Question

A hollow tube has a disk DD attached to its end as shown in the below figure. When air of density ρ is blown through the tube, the disk attracts the card CC. Let the area of the card be A and let v be the average air speed between the card and the disk. Calculate the resultant upward force on CC. Neglect the card’s weight; assume that v0<

Answer 1

Answer:

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Answer 2

Answer:

Apply Bernoulli’s equation just before it leaves the hollow tube, the total energy of the flow of air is

p1+ ½ ρv02 + ρgy1 = constant

Here, pressure of the flow of air just before it leaves the hollow tube is p1, speed of the air in the hollow tube is v0, and position of the flow of air from the ground before the air leaves the hollow tube is y1.

On the upper surface of the card CC, the total energy of the flow of air is

p2+ ½ ρv2 + ρgy2 = constant

Here, pressure of the flow of air above the car CC is p2, speed of the air above the card CC is v, and position of the flow of air above the card CC from the ground is y2.

The total energy for the flow of air remains the same when it flows from the hollow tube to the card CC.

Thus,

p1+ ½ ρv02 + ρgy1 = p2+ ½ ρv2 + ρgy2

The position of the flow of air is same for the same flow of air.

Thus,

y1 = y2

Also, it is given that v0<<v.

So, the speed in the hollow tube v0 can be neglected.

Insert y1 = y2   and neglecting v0 in the equation  p1+ ½ ρv02 + ρgy1 = p2+ ½ ρv2 + ρgy2

Gives

p1+ ½ ρv02 + ρgy1 = p2+ ½ ρv2 + ρgy2

p1 - p2 = ½ ρv2

Now, the resultant upward force on the card CC is

F = (p1-p2) A

Substitute ½ ρv2 for p1-p2 in the above equation gives

F = (p1-p2) A

 = ½ ρv2A

Therefore, the resultant upward force on the CC is ½ ρv2A.