TehniÄka termodinamika - Kemijsko-tehnoloÅ¡ki fakultet

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TEHNIČKA TERMODINAMIKA__________________________________________ b) Stanje 1 Stanje 2 T 1 = 278 T 2 = 308 K x 1 = 0.94 x 2 =1 p 1 =1.897 −1 h 1 = 358.5 kJ kg p 2 = 5.331bar −1 h 2 = h′′ 308 K = 392 kJ kg Stanje 3 Stanje 4 T 3 = 308 T4 = T0 = 278 K x 3 = 0 x 4 = 0.11 p 3 = 5.331bar −1 h 3 = h′ 308 K = 48.1kJ kg p4 = p0 = 1.897 bar h 4 = h3 = h308 ′ K = 48.1kJ kg ( h′′ − ) h ′ ′ 4 = h4 + x4 ⋅ 4 h4 h4 − h4′ 48.1 − 6.7 x 4 = = = 0.11 h4′′ − h4′ 381 − 6.7 −1 c) ε = w q 310.4 = 33.5 0 h = 9.27 −1 w = h1 − h2 = 358.5 − 392 = −33.5 kJ kg ili ε Q 0.465 = 5.018⋅10 0 h = = −2 W 9.27 −1 −3 −1 −2 −1 W = −33.5 kJ kg ⋅1.498 ⋅10 kg s = −5.018 ⋅10 kg s W = −5.018⋅10 −2 kW 322
_______________________________________________________________Rashladni strojevi Slika P7.4. Primjer 7.5. Za koliko se poveća rashladni učin rashladnog uređaja iz primjera 7.4. ako se u kondenzatoru kondenzat pothlađuje do temperature zraka? Koliki je tada stupanj rashladnog učina? Skicirati proces u T, s – dijagramu! Rješenje Slika P7.5. 323
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TehniÄka termodinamika - Kemijsko-tehnoloÅ¡ki fakultet

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