AP Physics B Fluid Dynamics 1. College Board Objectives. FLUID MECHANICS AND THERMAL PHYSICS A.Fluid...

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AP Physics BFluid Dynamics

College Board Objectives . FLUID MECHANICS AND THERMAL PHYSICSA. Fluid MechanicsB. Hydrostatic pressure

Students should understand the concept of pressure as it applies to fluids, so they can:

a) Apply the relationship between pressure, force, and area.

b) Apply the principle that a fluid exerts pressure in all directions.

c) Apply the principle that a fluid at rest exerts pressure perpendicular to any surface that it contacts.

Fluid mechanics, cont.

d) Determine locations of equal pressure in a fluid. e) Determine the values of absolute and gauge pressure for a particular situation.

f) Apply the relationship between pressure and depth in a liquid, P = g h

Fluid Mechanics, cont.

BuoyancyStudents should understand the concept

of buoyancy, so they can:a) Determine the forces on an object

immersed partly or completely in a liquid.

b) Apply Archimedes’ principle to determine buoyant forces and densities of solids and liquids.

Fluid Mechanics, cont.

Fluid flow continuityStudents should understand the

equation of continuity so that they can apply it to

fluids in motion.

Bernoulli’s equationStudents should understand Bernoulli’s

equation so that they can apply it to fluids in motion.

Equations :

P = F/ A

Sp.Gr. = ρ substance / ρwaterP = ρhg_____________________________ Ptotal = Patm + P liquidghpp 12

Homework : Chapter 9 Solids and Fluids Summaries 2 Examples per section 2 End of Chapter 9 problems per section 2 PROBLEMS IN THE COLLEGE BOARD https://apstudent.collegeboard.org/apcourse/ap-physics-b/exam-practice

10.1&2 Density & Specific Gravity1. The mass density of a substance is the mass of the substance divided by the volume it occupies:

unit: kg/m3

for aluminum 2700 kg/m3 or 2.70 g/cm3

mass can be written as m = V and

weight as mg = Vg

Specific Gravity: substance / water

2. A fluid - a substance that flows and conforms to the boundaries of its container.

3. A fluid could be a gas or a liquid; however on the AP Physics B exam fluids are typically liquids which are constant in density.

4. An ideal fluid is assumed

• to be incompressible (so that its density does not change),

• to flow at a steady rate, • to be nonviscous (no friction between the fluid

and the container through which it is flowing), and • flows irrotationally (no swirls or eddies).

5. Turpentine has a specific gravity of 0.9 . What is its density ?

Specific Gravity= substance / water

Ρsubstance = sp.gr. X ρwater =

0.9 X 1000kg/m3 = 900 kg / m312

• 6. A cork has volume of 4 cm3 and weighs 10-2 N . What is the specific gravity of cork ?

Fg = Weight = mg = 10-2 N = m (10 m/s2) m = 10-2 N / 10m/s2 = 10-3 kg

ρcork = m/v= 10-3 kg / 4cm3 ( 100 cm)3 / m3 = 250 kg / m3

sp. Gr. = ρsubstance /ρwater = 250kg /m3 / 1000kg /m3

sp. Gr. = 0.25 14

10.3 5. Pressure Any fluid can exert a force perpendicular to its

surface on the walls of its container. The force is described in terms of the pressure it exerts, or force per unit area:

Units: N/m2 or Pa (1 Pascal*) dynes/cm2 or PSI (lb/in2)

1 atm = 1.013 x 105 Pa or 15 lbs/in2

*One atmosphere is the pressure exerted on us every day by the earth’s atmosphere.

6. The pressure is the same in every direction in a fluid at a given depth.

7.Pressure varies with depth.

P = F/A = mg/A = ρVg/A

P = F = Ahg so P = gh

8. A FLUID AT REST EXERTS PRESSURE PERPENDICULAR TO ANY SURFACE THAT IT CONTACTS. THERE IS NO PARALLEL COMPONENT THAT WOULD CAUSE A FLUID AT REST TO FLOW.

