Question

Air enters an adiabatic nozzle at 300 kPa and 500 K with a velocity of 10 m/s. It leaves the nozzle at 100 kPa with a velocity of 180 m/s. The specific heat of air \(C_{\text{p}} = 1008\text{ J/kg}\cdot\text{K}\). The exit temperature of air is

• A. 516 K
• B. 532 K
• C. 484 K
• D. 468 K

Hint:

Always ensure units are consistent.
Velocity squared ($V^2$) yields units of $\text{m}^2/\text{s}^2$, which is equivalent to $\text{J/kg}$.
Using $C_{\text{p}}$ in $\text{J/kg}\cdot\text{K}$ (rather than $\text{kJ/kg}\cdot\text{K}$) ensures correct dimensional consistency without conversion errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The goal is to calculate the exit temperature of air passing through an adiabatic nozzle, given the inlet temperature, inlet velocity, and exit velocity, along with the specific heat of the air.

Step 2: Key Formula or Approach:
For an adiabatic, steady-flow nozzle with no work transfer and negligible potential energy changes, the Steady Flow Energy Equation (SFEE) simplifies to:
\[ h_{1} + \frac{V_{1}^{2}}{2} = h_{2} + \frac{V_{2}^{2}}{2} \] For an ideal gas like air, the enthalpy change can be expressed in terms of specific heat at constant pressure (\(C_{\text{p}}\)):
\[ C_{\text{p}} T_{1} + \frac{V_{1}^{2}}{2} = C_{\text{p}} T_{2} + \frac{V_{2}^{2}}{2} \] Rearranging to solve for the exit temperature \(T_{2}\):
\[ T_{2} = T_{1} - \frac{V_{2}^{2} - V_{1}^{2}}{2 C_{\text{p}}} \]

Step 3: Detailed Explanation:

• Identify the given parameters from the problem statement:
Inlet temperature, \(T_{1} = 500\text{ K}\).
Inlet velocity, \(V_{1} = 10\text{ m/s}\).
Exit velocity, \(V_{2} = 180\text{ m/s}\).
Specific heat, \(C_{\text{p}} = 1008\text{ J/kg}\cdot\text{K}\).

• Calculate the kinetic energy change term:
\[ V_{2}^{2} - V_{1}^{2} = 180^{2} - 10^{2} = 32400 - 100 = 32300\text{ m}^{2}/\text{s}^{2} \]
• Substitute these values into the rearranged SFEE:
\[ T_{2} = 500 - \frac{32300}{2 \times 1008} \] \[ T_{2} = 500 - \frac{32300}{2016} \] \[ T_{2} \approx 500 - 16.02 = 483.98\text{ K} \] Rounding to the nearest whole number yields \(484\text{ K}\).

Step 4: Final Answer:
The exit temperature of the air is \(484\text{ K}\).