FLUIDS 1.pdf

Fluid

1. Introduction

2. Physical Quantities

3. Kinematics

4. Forces & Universal Gravitation

5. Work, Energy and Power

6. Linear Momentum and Collision

7. Fluid Mechanics

8. Thermodynamics

FLUIDS

- any substance that does not

have definite shape and exhibits the phenomenon of flow.

-A fluid is a collection of

molecules that are randomly arranged and held together by weak cohesive forces and by forces exerted by the walls of a container.

FLUIDS

• Fluids (derived from “to flow”) include liquids and gases

• Liquids flow under gravity until they occupy the lowest possible regions of their containers

• Gases expand to fill their containers regardless of the container’s shape



Fluid Statics

– study of fluids at rest.



Fluid Dynamics

– study of fluids in motion.

Outline

• Part 1: Fluid Statics

• 1. Density and Specific Gravity • 2. Pressure in a Fluid

• 3. Pascal's Principle

• 4. Buoyancy and Archimedes’ Principle

Fluid Statics

• Study of Fluids at rest

• Includes density, pressure and buoyancy principles

• Very applicable in almost all the scientific and engineering fields. • Used to determine a lot of other

material characters.

• Considered in aspects that

1. Density and Specific Gravity

• Density

– Important property of a substance

– Ratio of mass to its volume • Density = ρ = mass = m • volume V

• The gram was originally defined as the mass of one cubic

centimeter of water, so the density of water served as a benchmark

1. Density and Specific Gravity

• Densities of substances vary with temperature

• The max. value of ρ_w occurs at 4o_C.

• A convenient unit of volume for fluids is the liter (L)

• 1 L = 103 _cm3 _{{cc} = 10}-3 _m3

Material Density (kg/m3)

air 1.2

water 1 x 103

aluminum 2.7 x 103

white dwarf stars 109

neutron stars 1018

• When an object’s density is

greater than that of water it will sink in water

• When its density is less it will float

• For floating object- the fraction of the volume of an object that is submerged in any liquid

equals the ratio of the density of the object to that of the liquid.

• Example: ice-0.92 g/cm3 ₍₉₂

percent of its volume is submerged)

1. Density and Specific Gravity

• The ratio of the density of the object to that of water is called specific gravity. It is unit-less! • s.g. = ρ_subs

• ρ_w

• If 0 < s.g. < 1.00 then it FLOATS • If s.g. > 1.00 then it SINKS

- is the ratio of the density of a material to a standard

density.

standard

density

For liquids: density of water at 4°C

1 g/cm3 _{or 1000 kg/m}3

For gases: density of oxygen at 0°C and pressure of 1 atm

1.43 kg/m3

 1. Rank the four materials in increasing densities as they

are submerged in water.

Example:

Spill it…

• A 200-ml flask is filled with water at 4o_{C. When the flask is heated}

a0t 80o_{C, 6g of water spill out.}

What is density of water 80o_C?

(Assume that the expansion of the flask is negligible.)

• Picture the problem

– The density of water at 80oC is ρ’

= m’/V (where V = 200ml = 200 cm3₎

Spill it…

1. Calculate the original mass of water in the flask at 4oC using ρ =1g/cm3_.

m = ρV = (1g/cm3_{) (200 cm}3_{) = 200 g}

2. Calculate the mass of water

remaining, m’, after 6g spills out m’ = m - 6g = 200g – 6g = 194g

3. Use this value of m’ to find ρ’ (density of water at 80o_C)

ρ’ = m’/V = 194g/200cm3

1. Density and Specific Gravity

A REMARK:

• Most solids and liquids expand only slightly when heated, and contract slightly when subjected in external pressure.

• Since these changes in volume are relatively small, we often treat the densities of solids and liquids as approximately independent of temperature and pressure.

