44.0 kg/kmol 30ºCThe mole fractions are Mm 29.56 kg/kmol y air = mf air = (0.9412) = 0.9606 M air 28.97 kg/kmol Mm 29.56 kg/kmol y C3H8 = mf C3H8 = (0.05882) = 0.03944 M C3H8 44.0 kg/kmolThe final pressure is expressed from ideal gas relation to be T2 T2 P2 = P1 r = (95 kPa )(9.5) = 2.977T2 (1) T1 (30 + 273.15) Ksince the final temperature is not known. Using Dalton’s law to find partial pressures, the entropies at theinitial state are determined from EES to be: T = 30°C, P = (0.9606 × 95) = 91.26 kPa ⎯ ⎯→ s air ,1 = 5.7417 kJ/kg.K T = 30°C, P = (0.03944 × 95) = 3.75 kPa ⎯ ⎯→ s C3H8 ,1 = 6.7697 kJ/kg.KThe final state entropies cannot be determined at this point since the final pressure and temperature are notknown. However, for an isentropic process, the entropy change is zero and the final temperature and thefinal pressure may be determined from Δs total = mf air Δs air + mf C3H8 Δs C3H 8 = 0and using Eq. (1). The solution may be obtained using EES to be T2 = 654.9 K, P2 = 1951 kPaThe initial and final internal energies are (from EES) u air ,1 = 216.5 kJ/kg u air , 2 = 477.1 kJ/kg T1 = 30°C ⎯ ⎯→ T2 = 654.9 K ⎯ ⎯→ u C3H8 ,1 = −2404 kJ/kg u C3H8 , 2 = −1607 kJ/kgNoting that the heat transfer is zero, an energy balance on the system gives q in + win = Δu m ⎯ ⎯→ win = Δu mwhere Δu m = mf air (u air,2 − u air,1 ) + mf C3H8 (u C3H8 ,2 − u C3H 8 ,1 )Substituting, win = Δu m = (0.9412)(477.1 − 216.5) + (0.05882)[(−1607) − (−2404)] = 292.2 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 13-3013-57 The moles, temperatures, and pressures of two gases forming a mixture are given. The mixturetemperature and pressure are to be determined.Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixtureas an ideal gas mixture. 2 The tank is insulated and thus there is no heat transfer. 3 There are no otherforms of work involved.Properties The molar masses and specific heats of CO2 and H2 are 44.0 kg/kmol, 2.0 kg/kmol, 0.657kJ/kg.°C, and 10.183 kJ/kg.°C, respectively. (Tables A-1 and A-2b).Analysis (a) We take both gases as our system. No heat, work, or mass crosses the system boundary,therefore this is a closed system with Q = 0 and W = 0. Then the energy balance for this closed systemreduces to E in − E out = ΔE system CO2 H2 0 = ΔU = ΔU CO 2 + ΔU H 2 2.5 kmol 7.5 kmol 0 =[mcv (Tm − T1 )]CO + [mcv (Tm − T1 )]H 200 kPa 400 kPa 2 2 27°C 40°CUsing cv values at room temperature and noting that m = NM,the final temperature of the mixture is determined to be (2.5 × 44 kg )(0.657 kJ/kg ⋅ °C)(Tm − 27°C) + (7.5 × 2 kg )(10.183 kJ/kg ⋅ °C)(Tm − 40°C) = 0 Tm = 35.8°C (308.8 K )(b) The volume of each tank is determined from ⎛ NRu T1 ⎞ (2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K) V CO 2 = ⎜ ⎜ ⎟ ⎟ = = 31.18 m 3 ⎝ P1 ⎠ CO 200 kPa 2 ⎛ NRu T1 ⎞ (7.