Question 9.PS.170: Supercharging of an engine is used to increase the inlet air......

Supercharging of an engine is used to increase the inlet air density so that more fuel can be added, the result of which is an increased power output. Assume that ambient air, 100 kPa and 27°C, enters the supercharger at a rate of 250 L/s. The supercharger (compressor) has an isentropic efficiency of 75%, and uses 20 kW of power input. Assume that the ideal and actual compressor have the same exit pressure. Find the ideal specific work and verify that the exit pressure is 175 kPa.
Find the percent increase in air density entering the engine due to the supercharger and the entropy generation.

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C.V.: Air in compressor (steady flow)

Cont:    \dot{ m }_{ in }=\dot{ m }_{ ex }=\dot{ m }=\dot{ V } / v _{ in }=0.29 \,kg / s

Energy:  \dot{ m }h_{ in }-\dot{ W }=\dot{ m }h_{ ex } \text { Assume: } \dot{ Q }=0

Entropy:    \dot{ m }s_{ in }+\dot{ S }_{ gen }=\dot{ m }s_{ ex }

Inlet state:      v _{\text {in }}= RT _{ in } / P _{\text {in }}=0.8614\,m ^3 / kg , P _{ r \text { in }}=1.1167

\begin{aligned}& \eta_{ c }= w _{ C \,s } / w _{ C \text { ac }}=-\dot{ W }_{ S }=-\dot{ W }_{ AC } \times \eta_{ c }=15 \,kW \\& – w _{ C s }=-\dot{ W }_{ S } / \dot{ m }=51.724\,kJ / kg , \quad- w _{ C \text { ac }}=68.966 \,kJ / kg\end{aligned}

Table A.7:

\begin{aligned}& h _{ ex\, s }= h _{ in }- w _{ Cs }=300.62+51.724=352.3 \,kJ / kg \\& \Rightarrow T _{\text {ex\, s }}=351.5 \,K , \quad P _{ r\,ex }=1.949\end{aligned}

P _{ ex }= P _{ in } \times P _{ r\,ex } / P _{ r \text { in }}=100 \times 1.949 / 1.1167= 1 7 4 . 5 \,k P a

The actual exit state is

\begin{gathered}h _{\text {ex \,ac }}= h _{\text {in }}- w _{ C \text { ac }}=369.6 \,kJ / kg \quad \Rightarrow T _{\text {ex \,ac }}=368.6 \,K \\v _{ ex }= RT _{ ex } / P _{ ex }=0.606 \,m ^3 / kg \\\rho_{ ex } / \rho_{\text {in }}= v _{\text {in }} / v _{\text {ex }}=0.8614 / 0.606= 1 . 4 2 \text { or } 4 2 \% \text { increase } \\\left. s _{\text {gen }}= s _{\text {ex }}- s _{\text {in }}=7.0767-6.8693-0.287 \ln (174 / 100)\right]= 0 . 0 4 8 4 \,k J / k g ~ K\end{gathered}

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