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Unformatted text preview: 14w fix Hwfiz kubkg MW #2 (-68 1"” 2/56 3.25 3’80 ['7'6 3’5‘ ‘ ('7‘! 2-] 3‘56 3-8q l-LZO Z_ 23 3'56 3-?5 ' 2-33 Hz 3_63 3-105 H 963? 4-5; i ‘l-éz 9“ 3. n n Ade E i— ll¢5 T (A; ~VLC .5 2,45 3—102 3-612; E: z_3L{ ave? 9'59 F. 1-50 5641064 Ms g 3’50 Wilma-cl 3 '52; P V = = PV=CPP8at@T Welect = V ° I = 12R W = IPdV W = m dev dh = cpdT m _ m = dmw Vx = Vf + X(Vg-Vf) ux = 11f + X(ug - uf) hx = hf + X(hg _ hf) incl: i out”; a W = 175-437 F = kx (linearsprz’ng) = «A d_T du = cvdT dX h=u+PV X: mm AKE=lm(V22_V12) APE=mg(zz—zl) mliquid + mvapor 2 cp = 0V + R Q—W+ 2351,01,. +V,.2 /2+gz,.) — 2,5100% +V92/2+gza) = dE/dt out 615 q-w=Au+Ake+Ape Q-W=AU+ Q—W=%(U+KE+PE) reversible As=s§—s§—RIn—El W=Mforng1 W=p1V1lnzg—forn=1 P1 1“ ’7 V1 As = cP 1n(Tz/Tl)—Rln(P2/ P1) . P1 _ Pr] V1 Vrl isentropic, variable Ideal gas, const prop — ' — — = — ' ‘ AS = CV mm m) +Rln(v2 ,Vl) P2 Pr; V2 Vr2 propertles, 1deal gas AS=Cp1n(T2/1‘1) incompressible p V k _2_ = -—1— isentropic, ideal gas p1 V2 W = H - QL _ Q (k—iyk k—l Q — Tr— heat % = = isentropic, ideal gas 1 1 2 (QH/QL)rev = reservoir COP = QH/Win (heat 7] = Wnet/QH (heat engine, power COP = QL/Win (refiig. systems) pump) cycle) 2 2 2 P 41 I V &_+Vi+zz=£1_+5_+zl 21+;:]—[Zz+%)=ngTO=HL HL=f—D-'2—— pg 2g pg 2g g one dimensionalflow: m = pVA Re = VDp v = p/p y IDEAL GAS CONSTANT: Ru= 0.08314 bar'm3/(kmol°K) Ru= 8.314 kJ/(kmol°K) Ru= 1.986 Btu/(lbmol°°R) Ru: 1545 ft'lbf/(lbmol°°R) AS=fa—Q— “I'ng dscv=2_Q—L+ minsin- s +3 T b dt J. T} out out gen f as out e15 AS = forreservoir 'fiob/ém/VL fi' 1 V [44/146 ‘///Cf‘ W330i: goégy 0‘4 #467 aka/(‘66, 0/ a 5LLéVW‘1V’VIC (rods/43 300 ire/ad ’fl‘f aflf‘f ELM/My @/ 5e“, , Assame baw mew/C DRESch LS Mr? pszq, Q/Lc/ v/fie area/re gray/{5 0/ LS /. 03>” MM \L “613$? Q) m cut an“ k/W " My“) 44,9, may)“ Sawmuwnc ¢fit§5w we @ \A, (a) To Quizwa "ii/mm 4 93th N. . I. ‘L [1va ~ 3: ~ H4 at» + (61-1? “:3- ) (32.1911 (30" SD paw/A; W x L W 591‘” (ca/.ux 32./:71/)/3oa) __éé_£‘fib_[{ inc/W ,.——-.-____ R) Check. CWUQ-t/falcw .kwlole‘s 1 Pat : 1L”! A ?r€§§cy‘€ : 32.179 112M 944/52 qfim’e .. 2 - _ . [991‘ [13,?!" ] [é—i = fistwga : fizi Eli-Hm“! QB; - / \ 1254 (Me A.on 42. [1 A42 5 /7‘/ Ink 2-; 7): Wig-7,0554 ,4 /67.Z?><300Jli W Mil/"‘7’ ’ M7 ~ “306) .. - fstcb . 