Question

Steam enters an adiabatic nozzle at \(2 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\) with a velocity of \(55 \mathrm{m} / \mathrm{s}\) and exits at \(0.8 \mathrm{MPa}\) and \(390 \mathrm{m} / \mathrm{s}\). If the nozzle has an inlet area of \(7.5 \mathrm{cm}^{2},\) determine (a) the exit temperature and (b) the rate of entropy generation for this process.

Short answer

Answer: The exit temperature is \(194.47^{\circ}\text{C}\), and the rate of entropy generation is \(0.02388 \, \text{kJ/s-K}\).

Step-by-step solution

List down the known parameters and equations

We are given the following data: 1. The initial pressure, \(P_1 = 2 \, \text{MPa}\) 2. The initial temperature, \(T_1 = 350^{\circ}\text{C}\) 3. The initial velocity, \(v_1 = 55 \, \text{m/s}\) 4. The final pressure, \(P_2 = 0.8 \, \text{MPa}\) 5. The final velocity, \(v_2 = 390 \, \text{m/s}\) 6. The inlet area of the nozzle, \(A_1 = 7.5 \, \text{cm}^2 = 0.00075 \, \text{m}^2\) We will use the expressions for the adiabatic process: 1. Continuity equation: \(A_1v_1 = A_2v_2\) 2. Energy equation: \(h_1+\frac{1}{2}(v_1^2)=h_2+\frac{1}{2}(v_2^2)\) 3. Entropy equation: \(s_2-s_1 = \int_1^2 R \frac{dT}{T}\)

Determine the initial enthalpy and entropy

Refer to steam tables for pressure \(P_1 = 2 \, \text{MPa}\) and temperature \(T_1 = 350^{\circ}\text{C}\) condition. We will obtain: 1. Initial enthalpy, \(h_1 = 3176.2 \, \text{kJ/kg}\) 2. Initial entropy, \(s_1 = 6.7806 \, \text{kJ/(kg }\cdot {\text{K})}\) 3. Gas constant, \(R = 0.4615 \, \text{kJ/(kg }\cdot {\text{K})}\)

Calculate the exit enthalpy and determine the exit temperature

We will use the energy equation to find the exit enthalpy \(h_2\): \(h_1+\frac{1}{2}(v_1^2)=h_2+\frac{1}{2}(v_2^2)\) \(h_2 = h_1+\frac{1}{2}(v_1^2)-\frac{1}{2}(v_2^2)\) Substituting the known values in the equation, we get: \(h_2 = 3176.2 + \frac{1}{2}(55^2) - \frac{1}{2}(390^2) \, \text{kJ/kg} = 2514.233 \, \text{kJ/kg}\) Now, we will use the steam table for the given pressure \(P_2 = 0.8 \, \text{MPa}\) and find the corresponding temperature for the exit enthalpy \(h_2 = 2514.233 \, \text{kJ/kg}\). After interpolation in the table, we find the exit temperature \(T_2 = 194.47^{\circ}\text{C}\). Answer (a): The exit temperature is \(194.47^{\circ}\text{C}\).

Calculate the exit entropy and find the rate of entropy generation

From the entropy equation, we have: \(\Delta s = s_2 - s_1 = \int_1^2 R \frac{dT}{T}\) Assuming the gas constant \(R\) is constant during the process, we can integrate to obtain: \(s_2 = s_1 + R \, \text{ln}\frac{T_2}{T_1}\) Now, substituting the known values, we get: \(s_2 = 6.7806 + 0.4615 \, \text{ln}\frac{194.47 + 273}{350 + 273} = 6.84 \, \text{kJ/(kg }\cdot {\text{K})}\) As this is an adiabatic process, the rate of entropy generation \(\dot{S}_{gen}\) can be calculated using the mass flow rate \(\dot{m}\) and the difference in entropy, \(\Delta s = s_2 - s_1\). First, we need to find the mass flow rate, which can be obtained from the continuity equation: \(\dot{m}=\frac{A_1v_1}{v_{1s}}\) Refer to the steam tables, we find the specific volume \(v_{1s}\) at \(2\, \text{MPa}\) and \(350^{\circ}\text{C}\): \(v_{1s} = 0.1024 \, \text{m}^3/\text{kg}\) Now, we can calculate the mass flow rate: \(\dot{m} = \frac{0.00075 \cdot 55}{0.1024}=0.402 \, \text{kg/s}\) Finally, we can calculate the rate of entropy generation: \(\dot{S}_{gen} = \dot{m} \Delta s = 0.402 (6.84 - 6.7806) = 0.02388 \, \text{kJ/s-K}\) Answer (b): The rate of entropy generation is \(0.02388 \, \text{kJ/s-K}\).