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In a De laval turbine, steam leaves nozzle with a velocity of 1300 m/s. The nozzle angle is 25° and the mean blade speed is 500 m/s. The inlet and outlet blade angles are equal. Find out work done per kg of steam if the blades of the turbine are frictionless.

Correct Answer: 678.205 kJ
C1 = 1300 m/s, α = 25°, Cbl = 500 m/s, θ = Ф Steps for Construction – Select a suitable scale and draw line LM corresponding to 500 m/s (Cbl). At M, cut a length at S corresponding to 1300 m/s (C1) at an angle of 25° (α). Join LS. Produce L to meet perpendicular from S at P. The inlet triangle is complete. By measurement – Cr1 = 872.81 m/s, θ = 39° Since the blades of the turbine are considered to be frictionless Cr2=Cr1 = 872.81 m/s Also, Ф = θ = 39° At L, cut a length at N corresponding to 872.81 m/s (Cr2) at an angle of 39°. Join MN. Produce M to meet perpendicular from N at Q. The outlet triangle is complete. By measurement – Cw1 = 1178.20 m/s and Cw2 = 178.21 m/s Work done per kg of steam = (Cw1 + Cw2)*Cbl = (1178.20 + 178.21)*500 = 678205 J or 678.205 kJ.

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