Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
Rješenje
a)
Stanje
p
bar
t
°C
v
m
3 kg
−1
h
−1
kJ kg
s
kJ kg
−1 K
−1
1 0.2 60.09 2.3014 958.11 2.9546 0.3
2 0.7206 90.67 2.3014 2661 7.4694 1
3 0.2 60.09 7.2088 2467.23 7.4694 0.94
x
Interpolacijom podataka iz tablica vrela voda i zasićena vodena para (s promjenom
tlaka) unutar intervala p = 0.196 bar i p = 0.245 bar, za traženi tlak p = 0.2
bar slijedi
t z
= 60.09 °C
.001017 m
3 kg
1
= 0
−
v′
.6689 m
3 kg
1
= 7
−
v′′
h′ =
−1
250.71kJ kg
−1
r = 2358 kJ kg
s′ =
s′′ =
0.8319 kJ kg
−1 K
−1
7.9076 kJ kg
−1 K
−1
.
( v ′′ − )
v ′
′
1 = v1
+ x1
⋅ 1 v1
3 −1
( 7.6689 − 0.001017) = 2.3014 m
v 1 = 0.001017 + 0.3 ⋅
kg
−1
h1 = h1′
+ x1
⋅ r1
= 250.71 + 0.3 ⋅ 2358 = 958.11kJ kg
−1
−1
( s ′′ − ′ ) = 0.8319 + 0.3 ⋅ ( 7.9076 − 0.8319) = 2.9546 kJ kg
s 1 = s1′
+ x1
⋅ 1 s1
K
3 −1
v2 = v1
= v′′
= 2.3014 m kg
258