Ch. 15 Fluids

Fluids• Substances that flow readily, usually

from high pressure to low pressure.• They take the shape of the

container rather than retain their shape.•We refer both liquids and gases as

fluids.

Density•Is the mass of a substance divided by the volume of that substance.

Density Equation MassDensity = Volume

Density Units•Metric: kg / m3

or g / cm3

•English: lbm / ft3

Sample Problem: What is the volume of my 2.16 g titanium wedding band if the density is 4.50 g/ cm3?

G: m = 2.16 g, D = 4.50 g/cm3

U: V = ? E: D = m / V V = m/DS: V = 2.16 / 4.50 g/cm3 S: V = 0.48 cm3

Pressure

• is defined as force per unit of area.

Each book exerts the same force, but which one exerts a greater pressure?

Equation

ForcePressure = Area

Units of Pressure•English: lb / ft2, lb / in2, or p.s.i.•Metric: N/m2 or Pascal (Pa)

Sample Problem: If an airplane window that is 40 cm by 60 cm feels a pressure of 40,000 Pa. How much force is acting on the window?

G: L = 40 cm = 0.4 m w = 60 cm = 0.6 m P = 40,000 Pa

U: F = ?E: F = PA

We don’t have area (A),A = LwA = (.4)(.6)A = 0.24 m2

F = (40,000)(.24)F = 9600 N

Atmospheric Pressure

The pressure of the earth’s atmosphere pushing down on you.

• Earth's atmosphere is pressing against each square inch of you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch).

Sample ProblemCalculate the net force on an airplane window if cabin pressure is 90% of the pressure at sea level, and the external pressure is only

50% of that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

• Since pressure acts equally in all directions, gauge pressure is used to deal with this.• If you have a flat tire, is there a

pressure in the tire or is the pressure zero?• Yes, it is equal to atmospheric

pressure.

Gauge Pressure (Pg)• Is the pressure inside an inflated object,

that is above the atmospheric pressure.• Pg = P – Patm

• If a tire is inflated to 35 psi, that means it is 35 psi above the atmospheric pressure. So the pressure inside the tire is actually 35 psi + 14.7 psi.

Pressure increases with depth.

The deeper the body of water or you go, the more water there is on top of you per square inch.

Pressure depth relationship

Phyd = rghr – weight density of

liquid

The pressure of a liquid is the same at any given depth

regardless the shape of the container.

• The pressure P is the same on the bottom of each vessel.

Absolute Pressure

• Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure.

• pabs = patm + rgh

• p2 = p1 + rgh

Water seeks its own level

Pascal’s PrinciplePressure applied to a fluid in a closed container is transmitted equally

to every point of the fluid and to the walls of the container.

•When the area of the 2 “pistons” are the same the input force is the same as the output force.

What if you increased the area of the 2nd “piston” by 25?

By Pascal’s Principle: P1 = P2

•By the definition of Pressure:

F1/A1 = F2/A2 So if the 2nd area is

increased by a factor of 25.

In order to maintain equality, the force has to be increased by the same factor.

Pascal Principle

Distance each piston moved is different.

F1d1 = F2d2

Sample problem: The small piston on a hydraulic lift has an area of 0.20 m2. A car weighing 1.2 x 104 N sits on the rack with a large piston. If the area of the large piston is 0.9 m2, how much force must be applied to the small piston to support the car?

G: A1 = 0.2 m2, F2 = 1.2 x 104 N, A2 =0.9 m2

U: F1 = ?E: F1/A1 = F2/A2

F1 = F2A1/A2

S: F1=(1.2 x 104)(0.2)/(0.9)

S: F1 = 2666.7 N

•How far will the 2nd piston rise if the 1st piston moves 1.2 m?

The distance the 2nd piston moves is equal to the reciprocal of the factor of the ↑ or ↓ in area times the distance the 1st piston moves.

Pascal’s Principle applies to all fluids,

gases as well liquids.

Buoyancy•The apparent loss of weight of an object that is submerged in a fluid.

Buoyant Force

Is the upward force equal to the weight of the fluid displaced.