7" Z{~SIW@ /91/ ml Act; a mass [fig 5W4" 4 rm M 3.5 ml 0 fl (amp-mséac/ 51F/M3 Qéoua. % 69 €545“, 3242 HS a fame 0/ flflfl 0V1 Me p/sle/L/ : v/flp a/masfldmvc flrnguc: As 25 52% / Zk£nwmw '? Made %@,(9A44& IJ WC EngAbx/L, / 146er Me, LH‘L‘) ‘rs I P A x} {i Q &/ (my 7 row W11; 7 CL Ada ' AF=§5€ \4/[i/1/lqe—Wigygas‘?:§e ‘3 1 e 1 w M3 “9 A (2.) E61 0‘, ecu/INN am 0‘; atoms ‘ 1303‘8 we, ,— Iv m$ I x OWEW C3) new +0 NWLM Jrovce -— ’1) we ' :%%,4W%% U/WL L0 Wm, ff: A f Lime?) Ava/fies ADM/v 3’ Mum; ?wé/W #3 fl mammc/n 6 05a] [IV/660,3 a”? M ayawk C! as“ 44:5 94 C's/wk» gray/17) 0/ @fiE/ WC/ 7416 QA/flgflfizd [fig/fill 5814/8844 #8 Z Gum‘s 672‘? "L 0 // fie /0 609/ l0 __———— /~3 /’Zo?¢5£q, o/g/gwwme 74245» aéswéz/c? flvzssu/F (I4 #6 We M 1/1,. 470w mi m 4% v’ém/c zs Anew @ flemcoutéc’o/ :1 (I’m/J A? =E’6‘N ‘ . 6 - I ’9’ a: P ~ 50041420 2 L25 >< ézuwém/Iga “ 925%; [32"7" £9) 5: 925, 001,,“ m6 k/ {#2 32.99 it.“ 797/115) : A 26/66; (A, «3: Palm «LAP = 17»? +1.25 am : /3a?0/ ’05(~W (,JLU “In C(rgcwice Be ’1 “We 8mg; Kiwi [WM er‘vc Ls Qowawz fr— fly! 0% fig QC¥/V/‘/QS M @A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power generation potential of this system is to be determined. Assumptions Water jet flows steadily at the specified speed and flow rate. Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses, and it can be converted to work entirely. Therefore, the power potential of the water jet is its kinetic energy, which is V2/2 per unit mass, and n'1V2 / 2 for a given mass flow rate: V2 (60 m/s.)2 lkJ/kg e e, =ke=—=— — =1.8kJ/k m h 2 2 (1000 mZ/sz) g Wmax = Emech = memech =(120 kg/s)(1.8 kJ/kg) lkw =216kW IkJ/s ti Therefore, 216 kW of power can be generated by this water jet at the stated conditions. The work required to compress a spring is to be determined. Analysis Since there is no preload, F = Irx. Substituting this into the work expression gives 2 2 2 k 2 2 F 1 i r l =ZOOlbf/inE1in)2_02 i =8.33lbf-ft Tc 2 12in 1Btu 778.169 lbf'ft = (8.331bf-ft)( ] = 0.0107 Btu classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from Qwoling = Qlights + Qpeople + Qheat gain where Q'lights =10 x 100 w = 1 kW Qpeople = 40x360 kJ/h = 4 kW gang,“ = 15,000 kJ/h = 4.17 kW Room 15,000 kJ/h 40 people Qcool Substituting, 10 bulbs Qcoolmg = 1+4 +4.17 = 9.17 kW Thus the number of air-conditioning units required is 9.17 kW ——.=1.83——-——>2 units 5 kW/unit 3-30 Complete the following table for H 2 0: T, °C P, kPa h, kJ / kg X Phase description five I 120.23 200 2046.0 0.7 Saturated mixture 21a L 140 361.3 1300 0.56 Saturated mixture 0‘?” 177.69 950 753.02 0.0 Saturated li uid -— My]! 80 500 334.91 — - - C0 ressed li uid ‘ 350 800 3161.7 — - - Sugerheated vaéor a; m, Cw: 6W @ *2!” ’1‘?“ god Ways": SF @ a X g 5’ y X it ‘\\“~..>:/,/ L ;-:-> M #200 a; M 523% 0"“ fliiji M ‘ ,. ’ , t 77mm, fl’g Q f‘fikfi“ é :k/ééa?Cé5§4%/) if = 500.10 awflskmmfi‘) j » 50910 + 15W, 33 M g'zo'aag 204%. xv,...w*"" N""“\.