• However, this is different from gases, in Phys 3, we always put gasses into standard conditions (Standard Temperature and

2. Pressure in a Fluid

• This force per unit area is called pressure P of the fluid:

A

F

2. Pressure in a Fluid

• Pressure [P] – Scalar

– Dual SI Unit:

• 1 N/m2 _{= 1 Pa (Pascal)}

– Other customary units are given in conversions

– 1 atm = 101325 Pa = 14.70 lb/in2

– 1 atm (atmosphere) is the pressure exerted by the atmosphere at sea level

Name Value(N/m2 = Pa)

1 pascal(Pa) 1

1 bar 1.00 x 105

1 atmosphere(atm) 1.013 x 105

1 mm Hg 1.33 x 102

1 torr 1.33 x 102

1 _lb/in2 _{6.89 x 10}3

• Which block exerts the greater force on the surface?

• Which block Exerts the greater pressure?

After a long Lecture, the daring

professor stretches out for a nap in a bed of nails, how is this

2. Pressure in a Fluid

• When a body is submerged in a fluid such as water, the fluid exerts a force

2. Pressure in a Fluid

• As any scuba diver knows, the pressure in a lake or ocean increases with depth.

• As any sky diver knows, the pressure of the atmosphere decreases with altitude.

• For water, density constant

2. Pressure in a Fluid

• We can prove this by considering a column of liquid (right) of height h

and cross-sectional area

A.

• To support the weight of the column, the pressure at the bottom must be

greater than the pressure at the top. The weight of the liquid column is

2. Pressure in a Fluid

• w = mg = ρVg = ρAhg • If P₀ is the pressure at the top

and P is the pressure at the bottom, the net upward force exerted by this pressure

difference is PA – P₀A.

• Setting this upward force equal to the weight of the column, we obtain:

• PA – P₀A = ρAhg

• P – P₀ = ρgh (ρ constant)

1. More dense liquid-> greater pressure

2. Pressure in a liquid at any location is exerted in equal amount in all direction

3. Pressure is depth dependent not volume dependent

Notes :

P = P

₀

+ ρgh

• 5. Liquid pressure acts

perpendicular to the sides of the container and increases with

2. Pressure in a Fluid

• LAKER EXERCISE:

• Find the pressure at a depth of 10 m below the surface of the Laguna

Lake if the pressure at the surface is 1 atm.

• Solution:

• P₀ = 1 atm = 101325 Pa • P = P₀ + ρ_wgh

• P = 101325 Pa + (103 _kg/m3_{) (9.8 m/s}2_{) (10 m)}

• P = 199325 Pa = 1.97 atm

A nurse administers medication

in a saline solution to a patient by infusion into the patient’s arm. The density of the

solution is 1 x 103 _kg/m3_{. And}

the pressure inside the vein is 2.4 x 103 _{Pa. How high must}

the container be hung?

p_liquid = p_vein

gh = 2.4 x 103 _Pa

solving for h:

h = p_vein = 2.4 x 103 _Pa

g (9.8m/s2)(1 x 103 kg/m3)

3. Pascal's Principle

• If we increase the pressure by

pressing down on the top surface with a piston, the increase in

pressure is the same throughout the liquid.

• This is known as Pascal's Principle, named after Blaise Pascal (1623-1662)

• “Pressure applied to an

3. Pascal's Principle

• A common application is the hydraulic lift shown right

• A small force F₁ on the small piston produces a change in pressure F₁/A₁ that is

transmitted by the liquid to the large piston.

• Since the pressures at the small and large pistons are the same, the forces are related by F₂/A₂ = F₁/A₁ ,

• The force on the large piston F₂ = (A₁/A₂)F₁ is much

3. Pascal's Principle

• PISTON BANG BANG

• The large piston in a hydraulic lift has a radius of 20 cm. What force must be applied to the small

piston of radius 2 cm to raise a car of mass 1500 kg?

• F₁ = PA₁

• PA₂ = F₂ = mg, so

• F₁ = PA₁ =>

• = (1500 kg) (9.81 m/s2) * (2cm/20 cm)2

• = 147 N

P= mg

A₂

mg

A₂ A1= mg

r₁2

2. Pressure in a Fluid

• Figure at the right shows water in a container with different shapes.