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(313 K) V H2 = ⎜ ⎟ = = 48.79 m 3 ⎜ ⎝ P1 ⎟ H⎠ 400 kPa 2Thus, V m = V CO2 + V H 2 = 31.18 m 3 + 48.79 m 3 = 79.97 m 3 N m = N CO2 + N H 2 = 2.5 kmol + 7.5 kmol = 10.0 kmoland N m Ru Tm (10.0 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(308.8 K) Pm = = = 321 kPa Vm 79.97 m 3PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 13-3113-58 [Also solved by EES on enclosed CD] The temperatures and pressures of two gases forming amixture in a mixing chamber are given. The mixture temperature and the rate of entropy generation are tobe determined.Assumptions 1 Under specified conditions both C2H6 and CH4 can be treated as ideal gases, and themixture as an ideal gas mixture. 2 The mixing chamber is insulated and thus there is no heat transfer. 3There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potentialenergy changes are negligible.Properties The specific heats of C2H6 and CH4 are 1.7662kJ/kg.°C and 2.2537 kJ/kg.°C, respectively. (Table A-2b). 20°CAnalysis (a) The enthalpy of ideal gases is independent of 9 kg/s C2H6pressure, and thus the two gases can be treated independently & &even after mixing. Noting that W = Q = 0 , the steady-flow 200 kPaenergy balance equation reduces to & & & 45°C CH Ein − E out = ΔEsystem 0 (steady) =0 4.5 kg/s 4 & & Ein = E out ∑m h = ∑m h & i i& e e 0 = ∑m h − ∑m h = m & &e e & i i C2H6 (he − hi )C H + mCH 4 (he − hi )CH & 0 = [mc (T − T )] + [mc p (Te − Ti )]CH 2 6 4 & p e & i C2H6 4Using cp values at room temperature and substituting, the exit temperature of the mixture becomes 0 = (9 kg/s )(1.7662 kJ/kg ⋅ °C )(Tm − 20°C ) + (4.5 kg/s )(2.2537 kJ/kg ⋅ °C )(Tm − 45°C ) Tm = 29.7°C (302.7 K )(b) The rate of entropy change associated with this process is determined from an entropy balance on themixing chamber, & & & & S in − S out + S gen = ΔS system 0 =0 & & & [m( s1 − s 2 )] C 2 H 6 + [m( s1 − s 2 )] CH 4 + S gen = 0 & S gen = [m( s 2 − s1 )] C 2 H 6 + [m( s 2 − s1 )] CH 4 & &The molar flow rate of the two gases in the mixture is & ⎛m⎞ & 9 kg/s N C2H6 = ⎜ ⎟ = = 0.3 kmol/s ⎝ M ⎠ C 2 H 6 30 kg/kmol & ⎛m⎞ & 4.5 kg/s N CH 4 = ⎜ ⎟ = = 0.2813 kmol/s ⎝ M ⎠ CH 4 16 kg/kmolThen the mole fraction of each gas becomes 0.3 y C2H6 = = 0.516 0.3 + 0.2813 0.2813 y CH 4 = = 0.484 0.3 + 0.2813Thus,PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 13-32 ⎛ T y Pm,2 ⎞ ⎛ T ⎞ ( s 2 − s1 ) C 2 H 6 = ⎜ c p ln 2 − R ln ⎜ ⎟ ⎟ ⎜ = ⎜ c p ln 2 − R ln y ⎟ ⎟ ⎝ T1 P1 ⎠ C2H6 ⎝ T1 ⎠ C2H6 302.7 K = (1.7662 kJ/kg ⋅ K) ln − (0.2765 kJ/kg ⋅ K) ln(0.516) = 0.240 kJ/kg ⋅ K 293 K ⎛ T y Pm,2 ⎞ ⎛ T ⎞ ( s 2 − s1 ) CH 4 = ⎜ c p ln 2 − R ln ⎜ ⎟ ⎟ = ⎜ c p ln 2 − R ln y ⎟ ⎜ ⎟ ⎝ T1 P1 ⎠ CH 4 ⎝ T1 ⎠ CH 4 302.7 K = (2.2537 kJ/kg ⋅ K) ln − (0.5182 kJ/kg ⋅ K) ln(0.484) = 0.265 kJ/kg ⋅ K 318 KNoting that Pm, 2 = Pi, 1 = 200 