Question 14.4: An axial turbine has the following data: ζ = 0.8, α2m = 60, ......

Given data

\zeta=0.8, \quad \alpha_3=0 , \quad \alpha_{\text{2m}}=60 \\ \Lambda_{\text{m}}=0.5 \quad Ca_{\text{m}}=200 \text{ m/s}

At any radius along the blade

\alpha_2=\tan^{-1}\left(\frac{C_{\text{u2}}}{C_{\text{a2}}} \right) , \quad \beta_2=\tan^{-1}\left(\frac{C_{\text{u2}}-U}{C_{\text{a2}}} \right) \\ \frac{r_{\text{h}}}{r_{\text{m}}} =\frac{2\zeta}{1+\zeta} =\frac{1.6}{1.8} =0.889 \\ \frac{r_{\text{t}}}{r_{\text{m}}} =\frac{2}{1+\zeta} =\frac{2}{1.8} =1.111

At mean section

C_{\text{a2}}=200 \text{ m/s} \\ C_{\text{u2}}=C_{\text{a2}} \times \tan \alpha_2=346.41 \text{ m/s} \\ C_{\text{u3}}=0

Moreover, since Λ = 0.5, then

\alpha_2=\beta_3=60^\circ, \quad \alpha_3=\beta_2=0 \\ U=C_{\text{u2}}=346.41 \text{ m/s}

The rotational speed at any radius is

U_{\text{i}}=\frac{r_{\text{i}}}{r_{\text{m}}} \times U_{\text{m}}, \quad \text{ thus} \\ U_{\text{h}}=307.92 \text{ m/s}\quad \text{and }\quad U_{\text{t}}=384.9 \text{ m/s}

Case (1)
Since C_{\text{u}2} = constant along the blade height, from Equation 14.57b

C_{\text{a}}\frac{\text{d}C_{\text{a}}}{\text{d}r} +C_{\text{u}}\frac{\text{d}C_{\text{u}}}{\text{d}r} +\frac{C^2_{\text{u}}}{r} =0     (14.57)

this design method is the general power case with N = 0.
The axial speed (C_{\text{a}2}) is calculated from Equation 14.59, which is expressed in state (2) as

\frac{C_{\text{a}}}{C_{\text{am}}}=\left\{1-\tan^2 \alpha_{\text{m}}\left(\frac{N+1}{N} \right)\left[\left(\frac{r}{r_{\text{m}}} \right)^{2N} -1\right] \right\}^{1/2} ,\quad N\neq 0 \quad \quad \quad (14.59) \\ C_{\text{a2}}=C_{\text{am}}\left[1-2\tan ^2\alpha_{\text{m}} \ln\frac{r}{r_{\text{m}}} \right] ^{1/2}

Since \alpha_3 = 0, from the same Equation 14.59 but applied at state (2), C_{\text{a}}3 = constant from blade hub to tip. Moreover, since \alpha_3=0,C_{\text{u}3}=0 along the blade height. The angles \beta_2 \text{ and } \beta_3 are calculated from Equation 14.1c.

\frac{U}{C_{\text{a}}} =\tan \alpha_2-\tan \beta_2=\tan \beta_3-\tan \alpha_3       (14.1c)

The results are summarized in Table 14.4a.
A plot for the results is shown in Figure 14.19.
The velocity triangles are shown in Figure 14.20.

Case (2)

Since \frac{C_{\text{a2t}}}{C_{\text{a2h}}} =\zeta^{\sin^2\alpha_2}

\frac{C_{\text{a2t}}}{C_{\text{a2h}}} =\left(\frac{r_{\text{h}}}{r_{\text{t}}} \right) ^{\sin^2 \alpha_2}=\left(\frac{r_{\text{m}}}{r_{\text{t}}} \right) ^{\sin^2\alpha_2}\left(\frac{r_{\text{h}}}{r_{\text{m}}} \right) ^{\sin^2 \alpha_2}

and

\frac{C_{\text{a2t}}}{C_{\text{a2h}}} =\frac{C_{\text{a2t}}}{C_{\text{a2m}}} =\frac{C_{\text{a2m}}}{C_{\text{a2h}}}

This design method is the constant nozzle angle at state (2), or

C_{\text{a2}}=C_{\text{a2m}}\left(\frac{r_{\text{m}}}{r} \right) ^{\sin^2 \alpha_2}, C_{\text{u2}}=C_{\text{u2m}}\left(\frac{r_{\text{m}}}{r} \right) ^{\sin^2 \alpha_2},C_2=C_{2\text{m}}\left(\frac{r_{\text{m}}}{r} \right) ^{\sin^2 \alpha_2}

At state (3),\alpha_3=0 ,\quad C_{\text{u3}}=0

C^2_{\text{a3}}=C^2_{\text{a3m}}+2U_{\text{m}}C_{\text{u2m}}\left[1-\left(\frac{r}{r_{\text{m}}} \right)^{\cos^2 \alpha_2} \right]

It is deduced here that all the velocity components C_{\text{u2}},C_{a2}, \text{ and }C_{\text{a3}} are variables. The final results are given in Table 14.4b.
A plot for the results is shown in Figure 14.21.

The velocity triangles are shown in Figure 14.22.
Case (3)
Since

\frac{C_{\text{u2t}}}{C_{\text{u2h}}} =\zeta=\frac{r_{\text{h}}}{r_{\text{t}}}

or rC_{u2} = constant, this design method is called free vortex.
At the rotor outlet, the flow is also a free vortex, or rC_{\text{u3}} = constant.
The axial velocity is then constant at both states (2) and (3), or C_{\text{a2}}=C_{\text{a2m}}=C_{\text{a3}}=C_{\text{a3m}}=\text{constant}

The results are summarized in Table 14.4c.

A plot for the results is shown in Figure 14.23, while the velocity triangles are shown in Figure 14.24.