PROBLEM 10-99. (a) Calculate the total force of the atmosphere

acting on the top of a table that measures

(b) What is the total force acting upward on the underside of the table?

m. 2.9m 6.1

PROBLEM 10-99. (a) Calculate the total force of the atmosphere

acting on the top of a table that measures Patmosphere = 1.013 X 105 N/m2

(b) What is the total force acting upward on the underside of the table?

a. The total force of the atmosphere on the table will be the air pressure times the area of the table.

5 2 51.013 10 N m 1.6 m 2.9 m 4.7 10 NF PA (b) Since the atmospheric pressure is the same on the underside of the

table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table,

54.7 10 N

m. 2.9m 6.1

10. A vertical column made of cement has a base area of 0.5 m2

If its height is 2m , and the specific gravity of cement is 3 , how much pressure does this column exert on the ground ?

10. A vertical column made of cement has a base area of 0.5 m2

If its height is 2m , and the specific gravity of cement is 3 , how much pressure does this column exert on the ground ?

P = F / A = mg /A = ρVg /A = ρAhg/A = ρhgSp. Gr. = ρ cement / ρwater

ρcement = Sp. Gr. X 1000 kg / m3 = 3 X 1000 kg / m3 =

3000 kg/m3

P = ρh g = 3,000 kg/m3 X 2m X 10m/s2 = 60,000 Pa = 60k Pa 21

11. Atmospheric Pressure and Gauge Pressure

• The pressure p1 on the surface of the water is 1 atm, or 1.013 x 105 Pa. If we go down to a depth h below the surface, the pressure becomes greater by the product of the density of the water , the acceleration due to gravity g, and the depth h. Thus the pressure p2 at this depth is

h h hp2 p2 p2

p1 p1p1

ghpp 12

12. In this case, p2 is called the absolute(total) pressure -- the total static pressure at a certain depth in a fluid, including the pressure at the surface of the fluid

13. The difference in pressure between the surface and the depth h is gauge pressure

ghpp 12

Note that the pressure at any depth does not depend of the shape of the container, only the pressure at some reference level (like the surface) and the vertical distance below that level.

h h hp2 p2 p2

p1 p1p1

14. (a) What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? (b) What will be the pressure against the side of the pool near the bottom?

(a)The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool.

14. (a) What are the total force and the absolute pressure on the bottom of a swimming pool 22.0 m by 8.5 m whose uniform depth is 2.0 m? (b) What will be the pressure against the side of the pool near the bottom?

(a)The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool.

5 2 3 3 2

1.013 10 N m 1.00 10 kg m 9.80 m s 2.0 m

1.21 10 N m

1.21 10 N m 22.0 m 8.5 m 2.3 10 N

P P gh

(b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the

bottom,

5 21.21 10 N mP

CW : Hydrostatic Pressure • 1. What is the hydrostatic gauge pressure at a point 10m

below the surface of the ocean ? The specific gravity of seawater is 1.025.

• 1. What is the hydrostatic gauge pressure at a point 10m below the surface of the ocean ? The specific gravity of seawater is 1.025.

Gauge pressure = ρgh = P2 - P1 Sp. Gr. = ρsubstance / ρwater

ρocean = Sp. Gr. X 1000kg/m3 = 1.025 X 1000kg/m3 =

= 1025 kg/ m3

Gauge Pressure = ρgh = 1025 kg/m3 ( 10 m/s2) ( 10m) = = 102, 500 Pa 28

2. A swimming pool has a depth of 4 m . What is the hydrostatic gauge pressure at a point 1 m below the surface ?

Gauge Pressure = ρhg = 1000kg/m3 ( 1m) (10m/s2) = = 10,000 Pa

15. What happens to the gauge pressure if we double our depth below the surface of the liquid ? What happens to the total pressure ?

15. What happens to the gauge pressure if we double our depth below the surface of the liquid ? What happens to the total pressure ? Gauge pressure will double. Gauge Pressure = ρgh

Total Pressure = Patm + ρgh

It will increase based on ρgh and having Patm as constant

16. A flat piece of wood , of area 0.5m2, is lying at the bottom of a lake . If the depth of the lake is 30 m , what is the force on the wood due to pressure ? ( use Patm= 1 X 10 5 Pa)

F= P/A P total = P atm + ρgh = 1 X 105 Pa + (1000 kg /m3) ( 10 m/s2) ( 30m) = 4 X 10 5 Pa

Force = P A = 4 X 10 5 Pa X 0.5 m2 = 2 X 10 5 N

CW : Hydrostatic Pressure • 3. Consider a closed container , partially filled with a liquid of

density ρ = 1200 kg/m3 and a point X that’s 0.5 m below the surface of the liquid .

