In a single row impulse turbine, the absolute velocity of steam leaving the turbine is 900 m/s. The nozzle angel is 20°. The mean blade speed is 300 m/s. The inlet blade angle is equal to the outlet blade angle. Determine the angle made by the discharging steam with the tangent of the wheel at the exit of moving blade. Neglect friction on turbine blades.

Correct Answer: 51°
C1 = 900 m/s, α = 20°, Cbl = 300, θ = Ф, K = 1 Steps for construction – Select a suitable scale and draw line LM corresponding to 300 m/s (Cbl). At M, at an angle of 20° cut a length corresponding to 900 m/s at S. Join LS. Produce L to meet perpendicular from S at P. The inlet triangle is complete. By measurement – Cr1 = 626.55 m/s, θ = 29° Since the blades of the turbine are considered to be frictionless Cr2 = Cr1 = 626.55 m/s Also, Ф = θ = 29° At L, cut a length at N corresponding to 626.55 m/s (Cr2) at an angle of 29°. Join MN. Produce M to meet perpendicular from N at Q. The outlet triangle is complete. By measurement, β = 51°.