\ .._ _. SFW acts-u p b“: 535’ '3 Fig/‘5} 05:»4/0‘2/9 0%,: 2953.0 tié/kg 3%, q «4, g, 5‘7 W a Mg 7: :3; £522.? 3940c? [gag ‘ f auség fl/«c 3 F... T: 0C : 600 “"”“ r’ [,1 ’3 731,33 ~' T” 7 8’0 ?:5flf3m “I? k: 5?;3955 xfi‘ Uké‘7fl 6’5? :Luwl’btl" Ex?” 0/ : .. A ’7 ’_ K [A é é/Ql [Ufa/é}, 5’55 «w fiLfiL '2": Lia-ow: ( (€504, — 9:93; Iii/0% Wfi' Oio‘oxoz‘; ;:~:,/}¢Ca é: ~74“ Odaazozg ( 500 r was?) : 35/2. 9/ 7‘ @.00/0?97 C 415705,!) t (33% ?/% 6. {5597' :- 35ng {5376* flew-mg ; 3275 Li A 354 f.“ 333% E? a“ 5'3" [£17351 .7}? : 3,23: 'drqsuolwlel IEMQWI-eldlllnw 9H1 mlqo p112 Z/((€Z)J+ (D)J)Z{=I 91m [upgozsdml {anew} 913L115 91.11 Sugsn empeomd 9111 91211511111 'sepu uop‘efiam} [amaluy-qdpmm 9111 1rqu 0], pemonogr empeooxd 9111 s; 12% (smgod p) '5 J W 3-53 [Also solved by EES on enclosed CD] A piston-cylinder device contains a saturated liquid-vapor mixtugg of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 350°C. The initial temperature, the total mass of water, the final volume are to be determined, and the P- v diagram is to be drawn. Analysis (21) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure, Efidffim 7 Tgflégw = 170.43 “c (Table A-5) W' a a” «s (b) The total mass in this case can easily be determined by adding the mass of each phase, V 3 vf ‘0.001115m /kg‘j _/ fir, V VT 0 9 m3 7”] / “Wk—“AMT; P =_§_ _ = 3 74 kc i ,x r 3 v, _ 0.5404 m37k‘ga ’4 3.74 = 93.43 kg m, = mf +mg ;?9.69 (c) At the final state water is Superheated vapor, and its specific volume is } v2 = m,v2 = (93.43 kg)(0.3544 m3 /kg) = 33.1 m3 P2 = 800 kPa =0.3544 3/k T2=350°C v2 m g gym. Jr; waif/w, (Table A-6) Then, if L} f I ,1 y L “ If," 31:”: ‘1: x 1 31/3422 / l, ‘“’ agjfisfi/KF J; mfg H20 decaf gr»; 3' .1 . 9” f“ WNW P = {dig} [fluid :40 la: ‘5; Q ,4, 53:55, 7 r, ! 'L/l ‘ '1, @5404 Cal/flay dew/Ce (“ct/1116;,wa a [n73 Agy/c/ #490 Md Oacl M30: 0W LL10 M ércwlbww @ Wipes. “301' ‘5 ““rWle’ed 0:} (Enslaved m—Ll law? W Consider a constant pressure process (of water) T. 'C 57444 0.003155 11,391“ 28 T-V Diagram Tl Critical P05!“ 41? é?’ / ‘9 z I/ ,’ V I ‘31, §§’/ COMPRBSED ‘ ,I’ (I / LIQUID I """""" “ ‘1 ,’ REGION / . / SUPERHEATED VAPOR REGION LIQUID—VAPOR REGION 29 Solutions to activity #5 Class #9: February 13, 2008 Problems 3—97, 3-102 III-class problems: 3-92, 4-59 3-92 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1 , R = 0.4615 kPa-m3/kg-K, Tcr = 647.1 K, Pcr = 22.06 MPa Analysis (a) From the ideal gas equation of state, I 3 0 ‘ r v = g = ————(°'4615 H2531)? me K) = 0.09533 m3lkg (3.7% error) t T “t a (1)) From the compressibility chart (Fig. A-15), P 3.5MPa P =—=———=o.159 R PC, 22.06 MPa Z 0961 H20 T _L__723_K_112 ”‘ “If” R T 647.1K ' 450C or Thus, v = 2videal = (0.961)(0.09533 m3/kg) = 0.09161 m3lkg (0.4% error) (c) From the superheated steam table (Table A-6), P =35 MPa 3 T = 450°C }v= 0.09196 m [kg 0 a @/g¢yflln€ l) 0/ jyflfyfic’ooécl a’g 040! 