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Solution: from the Given values:let\begin{array}{l}P_{2}=P_{3}=4.5 \text { bar } \\P_{4}=0.04 \text { baN } \\P_{6}=P_{7}=15 \text { bar } P_{8}=0.04 \text { bar. }\end{array}where The super heater is not used a sit.Graph!Here from \mathrm{Hg} table (given) F\begin{array}{c}h_{3}=h g @ P_{3}=4.5=355.98 \mathrm{~kJ} / \mathrm{kg} \\s_{3}=S_{9} @ P_{3}=0.5397 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\end{array}\begin{array}{c}s_{4}=s_{3} \\s_{4}{ }^{x} p_{4}+x_{4} S_{f_{4}}=0.5397 \\x_{y}=\frac{0.5397-0.0808}{0.6925-0.0808}=0.7502\end{array}where h_{y}=h_{f_{4}}+x_{y} h_{f_{y}}=29.98+0.7502 \times(329.85 -29.98)\begin{array}{l}h_{4}=254.94 \mathrm{~kJ} / \mathrm{kg} \\h_{1}=h_{R} \otimes P_{1}=29.98 \mathrm{~kJ} / \mathrm{kg}\end{array}h_{1}=h f \otimes P_{1}=29.98 \mathrm{~kJ} / \mathrm{kg}\begin{array}{rl}\omega_{p_{1}}=\vartheta_{1} & d p=V_{f_{1}} \times\left(p_{2}-p_{1}\right)=76.5 \times 10^{-6} \times(4.5-0.04) \times 10^{2} \\\omega_{p_{1}}=0.03412 \mathrm{~kJ} / \mathrm{kg}\end{array}h_{2}=h_{1}+\omega_{p_{1}}=29.98+0.03412=30.014 \mathrm{~kJ} / \mathrm{kg}\begin{array}{l} n_{1}=n_{H g}=\left|\frac{W_{T}-W_{P}}{Q_{S}}\right|_{H g} \\\left(W_{T}\right)_{A g}=\left(h_{3}-h_{4}\right)=355.98-254.59=101.04 \mathrm{~kJ} / \mathrm{kg} \\\left(\omega_{P}\right)_{H g}=w_{P}=0.03412 \mathrm{~kJ} / \mathrm{kg} \\\left(Q_{S}\right)_{H g}=h_{3}-h_{2}=355.98-30.014 \\=3250966 \mathrm{~kJ} / \mathrm{kg} \\n_{1}=\frac{101.04-0.03412}{325.96}=0.3098\end{array}for the strem lycle:-From stream table \Rightarrow at P_{6}=P_{7}=15 bar\begin{array}{l}h_{7}=h_{g} @ P_{7}=2791 \mathrm{~kJ} / \mathrm{kg} . \\S_{7}=S_{9} @ P_{7}=6.443 \mathrm{~kJ} / \mathrm{kg} \mathrm{K} \\S_{8}=S_{78}+x_{B} S_{898}=S_{7}=6.443 \quad, \quad \frac{6.443-0.4224}{8.0510}=0.7478 \quad\left[\because P_{8}=4 \mathrm{kF}\right] \\\left.x_{8}=\frac{10}{}\right]\end{array}\begin{aligned}h_{8} & =h_{f 8}+x_{8} h_{f_{8}}=121.3 .9+0.7478 \times 2432.3 \\h_{8} & =1940.26 \mathrm{~kJ} / \mathrm{kg} \quad h_{5}=h_{f} p_{5}=121.39 \mathrm{~kg} / \mathrm{kg} \\\left(\omega_{p}\right)_{2} & =v_{5}\left(P_{6}-P_{5}\right)=0.001004 \times(15-0.04)^{2} \times 10^{2} \\& =1.5 \mathrm{~kJ} / \mathrm{kg} \\h_{6} & =h_{5}+w_{P_{2}}=122.89 \mathrm{~kJ} / \mathrm{kg}\end{aligned}\begin{array}{l}\left(w_{T}\right)_{\text {sream }}=\left(h_{7}-h_{8}\right)=279-1940.26=850.79 \mathrm{~kJ} / \mathrm{k} \\\left(w_{p}\right)_{\text {sream }}=w_{P_{2}}=1.5 \mathrm{~kJ} / \mathrm{kg} \\\left(Q_{\text {s }}\right)_{\text {sream }}=h_{7-h_{6}}=2791-122.89=2668.11 \mathrm{~kJ} /\end{array}\left(Q_{\text {s }}\right)_{\text {sream }}=h_{7-h_{6}}=2791-122.89=2668.11 \mathrm{~kJ} /\begin{array}{l}n_{2}=\frac{\left(w_{T}\right)_{s}-\left(u_{p}\right) s}{\left(Q_{s}\right)_{s}}=\frac{850.74-1.5}{2668.11} \\\therefore n_{2}=0.31829\end{array}(4.1) Quevall ffficiency of cyclec\begin{array}{l}n_{\theta}=\eta_{1}+n_{2}-\eta_{1} \eta_{2} \\n_{0}=0.3098+0.31829-0.3098 \times 0.31829 \\n_{0}=0.5294=52.9 \%\end{array}(4.2) Here f low through (\mathrm{Hq}) turbin is:-\begin{array}{c}m_{+9}^{\prime}\left(h_{4}-h_{1}\right)=m^{\prime} s .\left(h_{7}-h_{6}\right) \\m^{\prime}+1 g\left(254.94^{\prime}-29.98\right)=48000(2791-129.89) \\\therefore m^{\prime}+g=5,69,298.009 \mathrm{~kg} \mathrm{hr}_{\mathrm{s}}\end{array}(4.3) useful workdone is io\begin{array}{l}\omega_{N_{2} t}=\left(\dot{w}_{T}-\dot{w}_{P}\right)_{\text {Strean }}+\left(\dot{W}_{T}-\dot{W}_{P}\right)_{\mathrm{Hg}} \\\omega_{\text {Net }}=\frac{489 p 00}{3600}(850.74-1.5)+\frac{5,69,298.009}{3600}(101.04-0.03412) \\\therefore \dot{w}_{\text {Net }}=15972.917 \mathrm{kw} \\\therefore \dot{w}_{\text {Net }}=15.972 \mathrm{mw}\end{array}Thank Yuu please HITIKE AND Comment For AnY QuIRES ...