Question 14.1: Consider a three-stage axial flow turbine where the total in......

1. The sketch (Figure 14.14) illustrates the three-stage axial turbine having a constant mean radius.
2. To draw the velocity triangles, the angles ($\alpha_2,\alpha_3,\beta_2,\beta_3$) must be calculated first.
The appropriate governing equations are the temperature drop and degree of reaction.

From Equations 14.9c and 14.32a,

$\Delta T_{0\text{s}}=\frac{UC_{\text{a}}(\tan \alpha_2+\tan \alpha_3)}{C_{\text{p}}}=\frac{UC_{\text{a}}(\tan \beta_2+\tan \beta_3)}{C_{\text{P}}} \quad \quad \quad (14.9\text{c}) \\ \Lambda=\frac{C_{\text{a}}}{2U} (\tan \beta_3-\tan \beta_2) \quad \quad \quad (14.32\text{a}) \\ \tan \beta_3+\tan \beta_2=\frac{C_{\text{p}}\Delta T_0}{UC_{\text{a}}} \quad \quad \quad (1) \\ \tan \beta_3-\tan \beta_2=\frac{2\Lambda}{\phi} \quad \quad \quad (2)$

Now solving these two equations simultaneously results in the angles $\beta_2 \text{ and }\beta_3$ .
For Stage (1)
Since $T_{01}=1100 \text{ K and }\Delta T_{0\text{s}}=145 \text{ K}, T_{03}=955 \text{ K}$.
Also, as $\Lambda_{\text{m}}=\frac{1}{2} ,\alpha_2=\beta_3,\alpha_3=\beta_2$

$\tan \beta_3+\tan \beta_2=\frac{1148 \times 145}{340 \times 272} =1.8 \quad \quad \quad (\text{a}) \\ \tan \beta_3-\tan \beta_2=\frac{2 \times 0.5}{0.8} =1.25 \quad \quad \quad \text{(b)}$

Then $\beta_2=\alpha_3=15.376^\circ \text{ and } \beta_3=\alpha_2=56.746^\circ$

Stage (2)
The same results were obtained for stage (2) as both have the same temperature drop per stage, axial flow coefficient, and axial and rotational speeds.

$\beta_2=\alpha_3=15.376^\circ \\ \beta_3=\alpha_2=56.746^\circ$

Stage (3)
From Equation 14.9c,

$\tan \alpha_3+\tan \alpha_2=\frac{C_{\text{P}}\Delta T_{0\text{s}}}{UC_{\text{a}}}$

With

$\alpha_3=0^\circ \text{ and } \Delta T_{0\text{s}}=120 \text{ K} \\ \therefore \tan \alpha_2=\frac{C_{\text{P}}\Delta T_{0\text{s}}}{UC_{\text{a}}} \\ \therefore \alpha_2=56.125^\circ$

To calculate the angles $\beta_2 \text{ and }\beta_3$ , we again use Equations 1 and 2

$\tan \beta_3+\tan \beta_2=\frac{1148 \times 120}{340 \times 272} =1.4896$

Also

$\tan \beta_3-\tan \beta_2=\frac{2 \times 0.4042}{0.8} =1.0105$

Solving both equations, we get $\beta_2=13.471^\circ \text{ and }\beta_3=51.341^\circ$

Now, the velocity triangles can be drawn. Stages (1) and (2) have the same velocity triangle, while the third stage has zero exit swirl as shown in Figure 14.15.

3. The degree of reaction for the third stage can be calculated from Equation 14.32c, namely

$\Lambda=1-\frac{C_{\text{a}}}{2U} (\tan \beta_3-\tan \alpha_3) \quad \quad \quad (14.32\text{c}) \\ \Lambda=1-\frac{\phi}{2} (\tan \alpha_3-\tan \alpha_2)$

With $\alpha_3=0 \text{ and }\alpha_2=56.125^\circ$

Λ = 0.4042

4. The pressure ratio of any stage is calculated from Equation 14.14a.

$\pi_{\text{s}}=\frac{P_{01}}{P_{03}} =\frac{1}{(1-(\Delta T_{0\text{s}}/\eta_{\text{s}}T_{01}))^{\gamma/(\gamma-1)}} \quad \quad \quad (14.14) \\ \left(\frac{P_{03}}{P_{01}} \right) _{\text{stage}}=\left[1-\frac{\Delta T_{0\text{s}}}{\eta_{\text{s}}T_{01}} \right] ^{\gamma/(\gamma-1)}$

Stage (1) 
With $T_{01}=1100 \text{ K and }\Delta T_{0\text{s}}=145 \text{ K, then }T_{03}=955 \text{ K}$

$\left(\frac{P_{03}}{P_{01}} \right) =\left[1-\frac{145}{(0.938) \times (1100)} \right] ^{1.33/0.33}=0.5432$

Stage (2)
With $T_{01}=955 \text{ K and } \Delta T_0=145 \text{ K, then }T_{03}=810 \text{ K}$

$\left(\frac{P_{03}}{P_{01}} \right) =0.4935$

Stage (3)
With $T_{01}=810 \text{ K and }\Delta T_0=120 \text{ K}$

$\left(\frac{P_{03}}{P_{01}} \right) =0.50097$

5. Now to correlate the efficiencies of the first two stages to the temperature drop per stage, since

$\left(\frac{P_{03}}{P_{01}} \right) _{1}=\left(\frac{P_{03}}{P_{01}} \right) _2 \\ \left(1-\frac{\Delta T_{01}}{\eta_{\text{s1}}T_{01}} \right) ^{\gamma/(\gamma-1)}=\left(1-\frac{\Delta T_{02}}{\eta_{\text{s2}}(T_{01}-\Delta T_{01})} \right) ^{\gamma/(\gamma-1)} \\ \frac{\Delta T_{01}}{\eta_{\text{s1}}T_{01}} =\frac{\Delta T_{02}}{\eta_{\text{s2}}(T_{01}-\Delta T_{01})} \\ \frac{\eta_{\text{s2}}}{\eta_{\text{s1}}} =\frac{T_{01}\Delta T_{02}}{\Delta T_{01}(T_{01}-\Delta T_{01})}$

Similarly, assuming equal pressure ratios for all stages:

$\therefore \left(\frac{P_{03}}{P_{01}} \right) _1=\left(\frac{P_{03}}{P_{01}} \right) _3 \\ \left(1-\frac{\Delta T_{01}}{\eta_{\text{s1}}T_{01}} \right) ^{\gamma/(\gamma-1)}=\left[1-\frac{\Delta T_{03}}{\eta_{\text{s3}}(T_{01}-\Delta T_{01}-\Delta T_{02})} \right] ^{\gamma/(\gamma-1)} \\ \therefore \frac{\Delta T_{01}}{\eta_{\text{s1}}T_{01}} =\frac{\Delta T_{03}}{\eta_{\text{s3}}(T_{01}-\Delta T_{01}-\Delta T_{02})} \\ \frac{\eta_{\text{s3}}}{\eta_{\text{s1}}} =\frac{T_{01}\Delta T_{03}}{\Delta T_{01}(T_{01}-\Delta T_{01}-\Delta T_{02})}$