• At first glance, it might seem that the pressure in the

largest section of the

container would be greatest and that water would

therefore be forced to a greater height in smaller sections of the container. • The fact that this does not

happen is known as

2. Pressure in a Fluid

• HYDROSTATIC PARADOX

– Although water in the largest section weighs more than in the smaller sections, some of this weight is supported by the normal force exerted by the walls of the container. – In fact, the shaded portion of the water is completely

2. Pressure in a Fluid

• We can use the fact that the pressure difference is

proportional to the depth of a fluid to measure unknown pressures.

• Figure to the right shows a simple pressure gauge, the

open tube manometer (consisting of a U-tube

2. Pressure in a Fluid

• The top of the tube is open to the atmosphere at pressure P_at. The other end of the tube is at

pressure P, which is to be measured.

• The difference P – P_at is known as gauge pressure, P_gauge = ρgh,

where ρ is the density of the liquid in the tube.

• The pressure you measure in the tires of automobile is gauge

pressure. When the tire is entirely flat, P_gauge is zero.

2. Pressure in a Fluid

• The absolute pressure P, is

obtained by adding atmospheric pressure to the gauge pressure. • RECALL:

• P = P₀ + ρgh

• P_gauge = P – P_at = ρgh

2. Pressure in a Fluid

• Figure at right shows a U-tube mercury barometer.

• The top end of the barometer has been closed off and

evacuated so that the pressure there is zero.

• The other end is open to the atmosphere at pressure P_at.

• The pressure Pat is given by P_at = ρgh, where is the density

of mercury Recall:

2. Pressure in a Fluid

• At 0o_{C, the density of mercury is}

13.595 x 103 kg/m3. What is the height of the mercury column in a U-tube manometer if the pressure is 1 atm = 101325 Pa?

• Solution: • h = P/ρg

• = 101325 N/m2

• (13.595 x 103 _kg/m3 _*9.8m/s2₎

• = 0.760 m = 760 mm

• * In practice pressure is often

measured in mm of Hg (mmHg), a unit called torr, after Italian

2. Pressure in a Fluid

• More units

• 1 atm = 760 mmHg = 760 torr

• = 29.9 inHg = 101.325 kPa

• = 14.7 lb/in2

1. More dense liquid-> greater pressure

2. Pressure in a liquid at any location is exerted in equal amount in all direction

3. Pressure is depth dependent not volume dependent

Notes :

• 5. Liquid pressure acts

perpendicular to the sides of the container and increases with

A tank is filled with water to a depth of 1.5 m. What is the pressure at the bottom of the tank due to the water

alone?

P = gh

P = (1 x 103 _kg/m3_)(9.8m/s2_{)(1.5 m)}

= 1.5 x 104 _N/m2

A nurse administers

medication in a saline

solution to a patient by

infusion into the patient’s

arm. The density of the

solution is 1 x 10

kg/m

.

And the pressure inside

the vein is 2.4 x 10

Pa.

How high must the

container be hung?

p

= p

gh = 2.4 x 10

Pa

solving for h:

h = p

= 2.4 x 10

Pa

g (9.8m/s

)(1 x 10

kg/m

)

4. Buoyancy and Archimedes’

Principle

• A heavy object submerged in water is “weighed” by

suspending it using a spring

scale, the reading on the scale is

less than when the object is weighed in the air.

• This is because water exerts an upward force that partially

balances the force of gravity.

• This force is more evident when we submerge a piece of cork: • When completely submerged,

4. Buoyancy and Archimedes’

Principle

• The force exerted by a fluid on a body submerged in it is called Buoyant Force (F_B).

– It is equal to the weight of the fluid displaced by the body.

• ARCHIMEDES’ PRINCIPLE

– A body wholly or partially

submerged in a fluid is

buoyed up by a force equal to the weight of the displaced fluid.

