Question 13.3: For an axial compressor stage, using the vortex energy equat......
For an axial compressor stage, using the vortex energy equation
\frac{1}{r^2} \frac{d}{dr} \left(r^2C^2_{\text{u}}\right) +\frac{d}{dr} \left(C^2_{\text{a}}\right) =0
Prove that, if the flow is represented by a free vortex, rC_{\text{u}} = constant, then
1. C_{\text{a}}=\text{constant} \\ 2. \tan \alpha_{\text{1h}}=\left(\frac{1+\zeta}{2\zeta} \right) \tan \alpha_{1\text{m}} \\ 3. \tan \beta_{1\text{h}}=\left(\frac{2\zeta}{1+\zeta} \right) \tan \beta_{1\text{h}}+\frac{(3\zeta+1)(\zeta-1)}{2\zeta(1+\zeta)} \tan \alpha_{1\text{m}}where ζ is the hub-to-tip ratio, \alpha_1 is the angle between the inlet absolute velocity and the axial direction, \beta_1 is the angle between the inlet relative velocity and the axial direction, and subscripts (h) and (m) refer to the hub and mean sections, respectively.
1. For rC_{\text{u}} = constant
\therefore \frac{d}{dr} (rC_{\text{u}})=0
Then, from the vortex energy equation
\frac{1}{r^2} \frac{d}{dr} \left(r^2 C^2_{\text{u}}\right) +\frac{d}{dr} \left(C^2_{\text{a}}\right) =\frac{d}{dr} (C^2_{\text{a}})=0. \\ \text{Thus }C^2_{\text{a}}=\text{constant} \\ \therefore C_{\text{a}}= \text{constant}
A typical velocity triangle is illustrated in Figure 13.24.
2. Since \tan \alpha_{1\text{h}}=C_{\text{u1h}}/C_{\text{a}},\tan \alpha_{\text{1m}}=C_{\text{u1m}}/C_{\text{a}},\text{ and } C_{\text{a}}=\text{constant},
Then at state (1)
r_{\text{h}}C_{\text{u1h}}=r_{\text{m}}C_{\text{u1m}} \\ \frac{C_{\text{u1h}}}{C_{\text{u1m}}}=\frac{r_{\text{m}}}{r_{\text{h}}}
Then from Equation (1)
\therefore \tan \alpha_{\text{1h}}=\frac{r_{\text{m}}}{r_{\text{h}}} \tan \alpha_{\text{1m}} \\ \frac{r_{\text{m}}}{r_{\text{h}} }=\frac{r_{\text{h}}+r_{\text{t}}}{2r_{\text{h}}} =\frac{(r_{\text{h}}/r_{\text{t}})+1}{2(r_{\text{h}}/r_{\text{t}})} =\frac{1+\zeta}{2\zeta} \\ \tan \alpha_{\text{1h}}=\frac{1+\zeta}{2\zeta} \tan \alpha_{\text{1m}}
3. From the kinematical relations
\tan \alpha_{\text{1h}}+\tan \beta_{\text{1h}}=\frac{U_{\text{h}}}{C_{\text{a}}} \\ \tan \alpha_{{\text{1m}}}+\tan \beta_{\text{1m}}=\frac{U_{\text{m}}}{C_{\text{a}}} \\ \therefore \frac{\tan \alpha_{\text{1h}}+\tan \beta_{\text{1h}}}{\tan \alpha_{\text{1m}}+\tan \beta_{\text{1m}}} =\frac{U_{\text{h}}}{U_{\text{m}}} \\ \tan \alpha_{\text{1h}}+\tan \beta_{\text{1h}}=\frac{U_{\text{h}}}{U_{\text{m}}} (\tan \alpha_{\text{1m}}+\tan \beta_{\text{1m}}) \\ \tan \beta_{\text{1h}}=\frac{U_{\text{h}}}{U_{\text{m}}} (\tan \alpha_{\text{1m}}+\tan \beta_{\text{1m}})-\tan \alpha_{\text{1h}}
Since
\frac{U_{\text{h}}}{U_{\text{m}}} =\frac{r_{\text{h}}}{r_{\text{m}}} =\frac{2\zeta}{\zeta+1} \quad \text{and } \quad \tan \alpha_{\text{1h}}=\frac{1+\zeta}{2\zeta} \tan \alpha_{\text{1m}} \\ \therefore \tan \beta_{\text{1h}}=\frac{2\zeta}{\zeta+1} (\tan \alpha_{\text{1m}}+\tan \beta_{\text{1m}})-\frac{1+\zeta}{2\zeta} \tan \alpha_{\text{1m}} \\ \tan \beta_{\text{1h}}=\frac{2\zeta}{1+\zeta} \tan \beta_{\text{1m}}+\left[\frac{2\zeta}{\zeta+1} -\frac{1+\zeta}{2\zeta} \right] \tan \alpha_{1\text{m}} \\ \tan \beta_{\text{1h}}=\frac{2\zeta}{\zeta+1} \tan \beta_{\text{1m}}+\frac{4\zeta^2-(1+\zeta)^2}{2\zeta(1+\zeta)} \tan \alpha_{\text{1m}} \\ \tan \beta_{\text{1h}}=\frac{2\zeta}{\zeta+1} \tan \beta_{\text{1m}}+\frac{(3\zeta+1)(\zeta-1)}{2\zeta(1+\zeta)} \tan \alpha_{\text{1m}}
