TehniÄka termodinamika - Kemijsko-tehnoloÅ¡ki fakultet

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TEHNIČKA TERMODINAMIKA__________________________________________ 7.1. PROCESI U RASHLADNIM STROJEVIMA Održavanje potrebne, dovoljno niske temperature u nekoj sredini – hlađenje – moguće je uključivanjem lijevokretnog kružnog procesa, npr. Carnotovog, odnosno utroškom određenog mehaničkog rada. Bez utroška toga rada ne bi bilo moguće toplinu odvesti, tj. podignuti od sredine s nižom temperaturom sredini s višom temperaturom. Ako bi bilo moguće, to bi bilo protivno II. zakonu termodinamike, toplina bi sama od sebe prelazila s hladnijeg na toplije tijelo. Ukupni porast entropije bio bi negativan (∆s < 0) jer bi promjena entropije sredine kojoj se odvodi toplina q0 pri Th bila veća od promjene entropije toplije okoline kojoj se dovodi isto tolika količina topline q0 pri temperaturi T0 , slika 109.a, tj. bilo bi q0 q0 ⎛ 1 1 ⎞ q0 ∆s = −∆sh + ∆s0 = − + = q0 ⎜ − ⎟ = h 0 < Th T0 ⎝ T0 Th ⎠ T0 ⋅Th pošto je T h < T 0 . ( T − T ) 0 Slika 109. Lijevokretni Carnotov kružni proces 302
_______________________________________________________________Rashladni strojevi Međutim, u našim uvjetima su nemogući procesi pri kojima bi porast entropije cijelog sustava bio manji od nule, a nemoguće je da se neki proces, u promatranom slučaju rashladni proces, odvija od stanja veće vjerojatnosti k stanju manje vjerojatnosti. To bi bilo moguće ako bi se paralelno s njim odvijao i neki kompenzacijski proces čiji bi smjer protjecanja bio suprotan, tj. od stanja manje vjerojatnosti k stanju veće vjerojatnosti i kojim bi se upravo kompenzirao negativni porast entropije. Takvim kompenzacijskim procesom postiglo bi se da ukupna entropija, a to znači i vjerojatnost sustava poraste, tj. da je ∆s > 0 ili u da u najmanju ruku, u graničnom slučaju, ostane stalna, tj. ∆s = 0. Ako bi ∆s → 0, tada bi i ∆ s0 → ∆sh pa bi kompenzacijskim procesom trebalo dovesti okolini, prema slici 109.a, toplinu ∆ 0 q = 2 ' − 4' − 4 − 3' − 2' = T ⋅ ∆s . To znači da bi se od neke sredine odvodila određena količina topline q 0 s ciljem da se u njoj održi, dok je god to potrebno. Temperaturu T h , koja je niža od temperature okoline T 0 , bilo bi potrebno dovoditi okolini u najpovoljnijem slučaju (za ∆s = 0) toplinu q = q 0 + ∆q . Površina 1 – 2 – 1’ – 4’ – 1 ≡ ∆q = ( q0 + ∆q) − q0 = q − q0 = w . To pak znači da bi, za utrošak najmanjeg kompenzacijskog rada (za ∆s = 0), bilo potrebno između rashladne sredine temperature Th i temperature okoline T0 uključiti suprotni (lijevokretni) kružni proces, npr. Carnotov. Prema slici 109.a izgleda da bi to trebao biti desnokretni ciklus, ali se pri tome ne smije izgubiti iz vida da se promjene stanja 1 – 2 i 1’ – 4’ na toj slici odnose na izvore topline, promjena stanja 1 – 2 na rashladnu sredinu od koje se odvodi toplina q 0 , a 1’ – 4’ na okolinu kojoj se dovodi toplina q = q 0 + ∆q . Međutim, kružni proces obavlja radno tijelo, slika 109.b, kojem se dovodi toplina q 0 od sredine koja se hladi, proces 1 – 2, odnosno od kojeg se okolini predaje toplina q = q 0 + ∆q , proces 3 – 4. Koeficijenti hlađenja ili stupanj rashladnog učina, tj. odnos odvedene topline od hladnijeg tijela i utrošenog rada lijevokretnog Carnotovog kružnog procesa ima vrijednost, prema oznakama na slici 109.b 303
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