Question 6.DE.17: Steam enters the first row of a series of stages at a static......
Steam enters the first row of a series of stages at a static pressure of 10 bars and a static temperature of 300°C. The blade angles for the rotor and stator of each stage are: α_1 = 25°, β_1 = 60°, α_2 = 70.2°, β_2 = 32°.
If the blade speed is 250 m/s, and the rotor efficiency is 0.94, find the degree of reaction and power developed for a 5.2 kg/s of steam flow. Also find the static pressures at the rotor inlet and exit if the stator efficiency is 0.93 and the carryover efficiency is 0.89.
Using the given data, the velocity triangles for the inlet and outlet are shown in Fig. 6.34. By measurement, C_2 = 225 m/s, V_2 = 375 m/s, C_1 = 400 m/s, V_1 = 200 m/s.
Work done per unit mass flow:
W_t = (250) × (400 cos 25° + 225 cos 70.2°) = 1, 09, 685 J/kg
Degree of reaction [Eq. (6.25)]
Λ = \frac{V^2_2 – V^2_1} {2 \times W_t} = \frac{375^2 – 200^2} {(2) \times (1, 09, 685)} = 0.4587, or 45.87%
Power output:
P = \dot{m}W = \frac{(5.2) \times (1, 09, 685)} {1000} = 570.37 kW
Isentropic static enthalpy drop in the stator:
Δh_s^′ = \frac{C^2_1 – C_2^2} {η_s} = \frac{(400^2 – (0.89) × (225^2))} {0.93}
= 1, 23, 595 J/kg, or 123.6 kJ/kg
Isentropic static enthalpy drops in the rotor:
Δh_s^′ = \frac{W} {η_rη_s} = \frac{1, 09, 685} {(0.94) \times (0.93)}
= 1, 25, 469 J/kg, or 125.47 kJ/kg
Since the state of the steam at the stage entry is given as 10 bar, 300°C,
Enthalpy at nozzle exit:
h_1 – [Δh^′]_{\text{stator}} = 3051.05 – 123.6 = 2927.5 kJ/kg
Enthalpy at rotor exit:
h_1 – [Δh^′]_{\text{rotor}} = 3051.05 – 125.47 = 2925.58 kJ/kg
The rotor inlet and outlet conditions can be found by using the Mollier
Chart.
Rotor inlet conditions: P_1 = 7 bar, T_1 = 235°C
Rotor outlet conditions: P_2 = 5 bar, T_2 = 220°C