Question

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are \(4 \mathrm{MPa}, 500^{\circ} \mathrm{C}\), and \(80 \mathrm{m} / \mathrm{s}\), and the exit conditions are \(30 \mathrm{kPa}, 92\) percent quality, and \(50 \mathrm{m} / \mathrm{s}\) The mass flow rate of the steam is \(12 \mathrm{kg} / \mathrm{s}\). Determine (a) the change in kinetic energy, ( \(b\) ) the power output, and (c) the turbine inlet area.

Short answer

Answer: The change in kinetic energy is -2550 J/kg, the power output is 11.64 kW, and the turbine inlet area is 0.01098 m².

Step-by-step solution

Determine the enthalpies at inlet and outlet

To find the enthalpy at the inlet and outlet states, we first need to consult the superheated and saturated steam tables. We can find the specific volume, internal energy, and enthalpy corresponding to the given pressures and temperatures. At the inlet, \(P_1 = 4 \mathrm{MPa}\) and \(T_1 = 500^{\circ} \mathrm{C}\). From the superheated steam table, we get: \(h_1 = 3396 \frac{\mathrm{kJ}}{\mathrm{kg}}\) At the outlet, \(P_2 = 30 \mathrm{kPa}\) and has a quality of \(x_2 = 0.92\). From the saturated steam table, we find: \(h_{f_2} = 251.4 \frac{\mathrm{kJ}}{\mathrm{kg}}\) \(h_{fg_2} = 2393.3 \frac{\mathrm{kJ}}{\mathrm{kg}}\) Now, we can calculate the enthalpy at the outlet state: \(h_2 = h_{f_2} + x_2 h_{fg_2} = 251.4 + 0.92 * 2393.3 = 2424.83 \frac{\mathrm{kJ}}{\mathrm{kg}}\)

Calculate the change in kinetic energy

We can find the change in kinetic energy per unit mass between the inlet and outlet states using the given velocities: \(\Delta KE = \frac{1}{2}(V_2^2 - V_1^2) = \frac{1}{2}(50^2 – 80^2) = -2550\ \frac{\mathrm{J}}{\mathrm{kg}}\)

Calculate the power output

Since the turbine is adiabatic, we can apply the conservation of energy principle to find the power output per unit mass: \(W_s = -(\Delta h + \Delta KE) = -(h_2 - h_1) - 2550\) (note the negative sign as work is done by the system) \(W_s = -(2424.83 - 3396) - 2550 = 970.17\ \frac{\mathrm{J}}{\mathrm{kg}}\) Now, we can find the total power output by multiplying the power output per unit mass by the mass flow rate: \(P_{out} = m \cdot W_s = 12 \cdot 970.17 = 11642 \mathrm{W}\) or \(11.64 \mathrm{kW}\)

Determine the turbine inlet area

First, we need to find the specific volume at the inlet from the superheated steam table: \(v_1 = 0.0732\ \frac{\mathrm{m^3}}{\mathrm{kg}}\) Then using the mass flow rate and the velocity, we can find the turbine inlet area: \(A_1 = \frac{m \cdot v_1}{V_1} = \frac{12 \cdot 0.0732}{80} = 0.01098\ \mathrm{m^2}\) To summarize the results: (a) Change in kinetic energy: \(\Delta KE = -2550\ \frac{\mathrm{J}}{\mathrm{kg}}\) (b) Power output: \(P_{out} = 11.64\ \mathrm{kW}\) (c) Turbine inlet area: \(A_1 = 0.01098\ \mathrm{m^2}\)