Question

To control an isentropic steam turbine, a throttle valve is placed in the steam line leading to the turbine inlet. Steam at \(6 \mathrm{MPa}\) and \(600^{\circ} \mathrm{C}\) is supplied to the throttle inlet, and the turbine exhaust pressure is set at \(40 \mathrm{kPa}\). What is the effect on the stream exergy at the turbine inlet when the throttle valve is partially closed such that the pressure at the turbine inlet is 2 MPa. Compare the second-law efficiency of this system when the valve is partially open to when it is fully open. Take \(T_{0}=25^{\circ} \mathrm{C}\).

Short answer

Answer: Partially closing the throttle valve reduces the specific exergy at the turbine inlet from 967.91 kJ/kg to 393.81 kJ/kg, a decrease of 574.1 kJ/kg. The second-law efficiency of the system when the valve is partially open is 59.3%, compared to 100% when the valve is fully open.

Step-by-step solution

Determine the initial steam properties

Using steam tables, we find the specific enthalpy and entropy for the initial steam state of 6 MPa and 600°C: \(h_1 = 3625.3 \mathrm{\frac{kJ}{kg}}\) \(s_1 = 6.9028 \mathrm{\frac{kJ}{kg \cdot K}}\)

Determine the reference state properties

Using steam tables, we find the specific enthalpy and entropy for the reference state of 25°C: \(h_0 = 104.83 \mathrm{\frac{kJ}{kg}}\) \(s_0 = 0.3674 \mathrm{\frac{kJ}{kg \cdot K}}\)

Calculate the specific exergy for the initial state

Using the specific exergy formula, we compute the specific exergy for the initial steam state: \(e_1 = h_1 - h_0 - T_0(s_1 - s_0)\) \(e_1 = 3625.3 - 104.83 - (25 + 273.15)(6.9028 - 0.3674)\) \(e_1 = 967.91 \mathrm{\frac{kJ}{kg}}\)

Determine the properties for the partially closed valve state

Using steam tables, we find the specific enthalpy and entropy for the steam state after the valve is partially closed, with a pressure of 2 MPa: \(h_2 = 3051.2 \mathrm{\frac{kJ}{kg}}\) \(s_2 = 6.9028 \mathrm{\frac{kJ}{kg \cdot K}}\) (since the throttling process is isentropic)

Calculate the specific exergy for the partially closed valve state

Using the specific exergy formula, we compute the specific exergy for the partially closed valve steam state: \(e_2 = h_2 - h_0 - T_0(s_2 - s_0)\) \(e_2 = 3051.2 - 104.83 - (25 + 273.15)(6.9028 - 0.3674)\) \(e_2 = 393.81 \mathrm{\frac{kJ}{kg}}\)

Calculate the difference in specific exergies

Subtracting the specific exergy for the partially closed valve state from the initial state, we get: \(\Delta e = e_2 - e_1\) \(\Delta e = 393.81 - 967.91\) \(\Delta e = -574.1 \mathrm{\frac{kJ}{kg}}\)

Determine the second-law efficiency

The second-law efficiency, \(\eta_{II}\), is the ratio of actual work output (\(W_{actual}\)) to the ideal work output (\(W_{ideal}\)). Since the throttle valve is fully open in the initial state, we have no friction losses and the entropy is conserved. The actual work output is equal to the ideal work output, which is equal to the change in exergy: \(\eta_{II} = \frac{W_{actual}}{W_{ideal}} = \frac{-574.1}{967.91}\) \(\eta_{II} = 0.593\) The second-law efficiency of the system when the valve is partially open is 59.3%, compared to 100% when the valve is fully open. Closing the throttle valve reduces the available energy (exergy) of the system by decreasing the specific enthalpy of the steam while maintaining the same specific entropy, leading to lower efficiency in converting the exergy to work output.

Key concepts

Isentropic Process

An isentropic process is a theoretical or ideal process in which the entropy of the system remains constant. In other words, there is no entropy generation, and hence no heat transfer or irreversibilities present during the process. This concept plays a vital role in understanding thermodynamic cycles, particularly in the function of steam turbines, compressors, and nozzles. Isentropic processes are often used as a benchmark to compare real-life processes because they represent the most efficient transformation of energy. In the case of the throttle valve in our exercise, the isentropic assumption means that despite the reduction in pressure caused by the partially closed valve, the specific entropy of the steam remains unchanged, hence it retains the same disorder or randomness in the molecules at both the turbine inlet and the valve exit.

An isentropic process is depicted on the T-s diagram as a vertical line, indicating constant entropy. This characteristic allows us to evaluate the performance of devices like turbines, assuming they operate close to isentropic conditions. However, real processes always involve some irreversibilities, making them non-isentropic, which leads to less work extracted from a turbine, as demonstrated in the exercise. The throttling process, specifically, which is indeed an isentropic process, reduces the pressure without changing the entropy. However, it negatively impacts the exergy, which is a measure of the useful work potential of the system.

Second-Law Efficiency

The concept of second-law efficiency is crucial in thermodynamics to measure the performance of a system against the ideal, reversible operation dictated by the second law. It signifies the actual work output relative to the maximum possible work that could be achieved if the process were isentropic. Second-law efficiency is a more meaningful measure of efficiency than first-law efficiency since it considers the quality of energy, not just the quantity.

Mathematically, the second-law efficiency, often denoted as \(\eta_{II}\), is expressed as the ratio of the work output of an actual process (\(W_{actual}\)) to the work output that would be achieved by an ideal process (\(W_{ideal}\)) under the same constraints. For example, in our exercise, we witness a second-law efficiency of 59.3% when the valve is partially closed, signaling that the system has suffered a decline in its ability to convert heat into work due to the throttling. This is in contrast to the ideal case where the second-law efficiency can be 100%, meaning all the available energy could be converted into work. The drop in efficiency directly correlates with the loss of exergy due to the valve's partial closure.

Thermodynamic Properties

Thermodynamic properties such as enthalpy (\(h\)) and entropy (\(s\)) are intrinsic properties that describe the energy content and disorder within a system, respectively. They are pivotal in the analysis of thermodynamic processes and cycles. For instance, enthalpy represents the total energy of a system, which includes internal energy, pressure, and volume, while entropy measures the amount of energy in a system not available to do work.

In exergy analysis, these thermodynamic properties allow us to quantify the maximum useful work potential of a system. Exergy (\(e\)) itself is calculated using the formula \(e = h - h_0 - T_0(s - s_0)\), where \(h_0\) and \(s_0\) are the enthalpy and entropy at the reference or dead state, typically the environment, and \(T_0\) is the absolute environmental temperature. The reference state is a baseline to which the actual state of the system is compared. Understanding these properties, as presented in the steps of the exercise solution, is essential for performing an accurate exergy analysis. The reduction in enthalpy from the throttling valve results in a direct decrease in the exergy and consequently in the work output potential of the steam entering the turbine.