ch08_eng

ch08_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 8...

This preview shows pages 1–6. Sign up to view the full content.

SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 8 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Sonntag, Borgnakke and Wylen CHAPTER 8 SUBSECTION PROB NO. Concept-Study Guide Problems 140-146 Entropy, Clausius 147-148 Reversible Processes 149-158 Entropy Generation 159-163 Entropy of a Liquid or Solid 164-166 Entropy of Ideal Gases 167-173 Polytropic Processes 174-177 Rates or Fluxes of Entropy 178-179 Review problems 180-182 This problem set compared to the fifth edition chapter 8 set. New 5th SI New 5th SI New 5th SI 140 new 7 154 90 42 168 new 92 141 new 10 155 91 45 169 104 93 142 new 11 156 92 49mod 170 105 95 143 new 14 157 93 50 171 106 102 144 new 15 158 94 59 172 107 104 145 new 16 159 96 65 173 108 105 146 new 20 160 97 63 174 110 110 147 83 21 161 98 77 175 111 119 148 84 27 162 100 72 176 109 113 149 85 32 163 101 73 177 112 115 150 86 34 164 102 78 178 new 120 151 87 35 165 new 82 179 new 124 152 88 36 166 103 86 180 95 132 153 89 39 167 new 91 181 99 130 182 113 137
Sonntag, Borgnakke and Wylen Concept Problems 8.140 E Water at 20 psia, 240 F receives 40 Btu/lbm in a reversible process by heat transfer. Which process changes s the most: constant T, constant v or constant P? d s = dq T Look at the constant property lines in a T-s diagram, Fig. 8.5. The constant v line has a higher slope than the constant P line also at positive slope. Thus both the constant P and v processes have an increase in T. As T goes up the change in s is smaller. The constant T (isothermal) process therefore changes s the most. 8.141 E Saturated water vapor at 20 psia is compressed to 60 psia in a reversible adiabatic process. Find the change in v and T. Process adiabatic: dq = 0 Process reversible: ds gen = 0 Change in s: ds = dq/T + ds gen = 0 + 0 = 0 thus s is constant Table F.7.2: T 1 = 227.96 F, v 1 = 20.091 ft 3 /lbm, s 1 = 1.732 Btu/lbm R Table F.7.2 at 60 psia and s = s 1 = 1.732 Btu/lbm R T = 400 + 40 1.732 – 1.7134 1.736 – 1.7134 = 400 + 40 × 0.823 = 432.9 F v = 8.353 + (8.775 – 8.353) × 0823 = 8.700 ft 3 /lbm

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Sonntag, Borgnakke and Wylen 8.142 E A computer chip dissipates 2 Btu of electric work over time and rejects that as heat transfer from its 125 F surface to 70 F air. How much entropy is generated in the chip? How much if any is generated outside the chip? C.V.1 Chip with surface at 125 F, we assume chip state is constant. Energy: U 2 – U 1 = 0 = 1 Q 2 1 W 2 = W electrical in - Q out 1 Entropy: S 2 – S 1 = 0 = - Q out 1 T surf + 1 S 2 gen1 1 S 2 gen1 = Q out 1 T surf = W electrical in T surf = 2 Btu (125 + 459.7) R = 0.0034 Btu/R C.V.2 From chip surface at 125 F to air at 70 F, assume constant state. Energy: U 2 – U 1 = 0 = 1 Q 2 1 W 2 = Q out 1 - Q out 2 Entropy: S 2 – S 1 = 0 = Q out1 T surf - Q out2 T air + 1 S 2 gen2 1 S 2 gen2 = Q out2 T air - Q out1 T surf = 2 Btu 529.7 R - 2 Btu 584.7 R = 0.000 36 Btu/R 70 F air 125 F Q air flow cb
Sonntag, Borgnakke and Wylen 8.143 E Two 10 lbm blocks of steel, one at 400 F the other at 70 F, come in thermal contact. Find the final temperature and the total entropy generation in the process? C.V. Both blocks, no external heat transfer, C from table F.2.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 45

ch08_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 8...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online