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). Data given: - At 2 MPa, 300 °C: \(h = 3024.2 \, \text{kJ/kg}\) - At 2 MPa, 400 °C: \(h = 3248.4 \, \text{kJ/kg}\) - At 1 kPa: \(h_f = 29.3 \, \text{kJ/kg}, h_{fg} = 2513.7 \, \text{kJ/kg}\)

Hint:

For nozzle problems, always apply steady-flow energy equation: \[ h_{in} + \frac{V_{in}^2}{2} = h_{out} + \frac{V_{out}^2}{2} \] and neglect velocity at inlet when stated.

Answer 1

Correct Answer: 1515

Step 1: Enthalpy after mixing.
Mass flow rates: \[ \dot{m}_1 = 3 \, kg/min = 0.05 \, kg/s, \dot{m}_2 = 2 \, kg/min = 0.0333 \, kg/s \] Weighted average enthalpy: \[ h_{mix} = \frac{\dot{m}_1 h_1 + \dot{m}_2 h_2}{\dot{m}_1 + \dot{m}_2} \] \[ h_{mix} = \frac{(3)(3024.2) + (2)(3248.4)}{5} = \frac{9072.6 + 6496.8}{5} = 3113.9 \, kJ/kg \]

Step 2: Exit enthalpy at 1 kPa with quality 0.77.
\[ h_{exit} = h_f + x h_{fg} = 29.3 + (0.77)(2513.7) = 29.3 + 1935.55 = 1964.9 \, kJ/kg \]

Step 3: Nozzle energy balance.
\[ h_{mix} - h_{exit} = \frac{V^2}{2 \cdot 1000} \] (1000 factor converts kJ/kg → kJ/kg kinetic energy). \[ 3113.9 - 1964.9 = \frac{V^2}{2000} \] \[ 1149 = \frac{V^2}{2000} \] \[ V^2 = 2298000 \Rightarrow V = 1516 \, m/s \] Final Answer:
\[ \boxed{1516.00 \, m/s} \]