Sonntag, Borgnakke and van Wylen
13.101 For the previous problem, find the specific heat transfer using Kay’s rule and the generalized charts. To do the EOS we need the gas constant, so from Eq.12.5 we get Mmix = ∑ yi Mi = 0.6 × 28.054 + 0.4 × 26.068 = 27.26 Rmix = 8.3145/27.26 = 0.305 kJ/kg K y M 0.6 × 28.054 cC2H4 = = = 0.6175, Mmix 27.26
cC2H4 = 1 - cC2H4 = 0.3825
CP mix = ∑ ci CP i = 0.6175 × 1.548 + 0.3825 × 1.699 = 1.606 kJ/kg K Kay’s rule Eq.13.86 Pc mix = 0.6 × 5.04 + 0.4 × 6.14 = 5.48 MPa Tc mix = 0.6 × 282.4 + 0.4 × 308.3 = 292.8 K 6 300 = 1.095, Tr1 = = 1.025 Reduced properties 1: Pr1 = 5.48 292.8 Fig. D.1:
(h*1 − h1) = 2.1 × RTc = 2.1 × 0.305 × 292.8 = 187.5 kJ/kg
Reduced properties 2: Fig. D.1:
Pr2 =
6 = 1.095, 5.48
Tr2 =
400 = 1.366 292.8
(h*2 − h2) = 0.7 × RTc = 0.7 × 0.305 × 292.8 = 62.5 kJ/kg
The energy equation gives * 1q2 = (h2 - h1) = (h2 − h2)
+ (h*2 − h*1) + (h*1 − h1)
= -62.5 + 1.606 (400 – 300) + 187.5 = 285.6 kJ/kg mix