• A. if the space above the surface of the liquid is vacuum , what is the absolute pressure at point X ?

• B. if the space above the surface of the liquid is occupied by a gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X?

• 35

CW : Hydrostatic Pressure • Consider a closed container , partially filled with a liquid of

• P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m)• = 6 X 10 3 Pa • B. if the space above the surface of the liquid is occupied by a

gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X?

CW : Hydrostatic Pressure • Consider a closed container , partially filled with a liquid of

• P absolute = P atm + ρgh = 0 + 1200 kg/m3 (10m/s2)( 0.5m)• = 6 X 10 3 Pa • B. if the space above the surface of the liquid is occupied by a

gas whose pressure is 2.4 X 104 Pa , What is the absolute pressure at Point X?

• P absolute = P1 +ρgh = 2.4 X 104 Pa + 6 X 10 3 Pa = 3X 104 Pa 37

10.5 Pascal’s Principle

• Pascal’s Principle - if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Applications: hydraulic lift and brakes

Pout = Pin

And since P = F/a

Fout = Fin

Aout Ain

Mechanical Advantage:

Fout = Aout

Fin Ain

17. Buoyancy and Archimedes’ PrincipleThis is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different.

18. The buoyant force is found to be the upward force on the same volume of water:

10-7 Buoyancy and Archimedes’ Principle

19. The net force on the object is then the difference between the buoyant force and the gravitational force.

20. If the object’s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water.

21. For a floating object, the fraction that is submerged is given by the ratio of the object’s density to that of the fluid.

This principle also works in the air; this is why hot-air and helium balloons rise.

22.A geologist finds that a Moon rock whose mass is 9.28 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock?

ρrock = mrock/ Vrock

ρwater = mwater / V water

Vrock = Vdisplaced water = Mdisplaced water /ρ water

= 9.28kg -6.18 kg / 1000 kg/m3

= 0.0031 m3

ρrock = mrock/ Vrock = 9.28 kg / .0031 m3

ρrock =2.99X 103 kg/m3

23. A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water.

Tension (T)

Fb = Fwater Buoyant Force mg

23. A crane lifts the 18,000-kg steel hull of a ship out of the water. Determine (a) the tension in the crane’s cable when the hull is submerged in the water, and (b) the tension when the hull is completely out of the water.

When the hull is submerged, both the buoyant force and the tension force act upward on the hull, and so their sum is equal to the weight of the hull. The buoyant force is the weight of the water displaced.

buoyant

hull waterbuoyant hull water sub hull water hull

hull hull

3 34 2 5 5

1.00 10 kg m 1.8 10 kg 9.80 m s 1 1.538 10 N 1.5 10 N

7.8 10 kg m

T F mg

mT mg F m g V g m g g m g

(b)When the hull is completely out of the water, the tension in the crane’s cable must be equal to the weight of the hull.

4 2 5 51.8 10 kg 9.80 m s 1.764 10 N 1.8 10 NT mg

24. A 5.25-kg piece of wood

floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink?

For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead.

0.50SG

A 5.25-kg piece of wood

floats on water. What minimum mass of lead, hung from the wood by a string, will cause it to sink?

34.For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead.

0.50SG

weight buoyant wood Pb wood water Pb water

wood Pb water waterwood Pb water water Pb wood

wood Pb Pb wood

F F m g m g V g V g

m mm m m m

wood woodPb wood wood

1 11 1 10.50

5.25 kg 5.76 kg11 11 1

SGm m m

10-8 Fluids in Motion; Flow Rate and the Equation of Continuity

25. If the flow of a fluid is smooth, it is called streamline or laminar flow (a).

Above a certain speed, the flow becomes turbulent (b). Turbulent flow has eddies; the viscosity of the fluid is much greater when eddies are present.