kPa and substituting, S gen = (9 kg/s )(0.240 kJ/kg ⋅ K ) + (4.5 kg/s )(0.265 kJ/kg ⋅ K ) = 3.353 kW/K &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 13-3313-59 EES Problem 13-58 is reconsidered. The effect of the mass fraction of methane in the mixture onthe mixture temperature and the rate of exergy destruction is to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input from the Diagram Window"{Fluid1$=C2H6Fluid2$=CH4m_dot_F1=9 [kg/s]m_dot_F2=m_dot_F1/2T1=20 [C]T2=45 [C]P=200 [kPa]}{mf_F2=0.1}{m_dot_total =13.5 [kg/s]m_dot_F2 =mf_F2*m_dot_total} m_dot_total = m_dot_F1 + m_dot_F2T_o = 25 [C]"Conservation of energy for this steady-state, steady-flow control volume is"E_dot_in=E_dot_outE_dot_in=m_dot_F1*enthalpy(Fluid1$,T=T1) +m_dot_F2 *enthalpy(Fluid2$,T=T2)E_dot_out=m_dot_F1*enthalpy(Fluid1$,T=T3) +m_dot_F2 *enthalpy(Fluid2$,T=T3)"For entropy calculations the partial pressures are used."Mwt_F1=MOLARMASS(Fluid1$)N_dot_F1=m_dot_F1/Mwt_F1Mwt_F2=MOLARMASS(Fluid2$)N_dot_F2=m_dot_F2 /Mwt_F2N_dot_tot=N_dot_F1+N_dot_F2y_F1=IF(fluid1$,Fluid2$,N_dot_F1/N_dot_tot,1,N_dot_F1/N_dot_tot)y_F2=IF(fluid1$,Fluid2$,N_dot_F2/N_dot_tot,1,N_dot_F2/N_dot_tot)"If the two fluids are the same, the mole fractions are both 1 .""The entropy change of each fluid is:"DELTAs_F1=entropy(Fluid1$, T=T3, P=y_F1*P)-entropy(Fluid1$, T=T1, P=P)DELTAs_F2=entropy(Fluid2$, T=T3, P=y_F2*P)-entropy(Fluid2$, T=T2, P=P)"And the entropy generation is:"S_dot_gen=m_dot_F1*DELTAs_F1+m_dot_F2*DELTAs_F2"Then the exergy destroyed is:"X_dot_destroyed = (T_o+273)*S_dot_gen 38 2000 mfF2 T3 Xdestroyed 36 [C] [kW] 0.01 95.93 20.48 34 1600 0.1 502.5 24.08 Xdestroyed [kW] 32 0.2 761.4 27 1200 T3 [C] 0.3 948.5 29.2 30 0.4 1096 30.92 28 F1 = C2H6 0.5 1219 32.3 800 26 F2 = CH4 0.6 1324 33.43 0.7 1415 34.38 mtotal = 13.5 kg/s 24 400 0.8 1497 35.18 22 0.9 1570 35.87 0.99 1631 36.41 20 0 0 0,2 0,4 0,6 0,8 1 mfF2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. 13-3413-60 A mixture of hydrogen and oxygen is considered. The entropy change of this mixture betweenthe two specified states is to be determined.Assumptions Hydrogen and oxygen are ideal gases.Properties The gas constants of hydrogen and oxygen are 4.124 and 0.2598 kJ/kg⋅K, respectively (TableA-1).Analysis The effective gas constant of this mixture is R = mf H2 R H2 + mf O2 RO2 = (0.33)(4.1240) + (0.67)(0.2598) = 1.5350 kJ/kg ⋅ K Since thetemperature of the two states is the same, the entropy change is determined from P2 150 kPa s 2 − s1 = − R ln = −(1.5350 kJ/kg ⋅ K ) ln = 2.470 kJ/kg ⋅ K P1 750 kPa13-61 A mixture of nitrogen and carbon dioxide is heated at constant pressure in a closed system. Thework produced is to be determined.Assumptions 1 Nitrogen and carbon dioxide are ideal gases. 