Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
p
T
( S − S ) = 293⋅
26.83 7861
0 1 2
=
p
( V − )
0 ⋅ 1 V2
2 ⋅ 8.314 ⋅ 473
3
V 1 =
= 53.47 m
147.1
2 ⋅ 8.314 ⋅ 293
3
V 2 =
= 49.68 m
98.07
kJ
( V −V
) = 98.07 ⋅ ( 53.47 − 49.68) 372
0 ⋅ 1 2
=
Wmax . = 9636 − 7861+
372 = 1403 kJ
kJ
Primjer 4.13.
3
U rezervoaru volumena 0.3 m nalazi se zrak pod tlakom od 4903.3 kPa i
te mperature 20 °C. Tlak i temperatura okoline su p 0 = 98. 07 kPa i t 0 = 20 °C.
Izračunati maksimalni rad koji se može dobiti od komprimiranog zraka u
rezervoaru.
Rješenje
⎛ p
⎜ 1
W max.
= p1
⋅V1
⋅ ln −1
+
⎝ p0
p0
⎞
⎟
p1
⎠
ili
⎛ 4903.3 98.07 ⎞
Wm
ax . = 4903.3 ⋅ 0.3 ⋅ ⎜ 2.3log −1
+ ⎟ = 4306 kJ
⎝ 98.07 4903.3 ⎠
W
max.
W
= U
1
max.
−U
= p
2
− T
0
⋅
( S − S ) + p ⋅ ( V −V
)
1
T =
2
1 = T2
→ U1
U 2
0
⋅
( V −V
) − T ⋅ ( S − S )
1
p2
T2 = T1
→ S1
− S2
= m ⋅ R ⋅ ln p 1
2
0
1
0
2
1
2
184