Question

A single-stage axial compressor, with a 50 % degree of reaction, runs at a mean blade speed of 250 m/s. The overall pressure ratio developed is 1.3. Inlet pressure and temperature are 1 bar and 300 K, respectively. Axial velocity is 200 m/s. Specific heat at constant pressure, \( C_p = 1005 \, {J/kg/K} \) and specific heat ratio, \( \gamma = 1.4 \). The rotor blade angle at the outlet is __________ degrees (rounded off to two decimal places).

Hint:

The rotor blade angle can be calculated using the formula \( \beta_2 = \tan^{-1} \left( \frac{U}{V_a} \right) \), where \( U \) is the blade speed and \( V_a \) is the axial velocity.

Answer 1

We are given the following:
Degree of reaction \( R = 50% = 0.5 \),
Mean blade speed \( U = 250 \, {m/s} \),
Overall pressure ratio \( \pi = 1.3 \),
Inlet pressure \( p_1 = 1 \, {bar} = 10^5 \, {Pa} \),
Inlet temperature \( T_1 = 300 \, {K} \),
Axial velocity \( V_a = 200 \, {m/s} \),
Specific heat at constant pressure \( C_p = 1005 \, {J/kg/K} \),
Specific heat ratio \( \gamma = 1.4 \).
Step 1: The total pressure rise is: \[ \Delta p = p_2 - p_1 = 1.3 \times 10^5 - 10^5 = 30000 \, {Pa} \] The pressure rise across the rotor is: \[ \Delta p_{{rotor}} = R \cdot \Delta p = 0.5 \cdot 30000 = 15000 \, {Pa} \] Thus, the pressure at the rotor outlet is: \[ p_2 = p_1 + \Delta p_{{rotor}} = 10^5 + 15000 = 115000 \, {Pa} \] Step 2: The rotor blade angle \( \beta_2 \) is: \[ \beta_2 = \tan^{-1} \left( \frac{U}{V_a} \right) = \tan^{-1} \left( \frac{250}{200} \right) \approx 21.8^\circ \] Thus, the rotor blade angle at the outlet is: \[ \boxed{21^\circ} \]