Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
Rješenje
m p
V p
= v 4.9 ′′ bar
5.3438
= = 14 kg
0.3817
v
′
VH2O
mH2O
= = v 4.9 ′ bar
0.0066
0.0010918
= 6.045 kg
( v ′′ − v′
) = .0010918 + ⋅ ( 0.3817 0.0010918)
1 = v1
+ x1
⋅ 1 1 0 x1
−
x1
=
m p
m p
+ mH
O
2
14
= = 0.6984
14 + 6.045
3 −1
( 0.3817 − 0.0010918) = 0.2669 m
v 1 = 0.0010918 + 0.6984 ⋅
kg
−1
h1 = h1′
+ x1
⋅ r1
= 636.8 + 0.6984 ⋅ 2111 = 2111.12 kJ kg
−1
−1
( s′′
− ′ ) = 1.8531+
0.6984 ⋅( 6.8283 −1.8531) = 5.33 kJ kg
s 1 = s1′
+ x1
⋅ 1 s1
K
v = konst.
t2
= 182 °C, tj. T 2 = 455K
3 −1
v2 = v1
= 0.2669 m kg
T
K
455
v’’
m
3 kg
−1
0.1856
x = 1 v 2 = v 455 ′′ K 0.2669
≠ 0. 1856
x > 1 v 2 > v 455 ′
K 0 .2669 > 0. 1856
0 < x < 1 v 2 < v 455 ′′ K 0 .2669 < 0. 1856
Vidljivo je da je
v 2 > v 455 ′′ K
, tj. konačno stanje je u području pregrijane pare.
244