2 The process is reversible.Properties The mole numbers of nitrogen and carbon dioxide are 28.0 and 44.0 kg/kmol, respectively(Table A-1).Analysis One kg of this mixture consists of 0.5 kg of nitrogen and 0.5 kg of carbon dioxide or 0.5 kg×28.0kg/kmol=14.0 kmol of nitrogen and 0.5 kg×44.0 kg/kmol=22.0 kmol of carbon dioxide. The constituentmole fraction are then N N2 14 kmol y N2 = = = 0.3889 N total 36 kmol N CO2 22 kmol y CO2 = = = 0.6111 N total 36 kmol 50% N2 50% CO2 QThe effective molecular weight of this mixture is (by mass) M = y N2 M N2 + y CO2 M CO2 120 kPa, 30°C = (0.3889)(28) + (0.6111)(44) = 37.78 kg/kmolThe work done is determined from 2 ∫ w = PdV = P2v 2 − P1v 1 = R(T2 − T1 ) 1 Ru 8.314 kJ/kmol ⋅ K = (T2 − T1 ) = (200 − 30)K M 37.78 kg/kmol = 37.4 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. 13-3513-62E The mass fractions of components of a gas mixture are given. This mixture is compressed in anisentropic process. The final mixture temperature and the work required per unit mass of the mixture are tobe determined.Assumptions All gases will be modeled as ideal gases with constant specific heats.Properties The molar masses of N2, He, CH4, and C2H6 are 28.0, 4.0, 16.0, and 30.0 lbm/lbmol,respectively (Table A-1E). The constant-pressure specific heats of these gases at room temperature are0.248, 1.25, 0.532, and 0.427 Btu/lbm⋅R, respectively (Table A-2Ea).Analysis We consider 100 lbm of this mixture. The mole numbers of each component are m N2 15 lbm N N2 = = = 0.5357 lbmol M N2 28 lbm/lbmol m He 5 lbm 15% N2 N He = = = 1.25 lbmol 5% He M He 4 lbm/lbmol 60% CH4 m CH4 60 lbm 20% C2H6 N CH4 = = = 3.75 lbmol M CH4 16 lbm/lbmol (by mass) m C2H6 20 lbm N C2H6 = = = 0.6667 lbmol 20 psia, 100°F M C2H6 30 lbm/lbmolThe mole number of the mixture is N m = N N2 + N He + N CH4 + N C2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 lbmolThe apparent molecular weight of the mixture is mm 100 lbm Mm = = = 16.12 lbm/lbmol N m 6.2024 lbmolThe constant-pressure specific heat of the mixture is determined from c p = mf N2 c p , N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.15 × 0.248 + 0.05 × 1.25 + 0.60 × 0.532 + 0.20 × 0.427 = 0.5043 Btu/lbm ⋅ RThe apparent gas constant of the mixture is Ru 1.9858 Btu/lbmol ⋅ R R= = = 0.1232 Btu/lbm ⋅ R Mm 16.12 lbm/lbmolThen the constant-volume specific heat is cv = c p − R = 0.5043 − 0.1232 = 0.3811 Btu/lbm ⋅ RThe specific heat ratio is cp 0.5043 k= = = 1.323 cv 0.3811The temperature at the end of the compression is ( k −1) / k 0.323/1.323 ⎛P ⎞ ⎛ 200 psia ⎞ T2 = T1 ⎜ 2 ⎜P ⎟ ⎟ = (560 R )⎜ ⎜ 20 psia ⎟⎟ = 982 R ⎝ 1 ⎠ ⎝ ⎠An energy balance on the adiabatic compression process gives win = c p (T2 − T1 ) = (0.5043 Btu/lbm ⋅ R )(982 − 560) R = 213 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

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