Fluids

• Fluids flow – conform to shape of container

– liquids OR gas

Density

• Defined as:

• Or:

• Specific gravity = / water at 4°C

– water at 4°C = 1.0 g/cm3 = 1000 kg/m3

– Specific gravity has no units

• density of air (1 atm, 0ºC): 0.00129 g/cm3 = 1.29 kg/m3

V

M

M V

Example 1

Find the volume of 20,000 N weigh water.

Pressure as Function of Depth

Static Fluid

(Fluid at rest)

P2A = P1A + mg

P2 = P1 + ρgh

For a given fluid, all points at same depth below surface are at the same pressure.

Example2: Deepest Fish

(a) Deepest fish ever sighted was on the floor of the Mariana's trench at 11,500m depth. What is the pressure there?

(b) What would be the inward force on a 20cm diameter circular of a submarine wall at that depth?

Which liquid container will have the highest pressure (at the black dot locations)?

(1)A, (2) B, (3) C, (4) D, (5) all the same.

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Consider the four points in the lake. The points are all at the same depth below the surface, but the depth of the bottom of the lake underneath the points varies. At

which point is the pressure the greatest? (1)A, (2) B, (3) C, (4) D, (5) all the same.

Pressure only depends on depth of point where pressure is being measured.

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Two beakers are filled with fluid. One is filled with water. The other is filled with a mixture of oil (specific gravity 0.8) and water to the same level. Which beaker

has the greatest pressure at the bottom of the beaker.

(1) The Water beaker

(2) The Oil and Water beaker

(3) Both the same

Pressure at depth where oil and water meet is lower in the oil and water mixture (oil less dense).

Increase in pressure from this level is same, since water in both columns.

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The following bulbs are connected by a fluid filled tube. The fluid is blue. How do the pressures in the two bulbs, P1 and P2, compare?

P1 P2

(1) P1 > P2

(2) P1 = P2

(3) P1 < P2

Pressure at level B greater than at level A in fluid

A

B

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Measuring Pressure Using a Column of Fluid

• Absolute Pressure, P

• Gauge Pressure = P – Po

• Po = Atmospheric Pressure

= 1.01 x 105 Pa

Po

Po

P

P=0

h

h

Manometer

Barometer

oP P gh

Height of column and density of fluid determine pressure difference.

Pressure sometimes given in inches of Mercury (Hg) 30.2”

Blood Pressure

120 mm of Mercury/80mm of MercuryGauge pressure (relative to atmosphere)

As you travel up a mountain, your ears pop. The figure below is a pitiful attempt to show the ear drum (in red). As you travel up a mountain, what will the ear drum do

between 'pops'? (1) Bow towards the middle ear

(2) Hold its shape

(3) Bow towards the outside

Pressure outside decreases.

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Example 3

A manometer is made of a solution that has a specific gravity of 4.15 and is used to measure the pressure of gas in a container as shown. The reading in manometer shows 650 mm.

(a) What is the gauge pressure?

(b) What is the absolute pressure?

The area of the left piston is 10mm2. The area of the right piston is 10,000mm2. What force must be exerted on the left piston to keep the 10,000-N car on the right

at the same height? (1) 10N

(2) 100N

(3) 10,000N

(4) 106N

(5) 108N

P = F1/A1 = F2/A2 F1 = (A1/A2) F2

F1 = (10 mm2 / 10,000 mm2) * 10,000N

Clicker! Hydraulic Car Lift

A1 = 10 mm2 A2 = 10,000 mm2

When an object is submerged in a fluid, there is a buoyant force. An object of 7N weigh submerged in a fluid is hanged by a rope, which has a tension of 5 N. What is the buoyant force?

(1) 1 N downward (2) 1 N upward (3) 2 N downward

(4) 2 N upward (5) 5 N downward (6) 5 N upward

(7) 12 N downward (8) 12 N upward

5N

7N

FB

Just another forceΣ FY = 0

5N – 7N + FB = 0FB = 7N – 5N

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Floating versus Completely Submerged

• Floating:FB = W VDISPLACED< VOBJECT

ρOBJ < ρFLUID

FB = ρFLUID VDISPLACED g

W = ρOBJ VOBJECT g

• Completely SubmergedVDISPLACED=VOBJECT

FB = ρ g VOBJECT

FB

FG

ρOBJ > ρFLUID

Object B is larger than A. If we put both objects into water, which of the following

statement is true? (1) Both A, and B will immerse into water.

