Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
plin
T 1 = 400 K
p 1 = 2.65 bar
3 3
V s = V1 + V2
= 700 + 800 = 1500 dm = 1.5 m
okolina
T
0 = T2
=
p
0 = p2
=
273K
1.013bar
( T − T ) = 3.177 ⋅ 0.82 ⋅ ( 400 − 273) 331
U1 − U 2 = ms
⋅ cv,
s ⋅ 1 2
=
⎛ T
p
T0 ⋅
ln
⎝
( ) ⎜
1
1
s1
− s2
= T0
⋅ m ⋅ c p , s ⋅ ln − ms
⋅ Rs
⋅
T
⎟ 2
p2
⎠
⎛
400
265 ⎞
T0 ⋅ 1 2 ⎜
⎟ =
⎝
273
101.3 ⎠
( s − s ) = 273 ⋅ 3.177 ⋅1.132
⋅ ln − 3.177 ⋅ 0.312 ⋅ ln 115
p
⋅
⎛
( ) ⎜
s s 2
V1
− V2
= 101.3 ⋅ 1. −
p ⎟ 2 ⎠
0 5
⎝
m
⋅ R
⋅T
⎛ 3.177 ⋅ 0.312 ⋅ 273 ⎞
p0 ⋅ 1 2 ⎜
⎟ −
⎝ 101.3 ⎠
( V − V ) = 101.3 ⋅ 1.5 −
= 118
Wmax . = 331−115
−118
= 98 kJ
⎞
⎞
kJ
kJ
kJ
Primjer 4.17.
a) Koliko se maksimalno može dobiti rada iz 2 m 3 plinske smjese masenog
sastava CO 2 = 25 % i N 2 = 75 %, temperature 300 K i tlaka 2.1 bar ako je
stanje okoline p0
= 1.1 bar i t 0 = 15 °C?
b) Kolika bi bila radna sposobnost (eksergija) ovako nastale plinske smjese?
−1
−1
C p , CO
K
2
Podaci: = 36.08 kJ kmol .
190