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I have attached a image to solve the problems of the questions I hope it helps you and if so please UPVOTE and RATE my answersFor Hig cydeh_{l}=355.98 \mathrm{~kJ} / \mathrm{kg}s_{l}=0.5397 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}s_{l}=s_{m}s_{m}=s_{n}+x_{m}\left(s_{g}-s_{n}\right)0.5397=0.0808+x_{m}(0.6925-0.0808)x_{m}=0.7502n_{m} - quality at m5\begin{array}{l}h_{m}=h_{n}+n_{m}\left(h_{g}-h_{n}\right) \\h_{m}=29.98+0.7502(329.85-29.98)=254.94 \mathrm{~kJ} / \mathrm{kg} \\h_{n}=29.98 \mathrm{~kJ} / \mathrm{kg} \\W_{P H g}=\text { bump work for Hg }=v_{n}(1 P)=76.5 \times 10^{-6}(4.5-0.04) \times 100 \\=8.03412 \mathrm{~kJ} / \mathrm{kg} \\h_{k}=h_{n}+w_{P H g}=29.98+0.03412=30.014 \mathrm{ks} / \mathrm{kg} \\\end{array}For stream cycle \quad h_{1}=2791 \mathrm{ks} / \mathrm{kg} \quad s_{1}=6.443 \mathrm{~kJ} / \mathrm{kg}s_{1}=s_{2}=6.443 \mathrm{~kJ} / \mathrm{kg}\begin{array}{l} s_{2}=s_{f}+n_{2} s_{2} \text { at } 0.0413 a r \\6-443= 8.42239+n_{2} \times 8.0510 \quad n_{2}=0.7478 \\h_{2}=h_{f}+n_{2} h_{g}=121.39+0.7478 \times(2432.3) \\=1940.264 \mathrm{~kJ} 1 \mathrm{~kg}\end{array}\begin{aligned}\left(w_{p}\right)_{\text {water }} & =v_{3} \times \Delta p \times 100=0.0010041 \times(15-0.04) \times 100=1.502 \mathrm{ks} 1 \mathrm{~kg} \\h_{y} & =h_{3}+\left(w_{p}\right)_{\text {water }}=121.39+1.502=122.892 \mathrm{~kJ} 1 \mathrm{~kg}\end{aligned}\begin{array}{l}\left(h_{4}\right)_{\mathrm{Hg}}=\left(h_{1}-h_{m}\right)=355.48-254.94=101.04 \mathrm{~kJ} / \mathrm{kg} \\\left(b_{p}\right)_{\mathrm{Hg}}=0.03412 \mathrm{~kJ} / \mathrm{kg} \\\left(h_{\text {net }}\right) \mathrm{Hg}_{g}=\left(\omega_{p}\right)-\left.\left(\omega_{p}\right)\right|_{\mathrm{Hg}}=101.006 \mathrm{~kJ} 1 \mathrm{~kg}\end{array}\theta_{\text {Heat supplied }}\left(\theta_{S}\right)=\left(h_{l}-h_{k}\right)=355.98-30.014=325.966 \mathrm{~kJ} / \mathrm{kg}\begin{array}{l}\left(\theta_{R}\right)_{H g}=\left(\theta_{S}\right)_{H_{20}} \Rightarrow \quad m_{\mathrm{Hg}} \times 224.96=\frac{48000}{3600} \times 2668.108 \\\hat{m}_{\mathrm{Hg}}=158.138 \mathrm{~kg} / \mathrm{JeC} \quad \dot{m}_{\mathrm{H}} 20=13.33 \mathrm{~kg} / \mathrm{sPC} \\\eta_{\text {overall }}=\frac{158.138 \times 101.006+13.33 \times 849.234}{158.138 \times 325.966}=0.5295 \text { or } 52.95 \% \\\end{array}Ans(ii) Fow twrough merury turbine m^{\prime g g}=158.138 \mathrm{~kg} / \mathrm{sec} or 569296.8 \mathrm{~kg} / \mathrm{hour}Ans(iii)\begin{array}{l}\text { Useful 100rk }=\dot{m}_{H_{g}} \times\left(w_{\text {net }}\right)_{\mathrm{Hg}}+\dot{m}_{\mathrm{H}_{20}} \times\left(w_{\text {net }}\right)_{H_{20}} \\\begin{aligned}=158.138 \times 161.006+13.33 \times 849.234 & =\begin{array}{l}27293.176 \mathrm{KW} \\27.293 \mathrm{MW}\end{array} \\& =\text { Ans }\end{aligned} \\\end{array}(iv) \quad h_{1}^{\prime}=3038.2 \mathrm{~kJ} / \mathrm{kg} \quad s_{1}^{\prime}=6.9198 \mathrm{~kJ} / \mathrm{kg}=s_{2}^{\prime}Ansat 0.04 \mathrm{Bar} s_{g}=8.0510 \mathrm{~kJ} / \mathrm{kgk}>s_{2}^{\prime} \rightarrow it is in wet Region\begin{array}{l}\left(\theta_{\mathrm{g}}\right)_{\mathrm{Hg}}=\dot{m}_{\mathrm{Hg}} \times 325.966=172.746 \times 325.966 \\\eta_{\text {oveall }}=\frac{\left(W_{\text {nel })_{\mathrm{Kg}}} \times \dot{m}_{\mathrm{Hg}}+\left(W_{\text {nel }}\right)_{\mathrm{H}_{20}} \times \dot{m}_{H_{20}}\right.}{\dot{m}_{H_{G}} \times \theta_{\text {SHg }}} \\\end{array}\begin{aligned} \eta_{\text {overall }} & =\left(\frac{84.84 \times 172.746+13.33 \times 837.968}{172.746 \times 325.966}\right)=0.4586 \\ & \approx 45.86 \% \text { Ans }\end{aligned} ...