Question

A simple ideal Rankine cycle operates between the pressure limits of \(10 \mathrm{kPa}\) and \(5 \mathrm{MPa}\), with a turbine inlet temperature of \(600^{\circ} \mathrm{C}\). The mass fraction of steam that condenses at the turbine exit is \((a) 6\) percent \((b) 9\)percent \((c) 12\) percent \((d) 15\) percent \((e) 18\) percent

Short answer

Answer: (b) 9%

Step-by-step solution

State Points of Rankine Cycle

First, let's identify the state points for this Rankine cycle. 1. Boiler exit (Turbine inlet): High-pressure, high-temperature state 2. Turbine exit: Low-pressure, mixed-phase state 3. Condenser exit (Pump inlet): Low-pressure, low-temperature (saturated liquid) state 4. Pump exit (Boiler inlet): High-pressure, low-temperature state

Determine Enthalpy Values for Each State Point

Next, we'll determine the enthalpy values for each state point. We have the following data: - Boiler exit (1): \(P_1 = 5\,\text{MPa}\) and \(T_1 = 600^{\circ}\mathrm{C}\) - Condenser exit (3): \(P_3 = 10\,\text{kPa}\) (saturated liquid) Using the steam tables, we can find the enthalpy values: \(h_1 = 3595.1\,\text{kJ/kg}\) (Turbine inlet) \(h_3 = 191.8\,\text{kJ/kg}\) (Condenser exit) Since the ideal Rankine cycle is assumed to be isentropic, we can use the isentropic relations to find \(h_2\): \(s_1 = s_2\) \(s_1 = 6.9026\,\text{kJ/(kg K)}\) (Using the steam table) We look up \(s_2\) in the steam table for low-pressure (\(10\,\text{kPa}\)) and find an appropriate quality value, \(x_2\). Interpolating in steam tables, we find \(x_2 = 0.91 \). Now we can find enthalpy at state 2: \(h_2 = h_{f2} + x_2(h_{g2} - h_{f2})\) \(h_{2} = 191.8\,\text{kJ/kg} + 0.91(2584.9 - 191.8)\,\text{kJ/kg}\) \(h_{2} = 2378.783\,\text{kJ/kg}\)

Calculate the Mass Fraction of Condensation at the Turbine Exit

Now, we can calculate the mass fraction of steam that condenses at the turbine exit. The mass fraction can be found from the quality at point 2. \(y = 1 - x_2 = 1 - 0.91 = 0.09\) (or 9%) So, the answer is \((b) 9\%\).