A mixing beater consists of three thin rods, each 10.6 cm long. The rods diverge

Question

A mixing beater consists of three thin rods, each 10.6 cm long. The rods diverge from a central hub, separated from each other by 120 degree, and all turn in the same plane. A ball is attached to the end of each rod. Each ball has cross-sectional area 3.90 cm^2 and is so shaped that coefficient of 0.580. The drag force on each ball is R = 1/2 D rho AV^2 where D is the coefficient, rho the density of the fluid, A the cross-sectional area, and v the speed of the moving through the fluid. Calculate the power input required to spin the beater at 1000 rev/min in Water The beater's taken out of the water and held in air. If the input power remains the same (it wouldn't, but if it did), what would be the new rotation speed?

Explanation / Answer

a)

For 1000 rpm = 16.67 rps, v = wr = 16.67*2(Pi)*Length of rod = 16.67*(6.28)*(10.6*10-2) = 11.09 m/s

R = 0.5 * 0.58 * 1000 * (3.9*10-4) * 11.092 = 13.9 N on one ball

Energy for one ball = R * (Circumference*no. of rounds) = 13.9*(2*3.14*(10.6*10-2)*16.67) = 154.24 W

Power = energy for 3 balls = 3*154.24 = 462.7 W

b)

Air density = 1.225 kg/m3

Rnew = (13.9/1000)*1.225 = 0.017 N

Energy for one ball = R * (Circumference*no. of rounds) = constant

Rnew * No. of roundsnew =  R * no. of rounds

0.017*rpsNew = 13.9*16.67 = 13630 rps = 817810 RPM

RPSnew = 13630 rps = 817810 RPM

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