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$$
\begin{array}{l}{\text { Air is expanded from } 2000 \mathrm{kPa} \text { and } 500^{\circ} \mathrm{C} \text { to } 100 \mathrm{kPa} \text { and } 50^{\circ} \mathrm{C} \text { . Assuming constant specific heats, }} \\ {\text { determine the change in the specific entropy of air. }}\end{array}
$$

$$
\Delta S_{air}=C_{p} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}
$$

$$
A-2 b \rightarrow C_ p=1.040 \mathrm{kJ} / \mathrm{kg} . \mathrm{k}
$$

$$
A-2 a \rightarrow R=0.287 kJ / kg . k
$$

$$
\rightarrow \Delta S_{\text {air }}=(1.040) \ln \frac{50+273}{500+273}-0.287 \ln \frac{100 \mathrm{kPa}}{2000 \mathrm{KPa}}
$$

$$
=\quad-0.0478 \quad kJ / kg \cdot k
$$

$$
\begin{array}{l}{\text { Argon is expanded in an isentropic turbine from } 2 \text { MPa and } 500^{\circ} \mathrm{C} \text { to } 200 \mathrm{kPa} \text { . }} \\ {\text { Determine the dutlet temperature and the work produced by this turbine per }} \\ {\text { unit mass of argon. }}\end{array}
$$

$$
A - 2 b \rightarrow C_ p=1.013 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}
$$

$$
A - 2a \longrightarrow R=0.287 \quad k J / \mathrm{kg} . \mathrm{k}
$$

$$
E_{\text {in }}-E_{\text {out }}=\Delta E_{\text {sys }}
$$

$$
Q_{\text {in }}-W_{\text {out }}=\Delta U=m\left(U_{2}-U_{1}\right)=0 \quad\left(T_{1}=T_{2}\right)
$$

$$
Q_{i n}=W_{ou t}
$$

$$
W_{o ut}=m R T \ln \frac{P_{2}}{P_{1}}=(1)(0.287)(127+275)* \ln (\frac{200}{100})
$$

\(∴ \quad W_{a v t}=79.6 \mathrm{kJ} \)

\(∴ Q_{i n}=79.6 \mathrm{kJ} \)

$$
\begin{array}{l}{\text { Air is expanded in an adiabatic nozzle during a polytropic process with } n=1.3 \text { . }} \\ {\text { It enters the nozzle at } 700 \mathrm{kPa} \text { and } 100^{\circ} \mathrm{C} \text { with a velocity of } 30 \mathrm{m} / \mathrm{s} \text { and exits at }} \\ {\text { a pressure of } 200 \mathrm{kPa} \text { . Calculate the air temperature and velocity at the nozzle exit. }}\end{array}
$$

$$
C_{P}=1.005
$$

$$
k=1.4 \quad(2 a)
$$

$$
Pv^{n}=\text { const. }
$$

$$
\rightarrow T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{\frac{n-1}{n}}
$$

$$
=373\left(\frac{200}{700}\right)^{\frac{0.3}{1.3}}=279 \mathrm{k}
$$

$$
\dot{m}_{1}=\dot{m}_{2}=\dot{m}
$$

$$
\dot E_{i n}-\dot{E}_{o u t}=\Delta\dot E_{sys}\longrightarrow \dot E_{in}=\dot E_{out}
$$

$$
\rightarrow \dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}\right)=\dot m\left(h_{2}+\frac{V_{2}^{2}}{2}\right)
$$

$$
h_{1}+\frac{V_{1}^{2}}{2}=h_{2}+\frac{V_{2}^{2}}{2}\rightarrow V_2=[V_{1}^{2}+2(h_1-h_2)]^\frac{1}{2}
$$

$$
\longrightarrow V_{2}=\left[v_{1}^{2}+2 C_{P}\left(T_1-T_{2}\right)\right]^{0.5}\rightarrow [30^{2}+2(1.005)(373-279]*1000)^{0.5}
$$

\( ∴ V_{2}=436 \mathrm{m/s} \)

 

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