Sect. 14.6: Bernoulli’s Equation

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!

Daniel Bernoulli 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. Bernolli’s Equation

Net work done on the segment: W = (P 1 – P 2 ) V. Part of this goes into changing kinetic energy & 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 – mgy 1. Work also equals change in energy. Combining: (P 1 – P 2 )V =½ mv ½ mv mgy 2 – mgy 1

Rearranging and expressing in terms of density: P 1 + ½  v  gy 1 = P 2 + ½  v  gy 2 This is Bernoulli’s Equation. Often expressed as P + ½  v 2 +  gy = constant When fluid is at rest, this is P 1 – P 2 =  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 Bernolli’s Equation

Applications of Fluid Dynamics 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

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

Atomizer 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 (1) Flow on the level  y 1 = y 2  (1) becomes: 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.” P TOP < P BOT  LIFT! A 1  Area of wing top, A 2  Area of wing bottom F TOP = P TOP A 1 F BOT = P BOT A 2 Plane will fly if ∑F = F BOT - F TOP - 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 = A 2 v 2 (Continuity) With (2) this  P 2 < P 1

Ventilation in “Prairie Dog Town” & in chimneys etc. 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 = 0 v 1 = 0.6 m/s, P 1 = 3.8 atm Diameter d 1 = 5.0 cm (r 1 = 2.5 cm). A 1 = π(r 1 ) 2 18 m up: y 2 = 18 m, d 2 = 2.6 cm (r 2 = 1.3 cm). A 2 = π(r 2 ) 2 v 2 = ? P 2 = ? Continuity: A 1 v 1 = A 2 v 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 + (½)ρ(v 2 ) 2 + ρgy 2  P 2 = 2.0 atm