Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
b)
Stanje 1 Stanje 2
T 1 = 278 T 2 = 308 K
x 1 = 0.94
x 2 =1
p 1 =1.897 −1
h 1 = 358.5 kJ kg
p 2 = 5.331bar
−1
h 2 = h′′ 308 K = 392 kJ kg
Stanje 3 Stanje 4
T 3 = 308 T4 = T0
= 278 K
x 3 = 0
x 4 = 0.11
p 3 = 5.331bar
−1
h 3 = h′ 308 K = 48.1kJ kg
p4 = p0
= 1.897 bar
h 4 = h3
= h308 ′ K = 48.1kJ kg
( h′′
− )
h ′
′
4 = h4
+ x4
⋅ 4 h4
h4
− h4′
48.1 − 6.7
x 4 = = = 0.11
h4′′
− h4′
381 − 6.7
−1
c)
ε
= w
q
310.4
=
33.5
0
h
=
9.27
−1
w = h1 − h2
= 358.5 − 392 = −33.5 kJ kg
ili
ε
Q
0.465
=
5.018⋅10
0
h
= =
−2
W
9.27
−1
−3
−1
−2
−1
W = −33.5 kJ kg ⋅1.498
⋅10
kg s = −5.018
⋅10
kg s
W = −5.018⋅10
−2
kW
322