ThermoSolutions-CHAPTER04

# ThermoSolutions-CHAPTER04 - 4-1 Chapter 4 ENERGY ANALYSIS...

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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 k P a m1 k ( N / m ) k N m 1 k 32 3 ¡ ¡ J 4-5 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). 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 N 2 130 kPa 120 q C 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 ¢o ¢ ¢ n P P n n n n The boundary work is determined from kJ 1.86 0 0 0 0 249 . 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

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4-2 4-6 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 ¿ ¾ ½ q ¿ ¾ ½ q v T P T P Q Steam 0.3 kg 1 MPa 400 q C Noting that pressure is constant during the process, the boundary work is determined from kJ 22.16 0 0 /kg m ) 23275 . 0 30661 . 0 kPa)( kg)(1000 3 . 0 ( ) ( 3 2 1 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 u 0 0 /kg m ) 30661 . 0 60 . 0 30661 . 0 kPa)( kg)(1000 3 . 0 ( ) 60 . 0 ( 3 1 1 mP W b The temperature at the final state is (Table A-5) C 151.8 q ° ¿ ° ¾ ½ u 2 3 2 2 /kg m ) 30661 . 0 60 . 0 ( MPa 5 . 0 T P 4-7 A piston-cylinder device contains nitrogen gas at a specified state. The final temperature and the boundary work are to be determined for the isentropic expansion of nitrogen. N 2 130 kPa 120 q C Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a) Analysis The mass and the final volume of nitrogen 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 2 4 . 1 2 4 . 1 3 2 2 1 1 m 08443 . 0 kPa) (100 ) m kPa)(0.07 (130 ¢o ¢ ¢ k k P P The final temperature and the boundary work are determined as K 364.6 /kg.K) kPa.m kg)(0.2968 07802 . 0 ( ) m 43 kPa)(0.084 (100 3 3 2 2 2 mR P T kJ 1.64 0 0 0 0 4 . 1 1 ) m kPa)(0.07 (130 ) m 43 kPa)(0.084 (100 1 3 3 1 1 2 2 k P P W b
4-3 4-8 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6) /kg m .71643 0 C 200 kPa 00 3 /kg m 0.60582 vapor Sat. kPa 300 3 2 2 2 3 kPa 300 @ 1 1 ¿ ¾ ½ q ¿ ¾ ½ v T P P g Analysis The boundary work is determined from its definition to be kJ 165.9 ¸ ¸ ¹ · ¨ ¨ © § ¡ 0 0 0 ³ 3 3 1 2 1 2 2 1 out , m kPa 1 kJ 1 /kg m 0.60582) 43 kPa)(0.716 kg)(300 (5 ) ( ) ( V mP P d P W b 2 1 P (kPa 300 Discussion The positive sign indicates that work is done by the system (work output).

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## This note was uploaded on 09/21/2009 for the course ME 311 taught by Professor Ferrenberg during the Fall '09 term at Nevada.

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ThermoSolutions-CHAPTER04 - 4-1 Chapter 4 ENERGY ANALYSIS...

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