Tehnička termodinamika - Kemijsko-tehnološki fakultet

TEHNIČKA TERMODINAMIKA__________________________________________
b)
( s − )
q1,2
= T ⋅ 2 s1
1
−1
1
2.8119 kJ kg
252 C
K −
s = s′
o =
t1=
Stanje 2, p2 = 4. 41 bar i 2 = 252 °C odgovara stanju pregrijane pare, tj.
t
−1
−1
s2 = 7.3312 kJ kg K
3 −1
v2 = 0.5408 m kg
−1
−1
h2 = 2985 kJ kg K
.
−1
( 7.3312 − 2.8119) = 2373 kJ
q 1,2
= 525 ⋅
kg
3
q2,3
= u3
− u2
−1
Stanje 3, v3 = v2
= 0.5408 m kg i p 3 = 2. 16 bar odgovara stanju mokre pare
jer je v v′′
.
v
3 < p 3 =2.16 bar
′
( v′′
− v′
) = .0010627 + ⋅ ( 0.8248 − 0.0010627) 0. 5408
3 = v3
+ x3
⋅ 3 3 0 x3
=
x
−1
h3 = h3′
+ x3
⋅ r3
= 515 + 0.655 ⋅ 2195 = 1953 kJ kg
3 = 0.655
( h − p ⋅ v ) − ( h − p ⋅ )
q2,3
= 3 3 3 2 2 v2
2
2
−1
( 1953 − 2.16 ⋅10
⋅ 0.5408) − ( 2985 − 4.41⋅10
⋅ 0.5408) = −910 kJ
q 2,3
=
kg
q3,4
= h3
− h4
= 1953 − h4
−1
−1
Stanje 4, s4 = s1
= 2.8119 kJ kg K i p4 = p3
= 2. 16 bar odgovara stanju
mokre pare jer je s < s p ′′ =2.16 bar .
s
′
4 4
( s ′′ − s′
) = .5562 + ⋅ ( 7.1021−1.5562) 2. 8119
4 = s4
+ x4
⋅ 4 4 1 x4
=
x 4 = 0.226
−1
h4 = h4′
+ x4
⋅ r4
= 515 + 0.226 ⋅ 2195 = 1011 kJ kg
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