Question

An insulated nozzle has an inlet cross-sectional area of 314 cm². Air flows through the nozzle with an inlet temperature of 300 K at a steady rate of 1.256 m³/s. The velocity at the exit is greater than that at the inlet by 210 m/s. Assume a constant \(C_p = 1.004\) kJ/kg-K. The temperature (in K) of air at the exit of the nozzle lies between

• A. 330 and 331
• B. 269 and 270
• C. 320 and 321
• D. 277 and 278

Hint:

In nozzle and diffuser problems, the change in kinetic energy is significant. Always remember to use the SFEE. Pay close attention to units, especially for energy (J vs kJ). Convert everything to base SI units before calculating to avoid errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves applying the Steady Flow Energy Equation (SFEE) to a nozzle. A nozzle is a device that increases the kinetic energy of a fluid at the expense of its internal energy and pressure. For an insulated nozzle with no work done, the SFEE simplifies to a balance between enthalpy and kinetic energy.
Step 2: Key Formula or Approach:
The Steady Flow Energy Equation for a nozzle is: \[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \] where \(h\) is the specific enthalpy and \(V\) is the velocity. Subscripts 1 and 2 refer to the inlet and exit, respectively.
Since the nozzle is insulated (\(Q=0\)) and does no work (\(W=0\)), these terms are omitted.
The change in enthalpy for an ideal gas is given by \(\Delta h = C_p \Delta T\), so \(h_2 - h_1 = C_p(T_2 - T_1)\).
Step 3: Detailed Explanation or Calculation:
Given values:
Inlet Area, \(A_1 = 314 \text{ cm}^2 = 314 \times 10^{-4} \text{ m}^2 = 0.0314 \text{ m}^2\).
Volume flow rate at inlet, \(\dot{v}_1 = 1.256 \text{ m}^3/\text{s}\).
Inlet temperature, \(T_1 = 300 \text{ K}\).
Velocity relation, \(V_2 = V_1 + 210 \text{ m/s}\).
Specific heat, \(C_p = 1.004 \text{ kJ/kg-K} = 1004 \text{ J/kg-K}\).
1. Calculate inlet velocity (\(V_1\)): \[ V_1 = \frac{\text{Volume flow rate}}{\text{Area}} = \frac{\dot{v}_1}{A_1} = \frac{1.256 \text{ m}^3/\text{s}}{0.0314 \text{ m}^2} = 40 \text{ m/s} \] 2. Calculate exit velocity (\(V_2\)): \[ V_2 = V_1 + 210 = 40 + 210 = 250 \text{ m/s} \] 3. Apply the SFEE to find the exit temperature (\(T_2\)): Rearranging the SFEE: \[ h_1 - h_2 = \frac{V_2^2 - V_1^2}{2} \] Substitute the enthalpy relation: \[ C_p(T_1 - T_2) = \frac{V_2^2 - V_1^2}{2} \] Now, solve for \(T_2\): \[ T_1 - T_2 = \frac{V_2^2 - V_1^2}{2 C_p} \] \[ T_2 = T_1 - \frac{V_2^2 - V_1^2}{2 C_p} \] Plugging in the values (ensure all units are in base SI: J, kg, m, s, K): \[ T_2 = 300 - \frac{(250)^2 - (40)^2}{2 \times 1004} \] \[ T_2 = 300 - \frac{62500 - 1600}{2008} \] \[ T_2 = 300 - \frac{60900}{2008} \] \[ T_2 = 300 - 30.328 \approx 269.67 \text{ K} \] Step 4: Final Answer:
The temperature of the air at the exit of the nozzle is approximately 269.67 K.
Step 5: Why This is Correct:
The calculated value of 269.67 K lies between 269 K and 270 K, which corresponds to option (B). The calculation correctly applies the principle of conservation of energy to the control volume of the nozzle.