Tehnička termodinamika - Kemijsko-tehnološki fakultet

_____________________________________________Prvi glavni zakon termodinamike
Rješenje
q 1,2
Slika P2.7.
29.31
Q1 ,2 = m ⋅ c p ⋅ ( T2
− T1
) = 0.5 ⋅ ⋅ ( 500 − 300) = 101. 1kJ
29
V
V
m ⋅ R ⋅T
( V − )
W1,2
= p ⋅ 2 V1
8.314
0.5 ⋅ ⋅ 300
29
1
3
1 = =
= 0.146 m
p1
294.2
m ⋅ R ⋅T
8.314
0.5 ⋅ ⋅ 500
29
2
3
2 = =
= 0.244 m
p2
294.2
( 0.244 − 0.146) 28. 83
W1 ,2 = 294.2 ⋅
=
U 2 −U1
= Q1,2
−W1,2
= 101.1 − 28.83 = 72.17 kJ
ili
20.93
∆U = m ⋅ cv ⋅ ∆T
= 0 .5 ⋅ ⋅ 200 = 72.17 kJ
29
Na povećanje unutarnje energije troši se
kJ
93