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# ch06_eng - SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 6...

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SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 6 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN

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Sonntag, Borgnakke and Wylen CHAPTER 6 SUBSECTION PROB NO. Concept-Study Guide Problems 139-143 Continuity and Flow Rates 144-146 Single Flow Devices 147-160 Multiple Flow Devices 161-165 Multiple Devices, Cycle Processes 166-169 Transient processes 170-174 Review Problem 175 New 5th SI New 5th SI New 5th SI 139 new 6 152 84 43 165 85 96 140 new 9 153 87 49 166 90 59 141 new 14 154 88 51 167 91 99 142 new 19 155 new 54 168 92 100 143 new 20 156 new 61 169 93 105 144 73 22 157 77 67 170 94 111 145 new 27 158 76 63 171 new 112 146 74 26 159 new 69 172 95 114 147 75 - 160 89 72 173 100 110mod 148 new 33 161 86 78 174 102 119 149 81 30 162 79 84 175 101 133 150 82 36 163 78 86 151 83 40 164 new 88
Sonntag, Borgnakke and Wylen Concept-Study Guide Problems 6.139 E Liquid water at 60 F flows out of a nozzle straight up 40 ft. What is nozzle V exit ? Energy Eq.6.13: h exit + 1 2 V 2 exit + gH exit = h 2 + 1 2 V 2 2 + gH 2 If the water can flow 40 ft up it has specific potential energy of gH 2 which must equal the specific kinetic energy out of the nozzle V 2 exit /2. The water does not change P or T so h is the same. V 2 exit /2 = g(H 2 – H exit ) = gH => V exit = 2gH = 2 × 32.174 × 40 ft 2 /s 2 = 50.7 ft/s

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Sonntag, Borgnakke and Wylen 6.140 E R-134a at 90 F, 125 psia is throttled so it becomes cold at 10 F. What is exit P? State 1 is slightly compressed liquid so Table F.5.1: h = h f = 105.34 Btu/lbm At the lower temperature it becomes two-phase since the throttle flow has constant h and at 10 F: h g = 168.06 Btu/lbm P = P sat = 26.8 psia 1 2 2 P v 1 T h = C h = C
Sonntag, Borgnakke and Wylen 6.141 E In a boiler you vaporize some liquid water at 103 psia flowing at 3 ft/s. What is the velocity of the saturated vapor at 103 psia if the pipe size is the same? Can the flow then be constant P? The continuity equation with average values is written m . i = m . e = m . = ρ A V = A V /v = A V i /v i = A V e /v e From Table F.7.2 at 103 psia we get v f = 0.01776 ft 3 /kg; v g = 4.3115 ft 3 /kg V e = V i v e /v i = 3 4.3115 0.01776 = 728 ft/s To accelerate the flow up to that speed you need a large force ( PA ) so a large pressure drop is needed. P i cb P e < P i

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Sonntag, Borgnakke and Wylen 6.142 E Air at 60 ft/s, 480 R, 11 psia with 10 lbm/s flows into a jet engine and it flows out at 1500 ft/s, 1440 R, 11 psia. What is the change (power) in flow of kinetic energy? m . KE = m . 1 2 ( V 2 e V 2 i ) = 10 lbm/s × 1 2 (1500 2 – 60 2 ) (ft/s) 2 1 32.174 (lbf/lbm-ft/s 2 ) = 349 102 lbf-ft/s = 448.6 Btu/s cb
Sonntag, Borgnakke and Wylen 6.143 E An initially empty cylinder is filled with air from 70 F, 15 psia until it is full. Assuming no heat transfer is the final temperature larger, equal to or smaller than 70 F? Does the final T depends on the size of the cylinder? This is a transient problem with no heat transfer and no work. The balance equations for the tank as C.V. become Continuity Eq.: m 2 – 0 = m i Energy Eq.: m 2 u 2 – 0 = m i h i + Q – W = m i h i + 0 – 0 Final state: u 2 = h i & P 2 = P i T 2 > T i and it does not depend on V

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Sonntag, Borgnakke and Wylen Continuity and Flow Rates 6.144 E Air at 95 F, 16 lbf/in.
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