Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
p0
⋅
⎛ n
⋅ R ⋅T
⋅ R ⋅T
( ) ⎜ s 1 s 2
V1
− V2
= 110 ⋅ −
⎟ ⎝ p1
p2
⎠
n
⎞
b)
p
⋅
⎛ 0.089 ⋅ 8.314 ⋅ 300
⎜
⎝ 210
0.089 ⋅ 8.314 ⋅ 288 ⎞
⎟
110 ⎠
( V − V ) = 110 ⋅
−
= 97
0 1 2
−
W
= 24 + 106 − 97
max .
=
e = h
e = h
33 kJ
( s − )
1 − h0
− T0
⋅ 1 s0
( s − )
1 − h2
− T0
⋅ 1 s2
−1
( T − T ) = 0.99 ⋅ ( 300 − 288) = 12 kJ
h1 − h2
= c p , s ⋅ 1 2
kg
kJ
T
⋅
⎛ T p
T0 ⋅
ln
⎝
( ) ⎜
1
1
s1
− s2
= T0
⋅ c p , s ⋅ ln − Rs
⋅
T
⎟ 2 p2
⎠
⎛
300
288
210 ⎞
110 ⎠
−1
( s − s ) = 288 ⋅ 0.99 ⋅ ln − 0.27 ⋅ ln = −38 kJ
0 1 2 ⎜
⎟ kg
⎝
−1
e = 12 + 38 = 50 kJ kg
⎞
Primjer 4.18.
Treba odrediti gubitak eksergije za 1 kg N 2 koji iz stanja p 1 = 7 bar i
T 1 = 770 K prelazi u stanje p2 = 2 bar i T2 = 320 K ako je stanje okoline 283 K.
Rješenje
⎛ T p
− ∆e
= h1 − h2
− T0
⋅
ln
⎝
− ∆e
= 1.065 ⋅
( ) ( ) ⎜
1
1
s1
− s2
= c p ⋅ T1
− T2
− T0
⋅ c p ⋅ ln − R ⋅
T
⎟ 2 p2
⎠
⎛
⎞
( 770 − 320) − 283 ⋅ 2.3 ⋅ 1.065 ⋅ log − ⋅ log ⎟
⎠
⎜
⎝
−1
− ∆e
= 317 kJ kg
770
320
8.314
28
7
2
⎞
192