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Question 50
Question Statement:
A centrifugal compressor requires 1800 kW of power to compress 10 kg/s of air entering at 300 K stagnation temperature. Taking specific heat $C_p = 1000$ J/(kg K), the stagnation temperature (in K) of air leaving the compressor is _________ K.
Official Answer Key:
480.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.50)**
**1. Intuitive Physical Setup & Given Parameters**:
- Power input: $W_{comp} = 1800\text{ kW} = 1.8 \times 10^6\text{ W}$
- Mass flow rate: $\dot{m} = 10\text{ kg/s}$
- Inlet stagnation temperature: $T_{01} = 300\text{ K}$
- Specific heat: $C_p = 1000\text{ J/(kg K)}$
**2. Step-by-Step Thermodynamic Derivation**:
1. **Steady Flow Energy Equation across Compressor**:
$W_{comp} = \dot{m} C_p (T_{02} - T_{01})$
2. **Solving for Stagnation Temperature Rise ($\Delta T_0$)**:
$\Delta T_0 = T_{02} - T_{01} = \frac{W_{comp}}{\dot{m} C_p} = \frac{1800 \times 10^3\text{ W}}{10\text{ kg/s} \times 1000\text{ J/(kg K)}} = \frac{1800000}{10000} = 180\text{ K}$
3. **Exit Stagnation Temperature ($T_{02}$)**:
$T_{02} = T_{01} + \Delta T_0 = 300\text{ K} + 180\text{ K} = 480\text{ K}$
The exit stagnation temperature is exactly **480 K**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **480.0 K** (Acceptable Range: [478, 482]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **480.0**
- **Key Formula**: $T_{02} = T_{01} + \frac{\dot{W}}{\dot{m} C_p}$.
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