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# ch04_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4...

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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

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Sonntag, Borgnakke and Wylen CHAPTER 4 SUBSECTION PROB NO. Concept-Study Guide Problems 117-122 Simple processes 123-128 Review Problems 129-130 Polytropic processes 131-134 Multi-step processes, other types of work 135-138 Rates of work 139-140 Heat Transfer Rates 141-143 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 117 new 1 131 new 50 118 new - 132 66 47 119 new 9 133 65 48 120 new 14 134 75 57 121 new 4 135 69 59 122 new 3 136 73 63 123 new 22 137 72 60 124 68 32 138 76 71 125 64 25 139 63 82 126 New 24 140 new 91 127 new 38 141 77 95 128 62 36 142 78 97 129 67 109 143 79 100 130 70 111
Sonntag, Borgnakke and Wylen Concept Problems 4.117 E The electric company charges the customers per kW-hour. What is that in english units? The unit kW-hour is a rate multiplied with time. For the standard English Eng. units the rate of energy is in Btu/h and the time is in seconds. The integration in Eq.4.21 becomes 1 kW-hour = 3412.14 Btu/h × 1 h = 3412.14 Btu Conversions are found in Table A.1 4.118 E Work as F x has units of lbf-ft, what is that in Btu? Conversions are found in Table A.1 1 lbf-ft = 1.28507 × 10 -3 Btu = 1 778 Btu 4.119 E A work of 2.5 Btu must be delivered on a rod from a pneumatic piston/cylinder where the air pressure is limited to 75 psia. What diameter cylinder should I have to restrict the rod motion to maximum 2 ft? W = F dx = P dV = PA dx = PA x = P π 4 D 2 x D = 4W π P x = 4 × 2.5 Btu π × 75 psia × 2 ft = 4 × 2.5 × 778.17 lbf-ft π × 75 × 144 (lbf/ft 2 ) × 2 ft = 0.339 ft

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Sonntag, Borgnakke and Wylen 4.120 E A force of 300 lbf moves a truck with 40 mi/h up a hill. What is the power? Solution: W . = F V = 300 lbf × 40 (mi/h) = 12 000 × 1609.3 × 3.28084 3600 lbf-ft s = 17 600 lbf-ft s = 22.62 Btu/s
Sonntag, Borgnakke and Wylen 4.121 E A 1200 hp dragster engine drives the car with a speed of 65 mi/h. How much force is between the tires and the road? Power is force times rate of displacement as in Eq.4.2 Power, rate of work W . = F V = P V . = T ω We need the velocity in ft/s: V = 65 × 1609.3 × 3.28084 3600 = 95.33 ft/s We need power in lbf-ft/s: 1 hp = 550 lbf-ft/s F = W . / V = 1200 × 550 95.33 lbf-ft/s ft/s = 6923 lbf

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Sonntag, Borgnakke and Wylen 4.122 E A 1200 hp dragster engine has a drive shaft rotating at 2000 RPM. How much torque is on the shaft? Power is force times rate of displacement as in Eq.4.2 Power, rate of work W . = F V = P V . = T ω We need to convert the RPM to a value for angular velocity ω ω = RPM × 2 π 60 s = 2000 × 2 π 60 s = 209.44 rad s We need power in lbf-ft/s: 1 hp = 550 lbf-ft/s T = W . / ω = 1200 hp × 550 lbf-ft/s-hp 209.44 rad/s = 3151 lbf-ft
Sonntag, Borgnakke and Wylen Simple Processes 4.123 E A bulldozer pushes 1000 lbm of dirt 300 ft with a force of 400 lbf. It then lifts the dirt 10 ft up to put it in a dump truck. How much work did it do in each situation?

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ch04_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4...

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