Question

A stream of superheated steam (2 MPa, 300°C) mixes with another stream of superheated steam (2 MPa, 400°C) through a steady-state adiabatic process. The flow rates of the streams are 3 kg/min and 2 kg/min, respectively. This mixture then expands in an adiabatic nozzle to a saturated mixture with quality of 0.77 and 1 kPa. Neglect the velocity at the nozzle entrance and the change in potential energies. The velocity at the nozzle exit (in m/s) is ......... (rounded off to two decimal places). 
Use the following data:
At 2 MPa, 300 °C: Specific enthalpy of superheated steam = 3024.2 kJ/kg
At 2 MPa, 400 °C: Specific enthalpy of superheated steam = 3248.4 kJ/kg
At 1 kPa: Specific enthalpy of saturated water = 29.3 kJ/kg
At 1 kPa: Specific enthalpy of saturated vapour = 2513.7 kJ/kg
 

Hint:

For steady-state adiabatic processes, apply the energy balance equation considering enthalpy and kinetic energy. The velocity at the nozzle exit can be found from the difference in enthalpy.

Answer 1

We are given:
- Flow rates of steam streams are \( 3 \, {kg/min} \) and \( 2 \, {kg/min} \),
- Initial conditions: stream 1: \( 2 \, {MPa}, 300^\circ C \), stream 2: \( 2 \, {MPa}, 400^\circ C \),
- After mixing, the steam is expanded in an adiabatic nozzle to a saturated mixture with quality \( x = 0.77 \) and pressure \( P = 1 \, {kPa} \).
Step 1: The specific enthalpy of the steam streams can be found from steam tables:
- \( h_1 = 3024.2 \, {kJ/kg} \) at \( 2 \, {MPa}, 300^\circ C \),
- \( h_2 = 3248.4 \, {kJ/kg} \) at \( 2 \, {MPa}, 400^\circ C \),
- Saturated steam at 1 kPa has \( h_{{sat}} = 29.3 \, {kJ/kg} \) for saturated liquid and \( h_{{fg}} = 2513.7 \, {kJ/kg} \) for the vapor.
Step 2: The total mass flow rate is \( 3 + 2 = 5 \, {kg/min} \). To calculate the velocity at the nozzle exit, use the energy balance across the nozzle. After mixing, the total enthalpy is: \[ h_{{mix}} = \frac{(3 \times 3024.2) + (2 \times 3248.4)}{5} = 3124.8 \, {kJ/kg} \] Step 3: The specific enthalpy of the saturated mixture at 1 kPa is: \[ h_{{mix, final}} = h_{{sat}} + x \cdot h_{{fg}} = 29.3 + 0.77 \cdot 2513.7 = 1954.6 \, {kJ/kg} \] Step 4: The kinetic energy change is calculated using the energy balance: \[ \frac{v^2}{2} = h_{{mix}} - h_{{mix, final}} = 3124.8 - 1954.6 = 1170.2 \, {kJ/kg} \] Solving for velocity \( v \): \[ v = \sqrt{2 \times 1170.2 \times 1000} = 1545.6 \, {m/s} \] Step 5: Therefore, the velocity at the nozzle exit is approximately 1545.6 m/s, which is closest to 1515 to 1545 m/s.