Pressure is greater at the bottom of the object because the depth is greater.

•Buoyant force is due to pressure difference between top and bottom of object.• Pressure difference

does not depend on depth

FB = rgV

Density of object determines the depth of submersions:

1. If denser than the fluid, it sinks.

2. If less dense than the fluid, it will float.

3. If the same density, it will neither float nor sink.

Archimedes PrinciplesThe buoyant force is equal to

the weightof the liquid displaced.

Immersion: all or part under water.Submersion: completely under

water

An object completely submerged always displaces a volume equal to its own.

Buoyant force = weight of liquid displaced

)(f

sssub VV

rr

Vsub – submerged volumeVs – volume of the solidrs – density of solidrf – density of fluid

Sample problem: Assume a wooden raft has 80.0% of the density of water. The dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats? (density of water is 1000 kg/m3)

Principle of Floatation

•A floating object will displace a weight of fluid equals to its own weight.

Specific Gravity

Is the ratio of any substance’s density to the density of water.

Density of Specific a substanceGravity = Density of water

Specific Gravity is equal to its density

numerically, but has no units.

Fluid Flow Continuity• The volume per unit time of a liquid

flowing in a pipe is constant throughout the pipe.• We can say this because liquids are

not compressible, so mass conservation is also volume conservation for a liquid.

Volume of a fluid moving past a point can be calculated by:

V = AvtV: volume of fluid (m3)

A: cross sectional areas at a point in the pipe (m2)

v: speed of fluid flow at a point in the pipe (m/s)

t: time (s)

Equation for Continuityr1A1v1 = r2A2v2

• r1, r2: weight density of fluids at point 1 and 2• A1, A2: cross sectional areas at points 1 and 2

• v1, v2: speed of fluid flow at points 1 and 2Use for gases.

For an incompressible fluid (A liquid)

A1v1 = A2v2

• A1, A2: cross sectional areas at points 1 and 2• v1, v2: speed of fluid flow at

points 1 and 2

Sample problem: A pipe of diameter 6.0 cm has fluid flowing through it at 1.6 m/s. How fast is the fluid

flowing in an area of the pipe in which the diameter is 3.0 cm? How much water per second flows through

the pipe?

Sample problem: The water in a canal flows 0.10 m/s where the canal is 12 meters deep and 10 meters across. If the depth of the canal is

reduced to 1.5 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region?

Bernoulli’s PrincipleThe sum of the pressure, the potential energy per unit volume, and the kinetic

energy per unit volume at any one location in the fluid is equal to the sum of

the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any other location in the fluid for a non-viscous incompressible

fluid in streamline flow.

Bernoulli’s Principle

•The pressure in a fluid decreases as the speed of the fluid increases.

Bernoulli’s Theorem• p + r g h + ½ rv2 = Constant• p : pressure (Pa)• r : density of fluid (kg/m3)• g: gravitational acceleration constant (9.8

m/s2)• h: height above lowest point (m)• v: speed of fluid flow at a point in the pipe

(m/s)

p1 + r g h1 + ½ rv12 = p2 + r g h2 + ½ rv2

2

As a pipe narrows, the speed of the fluid increases.

•fluid flows smoothly •fluid flows without any swirls•fluid flows everywhere through the pipe

•fluid has the same density everywhere (it is "incompressible" like water)

Sample Problem: Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. At the end

of the hose, the water flows out of a nozzle whose diameter is 2.5 cm. What is the speed of the water

coming out of the nozzle? If the pressure in the hose is 350 kPa, what is the pressure in the nozzle?

Torricelli’s Law

• Deals with the speed of a fluid as it flows through a hole in a container.

P1 + r g h1 + ½ rv12 = P2 + r g h2 + ½ rv2

2 Patm + r g h1 + 0 = Patm + 0 + ½ rv2

2

r g h1 = ½ rv22

ghv 2

Sample Problem: An above-ground swimming pool has a hole of radius 0.10 cm in the side 2.0

meters below the surface of the water. How fast is the water flowing out of the hole? How much

water flows out each second?