# ch12_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 12...

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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 12 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

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Sonntag, Borgnakke and van Wylen CHAPTER 12 SUBSECTION PROB NO. Concept Problems 134-141 Mixture Composition and Properties 142-144 Simple Processes 145-151 Entropy Generation 152-156 Air Water vapor Mixtures 157-168 Review Problems 169-170 Correspondence List The correspondence between the new English unit problem set and the previous 5th edition chapter 12 problem set. New 5th SI New 5th SI New 5th SI 134 new 4 146 84 39 158 new 78 135 new 5 147 85 40 159 94 85 136 new 6 148 new 43 160 new 80 137 new 7 149 89 51 161 95 86 138 new 8 150 90a 53 162 96 87 139 new 9 151 88 55 163 97 89 140 new 10 152 86 62 164 99 92 141 new 16 153 87 66 165 new 98 142 81 21 154 91 68 166 98 102 143 82 26 155 92 72 167 100 105 144 new 30 156 90b - 168 101 115 145 83 34 157 93 76 169 103 127 170 102 130
Sonntag, Borgnakke and van Wylen Concept Problems 12.134 E If oxygen is 21% by mole of air, what is the oxygen state (P, T, v) in a room at 540 R, 15 psia of total volume 2000 ft 3 ? The temperature is 540 R, The partial pressure is P O2 = yP tot = 3.15 psia. At this T, P: v = RT/P = 48.28 × 540 3.15 × 144 (ft-lbf/lbm R) × R (lbf/in 2 ) (in/ft) 2 = 57.48 ft 3 /lbm 12.135 E A flow of oxygen and one of nitrogen, both 540 R, are mixed to produce 1 lbm/s air at 540 R, 15 psia. What are the mass and volume flow rates of each line? For the mixture, M = 0.21 × 32 + 0.79 × 28.013 = 28.85 F o r O 2 , c = 0.21 × 32 / 28.85 = 0.2329 F o r N 2 , c = 0.79 × 28.013 / 28.85 = 0.7671 Since the total flow out is 1 lbm/s, these are the component flows in lbm/s. Volume flow of O 2 in is V . = cm . v = cm . RT P = 0.2329 × 48.28 × 540 15 × 144 = 2.81 ft 3 /s Volume flow of N 2 in is V . = cm . v = cm . RT P = 0.7671 × 55.15 × 540 15 × 144 = 10.58 ft 3 /s 12.136 E A flow of gas A and a flow of gas B are mixed in a 1:1 mole ratio with same T. What is the entropy generation per kmole flow out? For this each mole fraction is one half so, Eq. 12.19: S = - R _ (0.5 ln0.5 + 0.5 ln0.5) = + 0.6931 R _ = 0.6931 × 1.98589 = 1.376 Btu/lbmol-R

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Sonntag, Borgnakke and van Wylen 12.137 E A rigid container has 1 lbm argon at 540 R and 1 lbm argon at 720 R both at 20 psia. Now they are allowed to mix without any external heat transfer. What is final T, P? Is any s generated? Energy Eq.: U 2 – U 1 = 0 = 2mu 2 - mu 1a - mu 1b = mC v (2T 2 – T 1a – T 1b ) T 2 = (T 1a + T 1b )/2 = 630 R, Process Eq.: V = constant => P 2 V = 2mRT 2 = mR(T 1a + T 1b ) = P 1 V 1a + P 1 V 1b = P 1 V P 2 = P 1 = 20 psia S due to temp changes only , not P S = m (s 2 – s 1a ) + m (s 2 – s 1b ) = mC [ ln (T 2 /T 1a ) + ln (T 2 /T 1b ) ] = 1 × 0.124 [ ln 630 540 + ln 630 720 ] = 0.00256 Btu/R Ar Ar cb
Sonntag, Borgnakke and van Wylen 12.138 E A rigid container has 1 lbm CO 2 at 540 R and 1 lbm argon at 720 R both at 20 psia. Now they are allowed to mix without any heat transfer. What is final T, P? No Q, No W so the energy equation gives constant U U = 0 = (1 × 0.201 + 1 × 0.124) × T 2 - 1 × 0.201 × 540 - 1 × 0.124 × 720 T 2 = 608.7 R, Volume from the beginning state V = [ 1 × 35.10 × 540/20 + 1 × 38.68 × 720/20 ]/144 = 16.25 ft 3 Pressure from ideal gas law and Eq.12.15 for R P 2 = (1 × 35.10 + 1 × 38.68) × 608.7/(16.25 × 144) = 19.2 psia CO 2 Ar

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Sonntag, Borgnakke and van Wylen 12.139 E A flow of 1 lbm/s argon at 540 R and another flow of 1 lbm/s CO 2 at 2800 R both at 20 psia are mixed without any heat transfer. What is the exit T, P?
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## This note was uploaded on 10/28/2009 for the course ME 2103241 taught by Professor Jtt during the Spring '09 term at Chulalongkorn University.

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ch12_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 12...

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