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Question 17.SP.73: Repeat Problem 17.44 assuming a diffuser efficiency of 80%....

Repeat Problem 17.44 assuming a diffuser efficiency of 80%.

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\begin{aligned}& V =1000 \,km / h =277.8 \,m / s , \quad v _1=\frac{ RT }{ P }=\frac{0.287 \times 261.15}{40}=1.874 \,m ^3 / kg \\ & h _1=261.48\, kJ / kg , \quad P _{ r 1}=0.6862 \\& h _{ o 1}=261.48+277.8^2 / 2000=300.07 \,kJ / kg \\& \Rightarrow T _{ o 1}=299.7\,K , \quad P _{ ro 1}=1.1107\end{aligned}

Same as problem 17.44, except \eta_D=0.80. We thus have from 17.44

\begin{aligned}& \frac{ h _3- h _1}{ h _{ o 1}- h _1}=\frac{ h _3-261.48}{300.07-261.48}=0.8 \\& \Rightarrow h _3=292.35 \,kJ / kg , P _{ r 3}=1.0129 \\& P _{ o 2}= P _3=40 \times 1.0129 / 0.6862=59.04\, kPa \\& P _{ ro 2}= P _{ ro1 }=1.1107\end{aligned}

\begin{aligned}& h _2=300.07-100^2 / 2000=295.07 \,kJ / kg \Rightarrow T _2= 2 9 4 . 7 K , P _{ r 2}=1.0462 \\& P _2= P _{ o 2} P _{ r 2} / P _{ ro 2}=59.04 \times 1.0462 / 1.1107= 5 5 . 6\,k P a \\& v _2= RT _2 / P _2=0.287 \times 294.7 / 55.6=1.521 \,m ^3 / kg \\& A _1 / A _2=\left( v _1 / v _2\right)\left( V _2 / V _1\right)=(1.874 / 1.521)(100 / 277.8)= 0 . 4 4 4\end{aligned}

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