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In a Parson’s reaction turbine, the outlet angle of blade is 25°. Calculate the work done per second by the blades if the blade mean blade speed is 60 m/s and the absolute velocity of the steam leaving the nozzle is 120 m/s. The mass flow rate of steam is 0.1 kg per second.

Correct Answer: 945.12 W
Given, α = Ф = 25°, Cbl = 60 m/s, C1 = 120 m/s, ṁ = 0.1 kg/s Steps of ConstructioSelect a suitable scale and draw a line LM to represent Cbl (= 60 m/s). At point M, at an angle of 25° (α), cut a length MS to represent the velocity C1 (= 120 m/s). Join LS. Produce L to meet the perpendicular drawn from S at P. Thus inlet triangle is completed. For a Parson’s turbine: Ф = α = 25° and Cr2= C1 = 120 m/s At L, at an angle of 25° (Ф), cut a length LN to represent Cr2(= 120 m/s). Join MN. Produce M to meet the perpendicular drawn from N at Q. Thus outlet triangle is completed. By measurement: Cw1 = 108.76 m/s and Cw2 = 48.76 m/s Work done per second = ṁ (Cw1 + Cw2) Cbl = 0.1 (108.76 + 48.76) 60 = 945.12 W.

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