Question

Steam enters the turbine of a cogeneration plant at \(4 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\). One-fourth of the steam is extracted from the turbine at 1200 -kPa pressure for process heating. The remaining steam continues to expand to \(10 \mathrm{kPa} .\) The extracted steam is then condensed and mixed with feedwater at constant pressure and the mixture is pumped to the boiler pressure of 7 MPa. The mass flow rate of steam through the boiler is \(55 \mathrm{kg} / \mathrm{s}\). Disregarding any pressure drops and heat losses in the piping, and assuming the turbine and the pump to be isentropic, determine the net power produced and the utilization factor of the plant.

Short answer

Question: Determine the net power produced and the utilization factor of a cogeneration plant, given the following conditions: initial state at the entrance of the turbine at \(P_1 = 4 MPa\) and \(T_1 = 500^\circ C\), state at the extraction point is \(P_2 = 1.2 MPa\), state after the remaining steam expanded to low pressure is \(P_3 = 10 kPa\), and state after the mixture is pumped back to the boiler is \(P_5 = 7 MPa\). The mass flow rate of steam is 55 kg/s, and 1/4 of the steam is extracted at state 2. Assume isentropic turbine and pump operation.

Step-by-step solution

Determine the steam properties at the key states

Using the initial state (\(P_1 = 4 MPa\) and \(T_1 = 500^\circ C\)), we can find the enthalpy (\(h_1\)) and the entropy (\(s_1\)) at state 1. Then, using the isentropic condition for the turbine, we can also find the enthalpy at states 2 and 3. Lastly, we can find the enthalpy at state 4 as it is the same as state 2 (constant pressure mixing), and at state 5 after pumping.

Calculate the mass flow rate of extracted steam and the remaining steam after extraction

Let \(m_2\) be the mass flow rate of extracted steam and \(m_3\) be the mass flow rate of remaining steam after extraction. We know that \(1/4\) of the steam is extracted at state 2. Thus, we can calculate \(m_2\) and \(m_3\) as follows: \(m_2 = \frac{1}{4} \times 55kg/s = 13.75kg/s\) \(m_3 = 55 - 13.75 = 41.25kg/s\)

Calculate the work of the turbine, pump, and process heating

Now, we can calculate the work output/input for each process in the cycle: - Turbine work: \(W_t = (m_2(h_1 - h_2) + m_3(h_1 - h_3))\) - Pump work: \(W_p = m_5 (h_5 - h_4)\) - Process heating: \(Q_{ph} = m_2 (h_2 - h_4)\)

Calculate the net power produced in the plant

The net power produced is the difference between the turbine work output and the pump work input: \(W_{net} = W_t - W_p\)

Calculate the utilization factor of the plant

The utilization factor is the ratio of process heating to the sum of the net power produced and the process heating: \(UF = \frac{Q_{ph}}{W_{net} + Q_{ph}}\) Now, you have the net power produced and the utilization factor of the plant using the steps outlined above.