Question

Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at \(9 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) and expands to a pressure of 1.6 MPa. At this pressure, 35 percent of the steam is extracted from the turbine, and the remainder expands to \(10 \mathrm{kPa}\). Part of the extracted steam is used to heat the feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at \(1.6 \mathrm{MPa}\). It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. Assuming the turbines and the pumps to be isentropic, show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine the mass flow rate of steam through the boiler for a net power output of \(25 \mathrm{MW}\).

Short answer

#Answer# To calculate the mass flow rate of steam through the boiler, we need to evaluate the work output from each component and use the net power output of 25 MW. Following the steps outlined above, we calculate the turbine work output, pump work input, heat transfer in the feedwater heater, boiler, and condenser, and finally, the net power output. Then, we can determine the mass flow rate using: \(m = \frac{W_{net}}{\Delta W_{t} - \Delta W_{p}}\) Plug in the values obtained in the previous steps to get the mass flow rate of steam through the boiler.

Step-by-step solution

Find the enthalpy and entropy at each state

First, we will need to determine the enthalpy (h) and entropy (s) at each state in the cycle. Using the given pressure and temperature values, we can look up the enthalpy and entropy in a steam table for each state: 1. State 1: \(h_1 = 3230.9 \mathrm{kJ/kg}\) and \(s_1 = 6.9216 \mathrm{kJ/kg} \cdot \mathrm{K}\) 2. State 2: Since the turbine is isentropic, \(s_1 = s_2\). Looking up in a steam table, we get \(h_2 = 2762.4 \mathrm{kJ/kg}\) 3. State 3: Again, since the turbine is isentropic, \(s_2 = s_3\). Looking up in a steam table at 10 kPa, we get \(h_3 = 1917.3 \mathrm{kJ/kg}\) 4. State 4: We are given that the extracted steam leaves the process heater as a saturated liquid at 1.6 MPa. Looking up in a steam table, we get \(h_4 = 836.0 \mathrm{kJ/kg}\)

Calculate the work in the turbine and pumps

Now we will calculate the work output from the turbine and the work input to the pumps. Turbine work output: \(\Delta W_{t} = (h_1 - h_2) + (1 - 0.35)(h_2 - h_3)\) Pump work input: \(\Delta W_{p} = v_4 (P_{1} - P_{4})\)

Calculate the heat transfer in the feedwater heater, boiler, and condenser

We will now calculate the heat transfer in the feedwater heater, boiler, and condenser. Heat transfer in the feedwater heater: \(Q_{fwh} = 0.35 (h_2 - h_4)\) Heat transfer in the boiler: \(Q_{in} = h_1 - (h_4 + \Delta W_{p})\) Heat transfer in the condenser: \(Q_{out} = (1 - 0.35)(h_3 - h_4 + \Delta W_{p})\)

Calculate the net power output and mass flow rate

Now we can find the net power output, which is given as 25 MW. \(W_{net} = \Delta W_{t} - \Delta W_{p}\) Now, using the net power output, we can determine the mass flow rate through the boiler: \(m = \frac{W_{net}}{\Delta W_{t} - \Delta W_{p}} = \frac{25 \times 10^3 \mathrm{kW}}{\Delta W_{t} - \Delta W_{p}}\) Calculate the mass flow rate using the values obtained in the previous steps, and you will have the mass flow rate of steam through the boiler.