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Tehnička termodinamika - Kemijsko-tehnološki fakultet

Tehnička termodinamika - Kemijsko-tehnološki fakultet

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TEHNIČKA TERMODINAMIKA______________________________________________<br />

Primjer 2.12.<br />

1 kg dušika ekspandira politropski s eksponentom politrope 1.3. Početno<br />

stanje je 392.3 kPa i 120 °C. Ako dušik ekspandira do tlaka od 98.07 kPa, kolika<br />

je temperatura na kraju ekspanzije, izvršeni rad i izmijenjena toplina?<br />

Rješenje<br />

T<br />

n−1<br />

1.3−1<br />

⎛ p2<br />

⎞ n ⎛ 98.07 ⎞ 1.3<br />

2 = T1<br />

⎜<br />

⎟ = 393⎜<br />

⎟ =<br />

p1<br />

⎝ 392.3 ⎠<br />

⎝<br />

⎠<br />

285 K<br />

8.314<br />

⋅ 393<br />

m ⋅ R ⋅T1<br />

⎛ T2<br />

⎞ 28 ⎛ 285 ⎞<br />

W1 ,2 = ⋅ 1 = ⋅ ⎜1<br />

− ⎟ = 106.89<br />

1 ⎜ −<br />

⎟<br />

kJ<br />

n − ⎝ T1<br />

⎠ 1.3 −1<br />

⎝ 393 ⎠<br />

( T )<br />

Q1,2<br />

= m ⋅ cn ⋅ 2 − T1<br />

c<br />

n<br />

= c<br />

v<br />

n − κ 20.93 1.3 −1.4<br />

0.249 kJ kg<br />

−1 K<br />

−1<br />

= ⋅ = −<br />

n −1<br />

28 1.3 −1<br />

( 285 − 393)<br />

Q<br />

1 ,2<br />

= −0.249<br />

= 26.89 kJ<br />

Primjer 2.13.<br />

3<br />

Zrak volumena 3 m , tlaka 294.2 kPa i temperature 50 °C komprimira se<br />

3<br />

do 1176.8 kPa tlaka i volumena 1 m . Treba naći eksponent politrope, T 2 , Q,<br />

W i ∆U.<br />

Rješenje<br />

⎛<br />

⎜<br />

⎝<br />

⎞<br />

V n 1<br />

V<br />

⎟ =<br />

2<br />

⎠<br />

p<br />

p<br />

2<br />

1<br />

98

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