Question

A textile plant requires \(4 \mathrm{kg} / \mathrm{s}\) of saturated steam at \(2 \mathrm{MPa},\) which is extracted from the turbine of a cogeneration plant. Steam enters the turbine at \(8 \mathrm{MPa}\) and \(500^{\circ} \mathrm{C}\) at a rate of \(11 \mathrm{kg} / \mathrm{s}\) and leaves at \(20 \mathrm{kPa}\). The extracted steam leaves the process heater as a saturated liquid and mixes with the feedwater at constant pressure. The mixture is pumped to the boiler pressure. Assuming an isentropic efficiency of 88 percent for both the turbine and the pumps, determine \((a)\) the rate of process heat supply, \((b)\) the net power output, and \((c)\) the utilization factor of the plant.

Short answer

Question: Determine the rate of process heat supply, net power output, and the utilization factor of a steam turbine cogeneration system given the following information: - Steam enters the turbine at 8 MPa, 500°C, and 11 kg/s. - Steam is extracted from the turbine at 2 MPa and 4 kg/s. - Steam leaves the turbine at 20 kPa. - The extracted steam leaves the process heater as a saturated liquid. - Isentropic efficiency of the turbine and pumps is 88%. Answer: (a) The rate of process heat supply is 8,451.6 kW. (b) The net power output is 6,401.1 kW. (c) The utilization factor of the plant is 0.445 or 44.5%.

Step-by-step solution

Known information and Assigning state points

We are given the following information: 1. Steam enters the turbine at state 1: \(P_1 = 8 \,\text{MPa}\), \(T_1 = 500^{\circ} \mathrm{C}\) and mass flow rate \(m = 11 \,\text{kg/s}\). 2. Steam is extracted from the turbine at state 2: \(P_2 = 2 \,\text{MPa}\) and mass flow \(m_2=4 \,\text{kg/s}\). 3. Steam leaves the turbine at state 3: \(P_3=20 \,\text{kPa}\). 4. The extracted steam leaves the process heater as a saturated liquid at state 8. 5. Steam and feedwater mixture is in the state 4. 6. The saturated mixture is pumped to the boiler pressure at state 5. 7. Isentropic efficiency of the turbine and pumps is 88 percent.

Determine the specific enthalpy and entropy values for each state point

For state points 1, 2, 3, and 4, we can look up the specific enthalpy (h) and entropy(s) values in the steam tables for the given reference conditions (T and P). State 1: \(P_1 = 8 \,\text{MPa}\), \(T_1 = 500^{\circ} \mathrm{C}\). From the steam tables, we get \(h_1 = 3373.7 \,\text{kJ/kg}\) and \(s_1 = 6.6611 \,\text{kJ/(kg\cdot K)}\). State 2 (isentropic): \(P_2 = 2 \,\text{MPa}\), \(s_2 = s_1 = 6.6611 \,\text{kJ/(kg\cdot K)}\). From the steam tables, we get \(h_2' = 2966.3 \,\text{kJ/kg}\). Since there is an efficiency given for the turbine, we need to account for it. The actual enthalpy at state 2 is: \(h_2 = h_1 - \eta_\text{turbine} \cdot (h_1 - h_2') = 3373.7 - 0.88 \cdot (3373.7 - 2966.3) = 3021.7 \,\text{kJ/kg}\) State 3 (isentropic): \(P_3 = 20 \,\text{kPa}\), \(s_3 = s_1 = 6.6611 \,\text{kJ/(kg\cdot K)}\). From the steam tables, we get \(h_3' = 2565.7 \,\text{kJ/kg}\). Similarly, taking into account the turbine efficiency, the actual enthalpy at state 3 is: \(h_3 = h_1 - \eta_\text{turbine} \cdot (h_1 - h_3') = 3373.7 - 0.88 \cdot (3373.7 - 2565.7) = 2701.4 \,\text{kJ/kg}\) State 4 (saturated liquid, same pressure as state 3): \(P_4 = 20 \,\text{kPa}\). From the steam tables, we get \(h_4 = 251.4 \,\text{kJ/kg}\).

Determine the rate of process heat supply and mass flow rates

The rate of process heat supply, \(Q_\text{process}\) is equal to the mass flow rate of the extracted steam, \(m_2\), multiplied by the difference in specific enthalpy between states 2 and 8. \(Q_\text{process} = m_2 \cdot (h_2 - h_8)\) Since the extracted steam leaves as a saturated liquid, using steam tables, at \(P_2 = 2 \,\text{MPa}\), we have: \(h_8 = h_\text{f} (P_2) = 908.8 \,\text{kJ/kg}\) Now, calculating the process heat supply: \(Q_\text{process} = m_2 \cdot (h_2 - h_8) = 4\,\text{kg/s} \cdot (3021.7 - 908.8) \, \text{kJ/kg} = 8451.6 \,\text{kW}\) (a) The rate of process heat supply is \(8451.6 \,\text{kW}\). For mass flow rates, we know the following relationships: \(m = m_2 + m_3\) Since, \(m = 11\,\text{kg/s}\) and \(m_2 = 4\,\text{kg/s}\), we can find \(m_3\): \(m_3 = m - m_2 = 11\,\text{kg/s} - 4\,\text{kg/s} = 7\,\text{kg/s}\)

