The relative velocity of air leaving a straight radial impeller of a centrifugal compressor is 100 m/s. If the impeller tip speed is 200 m/s, for a slip free operation, the absolute velocity (in m/s) of air at the impeller exit is _________ (accurate to one decimal place).
Official Answer Key:
223.5
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.64 / AE Q.54)**
**1. Intuitive Physical Setup & Given Parameters**:
- Straight radial impeller blades: the blade angle at exit is $\beta_2 = 90^\circ$ (purely radial relative exit velocity).
- Relative velocity at exit: $W_2 = 100\text{ m/s}$ (in the radial direction, $C_{r2} = W_2 = 100\text{ m/s}$).
- Impeller tip speed: $U_2 = 200\text{ m/s}$.
- Slip-free operation: Slip factor $\sigma = 1.0 \implies C_{w2} = U_2 = 200\text{ m/s}$.
**2. Step-by-Step Mathematical Derivation**:
1. **Exit Velocity Components**:
- Radial component of absolute velocity:
$C_{r2} = W_2 = 100\text{ m/s}$
- Tangential (whirl) component of absolute velocity:
$C_{w2} = U_2 = 200\text{ m/s}$
2. **Resultant Absolute Exit Velocity ($C_2$)**:
Since the radial and tangential components are mutually perpendicular:
$C_2 = \sqrt{C_{r2}^2 + C_{w2}^2} = \sqrt{(100)^2 + (200)^2} = \sqrt{10000 + 40000} = \sqrt{50000}$
$C_2 = 100 \sqrt{5} \approx 100 \times 2.236068 = 223.61\text{ m/s}$
Rounding to one decimal place:
$C_2 \approx 223.6\text{ m/s}$
This falls perfectly in the center of the official acceptable interval **[222.0, 225.0]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **222.0 to 225.0 m/s**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **223.5** (Acceptable Range: **222.0 to 225.0**)
- **Key Formula**: For straight radial vanes with zero slip: $C_2 = \sqrt{W_2^2 + U_2^2} = \sqrt{100^2 + 200^2} \approx 223.6\text{ m/s}$.
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