Tehnička termodinamika - Kemijsko-tehnološki fakultet

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b)
−1
−1
s1 = s2
= s293 ′′ K = 0.5605 kJ kg K
x
( s ′′ − )
s ′
′
1 = s1
+ x1
⋅ 1 s1
0.5605 −
=
0.5777 −
( − 0.0687)
( − 0.0687)
1 =
0.973
−1
[ 145 − ( −18)
] = 141 kJ
h 1 = −18
+ 0.973 ⋅
kg
−1
h4′ = h3′
= h288 ′ K = 14.2 kJ kg
−1
q0 = 141 −14.2
= 126.8 kJ kg
−1
w = h1 − h2
= h1
− h293 ′′ K = 141 − 164 = −23 kJ kg
ε
= w
q
126.8
=
23
0
h
=
5.51
Slika P7.7.b
T = 288 K
T 0 = 253K
q 0 = h1
− h4′
( h ′′ − )
h ′
′
1 = h1
+ x1
⋅ 1 h1
327