Question 12.4: To examine how prewhirl reduces the bending stress for the b......
To examine how prewhirl reduces the bending stress for the blades of centrifugal compressors, consider the following case where the blade shape at tip of the eye is shown in Figure 12.23 for the two cases with prewhirl and without prewhirl.
(r_{\text{i}})_{\text{prewhirl}}=15 \text{ cm}, \quad (r_{\text{i}})_{\text{no prewhirl}}=5\text{ cm}, \quad t=0.4 \text{cm}, \quad A=0.4 \text{ cm}^2
If the bending moment M = 10.0 Nm, calculate the maximum stresses in both cases and the percentage of stress reduction when prewhirl is employed.
The stress in such curved beams is given by
\sigma=\frac{M\text{y}}{Ae(\text{R}-\text{y})}
where R = A/(∫ dA/r) is the radius of the neutral axis, which, for a rectangular cross section is expressed by
R=\frac{t}{\ln\left(1+\frac{t}{r_{\text{i}}} \right) } \quad \quad \text{and }\quad \quad e=\bar{r} -\text{R}
\bar{r} is the radius of the center of gravity of the section,
\bar{r}=r_{\text{i}}+\frac{t}{2} \\ r=\text{R}-\text{y}
r is any radius defining any point within the rectangular cross section.
Since the maximum stress occurs at the point having minimum radius, the critical point is at the concave side of the blades (r=r_{\text{i}}).
Calculation for the two cases will be tabulated here as follows:
The above results shows that prewhirl reduced the maximum stresses by a percentage of 13%.
| No prewhirl | With prewhirl | |
| \text{R}=\frac{t}{\ln\left(1+\frac{t}{r_{\text{i}}} \right) } | 5.1974 cm | 15.199 cm |
| e= \bar{r}-\text{R} | 2.565 \times 10^{-3} \text{ cm} | 0.001 cm |
| \text{y}_{\text{inner}}=R-r_{\text{i}} | 0.1974 cm | 0.199 cm |
| r_{\text{inner}}=R-\text{y} | 5.0 cm | 15.0 cm |
| \sigma=\frac{M\text{y}}{Ae(\text{R}-\text{y})} | 384.8 Mpa | 331.6 Mpa |