Tehnička termodinamika - Kemijsko-tehnološki fakultet

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1 – 2 adijabatska kompresija
( h′′
− )
h ′
′
1 = h1
+ x1
⋅ 1 h1
T 0 = 278K
t = 30 + ∆t
= 30 + 5 = 35 °C
T = 273 + 35 = 308 °C
Iz tablica za zasićenu paru SO 2 slijedi
m =
−1
−1
s1 = s2
= s308 ′′ K = 1.289 kJ kg K
( s′′
− )
s ′
′
1 = s1
+ x1
⋅ 1 s1
−1
−1
s′ 1 = s278 ′ K = 0.0255 kJ kg K
−1
−1
s′ 1′
= s278 ′′ K = 1.372 kJ kg K
−1
h′ 1 = h278 ′ K = 6.7 kJ kg
−1
h′ 1′
= h278 ′′ K = 381 kJ kg
s1
− s1′
1.289 − 0.0255
x 1 = =
= 0.94
s1′′
− s1′
1.372 − 0.0255
−1
( 381 − 6.7) = 358.5 kJ
h 1 = 6.7 + 0.94 ⋅
kg
−1
h4 = h3
= h308 ′ K = 48.1kJ kg
−1
q0 = h1
− h4
= 358.5 − 48.1 = 310.4 kJ kg
−1
0.465 kJ s
−3
−1
−3
−1
= 1.498 ⋅10
kg s = 1.498 ⋅10
⋅ 3600 = 5.39 kg h
−1
310.4 kJ kg
m
n =
M
=
5.39
− 1
3 − 1
= 0.0842 kmol h ⇒ 0.0842 ⋅ 22.414 = 1.88 m n h
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