Tehnička termodinamika - Kemijsko-tehnološki fakultet

_________________________________________________Drugi glavni zakon termodinamike
p
m
⋅ R
⋅T
8.314
1.29 ⋅ ⋅ 348
=
40
−3
70 ⋅10
Ar Ar 2
1 =
=
W
max.
= U
V
1
−U
2
− T
0
⋅
1333 kPa
( s − s ) + p ⋅ ( V −V
)
20.93 − 8.314
U1 − U 2 = m ⋅ cv ⋅ ( T1
− T2
) = 1.29 ⋅
⋅ ( 348 − 288) = 24. 41 kJ
40
1
T 2 = T 0
p 2 = p 0
⎛ T
p
T0 ⋅
p
ln
⎝
( ) ⎜
1
1
s1
− s2
= T0
⋅ m ⋅ c ⋅ ln − m ⋅ R ⋅
T
⎟ 2 p2
⎠
⎛ 20.93 348 8.314 13.32 ⎞
T0 ⋅ 1 2 ⎜
⎟ −
⎝ 40 288 40 1 ⎠
( s − s ) = 288 ⋅ 1.29 ⋅ ⋅ ln −1.29
⋅ ⋅ ln = 163. 16
⎛
p 0 ⋅
10
⎝
2
( ) ⎜ −3
2
V1
− V2
= 1⋅10
⋅ 70 ⋅ −
p ⎟ 2 ⎠
2
0
1
m ⋅ R ⋅T
⎛
8.314 ⎞
⎜ 1.29 ⋅ ⋅ 288 ⎟
p
40
0 ⋅ 1 2
−
⎜
2
1⋅10
⎟
⎝
⎠
2
( ) 1 10 ⎜ −3
V − V = ⋅ ⋅ 70 ⋅10
−
⎟ = 70
Wmax . = 24.41+
163.16 − 70 = 117.57 kJ
2
⎞
⎞
kJ
kJ
Primjer 4.16.
Koliko se maksimalno može dobiti rada iz novonastale smjese ako se
3
700 dm plina masenog sastava H2 = 2 %, CO 2 = 38 % i N 2 = 60 %,
3
temperature 127 °C i tlaka 2.65 bar pomiješa s 800 dm plina masenog sastava
CO = 25 % i N2 = 75 % iste temperature i tlaka. Stanje okoline je normalno.
−1
−1
C p , CO
K
2
Podaci: = 36.08 kJ kmol .
187