Question

5. A rigid tank initially contains 10 kg O2 at 200 kPa and 600 K. Now O2 is gradually cooled under constant volume until its temperature reaches 455 K. (18 Points) (a) Calculate the pressure of O2 at final state. (4 points) (b) Determine the boundary work. (6 points) (b) Calculate the heat transfer during this process. (8 Points)

MAE 320 - Thermodynamics + > e ecampus.wvu.edu/bbcswebdav/pid-6846897-dt-content-rid-82617141 1/courses/star50314.202005/MAE320-2020-Summer-HW04.pdf Q4 to Q6 should be submitted on-line. 4. A piston-cylinder system initially contains 0.25 kg of ideal gas at P,=8 bar and ¥;=0.02 m (State 1). After a polytropic expansion process described by PV1.2 = C, the pressure of gas in the cylinder drops to 2 bar (State 2). During this process, heat transfer between the system and environment occurs. The specific internal energy of the gas decreases by 55 kJ/kg. Ignoring the change in the kinetic energy and potential energy of this system, determine (12 points) (a) the boundary work done by the gas, in kJ (6 points) (b) the heat transfer during the process, in kJ (6 points) 5. A rigid tank initially contains 10 kg O, at 200 kPa and 600 K. Now O, is gradually cooled under constant volume until its temperature reaches 455 K.(18 Points) (a) Calculate the pressure of O2 at final state. (4 points) (b) Determine the boundary work. (6 points) (b) Calculate the heat transfer during this process. (8 Points) 6. A free piston-cylinder system initially contains 10 kg compressed liquid water at 200 kPa and 90 °C. A resister heater is now turned on to heat the water until the temperature of water reaches 150 °C. During this process, the heat loss from water to ambient air (25 °C) is 200 kJ. You can ignore the changes in potential energy and kinetic energy of water (20 Points). (a) Find the phase of water at the initial and final states. (5 Points) (b) Present this process in both P-V and T-V diagram. (5 points) (c) Determine the boundary work and electrical energy consumed during this process. (10 points)

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Answer 1

Qout cooling Rigid cylinder → To=455k } final condition Vi=vz constant volume P2 acht u diatomic gas r = 14 OL 02 m=lokg Pe = 200 kPa initial maloky condition T=book. P =200 kPa Ti = Gook y constant volume process an PIN P2 W TI Ta s pl P2 TI T2 200 P2 > Pz-piral produse 600 455 1P2 = 151667 kPa Ans@ work W12= p du since change in volume du= V₂-4 =0 Wizzo No boun indary work. (U2-Up) + Wiz of thermo dynamics R=0.5196 wlagle Cv -0.5196 rol 1.41 Cv = 1.299 Kilkg k = 10x1-299 (455-600) -1883.64 kj7 Ans (6) Boundary © Heat foarster Qout Qout = mer (T2-T 1st law & Qout Qout