(2) A will completely immerse, B will float.

(3) B will completely immerse, A will float.

(4) Both will float.

(5) None of the above is correct.

Density of A > density of waterDensity of B < density of water

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A2000 kg/m^3

B500 kg/m^3

Example 4: If ice has a density of 920 kg/m3, what percentage of an iceberg sticks out above the waterline?If the mass of the ice is 20 kg, what is the volume of the ice that is sticking out of water?

FBUOYANT = Weight

ρw g VDISPLACED = ρICE g VICE

ρw g VSUBMERGED = ρICE g VICE

VSUB / VOBJ = ρICE / ρw

VSUB / VOBJ = 0.92

So fraction not submerged is 1-0.92 or 0.08, 8%

An object of mass 1500 kg with a density of 2000 kg/m3 is put into water.(a) Will this object float?(b) What will be the weight of the object in water?

Example 5

Fill two identical glasses with water up to the rim. Now place ice in one (water will overflow into the sink). Now consider the two glasses, one with water, one with

water and ice. Which weighs more? (1) The glass without ice cubes

(2) The glass with ice cubes

(3) The two weigh the same

Ice less dense, but occupies more volume. Each cube displaces the weight of the cube in water.Think of the following: two beakers filled to the edge with water. Add an ice cube. The weight of the fluid lost over theedge equals the weight of the ice cube.

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An incompressible fluid is flowing through a pipe. At which point is the fluid

traveling the fastest?

(6) All points have the same speed

Smallest cross-sectional area, highest speed

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Bernoulli's Equation – Conservation of Energy

One way to look at it:

WNC = ΔKE + ΔPE

F*d (1/2)mv2 mgh

Divide by volume: ΔPressure (1/2)ρv2 ρgh

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

If height not changing much, higher speed leads to lower pressure

An incompressible fluid flows through a pipe. Compare the speed of the fluid at

points 1 and 2. (1) Greater at 1

(2) Greater at 2

(3) Both the same

(4) Not enough information

Area is smaller at 2, therefore fluid must travel faster in orderto conserve the volume flow rate.

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An incompressible fluid flows through a pipe. Compare the pressure at points 1

and 2. (1) Greater at 1

(2) Greater at 2

(3) Both the same

(4) Not enough information

P + (1/2) ρv2 + ρgh = constant(1/2) ρv2 is larger (higher v due to smaller A)ρgh is larger (higher)Therefore pressure must decrease at point 2

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

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Example 6Calculate the lift force of an airplane wing with a surface area of 12.0m2. Assume

the air above the wing is traveling at 70.0 m/s and the air below the wing is traveling at 60.0m/s. Assume the density of the air to be constant at 1.29 kg/m3.

FLIFT = FUP - FDOWN

= PBELOW A – PABOVE A

= (ΔP) A

From Bernoulli, find (ΔP) assuming Δy negligible.

PA + (1/2)ρvA2 + ρgyA = PB + (1/2) ρvB

2 + ρgyB

(ΔP) = PB – PA = ((1/2)ρvA2 – (1/2)ρvB

2) + (ρgyA – ρgyB)

(assume ΔPE negligible – if yA-yB = 10cm, only 1.26 Pa)

ΔP = (1/2) (1.29)(702 – 602 ) = 838.5 Pa

FLIFT = (838.5Pa)(12m2) = 1.01 x104 N

The Water TankWater is leaving a tank at a speed of 3.0 m/s. The

tank is open to air on the top. What is the height of the water level above the spigot?

Assume the area of the tank is much larger than the area of the spigot.

h = 0.459m

P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv2

2 + ρgy2

P0 + (1/2)ρv12 + 0 = P0 + (1/2)ρv2

2 + ρgh

Since A2 >> A1, v1 >> v2

(1/2)ρv12 = (essentially 0) + ρgh