Chapter 13

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Chapter 13

T . Norah Ali Al moneef

THE MECHANICS OF NONVISCOUS FLUIDS

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Steady/Unsteady flow• In steady flow the velocity of particles is constant with time• Unsteady flow occurs when the velocity at a point changes with

time. Turbulence is extreme unsteady flow where the velocity vector at a point changes quickly with time e.g. water rapids or waterfall.

• When the flow is steady, streamlines are used to represent the direction of flow.

• Steady flow is sometimes called streamline flow


• Streamlines never cross. A set of streamlines can define a tube of flow, the borders of which the fluid does not cross

Turbulent Flow Turbulent flow is an extreme kind of unsteady flow and occurs when there are sharp obstacles or bends in the path of a fast-moving fluid.

In turbulent flow, the velocity at a point changes erratically from moment to moment, both in magnitude and direction.

3T . Norah Ali Al moneef

Viscous/Non viscous flow• A viscous fluid such as honey does not flow readily, it has a large viscosity.• Water has a low viscosity and flows easily.• A viscous flow requires energy dissipation.• Zero viscosity – requires no energy. with no dissipation of energy. Some

liquids can be taken to have zero viscosity e.g. water.• An incompressible, non viscous fluid is said to be an ideal fluid

T . Norah Ali Al moneef 4

13. 2Stream flow



The volume flow rate, Q is defined as the volume of fluid that flows past an imaginary (or real) interface.

A v represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q


• If the fluid is incompressible, the density remains constant throughout

• Av represents the volume of fluid per second that passes through the tube and is called the volume flow rate Q

2211 vAvA AvQ

This just means that the amount of fluid moving in any “section of pipe” must remain constant.

•If the area is reduced the fluid must speed up!

tAvxAV

• Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter


The product of the area and the fluid speed at all points along a pipe is constant for an incompressible fluid.


Q: Have you ever used your thumb to control the water flowing from the end of a hose?

A: When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases.

This kind of fluid behavior is described by the equation of continuity.

Example: Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius of 0.305 m. How much oil flows in 1 day?


• A river flows in a channel that is 40. m wide and 2.2 m deep with a speed of 4.5 m/s.

• The river enters a gorge that is 3.7 m wide with a speed of 6.0 m/s. • How deep is the water in the gorge?

The area is width times depth. A1 = w1d1

Use the continuity equation. v1A1 = v2A2

v1w1d1 = v2w2d2

Solve for the unknown d2.

d2 = v1w1d1 / v2w2 =(4.5 m/s)(40. m)(2.2 m) / (3.7m)(6.0 m/s) = 18 m

example

example


example


If pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge andpipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe?

example


example


example

example


Example:The volume rate of flow in an artery supplying the brain is 3.6x10-6 m3/s. If the radius of the artery is 5.2 mm, A) determine the average blood speed.

B) Find the average blood speed if a constriction reduces the radius of the artery by a factor of 3 (without reducing the flow rate).


Example:

Example,decrease in area of stream of water coming from tap.

So v2 > v1 P2< p1


if the cross-section area, A, is 1.2 x 10-3m2 and the discharge, Q is 24 l / s , then the mean velocity,

if the area A 1 = 10 x10-3 m2 and A 2 = 3 x10-3 m 2 and the upstream mean velocity, 1 = 2.1 m / s , then the downstream mean velocity

Example

Example


• The figure shows 3 straight pipes through which water flows. The figure gives the speed of the water in each pipe.

• Rank them according to the volume of water that passes through the cross-sectional area per minute, greatest first. 6

Example:

sameExample:

• Water flows smoothly through the pipe shown in the figure, descending in the process.Rank the four numbered sections of pipe according to

• (a) the volume flow rate Q through them• (b) the flow speed v through them• (c) the water pressure p within them,

greatest first. a)Same b) 1,2,3,4 c ) 4,3,2,1

18

1)

2)

3)


P = 73,500 N/m2

PressurePressure is the ratio of a force F to the area A over which it is applied:

Pressure ; Force FPArea A

A = 2 cm2

1.5 kg2

-4 2

(1.5 kg)(9.8 m/s )2 x 10 m

FPA

example

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

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?(a) The total force of the atmosphere on the table will be the air pressure times the area of the

table.

The greater the force, the greater the pressure is. The greater the area, the smaller the force is.


example

T . Norah Ali Al moneef 21T . Norah Ali Al moneef

ExampleThe mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress.

