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__________________________________Tehnička postrojenja za provedbu kružnih procesa Motori s unutarnjim izgaranjem danas su najčešći strojevi za proizvodnje mehaničke energije (pogonski strojevi). Ocjenjuje se da njima pripada više od 90 % ukupne snage svih pogonskih strojeva koji energiju goriva i potencijalnu energiju vode transformiraju u mehaničku energiju. Uzrok tako širokoj primjeni treba tražiti u činjenici što se mogu ekonomično izgraditi motori u širokom rasponu snage i što je vrlo jednostavno osigurati opskrbu pogonskim gorivom, bez posebnih vodova i priključaka. Motori s unutarnjim izgaranjem danas se u prvom redu ugrađuju u prometna sredstva (u osobne i teretne automobile, motocikle, zrakoplove, brodove, lokomotive), ali je njihova uporaba znatna i u poljoprivredi (pokreću traktore, vršilice i sl.), u građevinarstvu ( pokreću rovokopače, miješalice, dizalice i sl.) i u vojsci (pokreću tenkove, zrakoplove, prijevozna sredstva na kopnu i moru). Motori s unutarnjim izgaranjem izvode se danas do snage od oko 34 do 40 MW. 6.2.3. Procesi u kompresoru U kompresoru se tlači zrak ili neki drugi plin, na što se troši mehanički rad, za razliku od pogonskih strojeva u kojima se ekspanzijom radnog tijela dobiva mehanički rad. Na slici 105. prikazan je teorijski rad jednocilindričnog idealnog klipnog kompresora, tj. takvog kompresora kod kojeg u cilindru ne postoji tzv. štetni prostor i kod kojeg se zanemaruju gubici energije. Međutim, u cilindru stvarnog kompresora postoji štetni prostor, slika 106. gdje uvijek ostaje određena količina stlačenog plina koji se širi (krivulja 3 – 4), miješa se sa svježim plinom i u isto vrijeme dovodi u cilindar. Štetni prostor smanjuje proizvodnost kompresora, a nas zanima teorijski rad idealnog kompresora. Ovdje crte 0 – 1 i 2 – 3 predstavljaju proces punjenja, odnosno pražnjenja kompresora, dok krivulja 1 – 2 predstavlja proces kompresije plina u njemu. Teorijski rad kompresora, odnosno onaj rad koji je teorijski potrebno utrošiti u kompresoru, prikazan je na slici 105. površinom 0 – 1 – 2 – 3; za 1 kg plina i za v0 = v3 = 0 iznositi će 2 w′ = p ⋅ v2 + ∫ p ⋅ dv − p0 ⋅ v1 . 1 297
TEHNIČKA TERMODINAMIKA__________________________________________ Slika 105. Idealan rad idealnog kompresora (bez štetnog prostora) Slika 106. Idealni rad stvarnog kompresora (sa štetnim prostorom) 298
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