TEST2_Sol - 61:0(adiabaiSIC ‘ l 0 plgjobaa 22339“ M...

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Unformatted text preview: 61:0 (adiabaiSIC) ‘ l 0 plgjobaa 22339“ M: $2.75.: L 17,320 2- 1-2 1:502; 3? Nah? I -= l0 '1: d? 2% f f: ”1% hze 2782.3 Air/,9 (7’4 mug a 30% ?‘ . A! (ii) (10 gamma the exit velocity of steam. 25' 8? Name: éfiMéh Page U4 ME 3720B Intro Fluid Thermo Eng — Spring 2008_ - Test #2 (February 22, 08:05 — 08:55 am, open textbook only) You do not need tofoliow any specific procedure. Use a hand-held calculator. Use mar ins to write down our derivations elearl Problem 1. (20 pt) Steam enters an adiabatic nozzle operating at steady state at 10 bar and 320°C, with a velocity of 10 W5. The inlet area is 1 m2 . The state of steam at the exit is p2 = 3 bar and T2 =160°C . Neglect the change in potential energy. (i) (10 pt) Calculate the mass flow rate. Answer: 27' 3 [kg/5] Answer: é [mfs] 0-: WlCAI'A-z.) "1L azfl-(lztz-ZL Name: épL E 1% Page 2;“! Problem 2. (30 pt) A rigid tank contains 1 kg of water at 2.0 MPa and 400°C. Because of the heat transfer to the surroundings at Tm = 300 K , the pressure of water inside the tank decreases to 500 kPa. (10 pt) Calculate the energy transfer by heat. $31.6 [H] Answer: .9547?!“ a _— i "I (ii) (10 pt) What is the entropy change of water during the process? Answer: ”'3‘ 2 ? t/ 8 [le1(] (iii) (10 pt) Calculate the entropy generation during the process. Answer: /‘ 8 352 [kaK] 05% "= dag-LEFT + 45% assume Name: épM 2% Page 3M Problem 3. (20 pt) Air is expanded in a piston-cylinder assembly from 400 kPa and 300 K to 100 kPa in an internally reversible and isothermal process. Model air as an ideal gas with a constant specific heat cp = 1.005 kakg - K and R = 0.287 ldfkg - K . (i) (10 pt) Determine the entropy change per unit mass of air. Answer: 0 ' 3 q; C? [kJr’kg-K] (i) (10 pt) Determine the energy transfer by heat per unit mass of air. Clearly indicate the direction of heat transfer. & Answer: I {q ”/1 [kakg] Direction: MM 557%?» "" T 0FPD§H€3 drmrfi‘dm 1‘0 1%t 9365de wahflvkm Problem 4. (30 pt) Air is compressed by an adiabatic compressor from 1 bar and 300 K. to a pressure of 8 bar at a steady-state mass flow rate of l kgfs. The isentropic efficiency of the compressor is 0.8 (80%). Changes in the kinetic and potential energy are negligible. Model air as an ideal gas with a constant specific heat 5;: Z1.005 kag-K and R=0.287 kakg-K. Name: é faith Page 4f4 1’31 '2 1 ga [5' (r) (10 pt) Determme the exrt temperature of arr. \ Answer: Ad 35’ / [K] H1111“- I “k 1 [IE—Ala: 7;*{=a-9 Wow 1 21,85“ Z Ara, 71—7; state 2r ' P P1; Sat-‘5'; 7'3"" (755 ‘773/0'5’ +77 0’- cpfig—‘l‘ 'Rfir}; "-1 {094/ K 735-— - 770E: Tile; $443.3 k (ii) (10 pt) Calculate the required power input to the compressor. Sketch the T-s diagram and clearly indicate state 1 and state 2. Answer: 3&5 '{ [kW] WG-2 W641'742— ”QCT" 7;)1: /X/:oan/é¢/- -.?&é) (iii) (10 pt) Calculate the rate of the entropy generation c'rcv. Answer: 0', #5? [kaK] ...
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