Tehnička termodinamika - Kemijsko-tehnološki fakultet

______________________________________________________Stvarni ili realni plinovi i pare
3 −1
v1 = v′′ = 0.1947 m kg
−1
h1 = h′′ = 2778 kJ kg
p = konst.
p2 = p1
= 10 bar
3 −1
v2 = v′ = 0.001127 m kg
−1
h2 = h′ = 763 kJ kg
.
−1
( v − ) = 1000 ⋅ ( 0.001127 − 0.1947) = −195 kJ
w 1,2
= p ⋅ 2 v1
kg
b)
−1
q1,2
= h2
− h1
= 763 − 2778 = −2015 kJ kg
u = u − u = u′
− u ′′
∆ 2 1
( h − p ⋅ v ) − ( h − p ⋅ v ) = ( h′
− p ⋅ v′
) − ( h ′′ − p ⋅ v )
∆ 2 2 2 1 1 1 2
1
u =
′′
∆u
= h′
− h ′′ − p ⋅
( v′
− v ′′ ) = 763 − 2778 −1000
⋅ ( 0.001127 − 0.1947)
−1
∆u = −2209 kJ kg
c)
Slika P5.10.
253