Question

Air at \(500 \mathrm{kPa}\) and \(400 \mathrm{K}\) enters an adiabatic nozzle at a velocity of \(30 \mathrm{m} / \mathrm{s}\) and leaves at \(300 \mathrm{kPa}\) and \(350 \mathrm{K}\) Using variable specific heats, determine ( \(a\) ) the isentropic efficiency, \((b)\) the exit velocity, and \((c)\) the entropy generation.

Short answer

Based on the given problem and its step-by-step solution, here is a short answer question: Q: An air is flowing through an adiabatic nozzle with specific heat at constant pressure (Cp) equals 1.005 kJ/kgK, specific gas constant (R) equals 0.287 kJ/kgK and it undergoes an isentropic process. If the initial temperature is 400 K, initial velocity is 30 m/s, final temperature is 350 K and process pressure changes from 500 kPa to 300 kPa. Calculate the isentropic efficiency, exit velocity, and entropy generation. A: (a) The isentropic efficiency is 0.429. (b) The exit velocity is 30 m/s. (c) The entropy generation is -0.067 kJ/kgK.

Step-by-step solution

Find actual exit enthalpy

To find the actual exit enthalpy, we can use the ideal gas equation and the variable specific heat relation: \(h_{2, actual} = h_1 + \frac{1}{2} v_1^2\) Since air can be assumed as an ideal gas, we can use \(h = C_p T\), where \(C_p\) is the specific heat at constant pressure. For given initial conditions, we have: \(h_1 = C_p T_1 = (1.005 \,\text{kJ/kgK})(400 \,\text{K}) = 402 \,\text{kJ/kg}\) And the exit enthalpy: \(h_{2, actual} = 402 \,\text{kJ/kg} + \frac{1}{2}(30 \,\text{m/s})^2 = 402 + 450 = 852 \,\text{kJ/kg}\)

Find ideal exit enthalpy

To find the ideal exit enthalpy, we can use the isentropic relation. For an isentropic process, \(\frac{h_2^\text{ideal}}{T_2} = \frac{h_1}{T_1}\) where \(h_2^\text{ideal}\) is the ideal exit enthalpy. We can now find the ideal exit enthalpy: \(h_2^\text{ideal} = \frac{h_1 T_2}{T_1} = \frac{402 \,\text{kJ/kg} \cdot 350}{400} = 367.05 \,\text{kJ/kg}\)

Calculate isentropic efficiency

The isentropic efficiency can be calculated as: \(\eta_{isentropic} = \frac{h_2^\text{ideal} - h_1}{h_{2, actual} - h_1}\) \(\eta_{isentropic} = \frac{367.05 - 402}{852 - 402} = 0.429\)

Calculate exit velocity

We can use the energy equation with the enthalpy difference to calculate the exit velocity: \(h_2^\text{actual} = h_1 + \frac{1}{2} v_2^2\) \(v_2 = \sqrt{2(h_2^\text{actual} - h_1)}\) \(v_2 = \sqrt{2(852 - 402)} = \sqrt{900} = 30 \,\text{m/s}\)

Calculate entropy generation

The entropy generation for the process can be calculated using the given temperatures and pressures: \(\Delta s = C_p \ln{\frac{T_2}{T_1}} - R \ln{\frac{P_2}{P_1}}\) where \(R = 0.287 \,\text{kJ/kgK}\) is the specific gas constant for air. Then, \(\Delta s = (1.005 \,\text{kJ/kgK})\ln{\frac{350 \,\text{K}}{400 \,\text{K}}} - (0.287 \,\text{kJ/kgK})\ln{\frac{300 \,\text{kPa}}{500 \,\text{kPa}}}\) \(\Delta s = -0.067 \,\text{kJ/kgK}\) The negative sign indicates that the entropy decreases, which is consistent with an adiabatic nozzle. Result: (a) The isentropic efficiency is 0.429. (b) The exit velocity is 30 m/s. (c) The entropy generation is -0.067 kJ/kgK.