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Exam 2 with Solutions - Thermodynamics | ENME 232, Exams of Thermodynamics

Material Type: Exam; Professor: Hines; Class: Thermodynamics; Subject: Engineering, Mechanical; University: University of Maryland; Term: Spring 2011;

Typology: Exams

2014/2015

Uploaded on 02/08/2015

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Download Exam 2 with Solutions - Thermodynamics | ENME 232 and more Exams Thermodynamics in PDF only on Docsity! Name: 1. Consider a ikg mass of water operating as the working fluid in the cycle shown below. a. Fillin the property table. b. Sketch and label the cycle on P-v diagram and a T-s Diagram. c. Calculate the cycle efficiency. d. Isthisa Carnot cycle? Give 2 quantifiable reasons. e. Calculate the entropy production o of the cycle as a whole. f. Does the cycle satisfy the Clausius Inequality? ' + 1 pee enn : 1 Saturated ' Vo vapor 1 1 0.7 MPa 4 O iP ' t 1 | Boiler 90% quality \ Ve ‘ z 5 oO 3 2okPa Gmeeenenernh 1@s eee + Saturated | \ : [ : \ 1 i liquid Condenser _. ne : +f ov mre a | - \ bor 4 3 : 110% quality 1 Lo : Le ann (ooo eee ee eee i ; yo s P (kPa) T (°C) h {ki/kg) s (kJ/kgk) x 4 700 tse 16S ©9722 992% ° |e 2 Foo oe 27ers = TBO oes 3 . a \ ; ZO the GOOG, 2373.87 WPL q \ calcolated 20 L $8723 SH sue wt * Jie? Gitte w TSat tos Bley Give Q at 2okK R&R hz het x hg- he\=2sl4 + x(26c9.7-231-4) THK S = s+ L(Sq~Se) = .8320 +x (7 BES = E52) yenq he 2373-67 B, So 7.2" Use veel We 487-23 s< 34 eye ©} wWordfen Cheha) tha hi) (2%62-5~ 231387) + (487.23-69 7. iz) Qaim hz-hy (27@3.S- 297.22) 2 BiCB2R ~ 387-63— 299-99. OBEF 2OOE. & Fe - bo (1) ;: _ = _ le - _ (F31GOOT)_ @ Mo Mreret = | Ty! (ties) |< Nee VN ocnst — Zprocesses 2-3 avd 14 ofe nto {Sentepic. sh nct adiaboic © §Q ame = COCERS _ hg ~he so (@SrZ7Z_ Gear 272 7 4.07s -34ESS- Ky ==. = .48 Kyk © yes Oo 7 392 i fone Name: 3. Consider the regenerative cycle using steam as the working fluid as shown below. Steam leaves the boiler and enters the turbine at 4MPa, 400°C. After expansion to 400kPa some of the steam is extracted from the turbine for the purpose of heating the feedwater in an open feedwater heater. The pressure in the open feedwater heater is 400kPa and the water leaving it is saturated liquid at 400kPa. The steam not extracted expands to 10kPa. a. Compiete the property table below. Sketch and labei the T-s Diagram for this cycle. Determine the cycle efficiency. If pump 2 is damaged and now has an isentropic efficiency of 0.8, how much additional work is needed in the pump, in kI/kg over the isentropic case? 2 ao — } ‘Steam _generatoe| P (kPa) T (°C) / h (kJ/kg) s (kJ/kgK) x 1 4 fo 400 3213.6 6.7690 000 nee 2 2685.6 .9752 FOO 4 [43G 3 2193.P sens? 1O +I 4d:St 193. ‘ 10 4581 14 83M .b4q3* S 5 GOO 1GZ.22BD 4} — 6 400 143-6 604.74 1.7766 Ss 7s 4000 ~ 608.6 { — nen (damaged 4OO 3 pump) A Bien \sect 5) . Oo os ese © has het A( hy he) * Se +x(S9-Se) = 141834 181 SOCK. 7- I-83) 6-169 = 6443 + 46.1502 -,6493) = 2143.4 fe - S A BIS gD ‘ea ' y — On hs & hg+ O4(Ps-Py ) = IGL@% + Ccloloz(doc~lo WW.) . » M222 ky OO @ OF CV hry + (-y)As =hg ye herbs . of: T#- 142-22 h2-he 2608S ~1 92: 2Z minchert y = (6D Win a ny = 83S TT) Oz Py ez “y ~ i Drei? Sette ed Mest es Hy Qwm 7 = hirhe) +(Fgyhehs)e(tg\hy-ha)+ beh) hy ~h 7 = 47643 . 37.53% tos ‘(om A) ‘Ape = he-has he- has he-hy a= he —(he-h7s) > O04 ]4 —(GOA-4 74¢- 608.6) 5 COV. 56 ee, 38 Peay? =(he-hq)= = -~3.86 Hey Oe =~J, 82 Ve | D0 Vey CourseSmart - Instructors - Print Page 2 of 2 Book: Fundamentals of Engineering Thermodynamics, 7th Edition Page: B93 lation of the law or N HINES part of ary book may be reproduced of transmitted by any means without the publisher's prior permission. Use (other than qualified fair use) in Terms of Service is prohibited, Violatars will be prosecuted to the fuil extent of the Tables in Sf Units 893 Envoy barks + K Sat Sat Liquid a 1 Press. bat _ Prohleunt! 3 ~~ Pretleunt | 1.0259 | 7.6700 | 0.40 1.0910 | 7.5939 0.50 44653 | 7.5320 | 0.60 aigig | 7.4797 | 0.70 1.2329 | 7.4346 | 0.80 1.2695 | 7.3949 0.56 13026 |'7.359a | 1.00 1.4336 | 7.2233 | 150 xggor | zaz7e |) 2.00 16072 | 7.0527 | 2.50 16718 | 69919 | 3.00 1.7275 | 6.9405 3.50 ‘y + 17766 | 6.8959 | 4.00 = fre fle unt 3 18207 | 6.8565 | 450 18607: | 6.8212 5.00 1.9312 | 6.7600 6.00 Kron tceroiy 700 -- Oo 20462 | BORE] “B60 Pr blow tl { 2.0946°| 6.6226 | 9.00 21987 | 65863 | 0.0 23150 | 6aas8 | 15.0 2.4a7s | 63009 | 20.0 255a7 | 6.2575 | 25.0 2.6457 | 6.1869 | 30.0 2.7253 | 6.2253 | 35.0 27964:| 6o7er | 40.0 2.8610.| 6.0199 | 45.0 2202 | 5.9734 | 50.0 30267 | 5.8892 | 60.0. pian | 5.8133 | 70.0 3.2068 | 5.7432 | 80.0 3.2858 | 5.6772 | 90.0 33596 | 5.61a1 | 100. 3.4295 | 5.5527 | 110. hitp://instructors.coursesmart.com/print?xmlid=9780470495902/892 &pagestoprint=2 2/25/2011
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