ch05_eng

# ch05_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 5...

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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 5 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

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Sonntag, Borgnakke and Wylen CHAPTER 5 SUBSECTION PROB NO. Concept-Study Guide Problems 139-144 Kinetic and Potential Energy 145-147 Properties from General Tables 148-150 Simple Processes 151-157 Multistep Processes and Review Problems 158-162, and 182 Solids and Liquids 164-167 Ideal Gas 168-172, and 163 Polytropic Processes 173-178 Energy equation in Rate Form 179-181 New 5th SI New 5th SI New 5th SI 139 new 1 154 new 44 169 121 95 140 new - 155 112 50 170 new 89 a 141 new 7 156 115 48 171 125 101 142 new 8 157 111 51 172 130 93 143 new 12 158 110 126 173 129 112 144 new 17 159 109 64 174 123 107 145 new 22 160 113 129 175 127 104 146 102 20 161 114 62 176 new 106 147 103 21 162 118 128 177 131 114mod 148 104 mod 32 163 124 85 178 132 115 149 105 mod 30 164 119 78 179 135 122 150 104 mod - 165 new 77 180 new 125 151 107 37 166 120 76 181 136 117 152 108 38 167 new 81 182 134 138 153 106 39 168 122 97
Sonntag, Borgnakke and Wylen Concept Problems 5.139 E What is 1 cal in english units, what is 1 Btu in ft-lbf? Look in Table A.1 for the conversion factors under energy 1 Btu = 778.1693 lbf-ft 1 cal = 4.1868 J = 4.1868 1055 Btu = 0.00397 Btu = 3.088 lbf-ft 5.140 E Work as F x has units of lbf-ft, what is that in Btu? Look in Table A.1 for the conversion factors under energy 1 lbf-ft = 1.28507 × 10 -3 Btu 5.141 E A 2500 lbm car is accelerated from 25 mi/h to 40 mi/h. How much work is that? The work input is the increase in kinetic energy. E 2 – E 1 = (1/2)m [ V 2 2 - V 2 1 ] = 1 W 2 = 0.5 × 2500 lbm [40 2 – 25 2 ] mi h 2 = 1250 [ 1600 – 625 ] lbm 1609.3 × 3.28084 ft 3600 s 2 1 lbf 32.174 lbm ft/s 2 = 2 621 523 lbf-ft = 3369 Btu

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Sonntag, Borgnakke and Wylen 5.142 E A crane use 7000 Btu/h to raise a 200 lbm box 60 ft. How much time does it take? Power = W . = F V = mg V = mg L t F = mg = 200 32.174 32.174 lbf = 200 lbf t = FL W . = 200 lbf × 60 ft 7000 Btu/h = 200 × 60 × 3600 7000 × 778.17 s = 7.9 s Recall Eq. on page 20: 1 lbf = 32.174 lbm ft/s 2 , 1 Btu = 778.17 lbf-ft (A.1)
Sonntag, Borgnakke and Wylen 5.143 E I have 4 lbm of liquid water at 70 F, 15 psia. I now add 20 Btu of energy at a constant pressure. How hot does it get if it is heated? How fast does it move if it is pushed by a constant horizontal force? How high does it go if it is raised straight up? a) Heat at 100 kPa. Energy equation: E 2 – E 1 = 1 Q 2 1 W 2 = 1 Q 2 – P(V 2 – V 1 ) = H 2 – H 1 = m(h 2 – h 1 ) h 2 = h 1 + 1 Q 2 /m = 38.09 + 20/4 = 43.09 Btu/lbm Back interpolate in Table F.7.1: T 2 = 75 F (We could also have used T = 1 Q 2 /mC = 20 / (4 × 1.00) = 5 F) b) Push at constant P. It gains kinetic energy. 0.5 m V 2 2 = 1 W 2 V 2 = 2 1 W 2 /m = 2 × 20 × 778.17 lbf-ft/4 lbm = 2 × 20 × 778.17 × 32.174 lbm-(ft/s) 2 /4 lbm = 500 ft/s c) Raised in gravitational field m g Z 2 = 1 W 2 Z 2 = 1 W 2 /m g = 20 × 778.17 lbf-ft 4 lbm × 32.174 ft/s 2 × 32.174 lbm-ft/s 2 lbf = 3891 ft

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Sonntag, Borgnakke and Wylen 5.144 E Air is heated from 540 R to 640 R at V = C. Find 1 q 2 ? What if from 2400 to 2500 R? Process: V = C Æ 1 W 2 = Ø Energy Eq.: u 2 u 1 = 1 q 2 – 0 Æ 1 q 2 = u 2 u 1 Read the u-values from Table F.5 a) 1 q 2 = u 2 u 1 = 109.34 – 92.16 = 17.18 Btu/lbm b) 1 q 2 = u 2 u 1 = 474.33 – 452.64 = 21.7 Btu/lbm case a) C v 17.18/100 = 0.172 Btu/lbm R, see F.4 case b) C v 21.7/100 = 0.217 Btu/lbm R (26 % higher)
Sonntag, Borgnakke and Wylen Kinetic and Potential Energy 5.145 E Airplane takeoff from an aircraft carrier is assisted by a steam driven piston/cylin-der with an average pressure of 200 psia. A 38 500 lbm airplane

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