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Solution Manual For Thermodynamics And Heat Power, 6th Edition

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  1. Page 1
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    CuarreR 1 EE Ee eee : THE FROBLEMS IN SECTION .4 ARE INTENDED 70 PROVIDE A REVIEW OF ARITHMETIC, ALGEBRA, AND TRIGONIIDET RIC OPERATIONS, 11, (3.70) +e. 2D _ su L2, (78 70)(26.0)%.80) = YP, 47L 13. 260° = &7doo : J 260% = dosss... ) x4 /.5, WTAE (¢z yes ) Radv as / 7.6. (33355) = — [86S 47 a) 1L3swz2s® = osy9d... 4.) 3.7 sm (7) = 3.7 sm (#°) =2378.... /8 a.) (5¢ kT )eos 160° = — S.262.. LT 4) (9.7 L7 Yeos Z- PJ eos 11.25" = 8925. gu /12

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    Caper 1 eee THE FROBLEMS IN SECTION 1.4 Apc INTENDED TO PROVIDE A REVIEW OF ARITHMETIC , ALGEBRAL, AND TR/IGONIIET RIC OPERATIONS , 41 (370X420. —— = £8... . 12. (1870)(26.0)%.80) = 474,47 13. 260° = &7g00 vu LH 260% = Hosss.. - 2.6 1.5. (62.7 (50) = G XE FTE / 16. (333545) = /66S™ 47 a) 13sm2s° = osy9d.. 4) STsm (FE) = Sy sm (#0) = 2378... "8 2) (5e kTVess 160° = — S262. AT 4) (3.7 LZ eos z= G,/ eos 71.257 = 8925. gu 12

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    Class Topics Readings Practice Home Problems Work 32 Problem Session 12.17,12.18 33 Review 34 Examination Three 35 Conduction Heat Transfer 15.1 15.2, 15.6, 15.2 15.10 36 Convection Heat Transfer 15.2 15.12,15.14 37 Conduction/Convection 15.3 15.20,15.24 15.28 38 Forced and Free Convection 15.4 15.30, 15.36 15.21 15.5 15.40, 15.39 39 Radiation Heat Transfer 15.6 15.44, 15.50 40 Heat Exchangers 15.7 15.54, 15.55 41 Useful Work 8.1,8.2 8.2,8.8 15.31 42 Availability and Free Energy 8.3, 8.4 8.10, 8.12 15.49 43 Review 8.13,8.16 8.9 44 Examination Four Aesonkkokok dokolekofle softest sk flecks sles ok ok kk sk ok ses fe sk fs ks sk ees fs lle / / + StudyXY

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    19. a) 648 Log (37.6) = /0.2072 .... 6) (0.2 kvm) (37000) = 2.1037. kN-m “7 1/06 & = 5.4739.... —~20.0 -9 eé = (2.06/.. ) x /O = = 4£8/0.... 1 Sove For Po: SP+/7= 22 cos 28° . 3P=22cos28°-/7 : pos (zz cos 28° /7)/3 ) P= 0.808... ps: 112 Sorve foe xX; = 324 x= 329% = g5¢8- Ft /./3 Sorve Fox VV: V% 2v= 265m THIS Is A uA o&47ic EouATeN AND -2 * z “ y= "“RITVR)-Y (1) zs) cz THE SoLU7/60) |/=-/7.3.. /S MT A BeXHL SoLuTrom /3

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    114, Sowve For T°: | 2/3/15 C= 27.600 ~ 0.0037 0.0037 = Z7 €00 ~Z27.3/8 =0.285s 7 = 0.285 fo.003 T = 9s°C 115 Sowve pVe=mRT For To we mAve . 7 = PY . mR 16 Rr xys 23 / : 7 , 3) he 2.3 2 PROBLEMS JN SECTION 1S PPovirs Some PRACTICE IN APPROXIMATING AREAS LUMIER CURVES AND USING TRAPEZDID Luis. LIT. Area under Curve = Aran oF Berge = BASE X HEIGHT = ca 3 4u (sm 06r*Y00 2) = 720 EN me B = [20 LN-m /4

