Question 13.5: Consider a free vortex design for a compressor stage having ......

Given data: \alpha_{1\text{m}}=30^\circ,\beta_{\text{1m}}=60^\circ, \text{ and }\zeta=0.6.

1. Free vortex design:
The data here are the same as in Example 2, except for the degree of reaction. Thus, the following same results are obtained for the inlet flow along the blade:

\alpha_{1\text{h}}=37.6^\circ, \quad \alpha_{\text{1t}}=24.8^\circ,\quad \beta_{\text{1h}}=43.9^\circ, \quad \beta_{\text{1t}}=67.5^\circ

It was proved in Example 2 that the degree of reaction may be expressed by the relation

\Lambda=\frac{\tan \beta_1+\tan \beta_2}{2(\tan \alpha_1+\tan \beta_1)}

Applying this equation at the mean section, then

\tan \beta_{2\text{m}}=2\Lambda_{\text{m}}(\tan \alpha_{\text{1m}}+\tan \beta_{1\text{m}})-\tan \beta_{\text{1m}} \\ \tan \beta_{2\text{m}}=2\times 0.6 \times 2.3094-1.732=1.0392 \\ \beta_{2\text{m}}=46.1^\circ \\ \tan \alpha_{\text{2m}}=\frac{U_{\text{m}}}{C_{\text{a}}} -\tan \beta_{\text{2m}}=1.2702 \\ \alpha_{\text{2m}}=51.78^\circ \\ \Lambda=\frac{\tan \beta_1+\tan \beta_2}{2(\tan \alpha_1+\tan \beta_1)} \\ \Lambda_{\text{h}}=\frac{\tan \beta_{\text{1h}}+\tan \beta_{\text{2h}}}{2(\tan \alpha_{\text{1h}}+\tan \beta_{\text{1h}})} =0.2905 \\ \Lambda_{\text{t}}=\frac{\tan \beta_{\text{1t}}+\tan \beta_{2\text{t}}}{2(\tan \alpha_{\text{1t}}+\tan \beta_{\text{1t}})} =0.7438

The variation of the angles from hub to tip is plotted in Figure 13.27.

2. Exponential design method:
Since a=U_{\text{m}}(1-\Lambda_{\text{m}})=100,\Lambda_{\text{m}}=0.6,

U_{\text{m}}=\frac{100}{0.4}250 \text{ m/s}

Moreover, with b = 40 m/s

\Delta T_0=\frac{2bU_{\text{m}}}{Cp} =\frac{2 \times 40 \times 250}{1005} =19.9 \text{ K} \\ C_{\text{u1}}=a-\frac{b}{R}

Since R_{\text{h}}=0.75, R_{\text{t}}=1.25,then

C_{\text{u1h}}=46.7 \text{ m/s}, \quad C_{\text{u1t}}=68 \text{ m/s}

Since

\frac{U_{\text{m}}}{C_{\text{am}}} =\tan \alpha_{\text{1m}}+\tan \beta_{\text{1m}} \\ C_{\text{am}}=\frac{250}{\tan 30+\tan 60} =108.25 \text{ m/s} \\ C_{\text{a1h}}^2=C^2_{\text{am}}-2[a^2 \ln R+\frac{ab}{R}-ab ] \\ C^2_{\text{a1h}}=(108.25)^2-2[(100)^2 \ln \frac{3}{4}+\frac{100 \times 40}{(3/40)}-100 \times 40 ]

C_{\text{a1h}}=121.7 \text{ m/s},\text{ similarly }C_{\text{a1t}}=94.1 \text{ m/s}

From Table 13.3

C^2_{\text{a2}}=C^2_{\text{a2m}}-2[a^2 \ell nR-\frac{ab}{R}+ab ]

C_{\text{u2t}}=132 \text{ m/s} \\ \tan \alpha_{\text{1h}}=\left(\frac{C_{\text{u1}}}{C_{\text{a1}}} \right) _{\text{h}}=0.384, \quad \alpha_{\text{1h}}=21^\circ \\ \tan \alpha_{2\text{h}}=\left(\frac{C_{\text{u}}2}{C_{\text{a2}}} \right) _{\text{h}}=0.93, \quad \alpha_{\text{2h}}=42.9 \\ \tan \alpha_{\text{1t }}=\left(\frac{C_{\text{u1}}}{C_{\text{a1}}} \right) _{\text{t}}=0.722, \quad \alpha_{\text{1t}}=35.85^\circ \\ \tan \alpha_{\text{2t}}=\left(\frac{C_{\text{u2}}}{C_{\text{a2}}} \right) _{\text{t}}=1.755, \quad \alpha_{\text{2t}}=60.32^\circ \\ U_{\text{h}}=\frac{3}{4}U_{\text{m}} =187.5 \text{ m/s} \quad \text{ and }\quad U_{\text{t}}=\frac{5}{4}U_{\text{m}} =312 \text{ m/s} \\ \left(\frac{U}{C_{\text{a1}}} \right) _{\text{h}}=\tan \alpha_{\text{1h}}+\tan \beta_{\text{1h}} \\ \tan \beta_{\text{1h}}=\frac{187.5}{121.7} -\tan21=1.1568

\tan \beta_{\text{2h}}=\left(\frac{U}{C_{\text{a2}}} \right) _{\text{h}}-\tan \alpha_{\text{2h}}=0.392

From both Tables 13.4 and 13.5 the following conclusions may be deduced:

1. The degree of reaction at hub for exponential design is much higher than the case of free vortex.
2. The rotor twist (\beta_1-\beta_2) for exponential design is less and much better than free vortex design.