1 http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm

web notes: Fluidslect3.pdf

buoyancy.pdf

surface.pdf – not examinable

Lecture 3 Dr Julia Bryant

Buoyancy and Archimedes’ principle Fluid statics

• What is a fluid? Density Pressure • Fluid pressure and depth Pascal’s principle • Buoyancy Archimedes’ principle

Fluid dynamics • Reynolds number • Equation of continuity • Bernoulli’s principle • Viscosity and turbulent flow • Poiseuille’s equation

2

Buoyancy •  When a solid object is wholly

or partly immersed in a fluid, the fluid molecules are continually striking the submerged surface of the object. "

•  The forces due to these impacts can be combined into a single force the buoyant force which counteracts the weight."

T = W - FB

W

FB

T

3

If Fb > Fg body floats. If Fb < Fg body sinks. A body floats in any

liquid with density ρfluid > ρbody

Fb Fb > Fg Fg

Fb < Fg

DEMO

4

How high will it float? - What fraction of an iceberg is under water?

Water expands on freezing by 10%. Density of ice is 0.9g/cm3 Fraction of iceberg submerged is ρice / ρ water= 0.9/1.0

Therefore 90% of the iceberg is submerged.

5 DEMO

How can we measure the force due to buoyancy?

6

m

•  When an object is immersed in a fluid, there is an upward buoyant force equal to the weight of the volume of fluid displaced by the object.

•  This applies to either full or partial immersion (i.e. a “sinking” or “floating” object)

T = W - FB

W

FB

T

Archimedes’ Principle

7

W a t e r d i s p l a c e d

Hung

T = W - FB

Weight

FB

Tension

Weight

FB Contact

FC = W - FB

Sunk

8

weight

buoyant force

Floating:

Floating: fully submerged

Water displaced

Some fish can remain at a fixed depth without moving by storing gas in their bladder.

Submarines take on or discharge water into their ballast tanks.

equals Water

displaced

9

If you accidently drop the bottle into the water (with the lid on), how much of the Kahlua would you have had to have drunk for the bottle to float back up to you? mg

FB

You are floating around a pool drinking a bottle of Kahlua.

10

How much of the Kahlua would you have had to have drunk for the bottle to float back up to you?

FB > mg

mg

FB

What is the volume of the bottle and it’s contents?

Full bottle has 350mls of liquid = 350cm3 = 3.5x10-4m3 with a specific gravity 1.15

Small amount of air in the top =πR2h= π x (1.5cm)2 x 2cm = 14 cm3 = 14 ml

How much glass is there? 375g (empty bottle) with density of 2500kg.m3 ==> V=1.5x10-4m3 =150cm3

Total volume = 350 + 14 + 150cm3 = 514cm3 =5.14x10-4m3

11

Liquid before any is drunk VKah= 350cm3 = 3.5x10-4m3

ρKah = 1150 kg.m-3 Air in the top Vair=14 ml =1.4x10-5 m3 Glass Vglass= 150cm3 with ρglass = 2500Kg.m3 Total volume =5.14x10-4m3

Buoyant force FB = ρwater V g = 1000 x 5.14x10-4 x 9.8 = 5.037 N mg

FB

mg = (ρair Vair + ρKah VKah + ρglass Vglass ) x g but some has been drunk

Vinside bottle = Vair + Vkah_before = 3.5x10-4 + 1.4x10-5 m3 = 3.64x10-4 m3

mg = [1.21 x (3.64x10-4 - VKah ) + 1150 x VKah + 2500 x 1.5x10-4] x 9.8 N = 0.00432 + 11258.14 VKah + 3.675 N

The bottle will float if FB > mg

5.037 > 0.00432 + 11258.14 VKah + 3.675 ==> VKah < 1.21x10-4 m3 or VKah < 121ml (equivalent mass to 139mls water)

12

What is wrong with the picture of the ship on the left?

13

Why can ships float?

A steel ship can encompass a great deal of empty space and so have a large volume and a relatively small density."

Weight of ship = weight of water displaced"

14

A 200 tonne ship enters a lock of a canal. The fit between the sides of the lock is so tight that the weight of the water left in the lock after it closes is much less than the ship's weight. Can the ship float?

15

Volume of water displaced. This volume is not necessarily the volume present.

Weight of ship = weight of water displaced

The buoyant force is equal to the weight of the water displaced, not the water actually present. The missing water that would have filled the volume of the ship below the waterline is the displaced fluid.

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h!ρF"

bottom"

ρo"

Why is buoyancy due to the weight of fluid displaced? An object floats because of the pressure difference between the top and bottom of the object.

F = (pbottom – ptop) A

F = (patm + ρF g h – patm) A

F = ρF g h A = ρF VF g

F = mF g = Weight of displaced fluid =>Buoyant force

A!top"

p = p0 + ρ g h

17

DEMO

Hydrometer can be used to measure fluid density. A hydrometer floats due to buoyancy. Higher fluid density => higher buoyant force

Hydrometer

FB = ρ V g

ρwater = 1000 kg.m-3

ρkerosene = 817 kg.m-3

18

If a hydrometer was a rod that had a length of 0.250 m, cross sectional area was 2.00×10-4 m2, and mass of 4.50×10-2 kg,

(a) How far from the bottom end of the rod should a mark of 1.000 be placed to indicate the relative density of the water (density 1.000×103 kg.m-3)?