We will deal with laminar flow.

26.The mass flow rate is the mass that passes a given point per unit time. The flow rates at any two points must be equal, as long as no fluid is being added or taken away.

Flow rate f = A v A = Cross-sectional area

v = flow speed

28. This gives us the equation of continuity:

(10-4a)

28. If the density doesn’t change – typical for liquids – this simplifies to . Where the pipe is wider, the flow is slower.

Flow speed is inversely proportional to the cross-sectional area or the square of the radius of the pipe

29. A pipe of non uniform diameter carries water. At one point in the pipe , the radius is 2 cm and the flow speed is 6 m/s .

a. What is the flow rate ?

b. What is the flow speed at a point where the pipe constricts to a radius of 1 cm ?

A pipe of non uniform diameter carries water. At one point in the pipe , the radius is 2 cm and the flow speed is 6 m/s .

a.What is the flow rate ?

f = Av = πr2v = π ( 2 X 10 -2m ) 2 (6m/s)

A pipe of non uniform diameter carries water. At one point in the pipe , the radius is 2 cm and the flow speed is

6 m/s .

a.What is the flow rate ?

f = Av = πr2v = π ( 2 X 10 -2m ) 2 (6m/s)

π ( 2 X 10 -2m ) 2 (6m/s) = π ( 1 X 10 -2m ) 2 v

v = 24 m/s

• 39. (II) A (inside) diameter garden hose is used to fill a round swimming pool 6.1 m in diameter. How long will it take to fill the pool to a depth of 1.2 m if water issues from the hose at a speed of

• 39. The volume flow rate of water from the hose, multiplied times the time of filling, must equal the volume of the pool.

inch-85

?sm 40.0

pool pool 5

2hose"hose hose 51

3.05 m 1.2 m 4.429 10 s

1m0.40 m s

1day4.429 10 s 5.1 days

60 60 24s

V VAv t

40. (II) What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 15 m?

40. Apply Bernoulli’s equation with point 1 being the water main, and point 2 being the top of the spray. The velocity of the water will be zero at both points. The pressure at point 2 will be atmospheric pressure. Measure heights from the level of point 1.

2 21 11 1 1 2 2 22 2

3 3 2 5 2

1 atm 2

1.00 10 kg m 9.8m s 15 m 1.5 10 N m

P v gy P v gy

P P gy

30. Bernoulli’s Equation

A fluid can also change its height. By looking at the work done as it moves, we find:

This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure goes down.

Bernoulli’s Equation

• P1 + ρgy1 + ½ ρv12 = P2 + ρgy2 + ½ ρv2

31. In the Figure below , a pump forces

Water at a constant flow rate through a pipe whose cross-sectional area , A gradually decreases. At the exit point A has decreased by 1/3 its value at the beginning of the pipe. If y =60cm and the flow speed of the water just after it leaves the pump is 1m/s , what is the gauge pressure at point 1?

P1 + ρgy1 + ½ ρv12 = PATM + ρgy2 + ½ ρv2

30. Bernoulli’s Equation

A fluid can also change its height. By looking at the work done as it moves, we find:

This is Bernoulli’s equation. One thing it tells us is that as the speed goes up, the pressure goes down.

31. In the Figure below , a pump forces

Water at a constant flow rate through a pipe whose cross-sectional area , A gradually decreases. At the exit point A has decreased by 1/3 its value at the beginning of the pipe. If y =60cm and the flow speed of the water just after it leaves the pump is 1m/s , what is the gauge pressure at point 1?

P1 + ρgy1 + ½ ρv12 = PATM + ρgy2 + ½ ρv2

P1 – PATM = ( ρgy2-ρgy1) + ½ ρv22 - ½ ρv1

Pgauge = (1000kg/m3)((10m/s2)(0.6m) + ½ (3v1)2 -v12) =

(1000kg/m3)((6 m2 / s2 +4 m2/s2 ) = 10,000 Pa

CW: Bernoulli’s Equation

• What does Bernoulli’s Equation tell us about a fluid at rest in a container open to the atmosphere?

