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Air enters the compressor of a Brayton refrigeration cycle at 100 kPa,260 K100\ \mathrm{kPa}, 260 \mathrm{~K}, and is compressed adiabatically to 300 kPa300\ \mathrm{kPa}. Air enters the turbine at 300 kPa,300 K300\ \mathrm{kPa}, 300 \mathrm{~K}, and expands adiabatically to 100 kPa100\ \mathrm{kPa}. For the cycle (a) determine the net work per unit mass of air flow, in kJ/kg\mathrm{kJ} / \mathrm{kg}. and the coefficient of performance if the compressor and turbine isentropic efficiencies are both 100%100 \%. (b) plot the net work per unit mass of air flow, in kJ/kg\mathrm{kJ} / \mathrm{kg}, and the coefficient of performance for equal compressor and turbine isentropic efficiencies ranging from 8080 to 100%100 \%.

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Answered 2 years ago
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First, fix each principal state (Table A-22).

State 1: T1=260Kh1=260.09kJkg,pr1=0.3405State 2: pr2s=pr1p2p1=2.5215h2s=356.31kJkgUsing the isentropic compressor efficiency:h2=h1+h2sh1ηca)ηc=100%h2=356.31kJkgb)ηc=80%h2=380.37kJkgState 3: T3=300 Kh3=300.19kJkg,pr3=1.3860State 4: pr4s=pr3p4p3=0.462h4s=218.97kJkgh4=h3(h3h4s)ηta)ηt=100%h4=218.97kJkgb)ηt=80%h4=235.21kJkg\begin{align*} \text{State 1: } T_1=260\, \mathrm{ K}\rightarrow h_1&=260.09\, \mathrm{\frac{kJ}{kg}},\, p_{r1}=0.3405 \\ \text{State 2: } p_{r2s}=p_{r1}\frac{p_2}{p_1}=2.5215\rightarrow h_{2s}&=356.31\, \mathrm{\frac{kJ}{kg}}\\ \text{Using the isentropic compressor efficiency:}\\ h_2&=h_1+\frac{h_{2s}-h_1}{\eta _c}\\ a) \eta _c=100\% \rightarrow h_2&=356.31\, \mathrm{\frac{kJ}{kg}}\\ b) \eta _c=80\% \rightarrow h_2&=380.37\, \mathrm{\frac{kJ}{kg}}\\ \text{State 3: }T_3=300\text { K}\rightarrow h_3&=300.19\, \mathrm{\frac{kJ}{kg}},\, p_{r3}=1.3860\\ \text{State 4: }p_{r4s}=p_{r3}\frac{p_4}{p_3}=0.462\rightarrow h_{4s}&=218.97\, \mathrm{\frac{kJ}{kg}}\\ h_4&=h_3-(h_3-h_{4s})\eta _t\\ a) \eta _t=100\% \rightarrow h_4&=218.97\, \mathrm{\frac{kJ}{kg}}\\ b) \eta _t=80\% \rightarrow h_4&=235.21\, \mathrm{\frac{kJ}{kg}}\\ \end{align*}

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