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A simple impulse turbine has a mean blade speed of 250 m/s. The nozzle angle is 20° and the steam velocity from the nozzle is 600 m/s. The turbine uses 3000 kg/h of steam. The blade outlet angle is 25°. Determine the amount of energy converted to heat by blade friction if the blade friction factor is 0.86.

Correct Answer: 30.50 kJ
Cbl = 250 m/s, α = 20°, C1 = 600 m/s, ṁ=3000 kg/h or 0.833 kg/s, K = 0.86, Ф = 25° Steps of Construction – Select a suitable scale and draw a line LM corresponding to 250 m/s (Cbl). At M, at an angle of 20° draw a line MS having a length of corresponding to 600 m/s (C1). Join L and S. The inlet triangle is completed. By measurement – Cr1 = 374.96 m/s Cr2 = K(Cr2) = 0.86(374.96) = 322.46 m/s To complete the outlet triangle, draw a line of length corresponding to 322.46 m/s (Cr2) i.e. LN at an angle of 25° at L. Join MN. Amount of energy converted to heat by blade friction = ṁ (Cr12-Cr22) = 0.833 (374.962-322.462) = 30499.92 J or 30.5 kJ.

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