Question

Consider an ideal steam regenerative Rankine cycle with two feedwater heaters, one closed and one open. Steam enters the turbine at \(10 \mathrm{MPa}\) and \(600^{\circ} \mathrm{C}\) and exhausts to the condenser at \(10 \mathrm{kPa}\). Steam is extracted from the turbine at 1.2 MPa for the closed feedwater heater and at 0.6 MPa for the open one. The feedwater is heated to the condensation temperature of the extracted steam in the closed feedwater heater. The extracted steam leaves the closed feedwater heater as a saturated liquid, which is subsequently throttled to the open feedwater heater. Show the cycle on a \(T-s\) diagram with respect to saturation lines, and determine \((a)\) the mass flow rate of steam through the boiler for a net power output of \(400 \mathrm{MW}\) and \((b)\) the thermal efficiency of the cycle.

Short answer

Short Answer: In the Rankine cycle with two feedwater heaters, we first visualize the process by drawing a T-s diagram with key stages like inlet of steam turbine, feedwater heaters, condenser outlet, pump exit, etc. We then determine enthalpies at each state using steam tables or ideal gas equations. Calculating the work done by the turbine and pump as well as the heat added in the boiler, we can derive the net work output and heat input into the system. These values help in calculating thermal efficiency of the cycle - the efficiency with which thermal energy is converted to work. Lastly, the mass flow rate of steam through the boiler is computed based on the given net power output. This step-by-step analysis helps in understanding the performance and efficiency of a Rankine cycle power plant.

Step-by-step solution

Draw the T-s diagram

To draw the \(T-s\) diagram for this Rankine cycle, plot the saturation dome and identify each state point on the diagram. The main state points are: 1. Inlet of the steam turbine (high pressure and temperature) 2. Extraction point for closed feedwater heater (1.2 MPa) 3. Extraction point for open feedwater heater (0.6 MPa) 4. Condenser outlet (saturated liquid) 5. Pump exit to open feedwater heater. 6. Open feedwater heater exit to the closed feedwater heater 7. Closed feedwater heater exit to the boiler These state points are on or within the saturation dome.

Determine enthalpies at each state

Now, we determine the enthalpies at each state point, which will help us carry out energy balance calculations later. For this, we need to apply steam tables and ideal gas relations. 1. Inlet of turbine: \(h_1 = h\text{(10 MPa, } 600^{\circ} \mathrm{C})\). 2. Extraction point for the closed feedwater heater: \(h_2 = h(\text{1.2 MPa, isentropic})\). 3. Extraction point for an open feedwater heater: \(h_3 = h(\text{0.6 MPa, isentropic})\). 4. Condenser outlet (saturated liquid): \(h_4 = h_{f}\text{(10 kPa})\). 5. Pump exit to open feedwater heater: \(h_5 = h_4 + v_4 \cdot (P_5 - P_4)\). 6. Open feedwater heater exit (0.6 MPa): \(h_6 = h_f(0.6 \text{MPa})\). 7. Closed feedwater heater exit (1.2 MPa): \(h_7 = h_f(1.2 \text{MPa})\).

Calculate heat input and work output for each process

Now, apply the first law of thermodynamics to determine the work output of the turbine (W_turbine), work input for the pump (W_pump), and heat input in the boiler (Q_boiler): - \(W_{turbine} = m_1 (h_1 - h_2) + (1 - y)m_1 (h_2 - h_3)\) indicates the work output of the turbine - \(W_{pump} = m_4(h_5 - h_4)\) is the work input for the pump system - \(Q_{boiler} = m_1 (h_1 - h_7)\) corresponds to the heat added to the cycle in the boiler

Calculate the net work output and heat input to the cycle

Use the information from Step 3 to find the net work output (\(W_{net}\)) and heat input (\(Q_{input}\)) to the cycle: - \(W_{net} = W_{turbine} - W_{pump}\) - \(Q_{input} = Q_{boiler}\)

Calculate the thermal efficiency

With the net work output and heat input, we can now calculate the thermal efficiency (\(\eta_{th}\)) of the cycle: - \(\eta_{th} = \dfrac{W_{net}}{Q_{input}}\)

Determine the mass flow rate of steam through the boiler

The given net power output is \(400 MW\). Using the net work output, we can calculate the mass flow rate (\(\dot{m}_1\)) through the boiler: - \(\dot{m}_1 = \dfrac{400 MW}{W_{net}}\) Now, we have determined \((a)\) the mass flow rate of steam through the boiler and \((b)\) the thermal efficiency of the cycle.