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Tehnička termodinamika - Kemijsko-tehnološki fakultet

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______________________________________________________Stvarni ili realni plinovi i pare<br />

−1<br />

q3,4<br />

= h4<br />

− h3<br />

= 1011 −1953<br />

= −942 kJ kg<br />

q 4 ,1 = 0<br />

q1,2<br />

− q2,3<br />

− q3,4<br />

2373 − 910 − 942<br />

η =<br />

=<br />

= 0.22 → 22 %<br />

q1,2<br />

2373<br />

c)<br />

−1<br />

−1<br />

s2 − s1<br />

= 7.3312 − 2.8119 = 45193 kJ kg K<br />

Primjer 5.17.<br />

U kotlu isparava voda pri tlaku 19.6 bar. Suha para odvodi se u parni stroj<br />

gdje adijabatski ekspandira do tlaka 0.0981 bar. Nakon toga para se kondenzira<br />

pri konstantnom tlaku, a onda kondenzat crpkom prebacuje u kotao.<br />

a) Koliki su entalpija i stupanj suhoće nakon adijabatske ekspanzije?<br />

b) Koliki je termički stupanj iskorištenja (zanemariti rad crpke za prebacivanje<br />

kondenzata)?<br />

c) Skicirati proces u p, v – i T, s – dijagramu!<br />

Rješenje<br />

a)<br />

−1<br />

−1<br />

2 = p 19.6 bar<br />

K<br />

2= =<br />

Stanje 2, p 2 =19. 6 bar i s s′′<br />

6.3476 kJ kg .<br />

−1<br />

−1<br />

Stanje 3, p 3 = 0. 0981 bar i s3 = s2<br />

= 6.3476 kJ kg K .<br />

( s ′′ − )<br />

s ′<br />

′<br />

( s3<br />

− s3′<br />

)<br />

( s ′′ − s′<br />

)<br />

3 = s3<br />

+ x3<br />

⋅ 3 s3<br />

6.3476 − 0.6443<br />

x 3 = =<br />

= 0.759<br />

3 3 8.1559 − 0.6443<br />

−1<br />

h3 = h3′<br />

+ x3<br />

⋅ r3<br />

= 190.3 + 0.759 ⋅ 2393 = 2007 kJ kg<br />

265

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