Textbook 2
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Textbook Example #2: Halliday and Resnick

 

An 80-cm2 plate of 500-g mass is hinged along one side. If air is blown over the upper surface only, what speed must the air have to hold the plate horizontal?

Problem #HR-75-P

 

Solution

(from the Halliday-Resnick companion solution book)

 

Apply Bernoulli's Principle to the top (1) and bottom (2) of the plate. These two values must be equal.

p1 + dV12/2 + dgy1 = p2 + dV22/2 + dgy2

(1)

Simplify since h(top) = h(under) for all practical purposes.

p1 + dV12/2  = p2 + dgy2

(2)

Since v(under) = 0 and v(top) = v,

p2 - p1 = dV2/2

(3)

Force = pressure * area.

F = A * dV2/2 = Weight

(4)

Plugging in:

Weight = mg = (0.5)(9.8) = 4.9 N (5)

1/2 (1.2)v2(80 * 10-4) = 4.9 (6)

v = 32.0 m/s.

(5)

 

Discussion

In general, the approach produces an answer that seems to conflicts with ordinary experience.

The he events on opposite sides of the glass plate do not appear to be in the same non-compressible, non-motivated closed flow stream so Bernoulli's Principle may not apply to link them. For example, if top and bottom were in the same stream, or even equivalent streams, then equation (3) would imply that the velocity of the air emerging from the device had zero velocity.

The answer might also conflict with ordinary experience. One can achieve a flow rate of about 32 m/s (about 70 mph) by blowing briskly. The conclusion (6), then, is that one can levitate a piece of glass about the size and shape of a small book simply by blowing briskly over it.

Another implication of this derivation is that parallel (non-expanding) flow across a plate can induce a pressure drop. This also conflicts with ordinary experience. For example, high velocity parallel-flow water passing through a garden hose does not flatten the hose.