Tehnička termodinamika - Kemijsko-tehnološki fakultet

_________________________________________________Drugi glavni zakon termodinamike
c)
R ⋅T2
0.265 ⋅ 734
p2 = = = 1706 kPa
v 2 0.114
κ
n −
q1,2
= cv ⋅ ⋅ 2 −
n −1
( T T )
1.5 −1.387
q 1,2
= 0.685 ⋅ ⋅
kg
1.5 −1
1
−1
( 734 − 303) = 66.72 kJ
q 2 ,3 = 0
q
−1
( T − ) = 0.950 ⋅ ( 303 − 313) = −9.5 kJ
q3,4
= c p ⋅ 4 T3
kg
p
68.69
4
−1
4,1
= p 1 ⋅ v1
⋅ ln = 120 ⋅ 0.669 ⋅ ln = −44.79 kJ kg
p1
120
q − q
η =
q
o
66.72 − 9.5 − 44.79
=
= 0.19 → 19 %
66.72
Primjer 4.12.
Koliko rada u najpovoljnijem slučaju se dobije od 2 kmola CO 2 stanja
p 1 =147.1 kPa i t 1 = 200 °C ako je stanje okoline p 0 98. 07 0 = 20
t °C?
Rješenje
W = U −U
− T ⋅ ( S − S ) + p ⋅ ( V −V
)
max.
1
2
0
1
2
( T − T ) = 2 ⋅ ( 35.08 − 8.314) ⋅ ( 473 − 293) 9636
U1 −U
2 = n ⋅ Cv ⋅ 1 2
= kJ
T
S p ⋅
1
1 − S2
= n ⋅ C ⋅ ln − n ⋅ R ln
T2
473
147.1
−1
S 1 − S2
= 2 ⋅ 35.08 ⋅ 2.3log − 2 ⋅ 8.314 ⋅ 2.3log
= 26.83 kJ K
293
98.07
0
p
p
1
1
2
2
183