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For \mathrm{Hg} cycle.\begin{array}{l}h_{l}=355.98 \mathrm{~kJ} / \mathrm{kg} \\s_{l}=0.5397 \mathrm{~kJ} / \mathrm{kgk} . \\s_{l}=s_{m} \\s_{m}=s_{n}+x_{m}\left(s_{g}-s_{n}\right) \\0.5397=0.0808+x_{m} x \\(0.6925-0.0808) . \\x_{m}=0.7502\end{array}x_{m} : quality at m\begin{array}{l}h_{m}=h_{n}+x_{m}\left(h_{g}-h_{n}\right) \\h_{m}=29.98+0.7502(329.85-29.98) \\h_{m}=254.94 \mathrm{~kJ} / \mathrm{kg} \text {, } h_{n}=29.98 \mathrm{~kJ} / \mathrm{kg} \text {. } \\W_{p_{H g}}=\text { loump work for } \mathrm{Hg}=v_{n}(\Delta p) \\=76.5 \times 10^{-6}(4.5-0.04) \times 100 \\=0.03412 \mathrm{~kJ} / \mathrm{kg} \text {. } \\h_{k}=h_{n}+w_{p_{\mathrm{Hg}}}=29.98+0.03412=30.014 \mathrm{~kJ} / \mathrm{kg} . \\h_{k}=30.014, k J 1 \mathrm{~kg} \\\end{array}For steam cycle h_{1}=2791.0 \mathrm{~kJ} / \mathrm{kg} s_{1}=6.443 \mathrm{~kJ} / \mathrm{lggk}s_{1}=s_{2}=6.443 \mathrm{~kJ} / \mathrm{kgk} \text {. }\begin{array}{l}s_{2}=s_{f}+x_{2} \text { sfg at } 0.04 \text { bar } \\\Rightarrow 6.443=0.42239+x_{2} \times 8.0510 \\x_{2}=0.7478 \\h_{2}=h_{f}+x_{2} h_{f g} \text { a } 0.04 \mathrm{bar} \\h_{3}=h_{f} \text { al } 0.043 a r \\h_{2}=121.39+0.7478 \times 2432.3 \\h_{3}=121.39 \mathrm{~kJ} / \mathrm{kg} \\h_{2}=1940.2641 \mathrm{cJ} / \mathrm{cg} \\W_{\text {Pwater }}=v_{3} \times(4 p) \times 100=0.0010041 \times(15-0.94) \times 100 \\W_{\text {proaler }}=1.502 \mathrm{~kJ} / \mathrm{kg} \\h_{4}=h_{3}+w_{\text {pwater }}=121.39+1.502 \\=122.892 \mathrm{~kJ} \mathrm{~kg} \\\end{array}\begin{array}{l}W_{T H}=h_{l}-h_{m}=355.98-254.94=101.04 \mathrm{~kJ} / \mathrm{kg} . \\w_{p r g}=0.03412 \mathrm{~kJ} / \mathrm{kg} \text {. } \\W_{\text {netrg }}=W_{\tau_{r g}}-W_{p_{r g}}=101.006 \mathrm{~kJ} / \mathrm{kg} \text {. } \\\end{array}Heat supplied \left(Q_{s}\right): Q_{s} H=h_{l}-h_{k}.\varphi_{\text {sHg }}=355.98-30.014=325.966 \mathrm{~kJ} 1 \mathrm{~kg} \text {. }Heat rejected \left(Q_{R}\right): Q_{R, r g}=h_{m}-h_{n}.\begin{array}{l}\left(\varphi_{R}\right)_{r g}=\left(\varphi_{S}\right)_{\mathrm{H}_{2} \mathrm{O}} \Rightarrow m_{r g} \times 224.96=\frac{48000}{3600} \times 2668.108 \\\dot{m}_{\mathrm{rg}}=158.138 \mathrm{~kg} / \mathrm{s}, \dot{m}_{\mathrm{H}_{2} 0}=13.33 \mathrm{~kg} / \mathrm{s} \\\end{array}(i)\begin{array}{l}\eta_{\text {ovenall }}=\frac{158.138 \times 101.006+13.33 \times 849.234}{158.138 \times 325.966} \\\eta_{\text {ovenall }}=0.5295=52.95 \%\end{array}(ii) Flow through mercury turbine \dot{m}_{\mathrm{rg}}=158.138 \mathrm{~kg} / \mathrm{s} or 569296.8 \mathrm{~kg} / \mathrm{hr}.\begin{array}{l}=158.138 \times 101.006+13.33 \times 849.234 \\=27293.176 \mathrm{kw} \text { or } 27.293 \mathrm{MW}\end{array}Ary(1 v)h_{1}^{\prime}=3038.2 \mathrm{~kJ} / \mathrm{kg} \quad s_{1}^{\prime}=6.9198 \mathrm{~kJ} / \mathrm{kg} k=s_{2}^{\prime}at 0.04 \mathrm{bar} s_{g}=8.0510 \mathrm{kJlkgk}. 78_{2}^{\prime}point 2 'is in wet region\begin{array}{l}6.9198=0.42239+x_{2}^{\prime} \times 8.05 \Rightarrow x_{2}^{\prime}=0.807 . \\h_{2}^{\prime}=121.39+0.807 \times 2432.3 \Rightarrow h_{2}^{\prime}=2084.256 \mathrm{KJ} / \mathrm{g}\end{array}\begin{array}{l}=837.968 \mathrm{~kJ} / \mathrm{kg} \\\end{array}\begin{array}{l}\eta_{\text {ovenay }}=\frac{84.84 \times 172.746+13.33 \times 837.968}{172.746 \times 325.966} \\=0.4586 \simeq 45.86 \% \text { Ans. } \\\end{array} ...