MAE0-W7A The homework i due Monday, November 4, 06. Each problem i worth the point indicated. Copying o the olution rom another i not acceptable. (). Multiple choice (0 point) a). Which tatement i invalid regarding the entropy generation (S gen )?. A). Entropy generation i alway a poitive quantity or zero. B). The value o entropy generation depend on the proce. C). The value o entropy generation i zero during an internal reverible proce. D). Entropy generation i a property o a ytem. E). None o the above b). The entropy o an iolated ytem increae during a proce. A). ometime B). alway C). never D). may decreae E). None o the above c). Which tatement i invalid regarding the Clauiu Inequality? A). For any thermodynamic cycle (reverible or irreverible), i alway le than or equal to zero. B). The equality in the Clauiu inequality hold or totally or jut internally reverible cycle. C). The Clauiu Inequality Statement only relect the irt law o thermodynamic. D). The Clauiu Inequality Statement relect the econd law o thermodynamic. E). None o the above d). Which tatement i invalid regarding the entropy? A). Entropy i a property. Like all other propertie, it ha a ixed value at a ixed tate. B). The entropy change o a ytem between two peciied tate i the ame whether the proce i reverible or irreverible. C). On a T-S diagram, the area under the proce curve repreent the heat traner or an internally reverible proce. D). Entropy i a path unction; and the entropy change i dependent on the path o a proce. E). None o the above
e). I an adiabatic compreor undergoe an internally reverible proce, we can conclude A). It obviouly i not an ientropic proce. B). Entropy generation (S gen ) i greater than zero. C). Entropy generation (S gen ) i zero D). eat may traner rom the compreor to the urrounding. E). The change in the entropy between the initial and the inal tate may not be zero. ). Which tatement i invalid? A). Entropy i a conerved property, and there exit a conervation o entropy principle. B). Procee can occur in a certain direction only, not in any direction. A proce mut proceed in the direction that complie with the increae o entropy principle. C). Entropy generation (S gen ) i a meaure o the magnitude o the irreveribilitie during a proce. D). Entropy may be tranerred by heat traner and ma traner. E). None o the above g). Which tatement i invalid? A). The T- diagram erve a a valuable tool or viualizing the econd law apect o procee and cycle. An ientropic proce appear a a vertical line egment on a T- diagram. B). The h- diagram i ueul analyi o adiabatic teady-low device, uch a turbine, compreor and nozzle. The horizontal ditance i a meaure o irreveribilitie aociated with the proce. C). The h- diagram i ueul analyi o adiabatic teady-low device. The vertical ditance h (between the inlet and the exit tate) on an h- diagram i a meaure o work. D). From the microcopic viewpoint, entropy i a meaure o molecular diorder or molecular randomne. E). According to the third law o thermodynamic, a pure crytalline ubtance at abolute zero temperature i in perect order. But it entropy never i zero.
. An inulated rigid tank contain 0 kg water. Initially it i a aturated liquid-vapor mixture o water with a quality o 0.5 at preure o 50 kpa. An electric heater inide i turned on and kept on until all the liquid vaporized. Determine the entropy change o the water during thi proce (0 point). From the team table (Table A-4 through A-6) P 00kPa v v + xv g x 0.5 + x v v at. vapor g 6.8649kJ/kg K 0.00+.06 Then the entropy change o the team become ( 0.5)(.694 0.00) + ( 0.5)( 6.0568) 0.44m /kg.868kj/kg K S ( ) ( 0kg)( 6.8649.868) kj/kg K 40.5kJ/K m O 0 kg W
