Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
Za stanje 3 slijedi
w2,3
= u2
− u3
2
−1
u2 = h2
− p2
⋅ v2
= 3032 −15.7
⋅10
⋅ 0.1618 = 2778 kJ kg
−1
−1
s3 = s2
= 6.886 kJ kg K
p 3 = 0.196 bar.
s 3 = s3′
+ x3( s3′′
− s3′
) = 0.8269
+ x3( 7.9139 − 0.8269) = 6. 886
v
′
x 3 = 0.855
( v ′′ − ′ ) = 0.0010169 + 0.855⋅
( 7.789 0.0010169)
3 = v3
+ x3
⋅ 3 v3
−
3 −1
v3 = 6.659 m kg
−1
h3 = h3′
+ x3
⋅ r3
= 249.2 + 0.855 ⋅ 2359 = 2266 kJ kg
2
−1
u3 = h3
− p3
⋅ v3
= 2266 − 0.196 ⋅10
⋅ 6.659 = 2133 kJ kg
−1
w2,3
= 2778 − 2133 = 645 kJ kg
c)
W
−1
wuk.
= 122 + 645 = 767 kJ kg
= 15 ⋅ 767
uk. = m ⋅ wuk.
=
11505kJ
Slika P5.15.c
262