Consider a three-stage axial flow turbine where the total inlet temperature is 1100 K, the flow coefficient is 0.8, and the mean rotational speed is constant for all stages and equal to 340 m/s.
Other data are given in the following table:
It is required to
1. Sketch the turbine shape.
2. Draw the velocity triangles at the mean sections for each stage (assume constant mean radius for all stages).
3. Calculate the degree of reaction at the mean section for the third stage.
4. Calculate the pressure ratio of each stage.
5. Prove that for equal pressure ratios of the three stages, the efficiencies must have the following relations:
\frac{\eta_{\text{s2}}}{\eta_{\text{s1}}} =\frac{T_{01}\times \Delta T_{02}}{\Delta T_{01}\times (T_{01}-\Delta T_{01})} \\ \frac{\eta_{\text{s3}}}{\eta_{\text{s1}}} =\frac{T_{01}\Delta T_{03}}{\Delta T_{01}\times (T_{01}-\Delta T_{01}-\Delta T_{02})}
Stage | Angles | Degree of Reaction \Lambda_{\text{m}} |
Efficiency \eta_{\text{s}} | Inlet Temperature (K) |
Temperature Drop [\Delta \ T_0 (K)] |
1 | \alpha_1=0 | 0.5 | 0.938 | 1100 | 145 |
2 | \alpha_1=\beta_2 | 0.5 | 0.938 | ? | 145 |
3 | \alpha_3=0 | ? | 0.94 | ? | 120 |
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})}
TABLE 14.3 Summary of results |
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Stages Parameters | Stage 1 | Stage 2 | Stage 3 |
T_{01} | 1100 | 955 | 810 |
\alpha_2 (degrees) | 56.746 | 56.746 | 56.126 |
\beta_3 (degrees) | 56.746 | 56.746 | 51.341 |
\alpha_3 (degrees) | 15.376 | 15.376 | 0 |
\beta_2 (degrees) | 15.376 | 15.376 | 13.471 |
\Lambda_{\text{m}} | 0.5 | 0.5 | 0.4042 |
P_{03}/P_{01} | 0.5432 | 0.4935 | 0.50097 |