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Tehnička termodinamika - Kemijsko-tehnološki fakultet

Tehnička termodinamika - Kemijsko-tehnološki fakultet

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_________________________________________________Drugi glavni zakon termodinamike<br />

3<br />

5 dm zraka t 1 = 100 °C i p1 = 0. 8 bar komprimira se politropski (n =<br />

1.35) na 35 bar. Kolika je promjena unutarnje energije?<br />

Rješenje<br />

m =<br />

p<br />

1<br />

R ⋅T<br />

( T )<br />

U v −<br />

⋅V<br />

1<br />

2 − U1<br />

= m ⋅ c ⋅ 2 T1<br />

1<br />

5<br />

0.8 ⋅10<br />

⋅ 5 ⋅10<br />

=<br />

8314<br />

⋅ 373<br />

29<br />

−3<br />

= 0.00374 kg<br />

T2<br />

T1<br />

⎛ p2<br />

⎞<br />

=<br />

⎜<br />

p<br />

⎟<br />

⎝ 1 ⎠<br />

n−1<br />

n<br />

n−1<br />

1.35−1<br />

⎛ p2<br />

⎞ n ⎛ 35 ⎞ 1.35<br />

T2 = T1<br />

⎜<br />

⎟ = 373⎜<br />

⎟ = 992 K<br />

⎝ p1<br />

⎠ ⎝ 0.8 ⎠<br />

20.93<br />

U 2 − U1<br />

= 0.00374<br />

=<br />

29<br />

( 992 − 373) 1. 67<br />

kJ<br />

Primjer 4.4.<br />

3<br />

5 kg vodika temperature 600 °C, volumena 1 m komprimira se izotermno<br />

na 3 puta manji volumen. Tada se toplina odvodi pri p = konst. dok se<br />

volumen ne smanji za 2 puta od volumena nakon izotermne kompresije. Odrediti<br />

veličine p, V, T, odvedenu toplinu i utrošeni rad.<br />

Rješenje<br />

8.314<br />

5 ⋅ ⋅ 873<br />

m ⋅ R ⋅T1<br />

p<br />

2<br />

1 = =<br />

= 18145 kPa = 181.45 bar<br />

V1<br />

1<br />

za T = konst.<br />

167

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