Bernoulli's Principle

      
Introduction

P₁  +  v₁²  + gy₁  = P₂  +  v₂²  + gy₂        Bernoulli's Equation 

or    P  + v²  + gy  =  Constant

If we apply this to the example to the left, P is the same because the top and the spigot are both exposed to atmospheric pressure.   Also, as long as the diameter of the can is much larger than the spigot, v₂ will be approximately zero.

v₁²  + gy₁  =  gy₂   

or

v₁  =  (  2g ( y₂  -  y₁ ) ) ^1/2   also called Torricelli's theorem

This special case was discovered a century earlier.

Another special case exist when there is no appreciable difference in height.

P₁  +  v₁²   = P₂  +  v₂²  

This equation shows why the pressure is high when the velocity is low and why the pressure is low when the pressure is high.

When a ping pong ball is placed over the opening to a hair dryer, the ball remains centered at all times even when the hair dryer is tilted.

Of course, if tilted sufficiently, the ball will drop.  The faster the air travels, the more the dryer can be tilted until the ball drops.

As the ball moves off center, the pressure will be less toward the center and more toward the outside.  This pushes the ball more toward the center.

It also explains how a wing works.  Since the distance traveled across the top is greater than the distance traveled under the wing, the air must travel faster over the wing.

Since the air is moving faster it has the lower press.  Since the pressure is less on the top of the wing, the wing has lift.

Bernoulli's principle also explain how a sailboat can sail into the wind.

The wind speed increases in the narrow space between the two sails.  Thus the atmospheric pressure behind the sail is greater than the reduced pressure in front of the sail.  The mainsail is set an an angle midway between the wind direction and the keel.  The net force on the sail acts nearly perpendicular to the sail.  This would tend to move the boat sideways.  The keel extends vertically downward beneath the water.  The water exerts a force on the keel nearly perpendicular to the keel.  The resultant of these two forces is almost directly forward.

A venturi tube is essentially a pipe with a narrow constriction.  A carburetor is an example of a venturi tube.  As the air speeds up as it enters the constriction, the pressure decreases.  This sucks the gasoline from the reservoir and puts it in the gaseous phase.  This makes it burn better in the cylinder. 

Modern cars use fuel injectors which use a pump to push the gasoline through an opening designed to separate it into a mist.  The smaller the gasoline droplets, the better it burns.

The chimney to the right also works partially because of Bernoulli's Principle.  As wind passes over the narrow opening at the top, the lower pressure sucks the smoke up the chimney.  It also rises because hot air is less dense than cold air.

Table of Contents

Problems

Table of Contents

Answers

1.	What is the lift due to Bernoulli's Principle on a wing of 70.0 m2 area if the air passes over the top and bottom surfaces at speeds of 340. m/s and 290. m/s respectively?
	Pb  +  vb2   = Pt  +  vt2                 Solve for Pb  -  Pt. Pb   -  Pt   =   vt2   -  vb2   Pb  -  Pt  =   x 1.00 x 10 3 kg m -3 x ( 340. m s -1 ) 2  -   x 1.00 x 10 3 kg m -3 x ( 290. m s -1 ) 2  Pb  -  Pt  =  1.58 x 107 N/m2   Force lift  =  1.58 x 107 N/m2  x  70.0 m2  =  1.10 x 10 9 N
2.	A can has an inside diameter of 12.2 cm.  (a)  How far below the water level do you want make a 0.0500 cm diameter hole to have a velocity of  2.25 m/s?  (b)  How far must the water drop for the speed to drop by 1/2?
	(a)  v1  =  (  2g ( y2  -  y1 ) ) 1/2                 Solve for y2  -  y1.     0.0500 cm is much smaller than 12.2 cm.                          v12         ( 2. 25 m s -1 )2  y2  -  y1  =  -------  =  ---------------------  =  0.258 m                      2g          2 x 9.80 m s -2       (b)                      v12         ( 1.13 m s -1 )2  y2  -  y1  =  -------  =  ---------------------  =  0.0651 m                      2g          2 x 9.80 m s -2      The water must drop from a difference of 0.258 m to a difference of 0.0651 m.   Thus the water must drop by    0.258 m  -  0.0651 m  or  0.193 m
3.	If the wind blows at 200. km/h over your house during Hurricane Bonnie, what is the net force on the roof if the area is 288 m2?
	200. km         1 h             1000 m -----------  x  ----------  x  ------------  =  55.6 m s -1      1 h            3600 s          1 km     Pb  +  vb2   = Pt  +  vt2                 Solve for Pb  -  Pt.    vb  =  0  Pb   -  Pt   =   vt2   -  vb2   Pb  -  Pt  =   x 1.00 x 10 3 kg m -3 x ( 55.6 m s -1 ) 2  -   x 1.00 x 10 3 kg m -3 x (  0 m s -1 ) 2  Pb  -  Pt  =  1.55 x 106 N/m2   Force lift  =  1.55 x 106 N/m2  x  288 m2  =  4.45 x 10 8 N
4.	Wind at 20.0 km/h blows over the 12.0 cm diameter opening of a chimney.  The box where the fuel burns is 1.00 m  x  1.50 m.  What is the pressure difference in the chimney?
	20.0 km         1 h             1000 m -----------  x  ----------  x  ------------  =  5.56 m s -1      1 h            3600 s          1 km     Pb  +  vb2   = Pt  +  vt2                 Solve for Pb  -  Pt.    vb  =  0  Pb   -  Pt   =   vt2   -  vb2   Pb  -  Pt  =   x 1.00 x 10 3 kg m -3 x ( 5.56 m s -1 ) 2  -   x 1.00 x 10 3 kg m -3 x (  0 m s -1 ) 2  Pb  -  Pt  =  1.55 x 104 N/m2  =  15.5 kPa

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