Question

Consider a steam power plant that operates on a reheat Rankine cycle and has a net power output of \(80 \mathrm{MW}\) Steam enters the high-pressure turbine at \(10 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) and the low-pressure turbine at \(1 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\). Steam leaves the condenser as a saturated liquid at a pressure of \(10 \mathrm{kPa} .\) The isentropic efficiency of the turbine is 80 percent, and that of the pump is 95 percent. Show the cycle on a \(T-s\) diagram with respect to saturation lines, and determine (a) the quality (or temperature, if superheated) of the steam at the turbine exit, \((b)\) the thermal efficiency of the cycle, and \((c)\) the mass flow rate of the steam.

Short answer

In summary, for this steam power plant operating on a reheat Rankine cycle: (a) The quality of the steam at the turbine exit (State 4) is 97.67%. (b) The thermal efficiency of the cycle is 11.20%. (c) The mass flow rate of the steam is 112.8 kg/s.

Step-by-step solution

Identify specific states in the Rankine cycle

We will denote the following specific states in the cycle: State 1: Steam entering the high-pressure turbine State 2: Steam leaving the high-pressure turbine State 3: Steam entering the low-pressure turbine State 4: Steam leaving the low-pressure turbine State 5: Exit of the condenser State 6: Exit of the pump

Determine the enthalpy of specific steps in the cycle

First, we determine the enthalpy of state 1 using steam tables: \(h_1 = h_{\mathrm{superheated}, P = 10 \, \mathrm{MPa}, T = 500^\circ \mathrm{C}} = 3372.7 \, \mathrm{kJ/kg}\) (Enthalpy at State 1) Since the turbine has an isentropic efficiency of 80 percent, we can determine the enthalpy of states 2 and 4 as: \(h_2 = h_1 - \eta_t (h_{1, \mathrm{isentropic}} - h_{2, \mathrm{isentropic}})\) \(h_4 = h_3 - \eta_t (h_{3, \mathrm{isentropic}} - h_{4, \mathrm{isentropic}})\) Where \(h_{1, \mathrm{isentropic}}, h_{2, \mathrm{isentropic}}, h_{3, \mathrm{isentropic}}, h_{4, \mathrm{isentropic}}\) are the enthalpy values of states 2 and 4 if they were isentropic (reversible adiabatic). We calculation below using steam tables: \(h_{2, \mathrm{isentropic}} = h_{\mathrm{saturated \, vapor}, P = 1 \, \mathrm{MPa}, s = s_1} = 2777.1 \, \mathrm{kJ/kg}\) Then, we find the enthalpy at state 2 considering the isentropic efficiency of the turbine: \(h_2 = 3372.7 - 0.8 (3372.7 - 2777.1) = 3216.5 \, \mathrm{kJ/kg}\) (Enthalpy at State 2) Now, we determine the enthalpy of state 3: \(h_3 = h_{\mathrm{superheated}, P = 1 \, \mathrm{MPa}, T = 500^\circ \mathrm{C}} = 3372.7 \, \mathrm{kJ/kg}\) (Enthalpy at State 3) \(h_{4, \mathrm{isentropic}} = h_{\mathrm{saturated \, vapor}, P = 10 \, \mathrm{kPa}, s = s_3} = 2039.5 \, \mathrm{kJ/kg}\) Then, we find the enthalpy at state 4 considering the isentropic efficiency of the turbine: \(h_4 = 3372.7 - 0.8 (3372.7 - 2039.5) = 2711.7 \, \mathrm{kJ/kg}\) (Enthalpy at State 4) For state 5 and state 6, we will consider the isentropic efficiency of the pump (95 percent): \(h_5 = h_{\mathrm{sat \, liquid}, P = 10\, \mathrm{kPa}} = 191.8 \, \mathrm{kJ/kg}\) (Enthalpy at State 5) \(v_5 = v_{\mathrm{sat \, liquid}, P = 10\, \mathrm{kPa}}\) (Specific volume at State 5) \(h_6 = h_5 + \frac{v_5 (10 \, \mathrm{MPa} - 10 \, \mathrm{kPa})}{\eta_p} = 191.8 + \frac{0.00102 (10 \times 10^3 - 10)\, \mathrm{kJ/kg}}{0.95} = 199.6 \, \mathrm{kJ/kg}\) (Enthalpy at State 6)

Calculate the cycle's net work output and heat input

The net work output equals both turbine works minus the pump work. We will also calculate the heat input from the boiler and reheater: \(W_{\mathrm{net}} = (h_1 - h_2) + (h_3 - h_4) - (h_6 - h_5)\) \(Q_{\mathrm{in}} = (h_1 - h_6) + (h_3 - h_2)\) Substituting the enthalpy values previously calculated: \(W_{\mathrm{net}} = (3372.7 - 3216.5) + (3372.7 - 2711.7) - (199.6 - 191.8) = 709.4 \, \mathrm{kJ/kg}\) (Net Work Output) \(Q_{\mathrm{in}} = (3372.7 - 199.6) + (3372.7 - 3216.5) = 6329.3 \, \mathrm{kJ/kg}\) (Heat Input)

Obtain the thermal efficiency of the cycle

Thermal efficiency (\(\eta_{\mathrm{thermal}}\)) can be calculated as the ratio of the net work output to the heat input: \(\eta_{\mathrm{thermal}} = \frac{W_{\mathrm{net}}}{Q_{\mathrm{in}}} = \frac{709.4}{6329.3} = 0.1120\) or \(11.20 \%\) (Thermal Efficiency)

Find the mass flow rate of the steam

Given that the net power output of the plant is \(80\, \mathrm{MW}\), we can determine the mass flow rate of the steam (\(\dot{m}\)) as: \(\dot{m} = \frac{W_{\mathrm{net,plant}}}{W_{\mathrm{net}}}\) Substituting the net power output and the net work output: \(\dot{m} = \frac{80 \times 10^6\, \mathrm{W}}{709.4\, \mathrm{kJ/kg}} = \frac{80 \times 10^3\, \mathrm{kJ/s}}{709.4\, \mathrm{kJ/kg}} = 112.8\, \mathrm{kg/s}\) (Mass Flow Rate) Now, we have the answers to all three parts of the question: (a) The quality of the steam at the turbine exit (State 4) can be determined using the enthalpy value we found: \(x_4 = \frac{h_4 - h_{\mathrm{sat \, liquid}, P = 10\, \mathrm{kPa}}}{h_{\mathrm{sat \, vapor}, P = 10\, \mathrm{kPa}} - h_{\mathrm{sat \, liquid}, P = 10\, \mathrm{kPa}}} = \frac{2711.7 - 191.8}{2584.6 - 191.8} = 0.9767\) or \(97.67 \%\) (Quality of the Steam) (b) The thermal efficiency of the cycle is \(11.20 \%\). (c) The mass flow rate of the steam is \(112.8\, \mathrm{kg/s}\).