The volume density of water at the normal condition (0oC and 1 atm) is 1000kg/m3. So the total mass of the water in the mattress is

Since the surface area of the mattress is 4.00 m2, the pressure exerted on the floor is

m

P

Therefore the weight of the water in the mattress is Wb) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor.

MWV kg31020.1300.000.200.21000

mg N43 1018.18.91020.1

AF

A

mg 3

4

1095.200.4

1018.1


13.3 Bernoulli’s Equation• Relates pressure to fluid speed and elevation• Bernoulli’s equation is a consequence of Work Energy Relation

applied to an ideal fluid• Assumes the fluid is incompressible and nonviscous, and flows

in a nonturbulent, steady-state mannerP2A P1A

area A


This work is negative because the force on the segment of fluid is to the left and the displacement is to the right. Thus, the net work done on the segment by these forces in the time interval


• States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline


Fluid At Rest In a Container :•Pressure in a continuously distributed uniform static fluid varies only withvertical distance and is independent of the shape of the container.• The pressure is the same at all points on a given horizontal plane in a fluid.• For liquids, which are incompressible, we have: pb + ρgh1 = patm + ρgh2

pb = patm + ρg (h2 - h1) = patm + ρgd

13.4 static consequences of Bernoulli's equation


h1

h2

d

Y a = y b = y c = yd

y A = y B = y C = y D

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Absolute Pressure and Gauge PressureThe pressure P is called the absolute pressure

Remember, P = Patm + ρ g hP – Patm = ρ g h ( so ρ g h is the gauge pressure)


As a fluid moves along if it changes speed or elevation then the pressure changes and vice versa.Bernoulli’s equation is really just conservation of energy for the fluid

• Bernoulli’s equation.• P + ½ ρv2 + ρg h = constant• Bernoulli’s equation shows that as the velocity of a fluid speeds up

it’s pressure goes down…..this idea used in airplane wings and frisbees (difference in pressure leads to upward force we call Lift)

Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphereUsual pressure gauges record gauge pressure. To calculate absolute pressure:

P = P atm + P gauge

, because it is the actual value of the system’s pressure.


PA – Patm = ρ g h

PA=PC

PA>PB

• P atm is atmospheric pressure =1.013 x 105 Pa• The pressure does not depend upon the shape of the

container

ghPP atm PA = PB = PC = PD


Variation of Pressure with Depth• If a fluid is at rest in a container, allportions of the fluid must be in static equilibrium• All points at the same depth must be at the same pressure– Otherwise, the fluid would not be inequilibrium

• Pressure changes with elevation•The pressure gradient in the vertical direction is

negative• The pressure decreases as we move upward

in a fluid at rest• Pressure in a liquid does not change due to

the shape of the container• The fluid would flow from the higher pressure

region to the lower pressure region•the pressure at a point in a fluid depends only on density, gravity and depth


example


The manometer

ghPPyygPP

atm

atm

)( 12

gygyPPgyPgyP

gyPPgyPP

PP

atmA

atmA

atmB

AB

A

12

21

2

1

Pressure Measurements:

• Manometers are devices in which one or more columns of a liquid are used

to determine the pressure difference between two points.

• U-tube manometer• One end of the U-shaped tube is open to the atmosphere• The other end is connected to the pressure to be measured


• Pressure at B is P=Patm+ρgh

Pressure at A = Pressure at B P A = P B = Patm+ ρgh

Pressure in a continuous static fluid is the same at any horizontal level so,

Pressure at A > Patm



Pressure at c =pressure at D

PC = PD

PC = PA + ρsgh

PB = Patm + ρ gh

PA + ρsgh = Patm + ρ gh

PA = Patm + ρ g h – ρsgh

Pblood =PA = Patm + ρ g h – ρsgh

Blood Pressure measurements by cannulation


Blood Pressure

• Blood pressure is measured with a special type of manometer called a sphygmomano-meter

• Pressure is measured in mm of mercury

Sphygmometer


Pressure with depthEstimate the amount by which blood pressure changes in an actuary in the foot P2 and in the aorta P1 when the person is lying down and standing up

Take density of blood = 1060kg/m3

PaPP

dingsPaghPP

downLying

412

12

104.135.18.91060

tan0


26.8 K P a


A fluid of constant density ρ = 960 kg / m3 is flowing steadily through the above tube. The diameters at the sections are d 1 =100 mm and d 2 = 80 mm . The gauge pressure at 1 is p1 = 200 k N/ m2 ,and the velocity here is u 1 = 5 m /s . We want to know the gauge pressure at section 2.

