Answer to Question #217587 in Mechanical Engineering for brian

Question #217587

A steam engine working on Rankine cycle operates between 1.96 MPa, 250ºC and 13.7 kPa. If engine consumes steam at the rate of 0.086 kg per second, determine Rankine cycle efficiency, neglecting pump work. Also, find Rankine cycle efficiency considering pump work.

A reheat cycle operates between 80 bar and 0.075 bar. Temperature of steam entering turbine is 600ºCand first stage of expansion is carried out till the steam becomes dry saturated. Subsequently steam is reheated up to the initial temperature at inlet. Neglecting pump work determine efficiency and specific steam consumption in kg/hp·hr.

Expert's answer


P_{1}=P_{2}=1.96 M P a ; t_{2}=250^{0} \Longrightarrow \\h_{2}=2904.9, S_{2}=6.559 From table A-1, For P_{3}=0.1 bar \\t_{s}=45.83^{\circ} C h_{f}=191.8 h_{f} g=2392.9 \\S_{f}=0.6493 S_{f} g=7.5018 Since \ From\ table A-1, For \ P_{3}=0.1 bar\\ t_{s}=45.83^{\circ} \mathrm{C} h_{t}=191.8 h_{f} g= 2392.9 S_{t}=0.6493 S_{f} g=7.5018\\ Since X_{3}=0.9, h_{3}=h_{f} 4+x_{3} h_{f} g =191.8+0.9(239.9)\\ =2345.4 k j / k g=191.8+0.9(239.9)= 2345.4 k j / k g\\ Also,\ since\ process\ 2- 3_{s} is isentropic, S_{2}\\ i.e., 7.1251=S_{f} g 4+X_{3} s S_{f} g 3 \\=0.6493+X_{3} s(7.5018)=0.6493+X_{3} s \therefore X_{3} s=0.863 \therefore h_{3} s=191.8+0.863(2392.9)= 2257.43 k j / k g

\eta_{R}=\frac{h_{2}-h_{3}}{h_{2}-h_{3 s}}=\frac{2904.9-2345.4}{2904.9-2257.43}=0.86\\ W_{p}=v \int d p=0.0010102(19.6-0.137) * 10^{3}=19.66 k J / k g \eta_{R}=\frac{2904.9-2345.4-19.66}{2904.9-2257.43}=0.83

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