1 sal engg collge andtechnology ahmedabad fluid dynamics by 130670106066

1

SAL ENGG COLLGE SAL ENGG COLLGE ANDTECHNOLOGY ANDTECHNOLOGY

AHMEDABADAHMEDABADFluid DynamicsFluid Dynamics

BY 130670106066BY 130670106066

2

College Board ObjectivesCollege Board Objectives II. FLUID MECHANICS AND THERMAL PHYSICSA. Fluid Mechanics1. 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.

3

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.Fluid mechanics, cont.

4

Fluid Mechanics, cont.Fluid Mechanics, cont.2. 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.

5

Fluid Mechanics, cont.Fluid Mechanics, cont.3. Fluid flow continuityStudents should understand the equation of

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

4. Bernoulli’s equationStudents should understand Bernoulli’s

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

6

10.1&2 Density & 10.1&2 Density & Specific GravitySpecific Gravity

• 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

V

m

7

Problem 10.5Problem 10.55. (II) A bottle has a mass of 35.00 g when empty

and 98.44 g when filled with water. When filled with another fluid, the mass is 88.78 g. What is the specific gravity of this other fluid?

5. Take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids.

fluid fluid fluidfluid

water waterwater

88.78 g 35.00 g0.8477

98.44 g 35.00 g

m V mSJ

m V m

SGfluid

8

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

• 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.

9

An An ideal fluidideal fluid is assumed 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).

10

10.3 Pressure10.3 PressureAny 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.

A

Fp

11

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

Pressure varies with depth.

P = F = Ahg so P = gh

A A

12

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.

13

PROBLEM 10-9PROBLEM 10-99. (I) (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?

9. (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

14

10.4 Atmospheric Pressure and Gauge 10.4 Atmospheric Pressure and Gauge PressurePressure

• 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

15

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

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

16

14.(II) (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

0

5 2

5 2 7

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

F PA

17

(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

18

10.5 Pascal’s Principle10.5 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

19

Problem 10-10Problem 10-10

10.(II) In a movie, Tarzan evades his captors by hiding underwater for many minutes while breathing through a long, thin reed. Assuming the maximum pressure difference his lungs can manage and still breathe is calculate the deepest he could have been. (See page 261.)

10.The pressure difference on the lungs is the pressure change from the depth of water

2

3 3 2

133 N m85 mm-Hg

1 mm-Hg 1.154 m 1.2 m

1.00 10 kg m 9.80 m s

PP g h h

g

20

10-7 Buoyancy and Archimedes’ Principle

This 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.

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

21


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

22


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.

23


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

24


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

25

22. (I) 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?

22. The difference in the actual mass and the apparent mass is the mass of the water displaced by the rock. The mass of the water displaced is the volume of the rock times the density of water, and the volume of the rock is the mass of the rock divided by its density. Combining these relationships yields an expression for the density of the rock.

rockactual apparent water rock water

rock

3 3 3 3rockrock water

9.28 kg1.00 10 kg m 2.99 10 kg m

9.28 kg 6.18 kg

mm m m V

m

m

26

24.(II) 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.

24.(a) 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

3 3

1

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

27

24. (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

28

34.(III) 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

1 1

F F m g m g V g V g

m mm m m m

water

wood woodPb wood wood

water

Pb Pb

1 11 1 10.50

5.25 kg 5.76 kg11 11 1

11.3

SGm m m

SG

29

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

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.

30

We will deal with laminar flow.

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.

This gives us the equation of continuity:


(10-4a)

31


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

32

10-9 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.

33

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?

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

.cm 0.2 2

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

1 min

V VA v r v v

t r t

34

• 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

2

pool pool 5

2hose"hose hose 51

2 8 "

5

3.05 m 1.2 m 4.429 10 s

1m0.40 m s

39.37

1day4.429 10 s 5.1 days

60 60 24s

V VAv t

t A v

35

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

1 sal engg collge andtechnology ahmedabad fluid dynamics by 130670106066

Documents