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Unformatted text preview: SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 3 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG • BORGNAKKE • VAN WYLEN Sonntag, Borgnakke and Wylen CHAPTER 3 SUBSECTION PROB NO. ConceptStudy Guide Problems 128132 Phase diagrams 133134 General Tables 135145 Ideal Gas 146148 Compressibility Factor 149, 157, 158 Review Problems 150156 Correspondence table The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and the SI number refers to the corresponding SI unit problem. New 5 th Ed. SI New 5 th Ed. SI 128 new 5 143 77E 53 129 new 7 144 new 62 130 new 9 145 79E 58 131 new 11 146 62E 69 132 new 17 147 new 65 133 new 23 148 69E c+d 70E d  134 61E 27 149 72E 81 135 68E ac 30 150 64E 113 136 68E df 30 151 new 74 137 new 40 152 81E 49 138 70E 36 153 new 99 139 73E 47 154 71E 95 140 74E 41 155 80E 61 141 new 44 156 83E 106 142 76E 51 157 65E 89 158 66E  Sonntag, Borgnakke and Wylen Concept Problems 3.128 E Cabbage needs to be cooked (boiled) at 250 F. What pressure should the pressure cooker be set for? Solution: If I need liquid water at 250 F I must have a pressure that is at least the saturation pressure for this temperature. Table F.7.1: 250 F P sat = 29.823 psia. 3.129 E If I have 1 ft 3 of ammonia at 15 psia, 60 F how much mass is that? Ammonia Tables F.8: F.8.1 P sat = 107.64 psia at 60 F so superheated vapor. F.8.2 v = 21.5641 ft 3 /lbm under subheading 15 psia m = V v = 1 ft 3 21.5641 ft 3 /lbm = 0.0464 lbm 3.130 E For water at 1 atm with a quality of 10% find the volume fraction of vapor. This is a twophase state at a given pressure: Table F.7.2: v f = 0.01 672 ft 3 /lbm, v g = 26.8032 ft 3 /lbm From the definition of quality we get the masses from total mass, m, as m f = (1 – x) m, m g = x m The volumes are V f = m f v f = (1 – x) m v f , V g = m g v g = x m v g So the volume fraction of vapor is Fraction = V g V = V g V g + V f = x m v g x m v g + (1 – x)m v f = 0.1 × 26.8032 0.1 × 26.8032 + 0.9 × 0.016 72 = 2.68032 2.69537 = 0.9944 Notice that the liquid volume is only about 0.5% of the total. We could also have found the overall v = v f + xv fg and then V = m v. Sonntag, Borgnakke and Wylen 3.131 E Locate the state of R134a at 30 psia, 20 F. Indicate in both the Pv and the Tv diagrams the location of the nearest states listed in the printed table F.10 T C.P. v 30 psia P C.P. v T 30 33.3 15.2 20 40 15.2 F 20 F 15 psia 33.3 psia Sonntag, Borgnakke and Wylen 3.132 E Calculate the ideal gas constant for argon and hydrogen based on Table F.1 and verify the value with Table F.4 The gas constant for a substance can be found from the universal gas constant from table A.1 and the molecular weight from Table F.1 Argon: R = R _ M = 1.98589 39.948 = 0.04971 Btu lbm R = 38.683 lbfft lbm R Hydrogen: R = R _ M = 1.98589 2.016 = 0.98506 Btu lbm R = 766.5 lbfft lbm R Recall from Table A.1: 1 Btu = 778.1693 lbfft Sonntag, Borgnakke and Wylen...
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 Fall '07
 Zaidi
 English, Dynamics, Thermodynamics, Vapor pressure, Water vapor

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