• Bernoulli’s Principle
(qualitative):
“Where the fluid velocity is high, the pressure is low, and where the velocity is low, the pressure is high.”
– Higher pressure slows fluid down. Lower pressure speeds it up!
• Bernoulli’s Equation
(quantitative).
– We will now derive it.
–
NOT a new law. Simply conservation of KE + PE
(or the Work-Energy Principle) rewritten in fluid language!

• 1700 – 1782
• Swiss physicist
• Published Hydrodynamica
– Dealt with equilibrium, pressure and speeds in fluids
– Also a beginning of the study of gasses with changing pressure and temperature

• As a fluid moves through a region where its speed and/or elevation above Earth’s surface changes, the pressure in fluid varies with these changes. Relations between fluid speed, pressure and elevation was derived by Bernoulli.
• Consider the two shaded segments
• Volumes of both segments are equal. Using definition work & pressure in terms of force & area gives: Net work done on the segment: W = (P
1
– P
2
) V .

• Net work done on the segment: W = (P
1
– P
2
Part of this goes into changing kinetic energy &
) V .
part to changing the gravitational potential energy.
• Change in kinetic energy:
ΔK = (½)mv
2
2 – (½)mv
1
2
– No change in kinetic energy of the unshaded portion since we assume streamline flow. The masses are the same since volumes are the same
• Change in gravitational potential energy:
ΔU = mgy
2
(P
1
– mgy
1.
Work also equals change in energy. Combining:
– P
2
)V =½ mv
2
2 - ½ mv
1
2 + mgy
2
– mgy
1

• Rearranging and expressing in terms of density:
P
1
+ ½ r v
1
2 + r gy
1
= P
2
+ ½ r v
2
2 + r gy
2
• This is Bernoulli’s Equation . Often expressed as
P + ½ r v 2 + r gy = constant
• When fluid is at rest, this is P
1
– P
2
= r gh consistent with pressure variation with depth found earlier for static fluids.
• This general behavior of pressure with speed is true even for gases
As the speed increases, the pressure decreases

• Streamline flow around a moving airplane wing
•
Lift is the upward force on the wing from the air
• Drag is the resistance
• The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal

• In general, an object moving through a fluid experiences lift as a result of any effect that causes the fluid to change its direction as it flows past the object
• Some factors that influence lift are:
– The shape of the object
– The object’s orientation with respect to the fluid flow
– Any spinning of the object
– The texture of the object’s surface

• The ball is given a rapid backspin
• The dimples increase friction
– Increases lift
• It travels farther than if it was not spinning

• A stream of air passes over one end of an open tube
• The other end is immersed in a liquid
• The moving air reduces the pressure above the tube
• The fluid rises into the air stream
• The liquid is dispersed into a fine spray of droplets

Water Storage Tank
P
1
+ (½)ρ(v
1
) 2 + ρgy
1
= P
2
+ (½)ρ(v
2
) 2 + ρgy
2
(1)
Fluid flowing out of spigot at bottom. Point 1
 spigot
Point 2
 top of fluid v
2

0 (v
2
<< v
1
)
P
2

P
1
(1) becomes:
(½)ρ(v
1
) 2 + ρgy
1
= ρgy
2
Or, speed coming out of spigot: v
1
= [2g(y
2
- y
1
)]
½ “Torricelli’s Theorem”

Flow on the level
P
1
+ (½)ρ(v
1
) 2 + ρgy
1
= P
2
+ (½)ρ(v
2
) 2 + ρgy
2
• Flow on the level  y
1
= y
2

(1) becomes:
(1)
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
(2) Explains many fluid phenomena & is a quantitative statement of
Bernoulli’s
Principle:
“Where the fluid velocity is high, the pressure is low, and where the velocity is low, the pressure is high.”

Application #2 a) Perfume Atomizer
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”
• High speed air ( v )

Low pressure ( P )

Perfume is
“sucked” up!

Application #2 b) Ball on a jet of air
(Demonstration!)
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”
• High pressure (
P ) outside air jet

Low speed
( v

0 ). Low pressure ( P ) inside air jet

High speed ( v )

Application #2 c) Lift on airplane wing
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”
A
1
P
TOP
< P
BOT
 LIFT!

Area of wing top, A
2

Area of wing bottom
F
TOP
= P
TOP
A
1
F
BOT
= P
BOT
Plane will fly if
∑F = F
BOT
- F
TOP
A
2
- Mg > 0 !

Sailboat sailing against the wind!
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”

“Venturi” tubes
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”
Auto carburetor

Application #2 e)
“Venturi” tubes
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”
Venturi meter: A
1 v
1
With (2) this

P
2
= A
2 v
2
< P
1
(Continuity)

P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”

Air is forced to circulate!

Blood flow in the body
P
1
+ (½)ρ(v
1
) 2 = P
2
+ (½)ρ(v
2
) 2 (2)
“Where v is high, P is low, where v is
low, P is high.”

Blood flow is from right to left instead of up (to the brain)

Example: Pumping water up
Street level: y
1 v
1
= 0.6 m/s, P
1
= 0
= 3.8 atm
Diameter d
1
( r
1
= 2.5 cm ).
= 5.0 cm
A
1
= π(r
1
) 2
(
18 m up: r
2 y
2
= 1.3 cm
= 18 m, d
2
). A
2
= 2.6 cm
= π(r
2
) 2 v
2
= ? P
2
= ?
Continuity: A
1 v
1
 v
2
= A
2 v
2
= (A
1 v
1
)/(A
2
) = 2.22 m/s
Bernoulli:
P
1
+ (½)ρ(v
1
) 2 + ρgy
1

P
2
= P
2
+ (½)ρ(v
2
) 2 + ρgy
2
= 2.0 atm