• Archimedes Principle works

4. Buoyancy and Archimedes’

Principle

• Archimedes (287-212 BC) has been tasked by King Hieron II to determine if the crown made for the king was made of pure gold or has been adulterated with some cheap metal such as silver. • As the story goes, Archimedes came up with the solution while

4. Buoyancy and Archimedes’

Principle

• What Archimedes found was a simple and accurate way to

determine the specific gravity of the crown (w/o destroying it~) which he could then compare with the specific gravity of gold! • The specific gravity of an object

is the weight of the object in air divided by the weight of an equal volume of water

• s.g = weight of object in air .

• weight of equal volume of water

• = w_o

4. Buoyancy and Archimedes’

Principle

• But according to Archimedes’ Principle, the weight of an equal volume of water equals the

buoyant force on the object when it is submerged.

• It is therefore equal to the loss in weight of the object when it is

weighed while submerged in water, thus

s.g. = weight of object in air . weight loss when submerged in water s.g. = wo

4. Buoyancy and Archimedes’

Principle

• The purity of King

Hieron’s crown could thus be determined by weighing the crown in air and then weighing it

again in water, and

CROWNING GLORY

• The specific gravity of gold is 19.3. If a crown made of pure gold weighs 8N in air, what will be its weight when it is

submerged in water?

• 1. The submerged weight equals weight in air minus weight in water

– w_sub = w_air/true - w_loss

• 2. Archimedes principle relates the weight loss to the weight (in air/true) and specific gravity:

s.g. = weight in air _ => wloss = wair => wloss = 8N = 0.415 N weight loss s.g. 19.3

4. Buoyancy and Archimedes’

Principle

• The measured weight F_s of an object submerged in a fluid is the difference between the true

weight w and the buoyant force F_B.

• F_s = w – F_B.

• If the density of the object is ρ, its volume is V, and the density of the fluid is ρ_f, the weight is w = ρgV, and the buoyant force is F_B = ρ_fgV.

• The measured weight is then

FLOATING CORK!!!

• A cork has a density of 200 kg/m3_{. Find the fraction of the}

volume of the cork that is submerged when the cork floats in water.

• Picture the Problem: Let V be the volume of the cork and V’ be the volume that is submerged when it floats. The weight of the cork is ρ_cgV, and the buoyant force due to the water is ρ_wgV’.

• 1. Since the cork is in equilibrium, the buoyant force equals the weight:

• ρ_cgV = ρ_wgV’ • 2. Solve for V’/V:

• V’ = ρ_c = 200 kg/m3 _{= 1}

• V ρ_w 1000 kg/m3 ₅

Archimedes’ Principle:

Totally Submerged Object

 The upward buoyant force is B = ρ_fluidgV_obj

 The downward gravitational force is w = mg = ρ_objgV_obj

 The net force is B – w = (ρ_fluid-ρ_obj)gV_obj

Depending on the

• The object is less dense than the fluid ρ_fluid>ρ_obj

• The object experiences a net

upward force

The

net force

is

B – w = (ρ

-ρ

)gV

 The object is more dense than

the fluid ρ_fluid<ρ_obj

 The net force is downward, so the

An aluminum block with a mass of 0.500 kg and a density of 2.7 x 103

kg/m3 _{is suspended from a string.}

What is the tension in the spring when it is completely immersed in water?

Draw and label all the forces acting on the object.

Application

T

W=mg _F

B

T + F

- W = 0

Compute for the volume of the fluid displaced:

V_fd = V_B = m_B/_B

V_fd = (0.5 kg)/(2.7 x 103 _kg/m3₎

V_fd = 1.9 x 10-4 _m3

Bouyant force is:

F_B = _fdV_fdg = (1 x103_)(1.9x10-4_)(9.8)

F_B = 1.86 N

From Newton’s 1st Law: T = W - F_B

T = mg - fdVfdg

= (0.5)(9.8) - 1.86 N

T = 3.04 N