Question 6.168: Steam enters a turbine steadily at 10 MPa and 550°C with a v......
Steam enters a turbine steadily at 10 \mathrm{MPa} and 550^{\circ} \mathrm{C} with a velocity of 60 \mathrm{~m} / \mathrm{s} and leaves at 25 \mathrm{kPa} with a quality of 95 percent. A heat loss of 30 \mathrm{~kJ} / \mathrm{kg} occurs during the process. The inlet area of the turbine is 150 \mathrm{~cm}^{2}, and the exit area is 1400 \mathrm{~cm}^{2}. Determine (a) the mass flow rate of the steam, (b) the exit velocity, and (c) the power output.
Steam expands in a turbine steadily. The mass flow rate of the steam, the exit velocity, and the power output are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible.
Properties From the steam tables (Tables A-4 through 6)
\left.\begin{array}{l} P_{1}=10 \mathrm{MPa} \\ T_{1}=550^{\circ} \mathrm{C} \end{array}\right\} \begin{gathered} {v}_{1}=0.035655 \mathrm{~m}^{3} / \mathrm{kg} \\ h_{1}=3502.0 \mathrm{~kJ} / \mathrm{kg} \end{gathered}
and
\left.\begin{array}{c} P_{2}=25 \mathrm{kPa} \\ x_{2}=0.95 \end{array}\right\} \begin{gathered} {v}_{2}={v}_{f}+x_{2} {v}_{f g}=0.00102+(0.95)(6.2034-0.00102)=5.8933 \mathrm{~m}^{3} / \mathrm{kg} \\ h_{2}=h_{f}+x_{2} h_{f g}=271.96+(0.95)(2345.5)=2500.2 \mathrm{~kJ} / \mathrm{kg} \end{gathered}
Analysis (a) The mass flow rate of the steam is
\dot{m}=\frac{1}{v_{1}} V_{1} A_{1}=\frac{1}{0.035655 \mathrm{~m}^{3} / \mathrm{kg}}(60 \mathrm{~m} / \mathrm{s})\left(0.015 \mathrm{~m}^{2}\right)=\mathrm{2 5 . 2 4} \mathrm{~ k g} / \mathrm{s}
(b) There is only one inlet and one exit, and thus \dot{m}_{1}=\dot{m}_{2}=\dot{m}. Then the exit velocity is determined from
\dot{m}=\frac{1}{v_{2}} V_{2} A_{2} \longrightarrow V_{2}=\frac{\dot{m} v_{2}}{A_{2}}=\frac{(25.24 \mathrm{~kg} / \mathrm{s})\left(5.8933 \mathrm{~m}^{3} / \mathrm{kg}\right)}{0.14 \mathrm{~m}^{2}}=\mathrm{1 0 6 3} \mathrm{m} / \mathrm{s}
(c) We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as
\begin{aligned} & \underbrace{\dot{E}_{\text {in }}-\dot{E}_{\text {out }}}_{\begin{array}{c} \text { Rate of net energy transfer } \\ \text { by heat, work, and mass } \end{array}}=\underbrace{\Delta \dot{E}_{\text {system }}{}^{\nearrow 0 \text { (steady) }} }_{\begin{array}{c} \text { Rate of change in internal, kinetic, } \\ \text { potential, etc. energies } \end{array}}=0 \\ & \dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\ & \dot{m}\left(h_{1}+V_{1}^{2} / 2\right)=\dot{W}_{\text {out }}+\dot{Q}_{\text {out }}+\dot{m}\left(h_{2}+V_{2}^{2} / 2\right) \quad(\text { since } \Delta \text { pe } \cong 0) \\ & \dot{W}_{\text {out }}=-\dot{Q}_{\text {out }}-\dot{m}\left\lgroup h_{2}-h_{1}+\frac{V_{2}^{2}-V_{1}^{2}}{2}\right\rgroup \end{aligned}
Then the power output of the turbine is determined by substituting to be
\begin{aligned} \dot{W}_{\text {out }} & =-(25.24 \times 30) \mathrm{kJ} / \mathrm{s}-(25.24 \mathrm{~kg} / \mathrm{s})\left\lgroup 2500.2-3502.0+\frac{(1063 \mathrm{~m} / \mathrm{s})^{2}-(60 \mathrm{~m} / \mathrm{s})^{2}}{2}\left\lgroup \frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^{2} / \mathrm{s}^{2}}\right\rgroup \right\rgroup \\ & =\mathrm{1 0 , 3 3 0} \mathrm{k W} \end{aligned}
