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Tehnička termodinamika - Kemijsko-tehnološki fakultet

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______________________________________________________Stvarni ili realni plinovi i pare<br />

6.86 6.7152<br />

Nakon izvršene interpolacije slijedi<br />

p 3 = 6.73bar<br />

d)<br />

Primjer 5.9.<br />

w<br />

( h − p ⋅ v ) − ( h − p ⋅ )<br />

1,2<br />

= u1<br />

− u2<br />

= 1 1 1 2 2 v2<br />

2<br />

2<br />

( 2941 −15.7<br />

⋅10<br />

⋅ 0.1482) − ( 2449 − 1.08 ⋅10<br />

1.4173)<br />

w 1,2<br />

=<br />

⋅<br />

−1<br />

w1,2<br />

= 397 kJ kg<br />

15 kg vodene pare stanja p1 = 9. 81bar i t 1 = 220 °C komprimira se<br />

⎛ v1<br />

⎞<br />

izotermno do stanja 2 ⎜v 2 = ⎟ .<br />

⎝ 3 ⎠<br />

a) Odredi odvedenu količinu topline i gustoću vodene pare u točki2.<br />

b) Skiciraj proces u p, v – i T, s – dijagramu.<br />

Rješenje<br />

a)<br />

Za p 1 = 9.81 bar i 1 = 220 °C iz tablica za pregrijanu paru slijedi<br />

t<br />

3 −1<br />

v1 = 0.2214 m kg<br />

−1<br />

h1 = 2875 kJ kg<br />

−1<br />

−1<br />

s1 = 6.7977 kJ kg K<br />

1 0.2214<br />

3 −1<br />

v 2 = v = = 0.0738 m kg<br />

3 3<br />

1 1<br />

−<br />

ρ 2 = = = 13.55 kg m<br />

3<br />

v 2 0.0738<br />

251

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