Question 15.2: A radial inflow turbine has the following characteristics an......

(a) Total-to-static efficiency
From Figure 15.5, with specific speed of \Omega_{\text{S}} = 0.58 rad and  \alpha_2 = 74°, then  \eta_{\text{ts}} = 0.85.

(b) Geometrical dimensions

\bullet D_2=\frac{60U_2}{\pi N} =\frac{60 \times 500}{\pi \times 40,000} =0.2387 \text{ m} \\ \bullet \frac{b_2}{D_2} =0.0873 \\ \bullet b_2=0.0873 \times 0.2387 \text{ m} \\ \bullet \frac{D_{3 \text{s}}}{D_2} =0.7 \\ \bullet D_{\text{3s}}=0.1671 \text{m} \\ \bullet \frac{D_{3\text{h}}}{D_{\text{3s}}} =0.4 \\ \bullet D_{\text{3h}}=0.06684 \text{ m} \\ \bullet D_{3\text{av}}=\frac{D_{\text{3h}}+D_{\text{3s}}}{2} =\frac{0.06684+0.1671}{2} \\ \bullet D_{\text{3av}}=0.11697 \text{ m}

(c) The outlet conditions

• From Equation 15.4, the specific work is

W=Cp(T_{01}-T_{03})=U_2 \ C_{\text{u2}}=U_2^2

Then 1148 \times (1400-T_{03})=(500)^2

T_{03}=1182.23\text{K}

• The specific speed \Omega_{\text{S}}=1.131 \sqrt{\frac{Q_3}{N \ D_2^3} }

0.58=1.131 \sqrt{\frac{Q_3}{(40,000/60)(0.2387)^3} }

Then Q_3=2.3845 \text{ m}^3/\text{s}.

Since

Q_3=C_{\text{a3}}(\pi/4)(D^2_{3S}-D^2_{3h})

Then

2.3845=C_{\text{a3}}(\pi/4)((0.1671)^2-(0.06684)^2) \\ C_{\text{a3}}=129.44 \text{ m/s}=C_3 \\ T_3=T_{03}-\frac{C^2_3}{2 \ Cp} =1182.23-\frac{(129.44)^2}{2 \times 1148} \\ T_3=1174.93 \text{ K}

• The static pressure is correlated to the total-to-static efficiency by the relation

\frac{P_3}{P_{01}} =\left[1-\frac{1}{\eta_{\text{ts}}}\left(1-\frac{T_{03}}{T_{01}} \right) \right] ^{(\gamma/\gamma-1)}=\left[1-\frac{1}{0.85}\left(1-\frac{1182.23}{1400} \right) \right] ^{(\gamma/\gamma-1)}=0.4455 \\ P_3=1.337 \text{ bar}

(d) Mass flow rate

\dot{m}=\rho_3Q_3=\frac{P_3}{RT_3} \times Q_3 \\ \dot{m}=\frac{1.337 \times 10^5}{287 \times 1174.93} \times 2.3845=0.9454 \text{ kg/s}