1/s@ cc 05mg Caffdew/ gas (’ t’w- KT) ’ . /- . I e 6 (351llm‘tle Z 5) (Mg/glib Cltw/ 5BR, 04“! IA ,- 3456 4 Lgévldd) ('9‘ 3-97 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1, R = 0.5182 kPa-m3/1(g-K, TCI = 191.1 K, Pcr = 4.64 MPa Analysis From the ideal gas equation, _( (5 (M‘s/ac, I P _._ Cad/k; v _______./’54‘53 T =T—i= 300K 1.5 =450K a, ' 2 1 V1 ( X ) J FE : T‘ V; v, From the compressibility chart at the initial state (Fig. A-15), T TR, =—1=flK— =1.570 Tcr 191.1 K 21 =0.88, VRl P PR1 =—1—fl'fl)i=1.724 P ' 4.64 MPa or At the final state, 2 — . Thus, T2 = P2v2 =3:vacr = 8000kPa (1.2)(191.1K) =406K 22R 22 Por 0.975 4640 kPa Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate. (film/Mam] flaw“ 5/7ch 3w/C “,5 @412] lo I Zgfi/J/ 1/ [tag ((4646chch 507- CV2,‘ LEVI), its/fig (Q 1:15:41 £56 5) (Mg/bay CAWA 3-102 Methane is heated in a rigid container. The final pressure of the methane is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state. Analysis ((1) From the ideal gas equation of state, T P2 = P1 —2—=(100kpa) 673K T1 293K = 229.7 kPa The specific molar volume of the methane is RT . - 3 - V1=V2 =LI=W=Z436msflqnol (b) The specific molar volume of the methane is RT . - 3 - V1 =V2 = u 1 = (8 314kPa m /km01K)(293K)=24-36m3/k1n01 Using the coefficients of Table 3-4 for methane and the given data, the Benedict-Webb-Rubin equation of state for state 2 gives RuTz C0 1 bRuTz—a aa c y -2 P =—+ B T — ———-— —+—+—+ 1+— ex — /v 2 V2 ORu 2 A0 T22 V2 V3 V6 V3132 V2 y ) _ (8.314)(673) + 24.36 24.362 24.363 6 [0.04260x8'314x673487.9%2.22:;10 ] 1 +0.003380x8.314x673 5.00 5.00x1.244x10_4 2.578x105 0.0060 + —6——+—fl 1+ 2 24.36 24.36 (673) 24.36 = 229.8 kPa )exp(—0.006O/24.362) 4—59 The internal energy changes for neon and argon are to be determined for a given temperature change. Assumptions At specified conditions, neon and argon behave as an ideal gas. Properties The constant-volume specific heats of neon and argon are 0.6179 kJ/kg'K and 0.3122 kJ/kg‘K, respectively (Table A-Za). Analysis The internal energy changes are Auneon = cVAT = (0.6179 kJ/kg ~ K)(180 — 20)K = 98.9 lekg Au = cVAT = (0.3122 kJ/kg -K)(180 — 20)K = 50.0 kJ/kg argon flat/Maul 742% 122 45 Méwgca% 20? fl» lwoc. HA (“law-cl Mae/mm) twill; will to M Le (£451er 2r AV ,7 5&9. W73 Z/’,Zl/ ///8/ CD Rewwlow M h Am‘l’Le' were Em «2 amt—M4 TWP H T7. , 41L: EVUMLT 1 CV87“ CUM— 't CVCWQJW >Tm A2, CV new ...
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