prvi i drugi zakon termodinamike (kombinovani ... - MASINAC.org

zbirka zadataka iz termodinamike strana 1PRVI I DRUGI ZAKON TERMODINAMIKE(KOMBINOVANI PROBLEMI)5/2/!U toplotno izolovanom rezervoaru nalazi se!211!lh!vode!)d x >5/29!lK0)lhL**!po~etne temperature!U x >3:4!L/!Komad bakra!)d Dv >1/49!lK0)lhL**!mase!51!lh, po~etne temperature!U Dv >469!L!i komadgvo`|a!)d Gf >1/57!lK0)lhL**!mase!31!lh, po~etne temperature!U Gf >454!L-!naglo se unesu u rezervoar savodom. U momentu uno{ewa u rezervoaru se ukqu~uje me{alica vode snage!711!X, koja radi dok se neuspostavi stawe termi~ke ravnote`e!U + >3::!L/!Odrediti:a) vreme rada me{aliceb) promenu entropije izolovanog sistema tokom navedenog procesaGf!Dv!I 3 P!!a)! prvi zakon termodinamike za proces u zatvorenom termodinami~komsistemu:! R 23 !>!∆V 23 !,!X U23 ! ⇒ ! XU23>V 2! − ! V 3! !! !! V = n ⋅ d ⋅ U + n ⋅ d ⋅ U + n ⋅ d ⋅ !2 x x x Dv Dv Dv Gf Gf UGf! V 2 = 211 ⋅ 5/29 ⋅ 3:4 + 51 ⋅ 1/49 ⋅ 469 + 31 ⋅ 1/57 ⋅ 454 >242182/3!lK!!! V = n ⋅ d+⋅ U + n ⋅ d+⋅ U + n ⋅ d ⋅+!3 x xDv DvGf Gf U! V 3 = 211 ⋅ 5/29 ⋅ 3:: + 51 ⋅ 1/49 ⋅ 3:: + 31 ⋅ 1/57 ⋅ 3:: >243388/7!lK!! !! X U23 !>!242182/3!−!243388/7!>!−2317/5!lK!!XU23− 2317/5! τ = =>3121/78!t!⋅−4− 711 ⋅21X U23!dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 2!b)!! ∆T TJ !>!∆T SU! ,!∆T P! >///!>5/58! LlK !! ∆T SU !>!∆T x !,!∆T Dv !,!∆T Gf !>///>9/58!−!3/85!−!2/37!!>!5/58! LlK !!!!+Ud+x ⋅ eUU3:: lK∆Tx= nx⋅ = nx⋅ d x mo = 211 ⋅ 5/29 ⋅ mo >9/58!∫!UU3:4 LUx+Ud+Dv ⋅ eUU3:: lK∆TDv= nDv⋅= nDv⋅ dDvmo = 51 ⋅ 1/49 ⋅ mo >−3/85!∫!UU469 LUDv+Ud+Gf ⋅ eUU3:: lK∆TGf= nGf⋅= nGf⋅ dGfmo = 31 ⋅ 1/57 ⋅ mo >−2/37!∫!UU454 LUGfxGfDv!5/3/!U kalorimetarskom sudu, zanemarqivog toplotnog kapaciteta, nalazi se te~nost polaznetemperature U U >3:1!L!)stalnog toplotnog kapaciteta D>2/36!lK0L*/!U sud je unet bakarni uzorak mase!n C >1/26!lh!i polazne temperature!U C >484!L/!Zavisnost specifi~nog toplotnog kapaciteta za bakardC−5Uod temperature data je izrazom:! ! = 1/498 + 1/76: ⋅21⋅ /Tokom uspostavqawa[ lK 0 ( lhL* )][ L]toplotne ravnote`e u kalorimetru, okolini stalne temperature!U p >394!L-!predato je!6% od koli~inetoplote koju je predao bakarni uzorak. Odrediti:a) temperaturu u kalorimetarskom sudu u trenutku uspostavqawa toplotne ravnote`eb) promenu entropije izolovanog sistema od polaznog stawa do stawa uspostavqawa toplotneravnote`e u