Question 13.8: Boeing 737–200 airplanes have the ability to take off and la......
Boeing 737–200 airplanes have the ability to take off and land from unpaved runways (Figure 13.39). Moreover, their engines are close to ground thus, more susceptible to ingestion of solid particles into its core. Solid particles will erode the compressor blades. An examination of one of the compressor blades revealed that erosion led to
1. Shortening of the chord by 10%
2. Increasing the incidence angle (i) from 4° to 8°
Other data of the compressor stage are
Stage total temperature at entry to the stage is T_{01} = 320 K
Stage total temperature rise = ΔT_{0\text{s}} = 25 K
For the new blade,
\frac{s}{c_{\text{new}}} =1.0 \quad \text{and}\quad \alpha_{1\text{new}}=50^\circ
For both new and eroded blade, \alpha_2=30^\circ \text{ and } s/h=0.25.
Calculate
1. Stage efficiency for both new and eroded blade
2. Stage pressure ratio for new and eroded blade
3. Comment!
Calculations will be arranged in Table 13.10.
Comments:
1. The stage efficiency has been decreased by 6%
2. The stage pressure ratio has been decreased by 1.4%
| TABLE 13.10 Properties of New and Eroded Compressor Stage | ||
| New Blade | Eroded Blade | |
| i | 4° | 8° |
| \alpha_1 | 50° | 54° |
| \alpha_2 | 30° | 30° |
| \alpha_{\text{m}}=\tan^{-1}\left[\frac{1}{2}(\tan \alpha_1+\tan \alpha_2) \right] | 41.5° | 44.33° |
| s/c | 1.0 | \frac{1}{0.9} =1.11 |
| s/h | 0.25 | 0.25 |
| C_{\text{D}_{\text{p}}} (from Figure 13.35) | 0.025 | 0.075 |
| C_{\text{L}} (from Figure 13.35) | 0.9 | 0.94 |
| C_{\text{DS}}=0.018C^2_{\text{L}} | 0.01458 | 0.0159 |
| C_{\text{DA}}=0.02\frac{s}{h} | 0.005 | 0.005 |
| C_{\text{D}}=C_{\text{D}_{\text{P}}}+C_{\text{DA}}+C_{\text{DS}} | 0.04458 | 0.0959 |
| \frac{\Delta P_{\text{th}}}{(1/2)\rho V_1^2} =1-\frac{\cos^2 \alpha_1}{\cos^2 \alpha_2} | 0.449 | 0.5393 |
| \frac{\bar{w}}{(1/2)\rho V_1^2} =\frac{C_{\text{D}}}{(s/c)\left(\cos^3 \alpha_{\text{m}}/ \cos^2 \alpha_1\right) } | 0.04384 | 0.08154 |
| \eta_{\text{b}}=1-\frac{\bar{w}/((1/2)\rho V_1^2)}{\Delta P_{\text{th}}/((1/2)\rho V_1^2)} | 0.90236 | 0.84878 |
| \eta_{\text{b}}=\eta_{\text{s}} | 0.9024 | 0.8488 |
| \pi_{\text{s}}=\left(1+\eta_{\text{s}}\frac{\Delta T_{0\text{s}}}{T_1} \right) ^{(\gamma/\gamma-1)} | 1.2693 | 1.2519 |
