Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
( s′′
− )
s ′
′
2 = s2
+ x2
⋅ 2 s1
( 7.3340 1.3268)
6.7223
= 1.3268 + x 2 ⋅ −
v
x 2 = 0.898
−1
h2 = h2′
+ x2
⋅ r2
= 426.5 + 0.898 ⋅ 2252 = 2449 kJ kg
= v′
x v
( v′′
− ′ ) = 0.0010448 + 0.898⋅
( 1.578 0.0010448)
2 2 + 2 ⋅ 2 1
−
b)
3 −1
v2 = 1.4173 m kg
Slika P5.8.
c)
−1
−1
s1 = s3
= s3,
′′ p ? 6.7223 kJ kg K
3= =
Iz tablica toj vrijednosti entropije odgovara
p
bar
s ′
kJ kg
−1 K
−1
5.88 6.7675
6.7223
250