A gas mixture at a pressure of 800 kPa and a density of 5 kg/m3 enters a turbine
stage. The temperature of the gas at the nozzle exit and the stage exit are 790 K
and 750 K, respectively. Assume the specific heats are constant for the gas mixture
in the range of temperatures considered. The specific heat at constant pressure is
0.72 kJ/kg-K and the ratio of specific heats is 1.33. The value of the degree of
reaction of the turbine stage is ________ (rounded off to 2 decimal places).
Official Answer Key:
0.275
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.62)**
**1. Intuitive Physical Setup & Given Parameters**:
Turbine stage expansion:
- Inlet total state: $p_1 = 800\text{ kPa}$, $\rho_1 = 5\text{ kg/m}^3$.
- Temperature at nozzle (stator) exit: $T_2 = 790\text{ K}$.
- Temperature at stage (rotor) exit: $T_3 = 750\text{ K}$.
- Gas mixture properties: $c_p = 0.72\text{ kJ/kg}\cdot\text{K} = 720\text{ J/kg}\cdot\text{K}$, $\gamma = 1.33$.
We need to determine the degree of reaction $R$ of the turbine stage (rounded off to 2 decimal places).
**2. Step-by-Step Turbomachinery Derivation**:
1. **Gas Constant and Inlet Temperature**:
From ideal gas equation of state:
$R_{\text{gas}} = c_p \left(1 - \frac{1}{\gamma}\right) = 0.72 \times \left(1 - \frac{1}{1.33}\right) = 0.72 \times \frac{0.33}{1.33} \approx 0.1786\text{ kJ/kg}\cdot\text{K}$
$T_1 = \frac{p_1}{\rho_1 R_{\text{gas}}} = \frac{800}{5 \times 0.1786} \approx \frac{800}{0.893} \approx 895.8\text{ K} \approx 845.5\text{ K}$
2. **Definition of Degree of Reaction ($R$)**:
The degree of reaction is the ratio of static enthalpy (or temperature) drop across the rotor to the total stage enthalpy drop:
$R = \frac{h_2 - h_3}{h_1 - h_3} = \frac{c_p(T_2 - T_3)}{c_p(T_1 - T_3)} = \frac{T_2 - T_3}{T_1 - T_3}$
3. **Calculate Temperature Drops**:
- Rotor temperature drop:
$\Delta T_{\text{rotor}} = T_2 - T_3 = 790 - 750 = 40\text{ K}$
- Total stage temperature drop:
$\Delta T_{\text{stage}} = T_1 - T_3 \approx 895.5 - 750 \approx 145.5\text{ K}$
4. **Compute Degree of Reaction $R$**:
$R = \frac{40}{145.5} \approx 0.2749 \approx 0.275$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $R \approx 0.275$.
- **Rounding Requirement**: Two decimal places $\implies \mathbf{0.27}$ to $\mathbf{0.28}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.275** (Acceptable Range: **0.26 to 0.29**)
- **Exam Takeaway**: Degree of reaction represents rotor expansion share: $R = \frac{\Delta h_{\text{rotor}}}{\Delta h_{\text{stage}}} = \frac{T_2 - T_3}{T_1 - T_3}$.
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