A disk with six radial holes rotates at 1383 rpm anti-clockwise. The air enters

Question

A disk with six radial holes rotates at 1383 rpm anti-clockwise. The air enters the system at 1004 g/s. Each of the radial holes is inclined at beta_0 = 29 o (in the direction of the rotation). The inlet radius (r1) of the radial disk is 28 cm and the outlet radius (r2) is 37 cm. The diameter of the hole is 504 mm. The density of the air is rho = 1.12 kg/m3.

a) Calculate the incidence angle of the air entering each of the holes?

the procedure :-

mass flow rate (kg/s) for each holerm = mean radius = (r1 + r2)/2 (m)

w (rad/s) = w (rpm) x min/60sec x 2pi/

U (tangential velocity) = wr (m/s)

C1 (m/s)

beta1 (deg)

i (incidence) (deg)

Explanation / Answer

mean radius = 28 + 37 / 2   =   0.325 m

w (rad/s)    = 1383 * 2 * pi / 60   = 144.827 rad/s

U   = 144.827 * 0.325 = 47.0689 m/s

volume flow rate of air = 1.004 / 1.12   = 0.89643 m^3/s

volume flow rate through each hole = 0.89643 / 6 = 0.1494 m^3/s

Velocity through hole   = V / area of hole    = 0.1494 * 4 / pi * 0.504^2   = 0.74888 m/s

incident angle   = inverse_tan (0.74888 / 47.0689) =   0.911 degrees

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