Tehnička termodinamika - Kemijsko-tehnološki fakultet

______________________________________________________Stvarni ili realni plinovi i pare
3 −1
v′ 2 = 0.001253 m kg
3 −1
v′′ 2 = 0.04953 m kg
−1
h′ 2 = 1088 kJ kg
−1
r2 = 1712 kJ kg
.
( v ′′ − )
v ′
′
2 = v2
+ x2
⋅ 2 v2
( 0.04953 0.001253)
0.00214
= 0.001253 + x 2 ⋅ −
x 2 = 0.0184
−1
h2 = h2′
+ x2
⋅ r2
= 1088 + 0.0184 ⋅1712
= 1119.5 kJ kg
2
2
−1
( 80.4 ⋅10
− 40.2 ⋅10
) = −239 kJ
q 1,2
= 1119.5 −1367
+ 0.00214 ⋅
kg
( − 239) = 892426
Q1,2
= m ⋅ q1,2
= 3734 ⋅ −
m p,
2
x2 = m
m p,2
0.0184 =
3734
m p, 2 = 0.0184 ⋅ 3734 = 68.71kg
3
V p, 2 = m p,2
⋅ v2′′
= 68 .71⋅
0.04953 = 3.4032 m
mv, 2 = m − m p,2
= 3734 − 68.71=
kJ
3665.29 kg
3
Vv, 2 = mv,2
⋅ v2′
= 3665 .29 ⋅ 0.001253 = 4.5926 m
V
V
v,2
p,2
4.5926 m
=
3.4032 m
3
3
= 1.35
255