Tehnička termodinamika - Kemijsko-tehnološki fakultet

______________________________________________________Stvarni ili realni plinovi i pare
Iz tablica ovom volumenu odgovaraju sljedeće vrijednosti p2
, t2,
h2
, s2
dobivene
interpolacijom unutar područja .408 m
3 1
= 2 kg
−
v′′
i .257 m
3 1
= 2 kg
−
v′′
.
Dakle, u stanju 2 imamo
b)
p 2 = 0.7206bar
t 2 t = 90. 76 °C
= z
−1
h2 = h′′ = 2661 kJ kg
−1
−1
s2 = s′′ = 7.4694 kJ kg K
p
3 = p1
=
0.2 bar
−1
−1
s3 = s2
= 7.4694 kJ kg K
.
( s ′′ − )
s ′
′
3 = s3
+ x3
⋅ 3 s3
( 7.9076 0.8319)
7.4694
= 0.8319 + x 3 ⋅ −
x 3 = 0.94
( v ′′ − )
v ′
′
3 = v3
+ x3
⋅ 3 v3
3 −1
( 7.6689 − 0.001017) = 7.2088 m
v 3 = 0.001017 + 0.94 ⋅
kg
−1
h3 = h3′
+ x3
⋅ r3
= 250.71 + 0.94 ⋅ 2358 = 2467.23 kJ kg
−1
−1
∆s1,2
= s2
− s1
= 7.4694 − 2.9546 = 4.5148 kJ kg K
c)
q
1,2
= u2
− u1
2
−1
u2 = h2
− p2
⋅ v2
= 2661 − 0.7206 ⋅10
⋅ 2.3014 = 2495.16 kJ kg
2
−1
u1 = h1
− p1
⋅ v1
= 958.11 − 0.2 ⋅10
⋅ 2.3014 = 912.08 kJ kg
−1
q1,2 = 2495.16 - 912.08 = 1583.07 kJ kg
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