Nozzle Exit Velocity Calculation

The problem requires calculating the exit velocity ($V_2$) of air flowing through a nozzle under specific conditions using thermodynamic principles.

Applying Steady Flow Energy Equation (SFEE)

For a steady-state, adiabatic process in a horizontal nozzle, assuming negligible potential energy changes and negligible inlet kinetic energy ($V_1 \approx 0$), the SFEE simplifies significantly. The equation per unit mass becomes:

$ h_1 + \frac{V_1^2}{2} \approx h_2 + \frac{V_2^2}{2} $

This simplifies to:

$ \frac{V_2^2}{2} \approx h_1 - h_2 $

Relating Enthalpy to Temperature

Air is treated as a perfect gas. For a perfect gas, the change in specific enthalpy ($h_1 - h_2$) is directly proportional to the temperature change ($T_1 - T_2$), given by:

$ h_1 - h_2 = C_p (T_1 - T_2) $

Substituting this into the simplified SFEE gives the formula for exit velocity:

$ V_2 = \sqrt{2 C_p (T_1 - T_2)} $

Numerical Calculation

Given the parameters:

• Specific heat at constant pressure, $ C_p = 1005 \text{ J/(kg K)} $
• Inlet temperature, $ T_1 = 300 \text{ K} $
• Exit temperature, $ T_2 = 296 \text{ K} $

First, find the temperature difference:

$ \Delta T = T_1 - T_2 = 300 \text{ K} - 296 \text{ K} = 4 \text{ K} $

Now, calculate the exit velocity $V_2$:

$ V_2 = \sqrt{2 \times 1005 \text{ J/(kg K)} \times 4 \text{ K}} $

$ V_2 = \sqrt{8040 \text{ J/kg}} $

$ V_2 \approx 89.67 \text{ m/s} $

The calculated velocity is approximately 89.67 m/s. The correct option is C.