Tehnička termodinamika - Kemijsko-tehnološki fakultet

_________________________________________________Drugi glavni zakon termodinamike
p 2 = 0.997 bar 8.314
⋅ 293
R ⋅T2
29
3 −1
v 2 = =
= 0.84 m kg
p 2
0.997 ⋅10
2
n − κ
Q2,3
= Cn
⋅ 3 2 v 3 T2
n −1
( T − T ) = C ⋅ ( T − )
155 . −14
.
1080 = T3 155 . −1
−
( 29.31 − 8.314) ⋅ ( 293)
T 3 = 482 K
p
3
R ⋅T
=
v
n−1
T2
⎛ V3
⎞
=
⎜
⎟
T3
⎝V2
⎠
1
1
⎛ T2 ⎞ n−1
v3
⎛ T2
⎞ n−1
⎜ ⎟ = → 3 = 2 ⋅ ⎜ ⎟
v v
⎝ T3
⎠ v2
⎝ T3
⎠
3
3
1
1.55− 1 =
⎛ 293 ⎞
3
v 3 = 0.84 ⋅ ⎜ ⎟ 0.34 m kg
⎝ 482 ⎠
−1
8.314
⋅ 482
p
29
3 =
= 406 kPa = 4.06 bar
0.34
T
p
4 = p3
=
4.06 bar
3 −1
v4 = v1
= 1.15 m kg
p
⋅ v
406 ⋅1.15
=
8.314
29
4 4
4 =
=
R
1628 K
−1
( T − ) = 29.31⋅
( 1628 − 482) = 33589 kJ
Q3,4
= C p ⋅ 4 T3
kmol
−1
( T − ) = ( 29.31 − 8.314) ⋅ ( 293 −1628) = −28030 kJ
Q4,1
= Cv
⋅ 1 T4
kmol
179