The mass flow rate in a supersonic wind tunnel is 2 kg/s when the stagnation
pressure and stagnation temperature are 1 MPa and 800 K, respectively. If the
stagnation pressure and stagnation temperature are changed to 3 MPa and 200 K,
the mass flow rate in the tunnel changes to _______ kg/s (answer in integer).
Official Answer Key:
12.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.51)**
**1. Intuitive Physical Setup & Given Parameters**:
Choked mass flow in a supersonic wind tunnel nozzle throat:
- Initial state 1: $p_{01} = 1\text{ MPa}$, $T_{01} = 800\text{ K}$, mass flow rate $\dot{m}_1 = 2\text{ kg/s}$.
- New state 2: $p_{02} = 3\text{ MPa}$, $T_{02} = 200\text{ K}$.
We need to determine the new mass flow rate $\dot{m}_2$ in $\text{kg/s}$ (integer).
**2. Step-by-Step Compressible Flow Derivation**:
1. **Choked Mass Flow Rate Equation**:
For isentropic sonic flow through a fixed nozzle throat area $A^*$:
$\dot{m} = \frac{p_0 A^*}{\sqrt{T_0}} \sqrt{\frac{\gamma}{R}} \left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma + 1}{2(\gamma - 1)}}$
Since nozzle throat area $A^*$ and gas properties ($\gamma, R$) are constant:
$\dot{m} \propto \frac{p_0}{\sqrt{T_0}}$
2. **Mass Flow Scaling Ratio**:
$\frac{\dot{m}_2}{\dot{m}_1} = \left(\frac{p_{02}}{p_{01}}\right) \sqrt{\frac{T_{01}}{T_{02}}}$
3. **Substitute Given Values**:
$\frac{p_{02}}{p_{01}} = \frac{3\text{ MPa}}{1\text{ MPa}} = 3$
$\frac{T_{01}}{T_{02}} = \frac{800\text{ K}}{200\text{ K}} = 4 \implies \sqrt{\frac{T_{01}}{T_{02}}} = \sqrt{4} = 2$
$\frac{\dot{m}_2}{\dot{m}_1} = 3 \times 2 = 6$
4. **Compute $\dot{m}_2$**:
$\dot{m}_2 = 6 \times \dot{m}_1 = 6 \times 2\text{ kg/s} = 12\text{ kg/s}$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $\dot{m}_2 = 12\text{ kg/s}$.
- **Rounding Requirement**: Nearest integer $\implies \mathbf{12}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **12.0** (Acceptable Range: **12 to 12**)
- **Exam Takeaway**: Sonic throat mass flow scales as $\dot{m} \propto p_0 / \sqrt{T_0}$. Tripling pressure and quartering temperature multiplies mass flow by $3 \times \sqrt{4} = 6$.
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