CW: Bernoulli’s Equation

• What does Bernoulli’s Equation tell us about a fluid at rest in a regular container open to the atmosphere?

Because the fluid is at rest , both v1 and v2 are zero and Bernoulli’s equation becomes

P1 + ρgy1 = P2 + ρgy2

P1 = Patm = 1X 105 Pa

P2 = Patm + ρg ( y1 – y2) Hydrostatic Pressure

36. (I) A 15-cm-radius air duct is used to replenish the air of a room every 16 min. How fast does air flow in the duct?

We apply the equation of continuity at constant density, Eq. 10-4b. Flow rate out of duct = Flow rate into room

2 room room

duct duct duct duct 22

to fill to fillroom room

9.2 m 5.0 m 4.5 m 3.1m s

60 s0.15 m 16 min

V VA v r v v

m 4.5m 5.0m 2.9

• 39. (II) A (inside) diameter garden hose is used to fill a round swimming pool 6.1 m in diameter. How long will it take to fill the pool to a depth of 1.2 m if water issues from the hose at a speed of

• 39. The volume flow rate of water from the hose, multiplied times the time of filling, must equal the volume of the pool.

inch-85

?sm 40.0

pool pool 5

2hose"hose hose 51

3.05 m 1.2 m 4.429 10 s

1m0.40 m s

1day4.429 10 s 5.1 days

60 60 24s

V VAv t

40. (II) What gauge pressure in the water mains is necessary if a firehose is to spray water to a height of 15 m?

40. Apply Bernoulli’s equation with point 1 being the water main, and point 2 being the top of the spray. The velocity of the water will be zero at both points. The pressure at point 2 will be atmospheric pressure. Measure heights from the level of point 1.

2 21 11 1 1 2 2 22 2

3 3 2 5 2

1 atm 2

1.00 10 kg m 9.8m s 15 m 1.5 10 N m

P v gy P v gy

P P gy

Visit the follow website from Boston University

• http://physics.bu.edu/~duffy/py105.html• For more information about (choose from left panel)

• Pressure; Fluid Statics• Fluid Dynamics• Viscosity

At the website complete the following:1. Read and record important equations and facts.2. For each equation write the quantity for each symbol3. Write the unit for each quantity (symbol ok)

Demonstrations to View• http://

www.csupomona.edu/~physics/oldsite/demo/fluidmech.html

Experiment : Density VS Buoyancy I. Purpose : To determine the density of water .To determine if 10 objects will float or sink.To determine the density of the 10 objects To calculate the weight of the 10 objects .To calculate the buoyant force exerted on the 10 objects when

submerged in water.

II. Materials :10 objects triple beam balance Water digital scale Cylinder

III.Data Table 1: Density of FluidWaterWater Mass in gMass in g V, ccV, cc D=M/VD=M/V

In g/ccIn g/cc

Trial 1Trial 1 10 mL10 mL

III. Data Table 2: Float or Sink

Name of the ObjectName of the Object WaterWater

III. Data Table 3 : Mass and Weight ObjectsObjects Mass, gMass, g Mass,KgMass,Kg Weight ,NWeight ,N

1.1. 22.2 22.2 .0222.0222 .222.222

4.4.5.5.

Data Table 4: Densities of ObjectsName Name Mass in gMass in g V, ccV, cc D=M/VD=M/V

In g/ccIn g/cc

10. 10.

Data Table 4: Buoyant Force VS Weight ObjectsObjects Volume of Volume of

object , ccobject , ccDensity of Density of Fluid Fluid

Buoyant Buoyant Force , NForce , N

WeightWeight

, N, NWhich is greater Which is greater Fw or FBFw or FB

IV. Calculations: Show all calculations of Densities of Fluids, Densities of Objects , Weights and Buoyant force.

V. Analysis and Calculations :Explain how the density of the fluid was verified.Explain how the density of objects were calculated.Explain how weight is calculated .Explain how the Buoyant force is calculated for each object .How does the density of the object relate to buoyancy ? ExplainHow will you prove that density and buoyancy are related .

AP Physics B Fluid Dynamics 1. College Board Objectives. FLUID MECHANICS AND THERMAL PHYSICS A.Fluid...

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