Sample Exam 2 (B) Solutions

Sample Exam 2 (B) Solutions - EML 3100 — Exam 2 ~ Spring...

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Unformatted text preview: EML 3100 — Exam 2 ~ Spring 2012 F ridaV March 23 KEY ‘ Name Wits” A lea?) UF~ID Problem 1 (20 points) Consider the general entropy balance equation for a closed system that undergoes a process fi'om state I to state 2. 2 A51—2 zf ?+Sgen,1—2 1 (A) How does the general entropy balance equation simplify if process .1 to 2 is reversible? {W _ 80 «Ash; " a: (B) How does the general entropy balance equation simplify if process 1 to 2 is adiabatic? ASI'Q “ SEMI-9 __.'«._.—“__ (C) How does the general entropy balance equation simplify if process 1 to 2 is isentropic? Problem 2 (20 points) Consider a compressor that receives ] kg/s of air at P1 = 100 kPa, and T1 = 300 K and delivers it to a nozzle at P2 = 1000 ld’a. The exit temperature of the nozzle is measured to be T3 = 500 K. The efficiency of the compressor is 775 = 75% while the nozzle is assumed to be isentropic. Assume the air behaves as an ideal gas and that the constant specific heat approximation is valid. First determine the temperature at state 2 if the compressor behaves isentropically, T25, then determine the actual temperature at state 2, T2. Also determine the actual work required by the compressong/C. ' C T25 : 39,9068“? k . ____.________________ l 1 . f“ Ale" w h—m' (R'TQ T _ \T - T) ‘795 a i ,1 : 30o“ wo'aq‘a 0:25 _ L Q T91: (92.27%qu M {Ms list oil/3 © we: myths.) Vy— (“0945M ~. U mm) (mm - 30% LL : 373.7(98907q w L 72: @7523 9:; WC: 373,3 le/ Problem 3 (20 points) A steady—state steam engine is used to generate 1075 kW of power output as shown in the figure below. The engine receives a 1 kg/s flow ofsaturated vapor water at P, = 800 kPa and a 1 kg/s flow of superheated vapor water at P2 = 600 kPa, and T2 = 300 °C. The water leaves the engine at P3 = 300 kPa, and T3 = 700 °C. Heat transfer of 3500 kW is added from a 1000 K reservoir and heat is lost to'a low-temperature reservoir at 300 K. . ® Determine the rate of heat transferred at 300 K, Q L, and the total rate of entrOpy production for the steam engine. 3915mm «inflates: 1000K t4 = 27M 13 "VG/ta 357667547“ I I 5‘ 2 (“@6097 W/‘fi‘k 1 1075kW \093 30(0le 301 = 7,3798 \A5 2 3%").10 Muss 300K 33 = 8.223261 sz‘wm Emmy I i? ( P Mlin‘ ‘i’VVJoiAQ 1L C9” 2 W303 +w +QL QL — Home WV r r t - €714 r C31: 33% + \M, 8, 44M; So i- ?H : W33: '3” TL ‘ r . e a} $80! : W333 “ WIS‘ “A”? SR + i ‘ 31:7 QL = (9 LLW__ + Votgg 3.930 . \ GOO Sgen : lv k” r QQU i 53% : t amsssses VV/k Problem 4 (20 points) Consider the following heat pump that is designed to provide heat at TH = 100 °C with the steady-state operating conditions shown in the figure. The low temperature reservoir is at 20 0C. (A) What is the external mechanical power that is needed to Operate the heat pump, Wm? (B) What is the coefficient of performance for heating, C OPH, for the device? (C) What is the theoretical maximum coefficient of performance of heating, C OPHMM , for the given reservoirs? (D) Are the operating conditions of the heat pump physically realistic? Why or why not? ® WMi ; (9H,, TH = 0C o Win : C‘lH’dL “loo My, da‘é L ‘0 0 <2 (<9 TH 3573 LS fihmgr = ; .——————— : Ltooq375 TH ‘1 80 gm ' 8.2”. J3: (3:9, +3594 * ’:H-' "A $94" ‘ TH To ‘L : sac YOU 373. 1‘5 g #315 —. _ o , o aqsvsva‘? W/k < 0 N07. (QWISDQ F‘- wm = (00 KW COP”: 5 _ detest COPHmmx _ Realistic? NO‘ LOPH Con/10+ ‘03 <~ COPCMNOT. 3539/, (Cl/mat [oe < O 9 Problem 5 (20 points) Consider the piston cylinder device with the given operating conditions as shown in the figure. Air inside the piston-cylinder device is heated from an external heat reservoir at TH = 475 K such that V2 = 1.25 V ,. Assume that the air behaves as an ideal gas. Do not, however, assume that the specific heat of air is constant in your analysis. ® First determine the final pressure P2 and the total change in entropy fi‘om state 1 to state 2. Given that the total amount of heat transferred during the process is 150 k], find the total entropy generation. ls the process reversible or irreversibler® ‘ @ 13/ f 200 KM T I = 309,19, T2 F 450,19 7 1 kg, Air A5 = sen-see,» «12. g /_ W0) TH = 475 K _/ i: (7,3173%53 {Lafittw ' L937) IQ/(aoo A5 = (0.357 ls‘3’7l3 mks/OHM) e rs. AS:_; )H, + SS¢M~Q l lg) __._'_-—- 332/‘1'1 1 k0‘35’72\ \ M75 ' ® SW“ SW“ >0 m Fmeees i5 r'fi‘amtrtv‘e" or some. We i’W‘R he'd“ P2 Z—(QLlO—LCL— V +6445le with. o fiM/J—e A514 2 - 367 a “37K V ch thfje A4 2 59971.14 =w r tare/Marge Reversible or Irreversible? ifffivfré {L016 ...
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