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Question 11.CSGP.180b: Do problem 11.24 with R-22 as the working fluid. A flow with......

Do problem 11.24 with R-22 as the working fluid.
A flow with 2 kg/s of water is available at 95°C for the boiler. The restrictive factor is the boiling temperature of 85°C. Therefore, break the process up from 2- 3 into two parts as shown in the diagram.

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\begin{gathered}-\dot{ Q }_{ AB }=\dot{ m }_{ H 2 O }\left( h _{ A }- h _{ B }\right)=2(397.94-355.88)=84.12 \,kW \\=\dot{ m }_{ R -22}(253.69-165.09) \Rightarrow \dot{ m }_{ R -22}=0.949 \,kg / s\end{gathered}

To verify that T _{ D }= T _3 \text { is the restrictive factor, find } T _{ C }.

\begin{aligned}& -\dot{ Q }_{ AC }=0.949(165.09-96.48)=65.11=2.0\left(355.88- h _{ C }\right) \\& h _{ C }=323.32 \,kJ / kg \Rightarrow T _{ C }=77.2^{\circ} C \text { OK }\end{aligned}

State 1:    40^{\circ} C , \quad 1533.5 \,kPa , v _1=0.000884 \,m ^3 / kg

CV Pump:      w _{ P }= v _1\left( P _2- P _1\right)=0.000884(4036.8-1533.5)=2.21 \,kJ / kg

CV: Turbine

\begin{aligned}& s _4= s _3=0.7918=0.3417+ x _4 \times 0.5329 \Rightarrow x _4=0.8446 \\& h _4=94.27+0.8446 \times 166.88=235.22 \,kJ / kg\end{aligned}

Energy Eq.:    w _{ T }= h _3- h _4=253.69-235.22=18.47 \,kJ / kg

Cycle:      w _{ NET }= w _{ T }- w _{ P }=18.47-2.21=16.26 \,kJ / kg

\dot{ W }_{ NET }=\dot{ m }_{ R 22} w _{ NET }=0.949 \times 16.26= 1 5 . 4 3 \,k W
101

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