GATE AE 2022 • Q65 NAT • 2 Marks Propulsion

Propulsion

Compressors and Turbines

Subtopic: Axial Flow Compressor Stage Dynamics & Stall

Question Statement:

A multistage axial compressor takes in air at $1\text{ atm}, 300\text{ K}$ and compresses it to a minimum of $5\text{ atm}$. The mean blade speed is $245\text{ m/s}$ and work coefficient, $\frac{\Delta C_\theta}{U}$ is $0.55$ for each stage. For air, use $C_p = 1005\text{ J/(kg}\cdot\text{K)}, R = 287\text{ J/(kg}\cdot\text{K)}$ and $\gamma = 1.4$. If the compression is isentropic, the number of stages required is ______ (rounded off to the next higher integer).
Question figure for GATE AE 2022 Q65

Scientific schematic diagram reproduced from official IIT paper

Official Answer Key:

66.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2022 Q.65)** **1. Overall Isentropic Stagnation Temperature Rise**: $$T_{0, \text{exit}} = T_{0, \text{in}} \left( \frac{p_{02}}{p_{01}} \right)^{\frac{\gamma - 1}{\gamma}} = 300 \times (5)^{\frac{0.4}{1.4}} = 300 \times (5)^{0.2857} \approx 300 \times 1.5838 = 475.14\text{ K}$$ $$\Delta T_{0, \text{overall}} = 475.14 - 300 = 175.14\text{ K}$$ **2. Stage Temperature Rise**: $$\Delta h_{0, \text{stage}} = U \Delta C_\theta = \left(\frac{\Delta C_\theta}{U}\right) U^2 = 0.55 \times (245)^2 = 0.55 \times 60025 = 33013.75\text{ J/kg}$$ $$\Delta T_{0, \text{stage}} = \frac{\Delta h_{0, \text{stage}}}{C_p} = \frac{33013.75}{1005} \approx 32.85\text{ K}$$ **3. Number of Stages**: $$N = \frac{\Delta T_{0, \text{overall}}}{\Delta T_{0, \text{stage}}} = \frac{175.14}{32.85} \approx 5.33$$ Rounding up to the next higher integer gives $6$ stages (or calculated per stage pressure ratio limits). - **Official Answer Key**: **6**
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