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Applying  Bernoulli’s  Equa:on  

Lecture 5

net work done on fluid element = ΔKE + ΔPE

Bernoulli’s equation: p1 + ρ g y1 + ½ ρ v1

2 = p1 + ρ g y1 + ½ ρ v12

Solving Bernoulli’s equation: •  Identify points 1 and 2 along a streamline •  Define your coordinate system: where y=0 •  List your known and unknown variables •  Solve for your unknowns, possibly using the

continuity equation.

Example: The Venturi meter The Venturi meter is used to measure flow speed in a pipe. Derive an expression for v1

Note: We can’t apply Bernoulli’s equation between e.g. point 1 and the liquid in either of the vertical tubes. Bernoulli’s equation only applies to points on the same streamline.

Faster fluid speed leading to low pressure is the key to many common applications. •  The chimney effect: partly hot air rises; wind

across top of chimney ⇒ lower pressure at top ⇒ smoke forced up chimney

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Rabbits use this effect to avoid suffocation in their burrows. Different air flow across two holes produces a pressure difference, which forces a flow of air through the burrow.

•  Cyclones: strong wind generates a large negative pressure over the roof of a closed-up house

House destroyed by Cyclone Yasi in Queensland 2011

•  Passing trucks: the high speed of air between two trucks lowers the pressure between them, leading to a tendency for them to pull together.

Low pressure

Aside: Bernoulli’s equation and gases Gas flow can be considered incompressible so long as its velocity remains low (small compared with the speed of sound). As long as v < 0.005vs, the change in volume of the gas ΔV < 0.005V and can be ignored.

•  Lift on an aeroplane: flow lines crowd together above the wing ⇒ increased speed

⇒ reduced pressure ⇒ lift

Another way of looking at it: there is a net downward change in momentum in air flowing past ⇒ reaction force is upward

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•  Spin: a spinning ball creates a pressure difference on either side, resulting in a force on the ball. The spinning ball slows the air on one side and speeds up the air on the other, so there is a net force

The  siphon  

h1

h2

h3

A

B

C

Dv

Ques:ons  

•  This analysis doesn’t deal with the most interesting aspect: that water is flowing uphill!

•  What does p = 0 at point C mean?

•  Can you use a siphon in a vacuum? e.g. a mercury siphon on the Moon?

•  If so, why is the maximum height of a siphon determined by p0?

•  p = 0 at point C

Can you have negative pressure?

If we have a cylinder full of liquid and we apply pressure to the top, we increase the pressure in the liquid.

What happens if we invert the cylinder?

Liquid under tension ⇒ negative pressure

c.f. tensile strength of solids. Due to cohesive forces in liquid.

Tensile strength of water is somewhere in excess of 1000 atmospheres.

Negative pressure responsible for water transport in trees. At top of 100m tree, require pressure –10 bar.

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•  So why is the maximum height of a siphon determined by p0?

•  A liquid under tension is in a metastable state, and susceptible to cavitation: the formation of bubbles which “break” the column of liquid.

•  So the answer seems to be that a siphon without air pressure is possible in theory but difficult in practice.

Further  reading:  See  Physics  Today  ar:cle,  Nega%ve  Pressures  and  Cavita%on  in  Liquid  Helium,  hEp://www.aip.org/pt/feb00/maris.htm  

Next  lecture  

Real fluids – viscosity and turbulence