Fluids

Fluids

solid liquid

gas

Fluid Mechanics

•A fluid is anything that flows: •Fluids can be a liquid or a gas

•e.g. liquid (water, mercury at room temp) gas (steam, air, CO2)

•Properties depend on temperature

Fluids

A container is needed to keep gas in a certain locationA solid has a rigid structure due to strong bondingA liquid has weak bondsA gas is free to move – molecules independent

Fluids - DensityThe density is the mass m of a substance divided by its volume V. {the density is defined as the mass of a unit volume}

SI units of density is kgm-3

So equal volumes of substances generally have different masses and therefore different densities.

V

m

Mass Density

Equal volumes of different substances generally have different masses. Characterise using density ()

Vm

volume

mass

Material Density (kg /m3) Aluminium 2700 Gold 19300 Ice 917 Water 1000 Oil 800 Air 1.29 Helium 0.179

Pressure (P)

Pressure is defined as the force acting perpendicular to a unit area. In other words, it is given by perpendicular force (F) to the surface divided by the area (A) over which the force acts.

SI unit of pressure:1 N/m2 = 1 Pa or 1 Pascal.

Atmospheric pressure 105 Pa

A

FP

© John Wiley and Sons, 2004

1. A cylinder of 4 cm radius is filled with water to the height of 10 cm. find the pressure at the bottom of the cylinder.

2. Cylindrical shape tube is filled with water. The pressure at the bottom is measured as 1 atm. Find the height of the water level.

3. If it is filled with mercury, find the height of the mercury level.

(Density of Hg-13600 kgm-3 Water-1000 kgm-3)

Examples:

Pressure with Depth in a static fluid

Pressure in a static fluid is proportional to depth.


Pressure with depthTake a column of liquid in equilibrium

Sum of forces is zero.

Relative to surface P1 = 0 soghPP

ghAAPAPF

ghAm

AhV

Vm

mgAPAPF

y

y

12

12

12

0

0

ghP

(A)

Examples:1. Find the pressure on an object of 10m depth

at sea. (density of sea water is given as 1030 kgm-3 and the atmospheric pressure is given as 105 Pa)

2. Find the pressure under 1m of,a. Waterb. Oil(density is 800 kgm-3)c. Hg(density – 13600kgm-3)

3 At what depth in the water the atmospheric pressure is doubled

Pascal’s Principle

Any change in pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the closed walls.

By applying force to a small area (F1) this pressure can be used in hydraulic systems to lift large masses e.g. car lift

1

212

1122

A

AFF

AFAFA

FP

Car lift

Problem:

1. If a downward force F1 = 50 N is applied to the small piston with area A1 = 100 cm2, what upward force F2 does the fluid apply to the large piston with area A2 = 1000 cm2?

Remember: Pressure is the same everywhere.

Problem 1A barber raises his customer’s chair by applying a force of 150N to a hydraulic piston of area 0.01 m2. If the chair is attached to a piston of area 0.1 m2, how massive a customer can the chair raise? Assume the chair itself has a mass of 5 kg.

AnswerTo solve this problem, first determine the force applied to the larger piston.

If the maximum force on the chair is 1500N, you can now determine the maximum mass which can be lifted by recognizing that the force that must be overcome to lift the customer is the force of gravity, therefore the applied force on the customer must equal the force of gravity on the customer.

If the chair has a mass of 5 kilograms, the maximum mass of a customer in the chair must be 148 kg.

Problem 2

A hydraulic system is used to lift a 2000-kg vehicle in an auto garage. If the vehicle sits on a piston of area 1 square meter, and a force is applied to a piston of area 0.03 square meters, what is the minimum force that must be applied to lift the vehicle?

Answer

Archimedes’ Principle

When a body is wholly or partially immersed in a fluid there is an up thrust which is equal to the weight of fluid displaced.


P2>P1 so the upward force exceeds the downward force!

The buoyant force FB = weight of the liquid displaced

displacedliquidofweightghA

Vm

densityliquid

volumehA

ghAAPPAPAPFB

1212


Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: so we can express that by the following equations.

If the buoyant force is strong enough to displace the force of gravity then the object will float in the fluid

fluidB WF gmgV fluidobjimmersed


An object of weight 100N is immersed in water. The deeper the object is the more fluid it displaces and the stronger the buoyancy force is. In (b) the buoyant force matches the weight of the cube so it floats!

fluidB WF



A solid square pine raft measures 4m by 4m by 0.3m thick. Determine if the raft will float in water if the density of the wood is 550kg/m3. Density of water is 1000kg/m3.



In the last example – compare the maximum possible buoyant force and the weight of the raft.

The comparison depends on the densities only!Any object solid throughout will float if the density of the material is less than the liquidWhat about a ship made of metal?It is not solid- hollow.. but has a large area to displace lots of water to balance its weight.

gVgV woodwaterwoodwood

Archimedes' Principle (Cont.)

Does Archimedes' principle depend on depth? No. Volume of

fluid displaced by submerged object is the same at any depth.

Buoyant force only depends on pressure differences Same at any depth

Sink or Float

An object more (less) dense than the fluid in which it is immersed will sink (float).What must you do to increase your buoyancy in water?

Sink or FloatAn object more (less) dense than the fluid in which it is immersed will sink (float).

What must you do to increase your buoyancy in water?

Decrease your weight density Life jacket increases volume

while adding very little weight

An object with a density equal to the surrounding fluid will neither sink nor float.

Fish (air sac) and submarines (ballast) can vary their densities

Crocodiles increase their weightdensity by swallowing stones.

The Principle of Flotation

How much water must be displaced for an object to float?Principle of flotation A floating object

displaces a volume of fluid with a weight equal to its own weight.