kalorimetru.!!a)oznake koje se koriste u daqem tekstu re{ewa:!!! R !! koli~ina toplote koju bakar predaje te~nosti u sudu( 23 ) C( R 23 ) p( R 23 ) Ukoli~ina toplote koju bakar predaje okolinikoli~ina toplote koju te~nost prima od bakra!! U S ! ! temperatura u kalorimetarskom sudu u trenutkuuspostavqawa toplotne ravnote`edipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 3!prvi zakon termodinamike za proces sa bakrom:( ) C−5R > [ ]! R 23 ,( 23 ) pn C ⋅ 1/498 + 1/76: ⋅21U ⋅ eU !! ( R 23 ) C,(23 ) p!!U S∫UC⎡3 3⎤−5US− UCn C ⎢S C + 1/76: ⋅21⎥ !! )2*!⎢⎣3 ⎥⎦R > ⋅ 1/498 ⋅ ( U − U )prvi zakon termodinamike za proces sa te~no{}u!( ) UU S! R 23 > ∫ D ⋅ eU > D ( U ⋅ US− UU)! ! ! ! ! )3*!UUUinterno razmewena toplota izme|u bakra i te~nosti:!!! − ! ! ! ! ! ! ! )4*!!( R 23 ) C> ( R 23 ) U[ ]! uslov zadatka:! ! ( R 23 ) p> / 16 ⋅ ( R 23 )C+ ( R23)1 ! ! )5*!!!! kombinovawem jedna~ina!)2*-!)3*-!)4*!j!)5*!dobija se:!! U S !>3:4/8!L-! >5736!K-! >−5736!K-! >−354!K!!b)!( R 23 ) U( R 23 ) C( R 23 ) p! ∆T TJ !>!∆T SU !,!∆T P !>!///>!2/2:!,!1/97!>!3/16! LK !p! ∆T SU !>!∆T U !,!∆T C !>!///>!26/96!−!25/77!>!2/2:! LK !!!!U S( )D U ⋅ eU US3:4/8 K∆TU== D U ⋅ mo = 2/36 ⋅ mo >26/96!∫!UU3:1 L∆TCUU= nCU SUU S.5( U) ⋅ eU ( 1/498 + 1/76: ⋅21U)d⋅ eU⋅= nC⋅=∫ U ∫U!UC⎡ US−5n ⋅ ⎢1/498⋅ mo + 1/76: ⋅21⋅ S⎣ UC( R23)p 354 K∆ TP= − = − >1/97! !U 394 LPUC( U − U )C⎤⎥ = /// = −25/77⎦KLdipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 4!5/4/!Meteor temperature!U>4111!L-!brzinom x>21!ln0t ule}e u ledeni breg temperature!U>384!L/!Masa meteora je!n n >21!lh-!a specifi~ni toplotni kapacitet!d n >1/9!lK0lhL/!Odrediti:a) koli~inu toplote koju meteor preda ledenom breguc*! promenu entropije izolovanog sistema koji ~ine meteor i ledeni breg!d*! masu otopqenog leda (toplota topqewa leda iznosi s>443/5!lK0lh)!!meteor x>21!ln0t!ledeni brega)!prvi zakon termodinamike za proces koji se de{ava sa meteorom:!x3− x2= n n n3 n2 + n ⋅ !3! R 23 !>!∆V 23 !,!X 23 !,!∆F l23 ! R n ⋅ d ⋅ ( U − U )4 321 ⋅2123 21 ⋅ 1/9 ⋅ 384 − 4111 − 21 ⋅ ⋅21! ( )( ) −4!b)!R23= >− 6/329⋅21lK!3lK! ∆Ttjtufn= ∆Tnfufps+ ∆Tmfe>///>−2:/286!,2:22/466>29:9/29! !LUn3384lK! ∆ Tnfufps= nn⋅ dnmo > 21 ⋅ 1/9⋅mo>!