ch02_eng

ch02_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 2...

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Unformatted text preview: SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 2 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN Sonntag, Borgnakke and Wylen CHAPTER 2 SUBSECTION PROB NO. Concept-Study Guide Problems 87-91 Properties and Units 92 Force, Energy and Specific Volume 93-96 Pressure, Manometers and Barometers 97-103 Temperature 104-105 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 6 th edition SI unit problem. New 5 th Ed. SI New 5 th Ed. SI 87 new - 97 43E 43 88 new 11 98 new 50 89 new 12 99 new 53 90 new 19 100 45E 70 91 new 20 101 46E 45 92 new 24 102 new 82 93 39E 33 103 48E 55 94 40E - 104 new 80 95 new 47 105 47E 77 96 42E 42 Sonntag, Borgnakke and Wylen Concept Problems 2.87 E A mass of 2 lbm has acceleration of 5 ft/s 2 , what is the needed force in lbf? Solution: Newtons 2 nd law: F = ma F = ma = 2 lbm 5 ft/s 2 = 10 lbm ft/s 2 = 10 32.174 lbf = 0.31 lbf 2.88 E How much mass is in 0.25 gallon of liquid mercury (Hg)? Atmospheric air? Solution: A volume of 1 gal equals 231 in 3 , see Table A.1. From Figure 2.7 the density is in the range of 10 000 kg/m 3 = 624.28 lbm/ft 3 , so we get m = V = 624.3 lbm/ft 3 0.25 (231/12 3 ) ft 3 = 20.86 lbm A more accurate value from Table F.3 is = 848 lbm/ft 3 . For the air we see in Figure 2.7 that density is about 1 kg/m 3 = 0.06243 lbm/ft 3 so we get m = V = 0.06243 lbm/ft 3 0.25 (231/12 3 ) ft 3 = 0.00209 lbm A more accurate value from Table F.4 is = 0.073 lbm/ft 3 at 77 F, 1 atm. Sonntag, Borgnakke and Wylen 2.89 E Can you easily carry a one gallon bar of solid gold? Solution: The density of solid gold is about 1205 lbm/ft 3 from Table F.2, we could also have read Figure 2.7 and converted the units. V = 1 gal = 231 in 3 = 231 12-3 ft 3 = 0.13368 ft 3 Therefore the mass in one gallon is m = V = 1205 lbm/ft 3 0.13368 ft 3 = 161 lbm and some people can just about carry that in the standard gravitational field. 2.90 E What is the temperature of 5F in degrees Rankine? Solution: The offset from Fahrenheit to Rankine is 459.67 R, so we get T R = T F + 459.67 = -5 + 459.67 = 454.7 R 2.91 E What is the smallest temperature in degrees Fahrenheit you can have? Rankine?...
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ch02_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 2...

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