Archimedes figured out that the meta center had to be determined which is a point where an imaginary vertical line (through the center of buoyancy) intersects another imaginary vertical line (through a new centre of buoyancy) created after the ship is displaced, or tilted, in the water.

How is it that an iron ship will float in water?

The center of buoyancy in a floating ship is the point in which all the body parts exactly balance each other and make each other float. In other words, the meta center remains directly above the center of buoyancy regardless of the tilt of the floating ship.When a ship tilts, one side displaces more water than the other side, and the center of buoyancy moves and is no longer directly under the center of gravity; but regardless of the amount of the tilt, the center of buoyancy remains directly below the meta center. If the meta center is above the center of gravity, buoyancy restores stability when the ship tilts. If the meta center is below the center of gravity, the boat is unstable and capsizes.


Problem 1.

1. A body weighs 450 g in air and 310 g when completely immersed in water. Find

(i) the loss in weight of the body(ii) the upthrust on the body(iii) the volume of the body

Answer 1

(i) loss in weight = weight in air - weight in water= 450 - 310= 140gf.(ii) upthrust = loss of weight= 140 gf(iii) weight of displaced water = upthrust = 140gfmass of displaced water = 140gAssumption: density of water is 1 gcm-3

Volume of displaced water =140/1=140cm3

Volume of the body = volume of displaced water = 140 cm3

2. The mass of a block made of certain material is 13.5 kg and its volume is 15 x 10-3 m3. Will the block float or sink in water? Give a reason for your answer.

Density of the block = 0.9 x 103 kg m-3

(density of water is 1 x 103 kg m-3)The block floats in water because its

density is less than that of water.

3. A hollow cubical box is 0.30 meters on an edge. This box is floating in a lake with 1/3 of its height beneath the surface. The walls of the box have a negligible thickness. Water is poured into the box. What is the depth of the water in the box at the instant the box begins to sink?

Fluids in motion

Flowing water or even wind is a moving fluid. It is important to be able to handle such phenomena.We use a few parameters to

describe the motion of the fluid particles

Steady or Unsteady flow (turbulence)Compressible or IncompressibleViscous or Non-viscous

Steady/Unsteady flow

In steady flow the velocity of particles is constant with timeUnsteady 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

Compressible/Incompressible fluids

Incompressible fluid the density remains constant with changing pressure.Most liquids can be modelled as incompressibleSome gasses cannot be said to be incompressible, therefore they are compressible.

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

Equation of continuity

The mass flow rate (ρAv)has the same value at every point along a tube that has a single entry and exit point for the fluid flow. For two positions

222111 vAvA If the area is reduced the fluid must speed up!


The mass flow rate (ρAv) unit is kg/s

secsec1sec

2

3222

kgmm

m

kgvA

kg

t

m

If the area is reduced the fluid must speed up!


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

Problem 1

Water runs through a water main of cross-sectional area 0.4 m2 with a velocity of 6 m/s. Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-sectional area of 0.3 m2.

Answer

Problem 2

Water enters a typical garden hose of diameter 1.6 cm with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.5 cm is attached to the end of the hose.

AnswerFirst, find the cross-sectional areas of the entry (A1) and exit (A2) sides of the hose.

Next, apply the continuity equation for fluids to solve for the water velocity as it exits the hose (v2).

Bernoulli’s PrincipleIn steady flow, the speed, pressure and elevation of an ideal fluid are related. Two situations.(1) Horizontal pipe, changing Area A. Pressure drop in thin pipe why? Fluid speeds – acceleration – force required

(2) Elevation change but same area A Lower fluid is under more pressure

Bernoulli’s Principle

In steady flow for a non-viscous incompressible fluid the density ρ, the pressure P, the fluid velocity v and elevation y are related through

2222

121

212

11 gyvPgyvP

Examples of Bernoulli’s Principle

(a) good plumbing, (b) bad plumbingIn (a) pressure difference between points A and B, water trap flushed out – sewer gas in houseIn (b) no pressure difference between points A and B using vent, water trap kept in place – no sewer gas in house


Air moves faster over top of wing, slower beneath. Leads to lower pressure over wing and uplift to raise plane!


Curve ball, air on one side of spinning ball faster than the other (leads to pressure difference!)

Example 1Water circulates throughout a house in a hot-water heating system. If water is pumped out at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm diameter pipe on the second floor 5.0 m above? Assume pipes do not divide into branches.

Answer 1

First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1.

v2 = (v1A1) / (A2)

= (v1π(r1)2) / (π(r2)2)

= (0.50 m/s)(0.020 m)2 / (0.013 m)2 = 1.2 m/s

To find pressure, we use Bernoulli’s equation:P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3) [ (0.50 m/s)2 – (1.2 m/s)2]P2 = (3.0 x 105 N/m2) – (4.9 x 104 N/m2) – (6.0 x 102 N/m2)

P2 = 2.5 x 105N/m2

Problem 1

What is the lift (in Newtons) due to Bernoulli’s principle on a wing of area 86 m2 if the air passes over the top and bottom surfaces at speeds of 340 m/s and 290 m/s, respectively?

Problem 2

Water at a gauge pressure of 3.8 atm at street level flows in to an office building at a speed of 0.06 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6cm in diameter by the top floor, 20 m above. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipe and ignore viscosity.

Problem 3The water supply of a building is fed through a main entrance

pipe 6.0 cm in diameter. A 2.0 cm diameter faucet tap positioned 2.0 m above the main pipe fills a 25 liter container in 30 s.(a) What is the speed at which the water leaves the faucet.(b) What is the gauge pressure in the main pipe.Assume that the faucet is the only outlet in the system

Technology

Fluids