−2:/286! !U4111L!c)mn2R236/329 ⋅21∆ T mfe = >U 384! R = n ⋅ ! ⇒!!23 m sm6>2:22/46! LlK !R236/329 ⋅21n m = >s 443/5m6>2681!lh!933dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 5!5/5/!U rezervoar sa n x >61!lh!vode!)d x >5/2:!lK0lhL*-!temperature!U x >394!L- uroni se zatvorena boca,na~iwena od ~elika)d • >1/59!lK0lhL*-!sa!W l >21!litara kiseonika (idealan gas). Masa ~eli~ne bocezajedno sa kiseonikom je!n • !,!n C >!31!lh/!Pre urawawa temperatura kiseonika i boce je!U l >U • >474!L-!a pritisak kiseonika u boci je!q l >26!NQb/!Sistem koji se sastoji od vode i boce sa kiseonikommo`e se smatrati izolovanim. Temperatura okolnog vazduha je!U p >3:4!L-!pritisak!q p >2!cbs-!azapreminski (molarni) udeo kiseonika u okolnom vazduhu je!32&/!Zanemaruju}i razmenu toplote saokolinom odrediti:a) temperaturu u sudu u trenutku uspostavqawa toplotne ravnote`eb) radnu sposobnost kiseonika u boci u trenutku postizawa toplotne ravnote`e!a)prvi zakon termodinamike za proces koji po~iwe urawawem boce a zavr{avase uspostavqawem toplotne ravnote`e;!!! R 23 !>!∆V 23 !,!X 23 ! ⇒! V 2 !>!V 3 !!! V = n ⋅ d ⋅ U + n ⋅ d ⋅ U + n ⋅ d ⋅ !2 x x x • • • l wL UL!+++V3 = nx⋅ d x ⋅ U + n•⋅ d•⋅ U + nl⋅ d wL ⋅ U !!!+ nx⋅ d x ⋅ Ux+ n•⋅ d•⋅ U•+ nl⋅ d wL ⋅ ULU =nx⋅ d x + n•⋅ d•+ nl⋅ d wL>///!!jedna~ina stawa idealnog gasa za kiseonik u boci na po~etku procesa:! q ⋅ W = n ⋅ S ⋅ U ! ⇒!lllhl)lnlq=S! n • !>! ( n + −!nl!>!31!−!2/6:!>29/52!lh!• n llhl⋅ W⋅ Ul726 ⋅21⋅21⋅21=371 ⋅ 474!! ! !!61 5/2: 394 29/52 1/59 474 2/6: 1/76 474U + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅= >397/7!L!61 ⋅ 5/2: + 29/52⋅1/59 + 2/6: ⋅ 1/76! !−4>2/6:!lh!napomena:!!! U + !−! temperatura u boci u trenutku uspostavqawa toplotne ravnote`e!!dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 6!b)jedna~ina stawa idealnog gasa za kiseonik u boci u trenutku uspostavqawatoplotne ravnote`e: q2 ⋅ Wl= nl⋅ Shl⋅ U2!nl⋅ ShL⋅ U22/6: ⋅ 371 ⋅ 397/7q2= =>22/96!NQb!W−421 ⋅21!! odre|ivawe pritiska kiseonika u okolnom vazduhu:!pql= sL⋅ qp>1/132!NQb!!!odre|ivawe radne sposobnosti kiseonika u boci u trenutku uspostavqawatoplotne ravnote`e:! ! Xnby= n ⋅ ( −∆v21+ Up⋅ ∆t2p− qp⋅ ∆w2p)!!!⎡⎛pU q ⎞⎤PL pXnby = nL⋅ ⎢dlw⋅ ( U2− UP) + UP⋅ ⎜dq ⋅mo− Sh⋅mo⎟ + qL⋅ ( w2− wp) ⎥ = 761⎢⎣U2q⎝2 ⎠⎥⎦lK !! !!Shl⋅ U2371 ⋅ 397/7n 4w2= =>!1/1174!! !q722/96 ⋅21lhwS2⋅ U371 ⋅ 3:4n 4 !hl pp = =>!4/7387!!p7ql1/132⋅21lh!! napomena:U delu zadatka pod b) radi lak{e preglednosti veli~ine stawa kiseonika uboci u trenutku uspostavqawa toplotne ravnote`e obele`ene su indeksom!2!!dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 7!5/6/!Radna materija u zatvorenom sistemu vr{i neki proces pri ~emu joj se u svakoj sekundi dovodi!R>4!