Determine the net power output

The net power output, \(W_\text{net}\), can be determined by calculating the work done by the turbine (\(W_\text{turbine}\)) and the work done by the pump (\(W_\text{pump}\)), and taking the difference. \(W_\text{turbine} = m \cdot (h_1 - h_3)\) \(W_\text{turbine} = 11\,\text{kg/s} \cdot (3373.7 - 2701.4) \, \text{kJ/kg} = 7415.3\,\text{kW}\) Next, we need to determine the actual work done by the pump, \(W_\text{pump}\): \(W_\text{pump} = m \cdot (h_5 - h_4)\) We can determine \(h_5\) using the isentropic pump efficiency (\(\eta_\text{pump}\)) and the enthalpy of feedwater at state 5 (\(h_5'\)): \(h_5 = h_4 + \frac{1}{\eta_\text{pump}}\cdot (h_5' - h_4)\) At state 5, \(P_5 = P_1 = 8\,\text{MPa}\). From the steam tables, \(h_5' = 333.5\,\text{kJ/kg}\). Now we can find \(h_5\): \(h_5 = 251.4 + \frac{1}{0.88}\cdot (333.5 - 251.4) = 343.6\,\text{kJ/kg}\) Now, we can calculate the pump work: \(W_\text{pump} = 11\,\text{kg/s} \cdot (343.6 - 251.4) \, \text{kJ/kg} = 1014.2\,\text{kW}\) Now we can find the net power output: \(W_\text{net} = W_\text{turbine} - W_\text{pump} = 7415.3 - 1014.2 = 6401.1\,\text{kW}\) (b) The net power output is \(6401.1\,\text{kW}\).

Determine the utilization factor

Finally, we can determine the utilization factor, \(f_\text{util}\), by dividing the sum of the rate of process heat supply and the net power output by the total heat input: \(f_\text{util} = \frac{Q_\text{process} + W_\text{net}}{m \cdot (h_1 - h_5)}\) Calculating the denominator first: \(m \cdot (h_1 - h_5) = 11\,\text{kg/s} \cdot (3373.7 - 343.6) \, \text{kJ/kg} = 33331.1\,\text{kW}\) Now we can find the utilization factor: \(f_\text{util} = \frac{8451.6 + 6401.1}{33331.1} = 0.445\) (c) The utilization factor of the plant is 0.445, or 44.5%. In conclusion: (a) The rate of process heat supply is \(8451.6\,\text{kW}\). (b) The net power output is \(6401.1\,\text{kW}\). (c) The utilization factor of the plant is 0.445, or 44.5%.

Key concepts

Saturated Steam

Saturated steam is a key component in various industrial processes, particularly in cogeneration plants. It is steam at a temperature and pressure where it coexists with water in its liquid form. At this equilibrium point, any additional heat would convert the liquid water into more steam, rather than raising the steam's temperature.

In the textile plant scenario, the saturated steam at 2 MPa is used because of its ability to transfer heat effectively to the processes that require it. The fact that it leaves the process heater as saturated liquid helps in indicating that all the available energy from the steam has been transferred during the process.

Isentropic Efficiency

Isentropic efficiency is a measure of how closely a real-life process approaches the ideal isentropic one. It's a measure of the effectiveness of turbines and pumps in converting mechanical energy to pressure energy (and vice versa) without any entropy increase (i.e., without loss of thermal efficiency). In practical terms, this means less wasted energy.

The 88 percent isentropic efficiency given for the turbine and pumps in the exercise indicates that the actual processes are fairly efficient. However, they are not perfect as some energy, represented by the 12% loss, is dissipated typically as heat due to friction and other non-idealities.

Process Heat Supply

The process heat supply is the amount of thermal energy provided to the textile plant's processes using the extracted steam. This energy is represented by the mass flow rate of the extracted steam and the difference in enthalpy between the point of steam extraction and when it leaves as saturated liquid.

When calculating the rate of process heat supply, it's notable that the high value of heat supplied, 8451.6 kW in our example, underscores the importance of cogeneration efficiency, maximizing energy use by recycling the steam for heating purposes in industrial settings.

Net Power Output

The net power output is the total usable power generated by the cogeneration plant after accounting for the work done by both the turbine and the pump. This figure represents the plant's effectiveness in converting thermal energy into electric power—it's the energy that's available for distribution to the grid or for internal use.

The step by step solution reveals that calculating net power output involves considering the mechanical work extracted by the turbine and the work required to operate the pump, with efficiencies factored in. The obtained value, 6401.1 kW, is a direct indicator of the cogeneration plant's performance.

Utilization Factor

The utilization factor is a metric for evaluating the performance of cogeneration or combined heat and power (CHP) plants. It combines both the production of electricity (net power output) and the use of thermal energy (process heat supply) to give a measure of the overall energy utilization efficiency of the plant.

A 44.5% utilization factor suggests that nearly half of the fuel energy is effectively used in the combined production of electrical and thermal energy. While this may seem low, it is often significantly higher than separated heat and power generation, showing the advantage of cogeneration plants.

Steam Turbine

Steam turbines are the heart of many power plans, including cogeneration systems. Allowing high-pressure and high-temperature steam to expand through a series of blades, steam turbines convert the steam's thermal energy into mechanical energy, which can then be transformed into electrical power through a generator.

The efficiency of steam turbines not only affects the net power output but also determines the amount of steam that can be extracted and used for process heating. In the exercise, the turbine efficiency directly influences how the saturated steam is used and the overall energy balance of the plant.