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    / /8. ABER NDEI CUBVE = ABA OF TEAPESOID A =£ (Base) sum or Two 5406s) A = £ (o0e-16C\e,@ ic * c@u0%) C,@ rode = 3.54 ,0/%/00 = 4 Ye ¢,@ IC 3.5+.0/x p50 = 34 “gc So As FEor)(ssrzc AT) yC = 3645 “Tey 119. Aen wvoere Cveve = A= Area viper: Cue WHERE yy WAR/ES INVES ELY WITH X AS sn ArreoK AY) A= C4 ar, WHewe (C= 50,000 26 J. 04 %= 50, 000 -4p So ’ A =(50, 000 2) Dn Hof = 9.3/4.7 #-lip AN ALTCRZATE AFPPROXINIATS SCOT ION CAA) BE OBrH/atD BY S/NG Smee 7BIASTOID ALSHS Simmer AS 180 APPENDIX Hs Ax.) /&

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    AS ONE EXAMPLE OF TMS, Sng 3 TRAFEEOIDAL ARSTS V,= lo#&3 7 250,000 HL» - V,= 2.0 A R= 257000 V, 23.04 R= /64e7 V, : 4.0 7, /2,500 THEN, THE Arr) CAN BE APPRIX/.#IT75D Ax FX 42) 404) Yug) / ERS (AT ~ L043) ee 41) Z L(4 7s 200 £2 +3 O)eéc7) +£ (1) 25/6 7) RUT Hl THIS FROSLeEM AA Also & Soren BY USING ComeuTsrt FRO§24.1 AREA AND pcre ComrPiTee L20 Aen UNDER CURVE = A = ABER UMD A Corre 4s 8/2" WHERE g- r, z= Vv n=/S 8B=C sn AEPNIIK AY. | /&

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    THEN ; A= ==(pV,-pV.) whees 215.0 ;a° B= 200 m4? V, = 100.0/m> - [%4 An oo As™ Pep v ) = 200(22) = 32s AND / A= roo xl00.0 ~ 34/4) 265/50) CL Hprn® T 632795 Len = 632795 ky in -_— 12] APProximArE AREA unpet CuRUE = 4 Arp A= Som or Smace TrAREECIDIL AREAS. = = (010% —,0/ #3) 1000 + Foo ZZ +4 0/17 -.0108 ){ Fo0+ $00) + 2/0/36 ~.0017 2 2 (Boo +700) + £ (ors ; 0130)(700+¢ 00) +%£ (oreo~.048Yeo0 +500) #L (‘020 — -0/e0)(s00 + #00) =g.) Hp Fr? /. 2 = 878.4 fr JAS 7

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    122. Arex Unoer. cvevs on J-s pucesm 4 AE Sum oF Tinea Aes . THIS CAN BE DONE USING SAme So OF CALCULATION AS IN ProG@eem /.2/ OC BY Using FROSrAM AREA Ann A INICED COMPUTER, . INCUT TS Tre PROGRAM ACEA wir Bs N=7 AND Y(1)= 3400 | x(1)=¢.7% Y(2)= 3500 , x(2)=¢.€1 Y(3) = 3oo ; X(3): 6.83) Y@): 3700 | X(4):¢.873 Y(5)=3800 | X(s):¢.904 Y(&) 23900 , X (6) = £.942 Y (PD =<000 , X (7): & G40 THE Bester 1s A= 665 /-23 Aren unpre curve on T-s Diacean=A AR Sum oF Trasces/0AL AreHs | THIS CAN BE DONE USING SAME Sor OF CALCULATIONS AS InN PLoBLEM /-2) or BY USING ProsZam AREA AD /%

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    A MICRO COMPUTER . INBUT To THE 206ran] AREA wil Bc j N= 3S AnD Y (1): 500 , XA): D4 Y (2): 600, X(2) 3.789 Y(2): 700, X(3): 39s¢ Y (4): 80, X(4)= %ooz Y (8): Goo, X(5): 40 THe esol 1s A= 3340s Bu L224 For p=20.SvV we Fup Pp AT vor From) [| Jo JO : P y 20.8 / 410 2 6/5 3 : 82.0 a 102.5 s 123.0 ¢ ) 43.8 7 - Jé4o & /8%5 7 205.0 /0 19

