Question 13.2: Consider a free vortex design for a compressor stage having ......

Since the degree of reaction is 50% at the mean section, then \alpha_{1\text{m}}=\beta_{2\text{m}} \text{ and } \alpha_{2\text{m}}=\beta_{1\text{m}}.
Since the hub-to-tip ratio is equal to 0.6, then at both stations (1) and (2),

r_{\text{t}}=\frac{5}{4} r_{\text{m}} \quad \quad r_{\text{h}}=\frac{3}{4}r_{\text{m}} \\ U_{\text{t}}=U_{\text{m}}\frac{r_{\text{t}}}{r_{\text{m}}} \quad \quad U_{\text{t}}=\frac{5}{4} U_{\text{m}} \\ U_{\text{h}}=U_{\text{m}}\frac{r_{\text{h}}}{r_{\text{m}}} \quad \quad U_{\text{h}}=\frac{3}{4} U_{\text{m}}

For a free vortex design, rC_{\text{u}} = Constant at stations 1 and 2

r_{\text{h}}C_{\text{uh}}=r_{\text{m}}C_{\text{um}}, \quad C_{\text{uh}}=\frac{r_{\text{m}}}{r_{\text{h}}} C_{\text{um}} \\ r_{\text{t}}C_{\text{ut}}=r_{\text{m}}C_{\text{um}}, \quad C_{\text{ut}}=\frac{r_{\text{m}}}{r_{\text{t}}} C_{\text{um}}

The swirl velocity components are

C_{\text{uh}}=\frac{4}{3} C_{\text{um}} \quad \quad C_{\text{ut}}=\frac{4}{5} C_{\text{um}}

Then, for both stations 1 and 2

C_{\text{u1h}}=\frac{4}{3} C_{\text{u1m}} \quad \quad C_{\text{u2h}}=\frac{4}{3} C_{\text{u2m}} \quad \quad U_{\text{h}}=\frac{3}{4} U_{\text{m}} \\ C_{\text{u1t}}=\frac{4}{5}C_{\text{u1m}} \quad \quad C_{\text{u2t}}=\frac{4}{5}C_{\text{u2m}} \quad \quad U_{\text{t}}=\frac{5}{4} U_{\text{m}} \\ \tan \alpha_{\text{h}}=\frac{C_{\text{uh}}}{C_{\text{a}}} =\frac{4}{3} \frac{C_{\text{um}}}{C_{\text{a}}} =\frac{4}{3}\tan \alpha_{\text{m}} \\ \tan \alpha_{\text{t}}=\frac{C_{\text{ut}}}{C_{\text{a}}} =\frac{4}{5}\frac{C_{\text{um}}}{C_{\text{a}}} =\frac{4}{5} \tan \alpha_{\text{m}} \\ \tan \beta_{\text{h}}=\frac{U_{\text{h}}-C_{\text{uh}}}{C_{\text{a}}} =\frac{3/4U_{\text{m}}-4/3C_{\text{um}}}{C_{\text{a}}} \\ \text{but}\frac{U_{\text{m}}}{C_{\text{a}}} =\tan \alpha_{\text{m}}+ \tan \beta_{\text{m}} \quad \text{and} \quad \frac{C_{\text{um}}}{C_{\text{a}}} =\tan \alpha_{\text{m}} \\ \therefore \tan \beta_{\text{h}}=\frac{3}{4} \tan \beta_{\text{m}}-\frac{7}{12} \tan \alpha_{\text{m}}

Similarly

\text{but}\frac{U_{\text{m}}}{C_{\text{a}}} =\tan \alpha_{\text{m}}+\tan \beta_{\text{m}} \text{ and }\frac{C_{\text{um}}}{C_{\text{a}}} =\tan \alpha_{\text{m}} \\ \tan \beta_{\text{t}}=\frac{U_{\text{t}}-C_{\text{ut}}}{C_{\text{a}}} =\frac{5/4U_{\text{m}}-4/5C_{\text{um}}}{C_{\text{a}}} \\ \therefore \tan \beta_{\text{t}}=\frac{9}{20}\tan \alpha_{\text{m}} +\frac{5}{4}\tan \beta_{\text{m}}

All the calculated angles (\alpha_1,\alpha_2,\beta_1,\beta_2) at the hub, mean, and tip sections are listed in Table 13.2.
From Equations 13.1a and 13.23e, the degree of reaction at any radius may be expressed as

\frac{U}{C_{\text{a1}}} = \tan \alpha_1+\tan \beta_1 \quad \quad \quad (13.1 \text{a}) \\ \Lambda=\frac{C_{\text{a}}}{2U}(\tan \beta_1+\tan \beta_2) \quad \quad \quad (13.23\text{e}) \\ \Lambda =\frac{\tan \beta_1+\tan \beta_2}{2(\tan \alpha_1+\tan \beta_1)}  

The difference in the angles (\beta_1-\beta_2) at different stations along the blade span represents the blade twist. The values of the degree of reaction and blade twist are listed also in Table 13.2. Thus, adopting free vortex design yields a highly twisted blade and a great variation in the degree of reaction from hub to tip. The degree of reaction is only 0.11 at the hub. As explained in Reference 5, because of the lower blade speed at the root section, more fluid deflection is required for a given work input, that is, a great rate of diffusion is required at the root section. It is, therefore, undesirable to have a low degree of reaction in this region, and the problem is aggravated as the hub-to-tip ratio is reduced.
The velocity triangles and the blade twist are shown in Figure 13.23.