Tehnička termodinamika - Kemijsko-tehnološki fakultet

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q
0 = h1′ − h4′
−1
h1′ = h′′ 278 K = 381 kJ kg
−1
h4′ = h3′
= h303 ′ K = 41 kJ kg
−1
q0 = 381 − 41 = 340 kJ kg
−3
Q0 = m ⋅ q0
= 1.498 ⋅10
⋅ 340 = 0.509 kW = 509 W
b)
Slika P7.6.b
q
0 = h1
′ − h4
−1
h1′ = h′′ 278 K = 381 kJ kg
−1
h4′ = h3′
= h308 ′ K = 48.1kJ kg
−1
q0 = 381 − 48.1 = 332.9 kJ kg
−3
Q0 = m ⋅ q0
= 1.498 ⋅10
⋅ 332.9 = 0.500 kW = 500 W
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