Axial gas turbine stage design calculation

Data: $U = 340$ m/s, $C_a = 250$ m/s, $\alpha_2 = 65^\circ$ (nozzle exit), $\alpha_3 = 0$, $T_{01} = 1200$ K, $\eta_{tt} = 0.88$. Stations: 1 nozzle inlet, 2 nozzle exit/rotor inlet, 3 rotor exit.

 Rotor inlet                 Rotor exit
      C2                         Vr3   C3=Ca
     /|\ Vr2                      \    |
    / | \                          \b3 |
   /65|b2\                          \  |
  +---U---+                       ---U-+

(a) Blade angles

$$C_{w2} = C_a\tan 65^\circ = 250 \times 2.1445 = 536.1\ \text{m/s}$$ $$\tan\beta_2 = \frac{C_{w2} - U}{C_a} = \frac{196.1}{250} \Rightarrow \beta_2 = 38.1^\circ$$

At exit $C_{w3} = 0$, so the relative whirl is $U$:

$$\tan\beta_3 = \frac{U}{C_a} = \frac{340}{250} \Rightarrow \beta_3 = 53.7^\circ$$

(b) Work and temperature drop

$$w = U(C_{w2} + C_{w3}) = 340 \times 536.1 = 182.3\ \text{kJ/kg}$$ $$\Delta T_0 = \frac{w}{c_p} = \frac{182.3}{1.148} = 158.8\ \text{K}$$

(c) Degree of reaction

$$R = \frac{C_a}{2U}(\tan\beta_3 - \tan\beta_2) = 1 - \frac{C_{w2}}{2U} = 1 - \frac{536.1}{680} = 0.212$$

(d) Pressure ratio

From $\eta_{tt} = \dfrac{\Delta T_0}{T_{01}(1 - (p_{03}/p_{01})^{(\gamma-1)/\gamma})}$:

$$\frac{p_{01}}{p_{03}} = \left[1 - \frac{\Delta T_0}{\eta_{tt}T_{01}}\right]^{-\gamma/(\gamma-1)} = \left[1 - \frac{158.8}{0.88 \times 1200}\right]^{-4.003} = 1.92$$

(e) Power

$$P = \dot m w = 25 \times 182.3 = 4557\ \text{kW} = 4.56\ \text{MW}$$

Stage loading $w/U^2 = 1.58$ and flow coefficient $C_a/U = 0.74$.

Answer: $\beta_2 = 38.1^\circ$, $\beta_3 = 53.7^\circ$; $w = 182.3$ kJ/kg, $\Delta T_0 = 158.8$ K; $R = 0.212$; $p_{01}/p_{03} = 1.92$; $P = 4.56$ MW.