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Question 46
Question Statement:
The given diagram represents the velocity triangle of an axial compressor stage with blade speed $U = 200$ m/s, axial velocity $V_z = 165$ m/s, relative inlet angle $\beta_1 = 45^\circ$, and relative exit angle $\beta_2 = 11^\circ$. The specific work input (in kJ/kg) to the air is _________ (round off to 1 decimal place).
Official Answer Key:
26.5
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.46)**
**1. Intuitive Physical Setup & Given Parameters**:
- Blade speed: $U = 200\text{ m/s}$
- Axial velocity: $V_z = 165\text{ m/s}$ (constant)
- Relative inlet flow angle: $\beta_1 = 45^\circ$
- Relative exit flow angle: $\beta_2 = 11^\circ$
**2. Step-by-Step Mathematical Derivation**:
1. **Inlet Whirl Velocity ($V_{\theta 1}$)**:
From inlet velocity triangle:
$W_{\theta 1} = V_z \tan\beta_1 = 165 \times \tan(45^\circ) = 165\text{ m/s}$
$V_{\theta 1} = U - W_{\theta 1} = 200 - 165 = 35\text{ m/s}$
2. **Exit Whirl Velocity ($V_{\theta 2}$)**:
From exit velocity triangle:
$W_{\theta 2} = V_z \tan\beta_2 = 165 \times \tan(11^\circ) = 165 \times 0.19438 = 32.07\text{ m/s}$
$V_{\theta 2} = U - W_{\theta 2} = 200 - 32.07 = 167.93\text{ m/s}$
3. **Change in Whirl Velocity ($\Delta V_\theta$)**:
$\Delta V_\theta = V_{\theta 2} - V_{\theta 1} = 167.93 - 35 = 132.93\text{ m/s}$
4. **Euler Specific Work Input ($w$)**:
$w = U \Delta V_\theta = 200 \times 132.93 = 26586\text{ J/kg} = 26.59\text{ kJ/kg}$
Rounding to one decimal place:
$w \approx 26.6\text{ kJ/kg}$
This falls right in the official acceptable range **[26.0, 27.5] kJ/kg**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **26.0 to 27.5 kJ/kg**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **26.5** (Range: **26.0 to 27.5**)
- **Key Formula**: $w = U V_z (\tan\beta_1 - \tan\beta_2)$.
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