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Tehnička termodinamika - Kemijsko-tehnološki fakultet

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TEHNIČKA TERMODINAMIKA__________________________________________<br />

p0<br />

⋅<br />

⎛ n<br />

⋅ R ⋅T<br />

⋅ R ⋅T<br />

( ) ⎜ s 1 s 2<br />

V1<br />

− V2<br />

= 110 ⋅ −<br />

⎟ ⎝ p1<br />

p2<br />

⎠<br />

n<br />

⎞<br />

b)<br />

p<br />

⋅<br />

⎛ 0.089 ⋅ 8.314 ⋅ 300<br />

⎜<br />

⎝ 210<br />

0.089 ⋅ 8.314 ⋅ 288 ⎞<br />

⎟<br />

110 ⎠<br />

( V − V ) = 110 ⋅<br />

−<br />

= 97<br />

0 1 2<br />

−<br />

W<br />

= 24 + 106 − 97<br />

max .<br />

=<br />

e = h<br />

e = h<br />

33 kJ<br />

( s − )<br />

1 − h0<br />

− T0<br />

⋅ 1 s0<br />

( s − )<br />

1 − h2<br />

− T0<br />

⋅ 1 s2<br />

−1<br />

( T − T ) = 0.99 ⋅ ( 300 − 288) = 12 kJ<br />

h1 − h2<br />

= c p , s ⋅ 1 2<br />

kg<br />

kJ<br />

T<br />

⋅<br />

⎛ T p<br />

T0 ⋅<br />

ln<br />

⎝<br />

( ) ⎜<br />

1<br />

1<br />

s1<br />

− s2<br />

= T0<br />

⋅ c p , s ⋅ ln − Rs<br />

⋅<br />

T<br />

⎟ 2 p2<br />

⎠<br />

⎛<br />

300<br />

288<br />

210 ⎞<br />

110 ⎠<br />

−1<br />

( s − s ) = 288 ⋅ 0.99 ⋅ ln − 0.27 ⋅ ln = −38 kJ<br />

0 1 2 ⎜<br />

⎟ kg<br />

⎝<br />

−1<br />

e = 12 + 38 = 50 kJ kg<br />

⎞<br />

Primjer 4.18.<br />

Treba odrediti gubitak eksergije za 1 kg N 2 koji iz stanja p 1 = 7 bar i<br />

T 1 = 770 K prelazi u stanje p2 = 2 bar i T2 = 320 K ako je stanje okoline 283 K.<br />

Rješenje<br />

⎛ T p<br />

− ∆e<br />

= h1 − h2<br />

− T0<br />

⋅<br />

ln<br />

⎝<br />

− ∆e<br />

= 1.065 ⋅<br />

( ) ( ) ⎜<br />

1<br />

1<br />

s1<br />

− s2<br />

= c p ⋅ T1<br />

− T2<br />

− T0<br />

⋅ c p ⋅ ln − R ⋅<br />

T<br />

⎟ 2 p2<br />

⎠<br />

⎛<br />

⎞<br />

( 770 − 320) − 283 ⋅ 2.3 ⋅ 1.065 ⋅ log − ⋅ log ⎟<br />

⎠<br />

⎜<br />

⎝<br />

−1<br />

− ∆e<br />

= 317 kJ kg<br />

770<br />

320<br />

8.314<br />

28<br />

7<br />

2<br />

⎞<br />

192

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