Introduction to Thermodynamics - Exam II Solved | ENGR 210, Exams of Thermodynamics

Download Introduction to Thermodynamics - Exam II Solved | ENGR 210 and more Exams Thermodynamics in PDF only on Docsity! ENGR 210 — Introduction to Thermodynamics~ Exam 2-16 Feb 12 - 50 minutes — NAME: de c [ SIGNATURE: By signing this document | attest that | have not used unauthorized material or technology during the test, nor have I received or provided aid to another student. There are-problems on both sides of these sheets. ALL work is to be shown on the exam sheet. An answer without work will not get full credit. Place answer in the space provided. A calculator, pen or pencil, eraser and ONLY are permitted. NOTE: Follow all course policies, in particular the one relating to significant digits TABLE Sue Saturated refrigerant-134a—Temperal Saturated refrigeront-134a—Pressure te H volume, volume, Speers Sat Sat. Sat Sal. Sat Temp., press., liquid, vapor, Press. temp., liquid, TC Poy Pav Me PRA i. "CY 40 51.25 0,0007084 0.36081 = gg. 36.98 0.0007098 56.86 0.0007084 0.32732 7) ~33.87 0,0007344 0.26929 62.95 0.007112 0.29751 gg. 31.13 0.000718 0.23753 69.55 0.007142 0.27099 a, _2a.65 0.0007243 0.21263 Jer! 0.007172 a2a711 |? . 100-2537 0.0007259 0.192b4 84.43 0.0007202 0.22880 = 9.007234 or66G 120. — 22,32 0.0007324 0.16212 <OL73 2007265 0.18926 140 ~18.77 6.0007383 0.14014 12137 0.000/297 0.27395 160 18.60 0.007437 0.12348 =22 121.72 0.007329 0.15995 180 —17.73 00007487 0.) 1041 Superheated refrigerant 1: roy u A 8 ee makg ite _aléxg Kuk K 2 yi eas 31121 709.12 4.89608 220.50 O3E04B 2z/.08 G.36a76 234.56 G.37BY3 241.92 €.39302 249.34 49705 25698 47102 804.77 0.43495 277.64 O.AC8sS 280.73 0.46289 286.9 .a7est 297.4) 0.49032 0.80420; ai92s, 2isis 23888 Dissat 21668 29980 Guores 22678 4749 poles 25095 26588 Geen Paid 26381 O83 248.79 272.17 o zee 2onm 2088 aen2s 72934 208 16 snl ‘nez6 puroe 2555 a0see 83499 eile i. 20 2, 120 3 720 | Total 7100 | . Page 1 of 5  ENGR 210 — Introduction to Thermodynamics— Exam 2-16 Feb 12 — 50 minutes Problem 1 (20 pts) Electric power is to be generated in a hydroelectric. power piant that receives water at a rate of 70.250 x 10° kg/s from an elevation of 65.3 m using a turbine-generator with an efficiency of 87.2 percent. When the frictional losses in the piping are disregarded, determine the electric power output. wa = 70 250 x0 Bls Power clectric GZ. Mw — BOw 9, et. t = Bry, 20,8725 = 4 20, ON Bw m ° 4 nec, eo We SP ba 20,872 Fre. 2souet hs x 40.332 Se [000 - paz | 444033 ua/.f = 3922574 kW os Q = BA BMW | og bets © Problem 2 (20 pts) A rigid tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains an ideal gas at 927°C. The other side is evacuated and has a volume ; twice the size of the part containing the gas. The partition is now removed and the gas expands to fill the entire tank. Heat is now applied to the gas until the pressure-equais the initial pressure. a @ . c me oe ® ° = espa G80". 2 6S.3 Wat = oar jm yep 0.872) Tatsvme a “{ I fei Determine the final temperature of the gas. Trina = 3330 " \ 3 Pe mm) po IN | ' | L Ve = Ne? 3N \ we Nea pag pve ret holds “ RL PNa P Ve @ 77 To Th 6 ; a T3= Ts Pie) (G24 2735) Y, = p200ns (3) = 3@ OSE - 27319 Page 2 of 5