(b) If the hydrometer sinks to a depth of 0.229m when placed into an alcohol solution. What is the density of the alcohol solution?

Hydrometer - example

Let h be the height to which it is submerged. FB = ρAhg upwards Fg= mg downwards ρAhg = mg

In water: h = m = 4.50×10-2 = 0.225m (ρA) (1.000×103)(2.00×10-4)

In alcohol solution: ρ  = m = 4.50×10-2 = 983 kg.m -3

(Ah) (2.00×10-4)(0.229)

h

0.250m

A

19

? A Two cups are filled to the same level. One cup has ice cubes floating on it. Which weighs more?"

B Two cups are filled to the same level. One of the cups has ice cubes floating in it. When the ice melts, in which cup is the level higher?

Cup 1 Cup 2

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? A Two cups are filled to the same level. One cup has ice cubes floating on it. Which weighs more?"

Cups weigh the same. • Weight of the ice cubes is equal to the buoyant force. "• The buoyant force is equal to the weight of the water displaced by the ice cubes. "• This means that the weight that the ice cubes add to the cup is exactly what an amount of water that is equal to that submerged volume of ice cubes would add. "

mg FB

Cup 1 Cup 2

21

? A Two cups are filled to the same level. One cup has ice cubes floating on it. Which weighs more?"

B Two cups are filled to the same level. One of the cups has ice cubes floating in it. When the ice melts, in which cup is the level higher?

Cup 1 Cup 2

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? B Two cups are filled to the same level. One of the cups has ice cubes floating in it. When the ice melts, in which cup is the level higher?

The level is the same. • The weight of the ice cubes is equal to the weight of the water that would fill the submerged volume of the cubes. "• When the cubes melt into the water the volume of melted water is exactly equal to the volume of water that the cubes were displacing."

Cup 1 Cup 2

23

A giant clam has a mass of 470 kg and a volume of 0.350 m3 lies at the bottom of a freshwater lake. How much force is needed to lift it at constant velocity?

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FT + FB"

FG"

a = 0"m"

m = 470 kg

Vclam = 0.350 m3

g = 9.8 m.s-2

ρwater = 1.0x103 kg.m-3

FT + FB = FG"

FT = FG - FB FT = m g - ρwater g Vdisplaced FT = m g - ρwater g Vclam FT = (470)(9.8) – (103)(9.8)(0.350) N

FT = 1.2×103 N

FT = ? N ΣF = 0

= force to lift 70 full 750ml Kahlua bottles in air!

25

A ring weighs 6.327×10-3 N in air and 6.033×10-3 N when submerged in water.

What is the volume of the ring? What is the density of the ring? What is the ring made of?

air

water

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Volume? density? What is the ring made of?

air water

FTair = FG FTwater + FB = FG

Archimedes’ Principle FB = weight of water displaced FB = ρF Vring g

VR = 3.0x10-8 m3 ρR = 2.2x104 kg.m-3 maybe gold

Decrease in weight due to buoyant force "FB = FGair – FGwater = 6.327×10-3 - 6.033×10-3 = 0.294x10-3 N "Find the volume of the ring:"Since the ring is fully submerged" VR = FB/ρF g =(0.294x10-3) / {(103 )(9.8)} m3 " = 3.00x10-8 m3"

Find the density of the ring:"ρR = mR / VR = Fgair /(g VR)=(6.327x10-3) / {(9.8)(3.0x10-8 )} kg.m-3 " = 21.5x103 kg.m-3"

What is the ring made of?"

Weighs 6.327×10-3 N in air and 6.033×10-3 N in water.

REYNOLDS NUMBER A British scientist Osborne Reynolds (1842 – 1912) established that the nature of the flow depends upon a dimensionless quantity, which is now called the Reynolds number Re.

Re = ρ v L / η

ρ density of fluid v average flow velocity over the cross section of the pipe L characteristic dimension η viscosity

As a rule of thumb, for a flowing fluid

Re < ~ 2000 laminar flow ~ 2000 < Re < ~ 3000 unstable laminar to turbulent flow Re > ~ 2000 turbulent flow

pascal 1 Pa = 1 N.m-2

newton 1 N = 1 kg.m.s-2 1 Pa.s = kg.m.s-2 . m-2 .s

Re = ρ v L / η [Re] ≡ [kg.m-3] [m.s-1][m] [Pa.s] ≡ kg x m x m x s2.m2 = [1] m3 s kg.m.s Re is a dimensionless number

Consider an IDEAL FLUID Fluid motion is very complicated. However, by making some assumptions, we can develop a useful model of fluid behaviour. An ideal fluid is

Incompressible – the density is constant Irrotational – the flow is smooth, no turbulence Nonviscous – fluid has no internal friction (η=0) Steady flow – the velocity of the fluid at each point is constant in time.

Velocity of particle is tangent to streamline"

Streamlines cannot cross

Streamlines"- in steady flow, a bundle of streamlines makes a flow tube.

Consider the average motion of the fluid at a particular point in space and time."

An individual fluid element will follow a path called a flow line."

Steady flow is when the pattern of flow lines does not change with time.

v