A binary-vapour cycle operates on mercury and steam. Saturated vapour at 4.5 bar is supplied to the mercury turbine, from which it exhaust at 0.04 bar. The mercury condenser generates saturated steam at 15 bar which is expanded in a steam turbine to 0.04 bar. 4.1 Determine the overall efficiency of the cycle. 4.2 If 50, 000 kg/h of steam flows through the steam turbine, estimate the flow through the mercury turbine. 4.3 Assuming that all processes are reversible, determine the useful work done in the binary vapour cycle for the specified steam flow. 4.4 If the steam leaving the mercury condenser is superheated to a temperature of 300°C in a superheater located in the mercury boiler, and if the internal efficiencies of the mercury and steam turbines are 0.85 and 0.87 respectively, calculate the overall efficiency of the cycle. The properties of saturated mercury are given below: P (bar) t (°C) hf hg sf sg vf vg (kJ/kg) (kJ/kgK) (m³/kg) 4.5 450 62.93 355.98 0.1352 0.5397 79.9 x10-6 0.068 0.04 216.9 29.98 329.85 0.0808 0.6925 76.5 x10-6 5.178

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Quss:- Givn:-P_{2}=P_{3}=4.5 \text { ban }P_{4}=0.04 \text { bar }\begin{array}{l}P_{6}=P_{7}=15 \text { bar } \\P_{8}=0.04 \text { bar }\end{array}Solv:-- When suparheater is not usd.\begin{array}{c}\text { From Hg table }(\text { giun }) \Rightarrow \quad h_{3}=h_{g} @ p_{3}=4.5=355.98 \mathrm{~kJ} / \mathrm{kg} \\x_{3}=s_{9} c P_{3}=0.5397 \mathrm{~kJ} / \mathrm{kg} \mathrm{K} \\S_{4}=s_{3} \\S_{f_{4} c P_{4}}+x_{4} S_{f_{4}}=0.5397 \\x_{4}=\frac{0.5397-0.0808}{0.6925-0.0808}=0.7502 \\h_{4}=h_{f, 4}+x_{4} h_{t g_{4}}=29.98+0.7502 \times(329.85-2998) \\h_{4}=254.94 \mathrm{~kJ} / \mathrm{kg}\end{array}\begin{array}{l}h_{1}=h_{f Q P_{1}}=29.98 \mathrm{~kJ} / \mathrm{kg} \\W_{P_{1}}=v_{1} d P=v_{f_{1}} \times\left(P_{2}-P_{1}\right)=76.5 \times 10^{-6} \times(4.5-0.09) \times 10^{2} \\W_{P_{1}}=0.03412 \mathrm{~kJ} / \mathrm{kg} \\h_{2}=h_{1}+W_{P_{1}}=29.98+0.03412=30.014 \mathrm{~kJ} 1 \mathrm{~kg}\end{array}#\begin{array}{l}n_{1}=n_{H g}=\left(\frac{W_{T}-W_{P}}{Q_{s}}\right)_{\mathrm{Hg}} \\\left(W_{T}\right)_{H g}=\left(h_{3}-h_{4}\right)=355.98-254.54=101.04 \mathrm{~kJ} / \mathrm{kg} \\\left(W_{P}\right)_{H g}=W_{P_{1}}=0.03412 \mathrm{~kJ} / \mathrm{kg} \\\left(Q_{S}\right)_{H g}=h_{3}-h_{2}=355.98-30.014=325.966 \mathrm{~kJ} / \mathrm{kg} \\n_{1}=\frac{101.04-0.03412}{325.966}=0.3098\end{array}For steam cycl:- Friom steam table \Rightarrow at P_{6}=P_{7}=15 ban\begin{array}{l}h_{7}=h_{g 0} p_{7}=2791 \mathrm{~kJ} / \mathrm{kg} \\S_{7}=S_{g} \otimes P_{7}=6.443 \mathrm{~kJ} / \mathrm{kg} \mathrm{K} \\S_{8}=S_{f 8}+x_{8} S_{f 98}=S_{7}=6.443 \\P_{8}=\underline{\underline{k R a}} \\x_{8}=\frac{6.443-0.4224}{8.0510}=0.7478 \\h_{8}=h_{f 8}+x_{8} h_{f 8}=121.39+0.7478 \times 2432.3 \\h_{8}=1940.26 \mathrm{~kJ} / \mathrm{kg} \quad