Question

Air is expanded in an adiabatic turbine of 90 percent isentropic efficiency from an inlet state of \(2800 \mathrm{kPa}\) and \(400^{\circ} \mathrm{C}\) to an outlet pressure of \(150 \mathrm{kPa}\). Calculate the outlet temperature of air, the work produced by this turbine, and the entropy generation.

Short answer

Calculating the isentropic exit temperature, we get: $$T_2s \approx 441.87 \mathrm{K} = 168.72^{\circ} \mathrm{C}$$ #tag_title#Step 2: Calculate the actual exit temperature#tag_content# Using the isentropic efficiency equation: $$\eta_s = \dfrac{c_p(T_1 - T_2)}{c_p(T_1 - T_{2s})}$$ Since the specific heat capacity \(c_p\) appears in both numerator and denominator, it cancels out. We can rearrange the equation to solve for the actual exit temperature \(T_2\): $$T_2 = T_1 - \dfrac{\eta_s (T_1 - T_{2s})}{1}$$ Given the isentropic efficiency \(\eta_s = 0.9\), we can plug in the values: $$T_2 = 673.15 - 0.9(673.15 - 441.87)$$ Calculating the actual exit temperature, we get: $$T_2 \approx 458.18 \mathrm{K} = 185.03^{\circ} \mathrm{C}$$ #tag_title#Step 3: Calculate the work output of the turbine#tag_content# We will use the work output equation to calculate the work produced by the turbine: $$W_t = m(h_1 - h_2)$$ Since for an ideal gas, \(h = c_p T\), the equation becomes: $$W_t = m c_p (T_1 - T_2)$$ We will use a value of \(c_p = 1.005 \mathrm{kJ/kg \cdot K}\) for air. Assuming a mass flow rate of 1 kg/s, we can plug in the values: $$W_t = 1 \cdot 1.005 (673.15 - 458.18)$$ Calculating the work produced by the turbine, we get: $$W_t \approx 215.97 \mathrm{kJ/kg}$$ #tag_title#Step 4: Calculate the entropy generation#tag_content# We will use the entropy generation equation to calculate the entropy generation: $$\Delta s = c_p \ln{\dfrac{T_2}{T_1}} - R \ln{\dfrac{P_2}{P_1}}$$ For air, we will use a value of \(R = 0.287 \mathrm{kJ/kg \cdot K}\). Plugging in the values: $$\Delta s = 1.005 \ln{\dfrac{458.18}{673.15}} - 0.287 \ln{\dfrac{150}{2800}}$$ Calculating the entropy generation, we get: $$\Delta s \approx 0.0506 \mathrm{kJ/kg \cdot K}$$ #Summary# The outlet temperature is approximately 185.03°C, the work produced by the turbine is approximately 215.97 kJ/kg, and the entropy generation is approximately 0.0506 kJ/kg·K.

Step-by-step solution

Calculate the isentropic exit temperature

We will use the isentropic relation for an ideal gas undergoing an adiabatic process to calculate the isentropic exit temperature \(T_2s\): $$T_2s = T_1 \left(\dfrac{P_2}{P_1}\right)^{(\gamma - 1)/\gamma}$$ Here, \(T_1 = 400^{\circ} \mathrm{C} = 673.15 \mathrm{K}\) and \(\gamma = 1.4\) for air (ideal gas). We can plug in the values: $$T_2s = 673.15 \left(\dfrac{150}{2800}\right)^{(1.4 - 1)/1.4}$$

Key concepts

Isentropic Process

In thermodynamics, an isentropic process is an idealized process that is both adiabatic and reversible, meaning there is no transfer of heat or matter into or out of the system, and it can be reversed without leaving any net change in the environment. An important characteristic of an isentropic process is that the entropy of the system remains constant throughout the entire process.

Regarding turbines, if they operated under genuinely isentropic conditions, there would be no entropy generation and the output work would be maximized. However, in reality, all processes involve some irreversibilities which lead to entropy generation and a decrease in work output. In our exercise, the turbine is 90 percent isentropic, signifying that it’s very efficient, yet not perfectly isentropic, thus, entropy will indeed be generated.

Ideal Gas Law

The ideal gas law is a fundamental relation between pressure (P), volume (V), temperature (T), and the number of moles (n) involving a gas, captured in the equation: \[PV = nRT\]where R is the universal gas constant. This equation is based on the assumption that gases consist of a large number of particles moving randomly, and the size of these particles is negligible compared to the distance between them.

When solving problems involving turbines and other thermodynamic systems, using the ideal gas law simplifies the calculations by treating the working fluid (in our case, air) as an ideal gas, ignoring the effects of intermolecular forces and the volume occupied by the gas particles themselves.

Entropy Generation

Entropy generation is a concept within the second law of thermodynamics that measures the irreversibility of a process. It signifies the amount of energy that has been dissipated, often as heat, and which cannot be converted into work. Entropy generation is always greater than or equal to zero, which aligns with the principle that total entropy, which is a measure of disorder, never decreases in an isolated system.

In practical systems like turbines, entropy is generated due to factors such as friction, turbulence, and other non-ideal effects. Recognizing and minimizing entropy generation is key to enhancing the efficiency of thermodynamic processes. The calculation of entropy generation is crucial in assessing the performance of the turbine in our exercise, as it quantifies the deviation from the ideal isentropic process.

Thermodynamic Work

Thermodynamic work in the context of turbines refers to the energy transferred from the system (the working fluid passing through the turbine) to its surroundings when the system expands against an external pressure. It’s a measure of the useful energy output that we can harness from the thermodynamic process. Work is represented by the symbol W and is most commonly measured in joules (J).

In an adiabatic turbine like the one described in the exercise, work is produced by the expansion of air. Maximizing this work output while minimizing energy loss is central to turbine efficiency. The work produced by the turbine is directly related to the temperature and pressure conditions at the inlet and outlet, and the specific characteristics of the working fluid. Using concepts such as the ideal gas law and principles of isentropic processes allows us to estimate the work output from these turbines quite accurately.