ch07_eng

ch07_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 7...

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Unformatted text preview: SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 7 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG • BORGNAKKE • VAN WYLEN Sonntag, Borgnakke and Wylen CHAPTER 7 SUBSECTION PROB NO. Concept-Study Guide Problems 92-96 Heat Engines and Refrigerators 97-100 Carnot Cycles and Absolute Temperature 101-110 Finite ∆ T Heat Transfer 111-114 Review Problems 115-117 Ideal Gas Carnot Cycles 118 This problem set compared to the fifth edition chapter 7 set and the current chapter 7 SI problem set. New 5th SI New 5th SI New 5th SI 92 new 2 101 55 40 110 70 63 93 new 3 102 56 44 111 59 80 94 new 5 103 58 47 112 61 75 95 new 7 104 60 48 113 66 73 96 new 15 105 63 51 114 62 61 97 54 20 106 64 60 115 67 84 98 new 22 107 65 72 116 71 87 99 new 30 108 68 - 117 72 91 100 57 26 109 69 62 118 73 79mod Sonntag, Borgnakke and Wylen Concept Problems 7.92 E A gasoline engine produces 20 hp using 35 Btu/s of heat transfer from burning fuel. What is its thermal efficiency and how much power is rejected to the ambient? Conversion Table A.1: 20 hp = 20 × 2544.4/3600 Btu/s = 14.14 Btu/s Efficiency: η TH = W . out /Q . H = 14.14 35 = 0.40 Energy equation: Q . L = Q . H- W . out = 35 – 14.14 = 20.9 Btu/s Q . H ⇒ Q . L ⇒ W . out ⇒ Sonntag, Borgnakke and Wylen 7.93 E A refrigerator removes 1.5 Btu from the cold space using 1 Btu work input. How much energy goes into the kitchen and what is its coefficient of performance? C.V. Refrigerator. The energy Q H goes into the kitchen air. Energy Eq.: Q H = W + Q L = 1 + 1.5 = 2.5 btu COP: β = Q L W = 1.5 / 1 = 1.5 The back side of the refrigerator has a black grille that heats the kitchen air. Other models have that at the bottom with a fan to drive the air over it. 1 2 Air in, 3 Air out, 4 7.94 E A window air-conditioner unit is placed on a laboratory bench and tested in cooling mode using 0.75 Btu/s of electric power with a COP of 1.75. What is the cooling power capacity and what is the net effect on the laboratory? Definition of COP: β = Q . L / W . Cooling capacity: Q . L = β W . = 1.75 × 0.75 = 1.313 Btu/s For steady state operation the Q . L comes from the laboratory and Q . H goes to the laboratory giving a net to the lab of W . = Q . H- Q . L = 0.75 Btu/s, that is heating it. 7.95 E A car engine takes atmospheric air in at 70 F, no fuel, and exhausts the air at 0 F producing work in the process. What do the first and the second laws say about that? Energy Eq.: W = Q H − Q L = change in energy of air. OK 2 nd law: Exchange energy with only one reservoir. NOT OK . This is a violation of the statement of Kelvin-Planck. Remark: You cannot create and maintain your own energy reservoir. Sonntag, Borgnakke and Wylen 7.96 E A large stationary diesel engine produces 20 000 hp with a thermal efficiency of 40%. The exhaust gas, which we assume is air, flows out at 1400 R and the intake is 520 R. How large a mass flow rate is that if that accounts for half the Q ....
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This homework help was uploaded on 02/13/2008 for the course ENG 322 taught by Professor Zaidi during the Fall '07 term at TCNJ.

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ch07_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 7...

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