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Thermo-Ch4 - 4-1 Chapter 4 ENERGY ANALYSIS OF CLOSED...

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4-1 Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Moving Boundary Work 4-1C It represents the boundary work for quasi-equilibrium processes. 4-2C Yes. 4-3C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case. 4-4C 1 kPa m 1 k(N /m ) m 1 kN m 1 kJ 3 2 3 = = = 4-5 Helium is compressed in a piston-cylinder device. The initial and final temperatures of helium and the work required to compress it are to be determined. Assumptions The process is quasi-equilibrium. Properties The gas constant of helium is R = 2.0769 kJ/kg K (Table A-1). Analysis The initial specific volume is /kg m 5 kg 1 m 5 3 3 1 1 = = = m V v 1 2 P (kPa) Using the ideal gas equation, 200 K 481.5 = = = K kJ/kg 0769 . 2 ) /kg m kPa)(5 (200 3 1 1 1 R P T v Since the pressure stays constant, 3 5 V (m 3 ) K 288.9 = = = ) K 5 . 481 ( m 5 m 3 3 3 1 1 2 2 T T V V and the work integral expression gives kJ 400 m kPa 1 kJ 1 m ) 5 kPa)(3 (200 ) ( 3 3 1 2 2 1 out , = = = = V V V P d P W b That is, kJ 400 = in , b W
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4-2 4-7E The boundary work done during the process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. P (psia) Analysis The work done is equal to the area under the process line 1-2: 2 1 500 Btu 111 = + = + = = 3 3 1 2 2 1 out , ft psia 5.404 Btu 1 .0)ft 2 (4.0 2 )psia 500 (100 ) ( 2 Area V V P P W b 100 2 4 V (ft 3 ) 4-8 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the polytropic expansion of nitrogen. Properties The gas constant for nitrogen is 0.2968 kJ/kg.K (Table A-2). N 2 130 kPa 120 ° C Analysis The mass and volume of nitrogen at the initial state are kg 07802 . 0 K) 273 20 kJ/kg.K)(1 2968 . 0 ( ) m kPa)(0.07 (130 3 1 1 1 = + = = RT P m V 3 3 2 2 2 m 08637 . 0 kPa 100 K) 273 /kg.K)(100 kPa.m kg)(0.2968 07802 . 0 ( = + = = P mRT V The polytropic index is determined from 249 . 1 ) m 37 kPa)(0.086 (100 ) m kPa)(0.07 (130 3 3 2 2 1 1 = → = → = n P P n n n n V V The boundary work is determined from kJ 1.86 = = = 249 . 1 1 ) m kPa)(0.07 (130 ) m 37 kPa)(0.086 (100 1 3 3 1 1 2 2 n P P W b V V
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4-3 4-9 A piston-cylinder device with a set of stops contains steam at a specified state. Now, the steam is cooled. The compression work for two cases and the final temperature are to be determined. Analysis (a) The specific volumes for the initial and final states are (Table A-6) /kg m 23275 . 0 C 250 MPa 1 /kg m 30661 . 0 C 400 MPa 1 3 2 2 2 3 1 1 1 = ° = = = ° = = v v T P T P Q Steam 0.3 kg 1 MPa 400 ° C Noting that pressure is constant during the process, the boundary work is determined from kJ 22.16 = = = /kg m ) 23275 . 0 30661 . 0 kPa)( kg)(1000 3 . 0 ( ) ( 3 2 1 v v mP W b (b) The volume of the cylinder at the final state is 60% of initial volume. Then, the boundary work becomes kJ 36.79 = × = = /kg m ) 30661 . 0 60 . 0 30661 . 0 kPa)( kg)(1000 3 . 0 ( ) 60 . 0 ( 3 1 1 v v mP W b The temperature at the final state is (Table A-5) C 151.8 ° = × = = 2 3 2 2 /kg m ) 30661 . 0 60 . 0 ( MPa 5 . 0
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