GATE AE 2024 • Q16 MCQ • 1 Mark Propulsion

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Turbojet, Turbofan & Turboprop Engine Performance

Question Statement:

For an ideal gas, the specific heat at constant pressure is 1147 J/kg K and the ratio of specific heats is equal to 1.33. What is the value of the gas constant for this gas in J/kg K?

Options:

  • (A) 284.6
  • (B) 1005
  • (C) 862.4
  • (D) 8314

Official Answer Key:

A

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2024 Q.16)** **1. Given Ideal Gas Thermodynamic Parameters**: - Specific heat at constant pressure: $c_p = 1147\text{ J/(kg}\cdot\text{K)}$ - Specific heat ratio: $\gamma = 1.33$ We need to find the specific gas constant $R$. **2. Thermodynamic Derivation**: By definition of the ratio of specific heats: $$\gamma = \frac{c_p}{c_v} \implies c_v = \frac{c_p}{\gamma}$$ Using Mayer's relation for an ideal gas: $$R = c_p - c_v = c_p - \frac{c_p}{\gamma} = c_p \left(1 - \frac{1}{\gamma}\right) = c_p \left(\frac{\gamma - 1}{\gamma}\right)$$ **3. Numerical Evaluation**: $$R = 1147 \times \left(\frac{1.33 - 1}{1.33}\right) = 1147 \times \frac{0.33}{1.33} = \frac{378.51}{1.33} \approx 284.594\text{ J/(kg}\cdot\text{K)}$$ Rounding off gives $284.6\text{ J/(kg}\cdot\text{K)}$. **4. Official Master Key**: - **Official Answer Key**: **A** (284.6) - **Exam Takeaway**: Mayer's relation links $c_p$ and $\gamma$ directly to $R$: $R = c_p \frac{\gamma - 1}{\gamma}$.
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