Question pls solve fast A binary vapor cycle operates on mercury and steam. Saturated mercury vapor at 4.5 bar is supplied to the mercury turbine, from which it exhausts at 0.04 bar. The mercury condenser generates saturated steam at 15 bar which is expanded in a steam turbine to 0.04 bar. (i) Determine the overall efficiency of the cycle. (ii) If 48000 kg/h of steam flows through the steam turbine, what is the flow through the mercury turbine ? (iii) Assuming that all processes are reversible, what is the useful work done in the binary vapor cycle for the specified steam flow (net power generated)? The properties of saturated mercury are given below: p (bar) +CC) 8. (kJ/kg) (kJ/kg K) 63.93 355.98 0.1352 0.5397 216.9 29.98 329.85 0.0808 0.6925 (m /kg) 4.5 450 79.9 X10 76.5 x 10- 0.068 5.178 0.04 12

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 ...