A mixing beater consists of three thin rods, each 10.4 cm long. The rods diverge

Question

A mixing beater consists of three thin rods, each 10.4 cm long. The rods diverge from a central hub, separated from each other by 120, 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 cm2 and is so shaped that it has a drag coefficient of 0.620. The drag force on each ball is R-D ? A v2 where D is the drag coefficient, ? the density of the fluid, A the cross-sectional area, and v the speed of the object moving through the fluid. 2 (a) Calculate the power input required to spin the beater at 1000 rev/min in water (b) The beater is 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? rev/min

Explanation / Answer

work in SI units (e.g. metres not cm). Note 1cm² = 10??m².
The value of angle (120°) is not used.
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a)

Circumference of motion = 2?r = 2? x 0.104 = 0.653m.
Since the angular speed is 1000rpm, each ball travels 1000x0.653 m in 60s.

Linear speed v = 1000x0.653/60 = 10.88m/s
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Density of water = 1000kg/m³

Drag force on each ball is
R = ½D?Av² = ½(0.620)(1000)(3.90x10??)(10.883)² = 14.319N

Total force on all 3 balls F = 3R = 3 x 14.319 = 42.957N
___________

Power required to overcome drag force is:
P = Fv = 42.957 x 10.88 = 467.372W
_______________________________________...

b) The density of air under normal conditions is about 1.2kg/m³. This is a factor 1000/1.2 = 833.3 trimes smaller than the density of water.

We are told to assume the power remains constant and we must also assume that in air, D remains 0.620.

There are various ways to do the calculation. Here is one method.

R = ½D?Av², F = 3R and P =Fv. Combining these gives:
P = 3 x ½D?Av² x v
. . = k?v³ where k is a constant as the same balls are used (k = (3/2)DA)

As the same power is supplied in wateer and in air, ?v³ must remain constant. Using proportion, we can say that since ? has decreased by a factor 833.3, then v³ must haved increased by a factor 833.3.

So v increases by a factor ?833.3 = 9.41

Angular speed is proportional to linear speed. So angular speed must increase by the samea factor, 9.41.

So the new angular speed in air is = 1000 x 9.41 = 9410rev/min

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