Question 17.CSGP.53: Air is expanded in a nozzle from a stagnation state of 2 MPa......
Air is expanded in a nozzle from a stagnation state of 2 MPa, 600 K, to a static pressure of 200 kPa. The mass flow rate through the nozzle is 5 kg/s. Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle.
Step-by-Step
\begin{aligned}P ^* & = P _{ o }\left\lgroup \frac{2}{ k +1} \right\rgroup^{ \frac{ k }{ k -1} }\\& =2 \times 0.5283=1.056 \,MPa \\T ^* & = T _{ o } \times \frac{2}{ k +1}=600 \times 0.8333=500 \,K \\v ^* & = RT ^* / P ^*=0.287 \times 500 / 1056 \\& =0.1359 \,m ^3 / kg\end{aligned}
\begin{aligned}& c ^*=\sqrt{ kRT ^*}=\sqrt{1.4 \times 1000 \times 0.287 \times 500}=448.2 \,m / s \\& A ^*= \dot{m}v ^* / c ^*=5 \times 0.1359 / 448.2= 0 . 0 0 1 5 2 ~ m ^2\end{aligned}
P _2 / P _{ o }=200 / 2000=0.1 \quad \Rightarrow \quad M _2^*=1.701= \mathbf{V} _2 / c ^*
Column with mach no. based on throat speed of sound.
\begin{aligned}& \mathbf{V}_2=1.701 \times 448.2=762.4\,m / s \\& T _2= T _{ o }\left( T _2 / T _{ o }\right)=600 \times 0.5176=310.56 \,K \\& v _2= RT _2 / P _2=0.287 \times 310.56 / 200=0.4456 \,m ^3 / kg \\& A _2=\dot{ m }v_2 / \mathbf{V} _2=5 \times 0.4456 / 762.4= 0 . 0 0 2 9 2 ~ m ^2\end{aligned}
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