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    THE AREA UNDE® THE CvBVvE p=2o.sv’ 1S APPEOXIMATED BY THE Sum OF TRAPE Zo)D AREAS | USINE THE ETD OF PROBLEM [2] OR Using FROS4N AREA AND A PERSONAL Come Re INPUT TO THE PROGAN] Could B& N= jo AND VALLES OF P Do JHE Y-vqLres AND For. THE X-vALues. THE eEsver /S AF So Ts USING CALCULUS : : 10 10 70 A= [pds = [fa « [zos vdv / ow» / / f 4609] ~ Latin) / A= J0/¥, 75 [125 FOR THE CHANGE IN INTEEIAL ENEESY OF A PE FecT GAs, AU, WE Hue N Au == D)6,ST = Area unver 4 n=1 Crews In ¢,~7 _ Dinceam. 26

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    12S Using §T oF 50 Legesss | we (ont) cALcutATE ©, AT T= 00 To T= Soo: ¢,=3.5L+.0346 7 cy= 7062 , T=, 00 a, 8.75 , T=/50 C= 108 , 7T=200 G= R22), T=RS0 : c,s/39Y, 7=300 cy /5e7, T=350 L&E I7Yo, T=400 c,c19./13, T=4450 . 820.86, 7=506 USING AREA AND 4 FPeESonAL COMPLTEE WITH INAUTS OF N=/0, Cy ~VALUVES FOR YAVALUES , AWD 7 Foi X-VAWES, THE BPeSwer ss A= 58574 “Ty USING CALcoLps : Soo Seo A= Seya7s [lsstr.copry 97 ed 152) p Soo = 3.567 + 2 (over?) /00 B 2

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    AU = §S7& “Tks /.26 AREA unos cues oF ove 2706 /S QIVEN IN ARRNDIX 4.4 a : As ZT (506% 10%) = est / ~Ve —_— THE SAME AREA CAN BE APPROXIMATD BY THE Sum OF TRAASEOIDAL AREAS. USING AELEA AMD A FEEsoNAL COMPUTER. WITH N= 30 AMD VALUES OF P FoR Y-VALUSS AMD V (Feom 10 To 300 /N INCREmMENIS OF JO) FOR X, VALUES or P ARE DETERMI Ji/D Feow) p= E7004) So THAT, AT V=/0, p= 8538; AT v=20, p= 603.7 AnD So on. [PROGRAM AREA NEEDS Tb BE REVISED 70 RUN THIS PRo&LeEM WiTH Jo POINTS, CHANGE LINE 100 70 READ : 100 DIM X(31) AND LINE Jo TO READ: 110 DIM YY (3) 22

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    1.2 THEN Ths COMPUTER oun Wye, (owr) GE A = TE788.2 (pemenomg —_—— [2 (58 & ON Bounp-ore oF PRLS) THE PROBLEMS IN SECTION 1,7 : ARE INTENDED TO PROVIDE PRACTICE IN USING ENGINEERING EQUATION soLver (£€s) [27 OPENING EES EQuaTion WINDOW Ano ENTEEING . {Problem 1-27} i x+2%y=3.4 X*2+y*2=4.5 : THAN ClLickNg CALCULATE ANP THs SOLVE GIVES © Unit Settings: [kJVICYIkPa)/[kg) [degrees] ) x =2.003 y =0.6985 No unit consistency or conversion problems were detected. 23 [- stuayv §

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    1.28 OPENING EES EQUATION W/m Dow AND ENTEZ IG : {Problem 1-28} $=3.458 . T's**1.4=4456 . THEN) CLICK CALCULATE ON The TOOLBAR, AND SOLVE EINES Unit Settings: [KJVICYIkPa)kg)/ [degrees] s =3.458 T =7845 No unit consistency or conversion problems were detected. [29 OPEN FES EQUATION Wonizou/ AND EnTeZ, ! {Problem 1-29} P*V**1.4=280 THEN CLICK TABLES oN TO0LBAR AND NEW PARAMETRIC. TA8LE T= = TC SET No. OF Runs To <0 , PUT P AND NUN vaersLes IN TpsLE - BY ctijcerng oN pp THe Alo, THEN) on Vv AND ADO, Crick 24

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