Quiz2SolnPC

Quiz2SolnPC - Note that 1 J = 1 kg m 2/s 2 and that 1 kPa =...

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------------------- --------- Name: Division: Quiz #2 ME 200 Prof. Lucht 27 Jan 2012 A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same initial state 1, PI = 100 kPa, t71 = 1.0 m 3 , VI = 400 kJ, and the same final state 2, P2 = 1000 kPa, t72= 0.1 m 3 , V 2 = 450 kJ. Process A: Constant volume process from state 1 to state la, where PIa = 10 bar, followed by a constant pressure process to state 2. Process B: P V= constant from state 1 to state 2. Neglect kinetic and potential energy changes. For each process, (a) sketch the process on a P V diagram, (b) calculate the work W l2 in kJ, and (c) calculate the heat transfer Q12 in kJ. Follow the solution procedure we have discussed. List Given, Find, Assumptions, Diagram and System Identification, Basic Equations, Solution, and Solution Check.
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Unformatted text preview: Note that 1 J = 1 kg m 2 /s 2 , and that 1 kPa = 1 kJ/m 3 • P(StoY\-cy(,'ttde; a:;5eJnhly =: /rOo r-f ~ Vi ),0 VV1 3 1lt". LfOO !: T p, :0-\fz ~ (2; Wi '] V 2-;: 4:<:0 jeT t"L:: ( (() 0 0 II-01' c" \J tJ ~u J41 e P roce~s A : { 0 Y1 S-t. . COYl$tl 1 t __--' \ ----------~ SDI~~()iA-'-@ Foo( p('oc.~$<; 11 ~ (0 2-&--------a I C( w -:::0 1.1 I ~ W 1 -/ a. l/V I ti 2- -I 2. .. A = p2-(d2--VI~) ~ COOO J;!3)(a, 1-/.0 ~"SJ ._ Y c@ ~ -qOO ~T ~tt-C_qOO kiJ QJ"2Jt"" W , 2-A+ V2.--U I l?f;3~l=r J6) :: f qOO)+ (q~(YLfOO) tr ~ ~'-r5 7 ~,Zpdlf f V-==C ~ Pi (II 5 2-C d: -= C) VI (~ ) = 10 I \I, I n(!jr ) I v \ , \71 \JV,2~ -~O ~3!/IOfh7) I~ (f~l) ~ -230.'3 kTl ~--~~~'~~ ---------------~ /?r3 ~ W'l-'6 + Vz--u; -:: ~·£-30~-t50 --/~o.o? k3 2 --------.. ------------~~------~ procpss WI 'LrS = +~...
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This note was uploaded on 02/22/2012 for the course ME 200 taught by Professor Gal during the Spring '08 term at Purdue.

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Quiz2SolnPC - Note that 1 J = 1 kg m 2/s 2 and that 1 kPa =...

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