Question 15.3: An IFR turbine with 12 vanes is required to develop 229.6 KW......
An IFR turbine with 12 vanes is required to develop 229.6 KW from a supply of hot gases available at stagnation temperature of 1000 K and a flow rate of 1 kg/s. It has the following data:
\frac{C_{\text{a3}}}{U_2} =0.25, \quad \zeta=0.4, \quad \frac{r_{\text{3s}}}{r_2} =0.7, \quad \text{and} \quad \frac{W_{\text{3m}}}{W_2} =2.0
Total-to-static efficiency 0.85
Static pressure at rotor exist is 100 kPa.
Nozzle enthalpy loss coefficient is \lambda_{\text{N}} = 0.06.
Use the optimum efficiency design method.
Determine
1. Absolute and relative flow angles at rotor inlet (\alpha_1,\beta_2)
2. Overall pressure ratio P_{01}/P_3
3. Rotor tip speed (U_2)
4. Ratio of relative velocity W_{\text{3s}}/W_2 at shroud
5. Diameter and width of rotor, and its speed of rotation
Whitefield equation
\cos^2 \alpha_2=\frac{1}{n} =\frac{1}{12} =0.08333 \\ \alpha_2=73.22^\circ
Also
\beta_2=2(90-\alpha_2)=2(90-73.22)=33.56^\circ \\ \Delta T_0=\frac{W}{Cp} =\frac{229.6}{1.148} =200 \text{ K} \\ \frac{P_{01}}{P_3} =\frac{1}{(1-(\Delta T_0/\eta_{\text{ts}}T_{01}))^{(\gamma/\gamma-1)}} =\left(1-\frac{0.2}{0.81} \right) ^{-4}=2.924
From the specific power relation, with
\frac{C_{\text{u2}}}{U_2} =\cos \beta_2 \\ W=U_2C_{\text{u2}}=U^2_2 \cos \beta_2 \\ U_2=524.9 \text{ m/s} \\ \frac{r_{\text{3m}}}{r_{\text{3s}}} =\frac{r_{\text{3s}}+r_{\text{3h}}}{2} \frac{1}{r_{\text{3s}}} =\frac{1+\zeta}{2} =0.7 \\ \frac{r_{\text{3m}}}{r_2} =\frac{r_{\text{3m}}}{r_{\text{3s}}} \frac{r_{\text{3s}}}{r_2} =0.7 \times 0.7=0.49 \\ \cos \beta_{\text{3m}}=\frac{C_{\text{a3}}}{U_2} \frac{r_2}{r_{\text{3m}}} =\frac{0.25}{0.49} =0.5102 \\ \beta_{\text{3m}}=59.32^\circ \\ \cos \beta_{\text{3s}}=\frac{C_{\text{a3}}}{U_2} \frac{r_2}{r_{\text{3s}}} =\frac{0.25}{0.7} =0.3571 \\ \beta_{\text{3s}}=69.077^\circ \\ \frac{W_{\text{3s}}}{W_2} =\frac{W_{\text{3s}}}{W_{\text{3m}}} \frac{W_{\text{3m}}}{W_2} =\frac{\sec \beta_{\text{3s}}}{\sec \beta_{\text{3m}}} \times 2.0=2.702 \\ T_{03}=T_{01}-\Delta T_0=1000-200=800 \text{ K} \\ T_3=T_{03}-\frac{C_3^2}{2 \times Cp} =T_{03}-\left(\frac{C_{\text{a3}}}{U_2} \right)^2\frac{U^2_2}{2 \times Cp} =800-(0.25)^2\frac{(524.9)^2}{2 \times 1148} = 792.5 \text{ K} \\ \dot{m}=\rho_3 C_{\text{a3}}A_3=\left(\frac{P_3}{RT_3} \right) \left(\frac{C_{\text{a3}}}{U_2} \right) U_2 \pi\left(\frac{r_{\text{3s}}}{r_2} \right) ^2\left(1-\zeta^2\right) r^2_2 \\ 1.0=\left(\frac{10^5}{287 \times 792.5} \right) (0.25) \times 524.9 \times \pi \times 0.7^2 \times \left(1-0.4^2\right) r^2_2 \\ r_2^2=0.0134 \\ r_2=0.1158 \text{ m} \\ D_2=0.2316 \text{ m} \\ \omega =\frac{U_2}{r_2} =4532.8 \text{ rad/s} \quad (N=43285.2 \text{ rpm})
Also since
\dot{m}=\rho_2C_{\text{r2}}A_2=\rho_2C_{\text{r2}}\pi D_2b_2 \\ \frac{b_2}{D_2} =\frac{\dot{m}}{4 \pi \times \rho_2 \times C_{\text{r2}} \times r^2_2}
To obtain the density \rho_2, the following steps have to be followed:
C_{\text{u2}}=U_2 \cos \beta_2=437.4 \text{ m/s}
C_{\text{r2}}=\frac{C_{\text{u2}}}{\tan \alpha_2} =\frac{437.4}{3.3163} =131.89 \text{ m/s} \\ C_2=\frac{C_{\text{u2}}}{\sin \alpha_2} =\frac{437.4}{0.9574} =456.8 \text{ m/s} \\ T_2=T_{02}-\frac{C_2^2}{2Cp} =1000-\frac{(456.8)^2}{2 \times 1148} =909.1 \text{ K} \\ T^\prime_2=T_2-\frac{\lambda_{\text{N}}C_2^2}{2Cp} =909.1-\frac{0.06 \times (456.8)^2}{2 \times 1148} =903.64 \text{ K} \\ P_2=P_{01}\left(\frac{T^\prime_2}{T_{01}} \right) ^{(\gamma/\gamma-1)}=2.924 \times 10^5 \times \left(\frac{903.64}{1000} \right) ^4=1.9497 \times 10^5 \text{ Pa} \\ \frac{b_2}{D_2} =\frac{\dot{m}}{4 \pi \times (P_2/RT_2)\times C_{\text{r2}} \times r^2_2} =\frac{1.0}{4 \pi \times ((1.9496 \times 10^5)/(287 \times 954.5))\times 131.89 \times 0.0134} \\ = 0.06322 \\ b_2=0.0146 \text{ m}The geometry of radial turbine is shown in Figure 15.6.
