Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
b)
m s ⋅ R s ⋅ T
2 29.80 ⋅1.315
⋅ 308
p
2
=
=
= 1207 kPa
V
10
W
p m = p a − p
b
max.
= U
−U
= 1207 − 101.3 = 1106 kPa
− T
⋅
( S1
− S2
) + p0
⋅ ( V1
−V2
)
⋅ c ⋅ ( T − )
1 2 0
U1 − U 2 = ms
v,
s 1 T2
c
n
p, s = ∑ω
i ⋅ c p,
i = ω N ⋅ c
2 p,N
+ ω
2 He ⋅ c p,He
i=
1
29.31 8 20.93
1 −1
c
6 p, s = ⋅ + ⋅ = 3.439 kJ kg
− K
14 28 14 4
−1
−1
c v, s = 3.439 −1.315
= 2,124 kJ kg K
U
1
− U
2
= 29.80 ⋅ 2.124 ⋅
( 308 − 290) = 1139
⎛ T
p
T0 ⋅
1 2 0 ⎜ s p,
s
s s
ln
⎝ T2
p
1
1
( S − S ) = T ⋅⎜m
⋅ c ⋅ ln − m ⋅ R ⋅ ⎟ =
⎛
308
1207 ⎞
= 290 ⋅ ⎜ 29.80 ⋅ 3.439 ⋅ ln − 29.80 ⋅1.315
⋅ ln ⎟ = −26369
kJ
⎝
290
101.3 ⎠
p
0
⋅ V
1
Primjer 4.15.
⎛
p0 ⋅
10
⎝
⋅ R
⋅T
( ) ⎜ s s 2
V1
− V2
= 101.3 ⋅ −
p ⎟ 2 ⎠
⎛ 29.80 ⋅1.315
⋅ 290 ⎞
( − ) = 101.3 ⋅ ⎜10
−
⎟ = −10351
⎠
V
2
kJ
⎝ 101.3
W = 1139 + 26369 −10351
17157 kJ
m
max .
=
3
U boci volumena 70 dm nalazi se argon pod tlakom 11 bar i temperaturom
koja je jednaka temperaturi okoline. Ako se uslijed zagrijavanja, argon u
boci ugrije za 60 °C koliko bi rada u najpovoljnijem
slučaju mogli dobiti iz tog
plina nakon zagrijavanja? Stanje okoline je p 0 = 1 bar i t 0 = 15 °C.
Rješenje
186
⎞
kJ
2
⎞
⎟
⎠