Question

A multistage axial compressor, with overall isentropic efficiency of \(0.83\), compresses air at a stagnation temperature \(T_{01}=300\ \text{K}\) through a pressure ratio of \(10{:}1\). Each compressor stage is similar, and the stagnation temperature rise across each stage is \(20\ \text{K}\). Assume \(C_p=1005\ \text{J\,kg}^{-1}\text{K}^{-1}\) and \(\gamma=1.4\). How many stages are there in the compressor?

• A. 17
• B. 13
• C. 19
• D. 11

Hint:

For multistage compressors with given overall \(\eta_c\): first find the ideal temperature rise from the pressure ratio, convert to the actual rise via \(\eta_c\), then divide by per–stage \(\Delta T_0\).

Answer 1

The Correct Option is A

Step 1: Isentropic exit temperature for the given pressure ratio.
For an ideal (isentropic) compressor, \[ \frac{T_{02s}}{T_{01}}=\left(\frac{p_{02}}{p_{01}}\right)^{(\gamma-1)/\gamma} =10^{0.4/1.4}=10^{0.285714}\approx 1.93. \] Hence \[ T_{02s}\approx 1.93\times 300 \approx 579\ \text{K}. \] 

Step 2: Use overall compressor isentropic efficiency.
For a compressor, \[ \eta_c=\frac{T_{02s}-T_{01}}{T_{02}-T_{01}} \quad\Rightarrow\quad T_{02}=T_{01}+\frac{T_{02s}-T_{01}}{\eta_c}. \] Thus \[ T_{02}=300+\frac{579-300}{0.83} =300+\frac{279}{0.83}\approx 300+336.1 \approx 636.1\ \text{K}. \] So the actual total temperature rise is \[ \Delta T_{0,\text{actual}}=T_{02}-T_{01}\approx 336.1\ \text{K}. \] 

Step 3: Convert total rise to number of identical stages.
Given each stage adds \(20\ \text{K}\) in stagnation temperature, \[ N=\frac{\Delta T_{0,\text{actual}}}{20} =\frac{336.1}{20}\approx 16.8 \approx 17\ \text{stages}. \] \[ \boxed{17} \]