Question

In a certain stage of an impulse turbine, the nozzle angle is 20° with the plane of the wheel. The mean diameter of the ring is 2.8 meters. It develops 55 kW at 2400 rpm. Four nozzles, each of 10 mm diameters expand steam isentropically from 15 bar and 250°C to 0.5 bar. The axial thrust is 3.5 N. Calculate: 1. Blade angles at entrance and exit, and 2.power lost in blade friction​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

11.57N

Explanation:

Definition of an Impulse Turbine:

• In impulse turbines, the high-velocity steam jet created by the nozzle impinges on blades attached to a rotor.
• The blades adjust the direction of the steam flow without changing the pressure.
• It causes a change in momentum, and the force that results pushes the turbine rotor.
• In response turbines, the pressure drops quickly.
• Over the fixed and moving blades, there is a gradual drop in pressure. A number of wheels are attached to the rotating shaft.
• Fixed guideways are provided between such a pair of spinning wheels.

Solution:

• $T_1_s = 250^0 C, h_2_s = 2943.10 \frac{kJ}{kg}, a_i = 426.63 \frac{m}{s}$
• $\frac{b}{a_i}=\frac{cos\alpha_i}{2} \implies b = 200.45$
• $r_i= \frac{a_i sin \alpha_i}{sin\beta_i} = 248.25\\\implies r_e = 223.42 m/s\\\omega = 392.35 m/s\\f = 30.85 m/s\\F_A_X = 11.57N\\F_D_R_I_V_I_N_G = 147.13N\\Power = 29.492 kW\\\eta_b_l_a_d_e = 86.40\%$