h_{5}=h_{f \odot P_{5}}=121.39 \mathrm{~km} / \mathrm{s} \\\left(W_{P_{2}}=v_{5} \cdot\left(P_{6}-P_{5}\right)=0.001004 \times(15-0.04) \times 10^{2}=1.5 \mathrm{~kJ} / \mathrm{kg}\right. \\h_{6}=h_{5}+w_{p_{2}}=122.89 \mathrm{~kJ} / \mathrm{kg} \\\end{array}\begin{aligned}\left(W_{T}\right)_{\text {steam }} & =\left(h_{7}-h_{8}\right)=279-1940.26=850.74 \mathrm{~kJ} / \mathrm{kg} \\\left(W_{p}\right)_{\text {steam }} & =W_{P_{2}}=1.5 \mathrm{kJ1kg} \\\left(Q_{s}\right)_{\text {stram }} & =h_{7}-h_{6}=2791-\frac{122.89}{}=2668.11 \mathrm{~kJ} / \mathrm{kg} \\\eta_{2} & =\frac{\left(W_{T}\right)_{8}-\left(W_{p}\right)_{s}}{\left(Q_{s}\right)_{s}}=\frac{850.74-1.5}{2668.11} \\\eta_{2} & =0.31829\end{aligned}#4.1 Ovmall efficiency of cycli- -\eta_{0}=n_{1}+n_{2}-n_{1} n_{2}\begin{array}{l}\eta_{0}=0.3098+0.31829-0.3098 \times 0.31829 \\\eta_{0}=0.5294 \cong 52.94 \%\end{array}# (4.2) Flom through \mathrm{Hg} trabini- Enrgy balance for Mercury condinses.\begin{array}{l}w_{\text {igg }}\left(h_{4}-h_{1}\right)=m_{s}\left(h_{7}-h_{6}\right) \\m_{H g}(254.94-29.98)=50,000(2791-122.89) \\m_{i g}=593,018: 75 \mathrm{~kg} / h_{\pi}\end{array}#(4.3) Usful work done. -\dot{W}_{\text {Nat }}=\left(\dot{W}_{T}-\dot{w}_{p}\right)_{\text {sttem }}+\left(\dot{w}_{T}-\dot{w}_{P}\right)_{M g}\begin{array}{c}\dot{W}_{\text {Nat }}=\frac{50,000}{3600}(850.74-1.5)+\frac{593,018.75}{3600}(101.04-0.03412) \\\dot{W}_{\text {Nat }}=28.433 .43 \mathrm{~kW}=28.433 \mathrm{~mW}\end{array}Qvarall efficiany:-(4.4)\text { If } \begin{aligned}T_{7} & =300^{\circ} \mathrm{C} \quad, P_{7}=15 \mathrm{bba} \\\left(\eta_{\text {Hg }}\right)_{+}=0.85 ; & \left.; \eta_{s}\right)_{+}=0.87\end{aligned}Noutr h_{7}=3038.15 \mathrm{~kJ} / \mathrm{kg} \quad \therefore S_{7}=6.9208 \mathrm{~kJ} / \mathrm{ksK}\begin{array}{l}\text { Nom } S_{8}=S_{7}=6.9208 \\x_{8}=\frac{s_{8}-s_{78}}{s_{78}}=\frac{6.9208-0.4224}{8.05_{10}}=0.8 \\\end{array}h_{8}=121.39+0.8 \times 2432.3=206724 \mathrm{~kJ} / \mathrm{kg}(Nom)\begin{array}{l}\left(W_{T}\right)_{\text {steen }}=\left(h_{7}-h_{8}\right)=3038.15-2067.24=970.32 \mathrm{~kJ} / \mathrm{g} \\\left(W_{T}\right)_{\text {actura }}=0.87 \times 970.92=844.7 \mathrm{KJ} / \mathrm{kg} \\\left(W_{\text {Nut }}\right)_{t}=844.7-1.5=843.2 \mathrm{~kJ} / \mathrm{kg} \\\left(Q_{s}\right)_{\star}=\left(h_{7}-h_{6}\right)=3038.15-122.89=2915.26 \mathrm{~kJ} / \mathrm{ks} \\\eta_{S}=n_{2}=\left(\frac{W_{\text {Nat }}}{Q_{S}}\right)_{t}=\frac{844.7}{2915.26}=0.2897 \\\end{array}\begin{aligned}\#\left(\eta_{+}\right)_{H g} & =0.85 \\& \left(W_{N_{a t}}\right)_{H g}=0.85 \times 101.04-0.03412=85.849 \\\eta_{1} & =\left(\frac{W_{N_{\text {At }}}}{Q_{s}}\right)_{H_{H}}=\frac{85.849}{325.966}=0.2633\end{aligned}Nom\begin{array}{l}\eta_{0}=\eta_{1}+n_{2}-n_{1} n_{2}=0.2633+0.2857-0.2633 x \\\eta_{0}=0.4767\end{array}H PLS RATE the solun ...