Question 12.CSGP.124: Repeat Problem 12.31, but assume that the compressor has an ......

Repeat Problem 12.31, but assume that the compressor has an efficiency of 82%, that both turbines have efficiencies of 87%, and that the regenerator efficiency is 70%.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

a) From solution 12.31:    T _2= T _1\left\lgroup \frac{ P _2}{ P _1} \right\rgroup^{\frac{ k -1}{ k }}=300(6)^{0.286}=500.8 \,K

\begin{aligned}& – w _{ C }=- w _{12}= C _{ P 0}\left( T _2- T _1\right)=1.004(500.8-300)=201.6 \,kJ / kg \\& – w _{ C }=- w _{ SC } / \eta_{ SC }=201.6 / 0.82=245.8 \,kJ / kg = w _{ T 1} \\& = C _{ P 0}\left( T _4- T _5\right)=1.004\left(1600- T _5\right) \Rightarrow T _5=1355.2 \,K \\& w _{ ST 1}= w _{ T 1} / \eta_{ ST 1}=245.8 / 0.87=282.5 \,kJ / kg \\& = C _{ P 0}\left( T _4- T _{5 S }\right)=1.004\left(1600- T _{5 S }\right) \Rightarrow T _{5 S }=1318.6 \,K\end{aligned}

s _{5 S }= s _4 \Rightarrow P _5= P _4\left( T _{5 S } / T _4\right)^{\frac{ k }{ k -1}}=600\left(\frac{1318.6}{1600}\right)^{3.5}= 3 0 4 . 9 \,kPa

b)   P_6=100\, kPa , \quad s _{6 S }= s _5

\begin{aligned}& T _{6 S }= T _5\left\lgroup \frac{ P _6}{ P _5}\right\rgroup^{\frac{ k -1}{ k }}=1355.2\left\lgroup \frac{100}{304.9}\right\rgroup^{0.286}=985.2 \,K \\& w _{ ST 2}= C _{ P 0}\left( T _5- T _{6 S }\right)=1.004(1355.2-985.2)=371.5 \,kJ / kg \\& w _{ T 2}=\eta_{ ST 2} \times w _{ ST 2}=0.87 \times 371.5= 3 2 3 . 2\, k J / k g \\& 323.2= C _{ P 0}\left( T _5- T _6\right)=1.004\left(1355.2- T _6\right) \Rightarrow T _6=1033.3 \,K\end{aligned}

\dot{ m }=\dot{ W }_{ NET } / w _{ NET }=150 / 323.2= 0 . 4 6 4\,k g / s

c)      w _{ C }=245.8= C _{ P 0}\left( T _2- T _1\right)=1.004\left( T _2-300\right) \Rightarrow T _2=544.8 \,K

\begin{aligned}& \eta_{ REG }=\frac{ h _3- h _2}{ h _6- h _2}=\frac{ T _3- T _2}{ T _6- T _2}=\frac{ T _3-544.8}{1033.3-544.8}=0.7 \\& \Rightarrow T _3=886.8 \,K \\& q _{ H }= C _{ P 0}\left( T _4- T _3\right)=1.004(1600-886.8)=716\,kJ / kg \\& \eta_{ TH }= w _{ NET } / q _{ H }=323.2 / 716= 0 . 4 5 1\end{aligned}

Related Answered Questions