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Unformatted text preview: Chapter 7 Entropy Review Problems 7137 It is to be shown that the difference between the steadyflow and boundary works is the flow energy. Analysis The total differential of flow energy P v can be expressed as ( 29 ( 29 flow b flow b w w w w dP v dv P Pv d = = + = δ δ δ Therefore, the difference between the reversible steadyflow work and the reversible boundary work is the flow energy. 7138E An insulated rigid can initially contains R134a at a specified state. A crack develops, and refrigerant escapes slowly. The final mass in the can is to be determined when the pressure inside drops to a specified value. Assumptions 1 The can is wellinsulated and thus heat transfer is negligible. 2 The refrigerant that remains in the can underwent a reversible adiabatic process. Analysis Noting that for a reversible adiabatic (i.e., isentropic) process, s 1 = s 2 , the properties of the refrigerant in the can are ( 29 ( 29 /lbm ft 0.3518 0.01209 1.5408 0.2222 0.01209 2222 . 0364 . 2209 . 0364 . 0774 . psia 30 R Btu/lbm 0.0774 F 80 psia 120 3 2 2 2 2 1 2 2 F 80 @ 1 1 1 = + = + = = = = = = ⋅ = 2245 = = fg f fg f f v x v v s s s x s s P s s T P Thus the final mass of the refrigerant in the can is lbm 3.411 = = = /lbm ft 0.3518 ft 1.2 3 3 1 v V m 7109 R134 120 psia 80 ° F Leak Chapter 7 Entropy 7139 An insulated rigid tank is connected to a pistoncylinder device with zero clearance that is maintained at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into the cylinder. The final temperatures in the tank and the cylinder are to be determined. Assumptions 1 Both the tank and cylinder are wellinsulated and thus heat transfer is negligible. 2 The water that remains in the tank underwent a reversible adiabatic process. 3 The thermal energy stored in the tank and cylinder themselves is negligible. 4 The system is stationary and thus kinetic and potential energy changes are negligible. Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s 2 = s 1 . From the steam tables, ( 29 kJ/kg 2377.2 ) kJ/kg 2052.7 )( 0.9306 ( 466.94 /kg m 1.079 ) 0.001053 1.1593 )( 0.9306 ( 0.001053 9306 . 7897 . 5 4336 . 1 8213 . 6 . kPa 150 K kJ/kg 6.8213 kJ/kg 2561.2 /kg m 0.3749 . kPa 500 , 2 , 2 3 , 2 , 2 , 2 , 2 kPa 150 @ , 2 1 2 2 kPa 500 @ 1 kPa 500 @ 1 3 kPa 500 @ 1 1 = + = + = = + = + = = = = = = = = ⋅ = = = = = = = fg A f A fg A f A fg f A A sat A g g g u x u u v x v v s s s x T T mixture sat s s P s s u u v v vapor sat P C 111.37 The initial and the final masses in tank A are Thus, kg 0.696 0.371 1.067 kg 0.371 /kg m 1.079 m 0.4 kg 1.067 /kg m 0.3749 m 0.4 , 2 , 1 , 2 3 3 , 2 , 2 3 3 , 1 , 1 = = = = = = = = = B A B A A A A A A m m m v V m and v V m ( b ) The boundary work done during this process is ( 29 B B B B B out b v m P V P dV P W , 2 , 2 , 2 2 1 , = = = ∫ Taking the contents of both the tank and the cylinder to be the system,...
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
 Spring '08
 Chung

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