Assignment 1
Assignment 1 ! ! ! ! Problem 2.29 Nitrogen (N2) gas within a piston-cylinder assembly undergoes a process from p1 = 20 bar, V1 = 0.5 m3 to a state where V2 = 2.75 m3. The relationship between pressure and volume during the process is pV1.35 = constant. For the N2, determine (a) the pressure at state 2, in bar, and (b) the work, in kJ. KNOWN: N2 gas within a piston-cylinder assembly undergoes a process where the p-V relation is pV1.35 = constant. Data are given at the initial and final states. FIND: Determine the pressure at the final state and the work. ENGINEERING MODEL: (1) The N2 is the closed system. (2) The p-v relation is specified for the process. (3) Volume change is the only work mode. SCHEMATIC AND GIVEN DATA: pV 1.35= constant p1 = 20 bar, V1 = 0.5 m3 V2 = 2.75 m3 N2 ANALYSIS: (a) ଵ ܸଵ ൌ ଶ ܸଶ Ǥହ୫య ଵǤଷହ ଶ ൌ ሺʹͲሻ ቀଶǤହ୫య ቁ ĺ ଶ ൌ ଵ ቀభ ቁ ; n = 1.35. Thus మ ൌ ʹ (b) Since volume change is the only work mode, Eq. 2.17 applies. Following the procedure of part (a) of Example 2.1, we have W= మ మ ିభ భ ଵି ൌ = 1285.7 kJ ሺଶୠୟ୰ሻሺଶǤହ୫యሻ ିሺଶሻሺǤହሻ ଵఱ Ȁ୫మ ଵିଵǤଷହ ቚ ଵୠୟ୰ ቚቚ ଵ୩ ଵయ ή୫ ቚ Problem 2.34 Carbon monoxide gas (CO) contained within a piston-cylinder assembly undergoes three processes in series: Process 1-2: Constant pressure expansion at 5 bar from V1 = 0.2 m3 to V2 = 1 m3. Process 2-3: Constant volume cooling from state 2 to state 3 where p3 = 1 bar. Process 3-1: Compression from state 3 to the initial state during which the pressure-volume relationship is pV = constant. Sketch the processes in series on p-V coordinates and evaluate the work for each process, in kJ. KNOWN: Carbon monoxide gas within a piston-cylinder assembly undergoes three processes in series. FIND: Sketch the processes in series on a p-V diagram and evaluate the work for each process. SCHEMATIC AND GIVEN DATA: p (bar) CO 5 Process 1-2: Constant pressure expansion at 5 bar from V1 = 0.2 m3 to V2 = 1 m3. Process 2-3: Constant volume cooling from state 2 to state 3 where p3 = 1 bar. Process 3-1: Compression from state 3 to the initial state during which the pressure-volume relationship is pV = constant. 1 1 2 3 pV = constant ! 0.2 1 ENGINEERING MODEL: (1) The gas is the closed system. (2) Volume change is the only work mode. (3) Each of the three processes is specified. ANALYSIS: Since volume change is the only work mode, Eq. 2.17 applies. Process 1-2: Constant pressure processes: ଵఱ Τ୫మ ܹଵଶ ൌ ሺͷሻሺͳ െ ͲǤʹሻଷ ቚ ଵୠୟ୰ ܹଵଶ ൌ మ ܸ݀ൌ ଵ ሺܸଶ െ ܸଵ ሻ భ ቚቚ ଵ୩ ଵయ ή୫ ቚ ൌ ͶͲͲ (out) Process 2-3: Constant volume (piston does not move). Thus W23 = 0 V (m3) Problem 2.33 (Continued) Process 3-1: For process 3-1, pV = constant = p1V1 . Noting that V3 = V2, we get య య య మ ܹଷଵ ൌ భ ܸ݀ൌ భ ܸ݀ ൌ ܥቀ భ ቁ ൌ ሺଵ ܸଵ ሻ ቀ భ ቁ Inserting values and converting units !" Ǥଶ୫య ଵఱ Τ୫మ ܹଷଵ ൌ ሺͷሻሺͲǤʹଷ ሻ ቀ ଵ୫య ቁ ቚ ଵୠୟ୰ ଵ୩ ቚ ቚଵయ ή୫ቚ = -160.9 kJ (in) 1. The net work for the three process is Wnet = W12 + W23 + W31 = (+400) + 0 + (-160.9) = 239.1 kJ (net work is positive - out) PROBLEM 3.6 Determine the phase or phases in a system consisting of H2O at the following conditions and sketch the p-!!and T-!!diagrams showing the location of each state. (a) (b) (c) (d) (e) p = 10 bar, T = 179.9oC p = 10 bar, T = 150oC T = 100oC, p = 0.5 bar T = 20oC, p = 50 bar p = 1 bar, T = - 6oC (a) p = 10 bar, T = 179.9oC p Two-phase liquid-vapor mixture 10 bar T 179.9oC 10 bar 179.9oC (Table A-3) !! !! (b) p = 10 bar, T = 150oC p T<Tsat@p sub-cooled liquid . 10 bar 179.9oC 150oC !! 10 bar T . 179.9oC 150oC !! Problem 3.5 (Continued) (c) T = 100oC, p = 0.5 bar p<psat@T superheated vapor p . T 1.014 bar (Table A-2) 0.5 bar . 1.014 bar 100oC 100oC 0.5 bar !! !! (d) T = 20oC, p = 50 bar p>psat@T sub-cooled liquid p . T 50 bar 0.02339 bar (Table A-2) 50 bar . 0.02339 bar 20oC 20oC !! !! o (e) p = 1 bar, T = - 6 C p>psat@T solid (T is below the triple point temperature) p . T 1 bar 0.003689 bar (Table A-5) 1 bar 0.003689 bar -6 oC !! . -6 oC !! PROBLEM 3.7 PROBLEM 3.22 Ammonia, initially at 6 bar, 40oC, undergoes a constant volume process in a closed system to a final pressure of 3 bar. At the final state, determine the temperature, in oC, and the quality. Locate each state on a sketch of the T-! diagram. constant volume 6 bar T 1 Ammonia . 40oC 3 bar 2 . -9.24oC !! The initial state is in the superheated vapor region. From Table A-25, !1 = 0.24118 m3/kg. The system is a closed system (constant mass) and the volume is constant. Therefore, !2 = !1. From Table A-14 at !2 = 0.24118 m3/kg, the state is in the two-phase liquid-vapor region, and T2 = Tsat(3 bar) = -9.24oC The quality is x2 = ௩మ ି௩మ ௩ౝమ ି௩మ ൌ Ǥଶସଵଵ଼ିଵǤହଷଵ୶ଵషయ ǤସଵିଵǤହଷଵ୶ଵషయ = 0.5924 (59.24%) PROBLEM 3.36 PROBLEM 3.43
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