The tube is horizontal, with y1 = y2 so Bernoulli gives us the following equation for pressure at section 2:

P2 = 2 x10 5 + 960 { 52 – ( 7.8 )2} /2 = 182796.8 pa

example


An example of the U-Tube manometer Using a u-tube manometer to measure gauge pressure of fluid density ρ = 700 kg/m3, and the manometric fluid is mercury, with a relative density of 13.6. What is the gauge pressure if: h1 = 0.4m and h2 = 0.9m?

pB = pC pB = pA + ρgh1 pB = pAtmospheric + ρman gh2 We are measuring gauge pressure so patmospheric = 0 pA = ρman gh2 - ρgh1

a) pA = 13.6 x 103 x 9.8 x 0.9 - 700 x 9.8 x 0.4 = 117 327 N, b) pA = 13.6 x 103 x 9.8 x (-0.1) - 700 x 9.8 x 0.4 = -16 088.4 N,The negative sign indicates that the pressure is below atmospheric


Example: Fluid is flowing from left to right through the pipe. Points A and B are at the same height, but the cross-sectional areas of the pipe differ. Points B and C are at different heights, but the cross-sectional areas are the same.Rank the pressures at the three points, from highest to lowest.

A) A and B (a tie), CB) C, A and B (a tie)C) B, C, AD) C, B, AE) A, B, C

E) PA > PB > PC


Pressure Measurements:

• A long closed tube is filled with mercury and inverted in a dish of mercury

• Measures atmospheric pressure as ρ g h

Vacuum pressure = 0


For mercury, h = 760 mm. How high will water rise?

No more than h = patm/g = 10.3 m

h patm

p=0

atmatm

PP gh h

g

Vacuum!•Since the closed end is at vacuum, it does not exert any force.


We can set (assume the hole is on the ground or is where we measure height from). We also have 1 atm. So we have

v2h

example


example


P2 =p1 + ρ (v12 –v2

2 ) / 2 = 180X103 + 103X(22 – 182) = 20X 103 pa

example


example


If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium

• If the height doesn’t change much, Bernoulli becomes: y1 = y2

• Where speed is higher, pressure is lower.

• Speed is higher on the long surface of the wing – creating a net force of lift.

222

12

212

11 vPvP FL


example


1.5mWater flows through the pipe as shown at a rate of .015 m3/s. If water enters the lower end of the pipe at 3.0 m/s, what is the pressure difference between the two ends?

A2 = 20 cm2

5.18.9

21

21

21

21

/sm 5.720

015.0

1035.710

pp

pp10

3223

12

2

1

2

221

2

2

221

2

11

42

P

g

gg

AQ

AVQ

yyVV

yVyV

V

example


ExampleEstimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of the pool with a depth 5.0 m.

We first need to find out the pressure difference that is being exerted on the eardrum. Then estimate the area of the eardrum to find out the force exerted on the eardrum.

F

Since the outward pressure in the middle of the eardrum is the same as normal air pressure

Estimating the surface area of the eardrum at 1.0cm2=1.0x10-4 m2, we obtain

ghW Pa4109.40.58.91000

APP 0 N9.4100.1109.4 44

PP atm

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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

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

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.

Example. A diver is located 20 m below the surface of a lake ( = 1000 kg/m3). What is the pressure due to the water?

h = 1000 kg/m3

P = gh

The difference in pressure from the top of the lake to the diver is:

h = 20 m; g = 9.8 m/s2

3 2(1000 kg/m )(9.8 m/s )(20 m)P P = 196 kPa

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Example(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?

(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 20

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

Example:a) At what water depth is the pressure twice the atmospheric pressure?

b) What’s the pressure at the bottom of the 11-km-deep Marianas Trench, the deepest point

in the ocean?

Take 1 atm = 100 kPa & water = 1000 kg/m3 .

0p p g h 0p phg

3 2

2 3

1 100 10 / /

9.8 / 1000 /

atm Pa atm N m Pa

m s kg m

10 m

(a)

(b) Pressure increases by 1 atm per 10 m depth increment.

3 111 1010

atmp mm

1100 atm 110 MPa


•Air speeds up in the constricted space between the car & truck creating a low-pressure area. Higher pressure on the other outside pushes them together.

•Hold two sheets of paper together, as shown here, and blow between them. No matter how hard you blow, you cannot push them more than a little bit apart!

Applications of Bernoulli's Equation

Chapter 13

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Transcript of Chapter 13