⋅lK0t!toplote i odvodi zapreminski rad! X)lK0t*-!koji se u toku vremena mewa po zakonu:!!⋅X 12τ= 3/5 ⋅ !! ){b!!1! < τ ≤ 2 i *![ lX] [] i⋅X23>,3/5![ lX]! ){b!!τ!?!2!i*! !b*! odrediti brzinu promene unutra{we energije sistema-! ∆ V ⋅23 )lX*-!u trenutku!vremena!τ>1/7!i!b) odrediti promene unutra{we energije sistema-!∆V 23 !)lK*-!u toku prva dva ~asa![ lX]! - !X ⋅⋅i]τ-! [ !2!3!a)⋅⋅⋅! ∆ V23 = R23− X23= 4 − 3/5 ⋅ τ = 4 − 3/5 ⋅ 1/7 = 2/67 lX !b)223τ2! X12 = X() τ ⋅ eτ= 3/5 ⋅ τ ⋅ eτ= 3/5 = 2/3 lXi !∫ ∫3 1113323 =22! X = ∫ X() τ ⋅ eτ= 3/5τ3/5 lXi !! X p3 !>!X 12 !,!X 23 !>!4/7!lXi!>!23:71!lK!! !3! R = ∫ R⋅() τ ⋅eτ= 4 ⋅3= 7 lXi 32711 lK !13 =1! ∆V 13 >!R 13 !−!X 13 !>!9751!lK!!dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv

zbirka zadataka iz termodinamike strana 8!5/7/!Toplotne karakteristike neke radne materije zadate su zavisnostima: !! ! ! q! / !w!>!B! / !U!!! ! ! v!>!C! /! U!,!D! / !U 3 !,!E!gde je!B>!3:8!K0)lhL*-!C>7:8!K0)lhL*-!D>!1/196!K0)lhL 3 *-!E!>!dpotu-!a!q-!w-!U!j!u su veli~ine stawa uosnovnim jedinicama!TJ. Radna materija mewa svoje toplotno stawe kvazistati~ki adijabatski od stawa!2)q 2 !>!1/2!NQb-!U 2 >511!L*!do stawa!3!)U 3 >2451!L*/!!b*! izvesti jedna~inu kvazistati~ke adijabatske promene stawa radne materije u obliku:!q>g)U*!c*! odrediti pritisak radne materije u stawu!3!!b*!prvi zakon termodinamike za proces sa radnim telom (diferencijalni oblik)!! δr= ev + q ⋅ew! ! )2*!! !! diferencijal proizvoda:! ! e( q ⋅ w) = q ⋅ ew + w ⋅ eq !)3*!!! kombinovawem jedna~ina!)2*!i!)3) sa toplotnim karakteristikama radnematerije dobija se:! !! 1!>!ev!, e q w − w ⋅ ! ⇒! ev!,!e q ⋅ w >! w ⋅ eq ! !!3! e( C U + D ⋅ U + E) + e( B ⋅ U)( ⋅ ) eq( )eq⋅ = B ⋅ U ⋅ !qeq⋅ = B ⋅ U ⋅ ! !q!B + C + 3D ⋅ U eUB + C U! ⋅ =⋅ moB U qB U3! e( B U + C ⋅ U + D ⋅ U + E)!! !B+CU3D⋅mo+ ⋅( U−U2)B U 2 Bq = q 2 ⋅ f! ⇒!q = 1/2⋅217⋅ f3:8+7:8⋅mo3:83Deq ! ! + ⋅ ( U − U )U5113⋅1/196+3:8⋅( U−511)q= mo !22 Bq23:8+7:8U5113⋅1/196⋅mo+ ⋅( U−511)7q = 1/2⋅21⋅ f3:83:8!b)!!!!ako u izvedenu jedna~inu stavimo!U>U 3 >2451!K, kao i vrednosti za navedenekonstante!)B-!C!j!D*!dobija se:q3:8+7:83 f2451 3⋅1/196⋅mo+ ⋅( 2451−511)7= 1/2⋅21⋅3:8 511 3:8>:/9!NQb!dipl.ing. @eqko Ciganovi}{fmlp@fvofu/zv