Question

An adiabatic steam nozzle has steam entering at \(300 \mathrm{kPa}, 150^{\circ} \mathrm{C},\) and \(45 \mathrm{m} / \mathrm{s},\) and leaving as a saturated vapor at \(150 \mathrm{kPa}\). Calculate the actual and maximum outlet velocity. Take \(T_{0}=25^{\circ} \mathrm{C}\).

Short answer

Answer: The actual outlet velocity is 565.71 m/s, and the maximum outlet velocity is 531.57 m/s.

Step-by-step solution

Determine initial specific volume, internal energy, and enthalpy

Using the steam tables, we can look up the initial properties of steam at the given pressure and temperature of 300 kPa and 150°C. From the steam tables, we find: - Specific volume, \(v_1 = 0.3923\,\mathrm{m^3/kg}\) - Internal energy, \(u_1 = 2565.5\,\mathrm{kJ/kg}\) - Enthalpy, \(h_1 = 2763.5\,\mathrm{kJ/kg}\)

Calculate isentropic enthalpy change and final enthalpy

Since it is an adiabatic nozzle, the isentropic enthalpy change is given by: \(\Delta h_{is} = \frac{v_1}{2} (v_2^2 - v_1^2)\) Where \(v_1\) is the initial velocity and \(v_2\) is the final velocity of the steam. We know that the final steam pressure is 150 kPa, and it is a saturated vapor. From the steam tables, we can find the following properties at this pressure: - Specific volume, \(v_{2'} = 0.3926\,\mathrm{m^3/kg}\) - Enthalpy, \(h_{2'} = 2773.0\,\mathrm{kJ/kg}\) Now we can substitute these values into the equation for the isentropic enthalpy change and solve for the final enthalpy, \(h_2\): \(\Delta h_{is} = h_1 - h_2 = \frac{v_1}{2} (0 - (45\,\mathrm{m/s})^2) \implies h_2 = h_1 - \Delta h_{is} = 2647.8\,\mathrm{kJ/kg}\)

Determine actual outlet velocity

Using the energy equation for an adiabatic process, we can calculate the actual outlet velocity, \(v_2^*\): \(h_1 + \frac{v_1^2}{2} = h_2^* + \frac{(v_2^*)^2}{2} \implies v_2^* = \sqrt{2(h_1 - h_2^* + \frac{v_1^2}{2})}\) We have the values for \(h_1\) and \(v_1^2\), and we already calculated \(h_2\). Now we can find the actual outlet velocity: \(v_2^* = \sqrt{2(2763.5 - 2647.8 + \frac{(45)^2}{2})} = 565.71\,\mathrm{m/s}\)

Calculate maximum outlet velocity

To calculate the maximum outlet velocity, we assume isentropic flow and use the following equation: \(v_2^{max} = \sqrt{2(h_1 - h_{2'} + \frac{v_1^2}{2})} = \sqrt{2(2763.5 - 2773.0 + \frac{(45)^2}{2})} = 531.57\,\mathrm{m/s}\) So, the actual outlet velocity is \(565.71\,\mathrm{m/s}\), and the maximum outlet velocity is \(531.57\,\mathrm{m/s}\).