. A reverible heat pump deliver heat at a rate o 00 kj/ to warm a houe maintained at 4 o C. The exterior air, which i at 7 o C, erve a the cool reervoir (9 point). (a) Draw a carton to illutrate the heat pump device (b) Calculate the rate o entropy change o the hot and cool reervoir, repectively, a well a the entire heat pump ytem (c) Jutiy i thi heat pump atiie the increae o entropy principle Since the heat pump i completely reverible, the combination o the coeicient o perormance expreion, irt Law, and thermodynamic temperature cale give COP P, rev 7.47 T L / T (80 K) /(97 K) The power required to drive thi heat pump, according to the coeicient o perormance, i then 00 kw W net, in 7.7 kw COP 7.47 P, rev According to the irt law, the rate at which heat i removed rom the low-temperature energy reervoir i W 00 kw 7.7 kw 8.8 kw L net, in The rate at which the entropy o the high temperature reervoir change, according to the deinition o the entropy, i S T 00 kw 97 K.0kW/K and that o the low-temperature reervoir i L 7.7 kw S L.0kW/K T 80 K L The net rate o entropy change o everything in thi ytem i S S + S.0. 0 0 kw/k total L a it mut be ince the heat pump i completely reverible. 00 4 C 7 C P 4
4. A rubber bag initially contain kg o water at 60 o C and 0 bar. It undergoe an iothermal internally reverible expanion proce during which 000 kj i received by heat traner (4 point). (a) Determine the inal preure, in kpa (b) Calculate the work done during the proce, in kj (c) Draw a T- diagram, and locate the initial and inal tate in the T- diagram, label the temperature and entropy at the initial and inal tate. (a). or an iothermal internally reverible procee: Q TdS m Td mt ( ) At T 60 o C and P 000 kpa. It i compreed liquid, @T.946 kj/kg K and u 674.79 kj/kg (Table A-4). Q 700 +.946 + 5.0605kJ / kg K mt (60 + 7) At T 60 o C, < < g, hence it i a mixture, P 68. kpa (Table A-4). x g 5.0605.946 4.8066 0.65 u 674.79 + 0.65 89 u + xu g 909.4 (b) rom the irt law: Q W U m( u u) W Q m( u u ) 000 (909.4 674.79) 50. kj (c) ee the curve T ( o C) P 0 bar 60 o C.946 5.0605 (kj/kg K) 5
5. An inulated piton-cylinder ytem i initially illed with 0.0764 m o team at 00 kpa and 00 C. The team i now compreed in an internal reverible manner until the preure reache 4.5 MPa. (a) Determine the peciic entropy at the initial tate (8 point) (b) Calculate the total work input during thi proce, in kj (0 point). (a) Thi i an internal reverible adiabatic (i.e., ientropic) proce, and thu. It i an ientropic proce. The kinetic and potential energy change are negligible. The cylinder i well-inulated and thu heat traner i negligible. The thermal energy tored in the cylinder itel i negligible. 4 The proce i tated to be reverible. Analyi Thi i a reverible adiabatic (i.e., ientropic) proce, and thu. From the team table (Table A-4 through A-6), Alo, v 0.764m /kg P 00kPa u 65.0kJ/kg T 00 C 7.kJ/kg K P 4.5MPa u 76.4kJ/kg V m v 0.0764m 0.kg 0.764m /kg O 00 kpa (b) We take the content o the cylinder a the ytem. Thi i a cloed ytem ince no ma enter or leave. The energy balance or thi adiabatic cloed ytem can be expreed a W b, in U u m( u ) Subtituting, the work input during thi adiabatic proce i determined to be ( ) ( 0.kg)( 76.4 65.0) kj/kg 6.54kJ W b, in m u u 6
6. Steam i expanded in an adiabatic turbine with a ingle inlet and a ingle outlet in an internally reverible manner. At the inlet, the team i at MPa and 60 o C. The team preure at the outlet i 00 kpa. (a) Find the peciic entropy and quality o water at exit? (6 point) (b) Calculate the work per unit ma produced by the turbine (8 point). (c) Draw a T- diagram, chematically locate the initial and inal tate in the T- diagram and chematically how the proce (5 point). (a) The inlet tate propertie are P T MPa 60 C h 59.9 kj/kg 6.998 kj/kg K (Table A - 6) For thi ientropic proce, the inal tate propertie are (Table A-5) P 00 kpa 6.998 kj/kg K x h h g + x 6.998.08 0.997 6.056 h 47.5+ (0.997)(57.5) 58.9 kj/kg g (b) Thi i a teady-low proce. There i only one inlet and one exit, and thu m m m. We take the turbine a the ytem, which i a control volume ince ma croe the boundary. The energy balance or thi teady-low ytem can be expreed in the rate orm a mh W out mh m ( h h ) Subtituting, w out + W out h h (59.9 58.9) kj/kg 6.0 kj/kg T (c) See the T- diagram MPa 00 kpa 7