Question 12.181E: The R-410A in Problem 12.180E flows through a heat exchanger...

Kay’s rule Eq.12.84

Table F.1:

Reduced properties 1:    P _{ r 1}=\frac{200}{681.5}=0.293, \quad T _{ r1 }=\frac{539.67}{621.45}=0.868

Fig. D.1:      \left( h _{1}^{*}- h _{1}\right)=0.45 \times RT _{ c }=0.45 \times 0.02736 \times 621.45=7.65   Btu / lbm

Reduced properties 2:    P _{ r 2}=\frac{200}{681.5}=0.293, \quad T _{ r 2}=\frac{739.67}{621.45}=1.19

Fig. D.1:        \left( h _{2}^{*}- h _{2}\right)=0.25 \times RT _{ c }=0.25 \times 0.02736 \times 621.45=4.25   Btu / lbm

The energy equation gives

Table F.9.2:    q = h _{2}- h _{1}=176.26-126.34= 4 9 . 9 2   Btu / lbm

The main difference is in the value of specific heat, about 0.25 Btu/lbm-R at the avg. T, whereas it is 0.1925 Btu/lbm-R at 77 F.

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Eq.12.84 : \left(P_{c}\right)_{\operatorname{mix}}=\sum_{i} y_{i} P_{c i}, \quad\left(T_{c}\right)_{\operatorname{mix}}=\sum_{i} y_{i} T_{c i}