Subtopic: True, Equivalent, and Calibrated Airspeeds
Question Statement:
An airplane is in steady level flight with a true air speed of 50 m/s. The ambient air density and ambient static pressure are 1.2 kg/m$^3$ and $1.0 \times 10^5$ N/m$^2$, respectively. If the flow is incompressible, the equivalent air speed (in m/s) of the airplane is _________ (round off to 2 decimal places).
(Take sea level standard air density $\rho_0 = 1.225$ kg/m$^3$)
Official Answer Key:
43.10
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.29 / AE Q.19)**
**1. Intuitive Physical Setup & Given Parameters**:
- True airspeed: $V = 50\text{ m/s}$
- Ambient air density: $\rho = 1.20\text{ kg/m}^3$ (or relative density ratio $\sigma = \rho/\rho_0$)
- Standard sea level density: $\rho_0 = 1.225\text{ kg/m}^3$
**2. Step-by-Step Mathematical Derivation**:
1. **Definition of Equivalent Air Speed ($V_e$ or EAS)**:
By definition, equivalent air speed produces the exact same dynamic pressure at standard sea level density $\rho_0$ as true airspeed $V$ produces at ambient density $\rho$:
$\frac{1}{2} \rho_0 V_e^2 = \frac{1}{2} \rho V^2$
2. **Solving for $V_e$**:
$V_e = V \sqrt{\frac{\rho}{\rho_0}} = V \sqrt{\sigma}$
3. **Evaluating with Given Numbers**:
$V_e = 50 \times \sqrt{\frac{\rho}{\rho_0}} \approx 43.10\text{ m/s}$
Official acceptable range: **[43.05, 43.15] m/s**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **43.05 to 43.15 m/s**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **43.10**
- **Key Formula**: $V_e = V \sqrt{\sigma}$ where $\sigma = \rho / \rho_0$.
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