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# Fluid Mechanics - PowerPoint PPT Presentation

Fluid Mechanics. A fluid is a substance that has the ability to flow, and therefore, does not maintain a specific shape. It can either be a liquid or a gas . A valuable measurement of a fluid is its density. Density is mass per unit of volume. ρ = m / V ρ = density m = mass (kg)

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## PowerPoint Slideshow about 'Fluid Mechanics' - ross-wagner

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Presentation Transcript

A fluid is a substance that has the ability to flow, and therefore, does not maintain a specific shape. It can either be a liquid or a gas. A valuable measurement of a fluid is its density.

Density is mass per unit of volume.

ρ = m / V

ρ = density

m = mass (kg)

V = volume (m3)

Things with low densities float in things with higher densities.

Specific gravity = the ratio of a substance’s density to the density of water

0.50

0.73

0.82

1.00 (water)

1.34

Pressure is force per unit of area.

P = F / A

P = pressure (Pascal)

F = force (N)

A = surface area (m2)

Which book exerts more pressure on the table?

Why can she lay on a bed of nails and be struck with a sledgehammer without getting hurt?

Fluids exert pressure in all directions.

Forces acting in all directions on the cube are equal (assuming cube is very small)

Fluids exert forces that are perpendicular to the solid surfaces they are in contact with.

If they are at rest, they do NOT exert forces parallel to the solid surface.

As you go deeper, the amount of water above you increases exerting more force on you because of gravity.

P = F / A

F (cube of water) = mg

m = ρV = ρAh

F = ρAhg

P = ρAhg / A

P = ρgh

P = pressure (Pa)

ρ = density (kg/m3)

g = gravitational acceleration (m/s2)

h = distance from surface (m)

Furthermore,

∆P = ρg∆h

Assuming that the difference between the surface of the water in the tank and the household faucet is 30 m, calculate the difference in water pressure between the surface and the faucet.

A manometer is used to measure pressure. A fluid is pushed on by two pressures – the atmospheric pressure from the outside (P0) and the pressure being measured (P).

The fluid adjusts itself to an equilibrium point based on the pressure difference which produces a height difference (Δh).

The term ρg∆h is known as the gauge pressure. It is the difference between the atmospheric pressure and the pressure being measured (or the pressure relative to the atmospheric pressure)

∆P = ρg∆h

P - P0 = ρg∆h

Units for P  Pa (N/m2), mm Hg (torr), atm

Intravenous infusions are often made under gravity. Assuming the fluid has a density of 1.00 g/cm3, at what height should the bottle be placed so that the liquid pressure is 55 mm-Hg?

If the blood pressure is 18 mm-Hg above atmospheric pressure, how high should the bottle be placed so that the fluid just barely enters the vein?

Pascal’s Principle Assuming the fluid has a density of 1.00 g/cm

If an external pressure is applied to a confined fluid, that extra pressure is transmitted throughout the fluid

Pin = Pout

Fin / Ain = Fout / Aout

Fout / Fin = Aout / Ain

Archimedes’ Principle Assuming the fluid has a density of 1.00 g/cm

Why do people rehabilitate injuries by training in a pool?

There is less stress on their joints.

Why?

Buoyant Force – an upward force exerted by a liquid on an object submerged in the liquid

Picture shows an object submerged in a fluid. The buoyant force is the net force in the vertical direction (we know F2 is greater than F1 because of the greater fluid pressure)

Fb = F2 – F1

Since P = F / A

= P2A2 – P1A1

= A(P2 – P1)

Since P = Assuming the fluid has a density of 1.00 g/cmρgh

Fb = A(ρFgh2 – ρFgh1)

= ρFgA (h2 – h1)

= ρFgA∆h

= ρFgV

Since ρF= mF / V

Fb= mFg

Fb = buoyant force (N)

mF = mass of fluid displaced (kg)

g = gravitational acceleration (m/s2)

Archimedes’ Principle – The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object.

An object’s apparent weight is the weight that object experiences when submerged in a liquid

W ´ = W – Fb

W ´= apparent weight

W = actual weight

Fb = buoyant force

There is a buoyant force of:

Fb = W – W ’

14.7(9.8)-13.4(9.8)

= 12.7 N

(equal to the weight of the liquid displaced)

If this 1200-kg log was experiences when submerged in a liquidfully submerged, the water it would displace would have a greater mass than the log itself (2000 kg)

It rises until the fluid it displaces matches the weight of the log.

Fluids in Motion have a total mass of 68.0 kg.

If fluid flow (gas or liquid) is smooth as in figure (a), it is called streamline or laminar flow. In this flow, each particle follows a smooth path or streamline, not crossing paths with other particles.

Turbulent flow exists with the development of eddies, which are whirlpool-like circles along the path of the streamlines. In this flow, particles cross paths and lose a great deal of energy.

m / ∆t

m = mass (kg)

∆t = time interval (s)

Fluids typically flow through pipes which have varying diameters. When the diameter changes, the flow changes.

Since ρ = m / V

m / ∆t = ρV / ∆t

Since V = AΔl

m / ∆t = ρAΔl/ ∆t

m / ∆t = ρAv

(v = fluid velocity)

ρA1v1 = ρA2v2

Since ρ is constant,

A1v1 = A2v2

(Continuity Equation)

What area must a heating duct have if air moving 3.0 m/s along it can replenish the air every 15 minutes in a room of volume 300 m3? Assume the air’s density remains constant.

Bernoulli’s Principle along it can replenish the air every 15 minutes in a room of volume 300 m

• Ever wondered…

• Why a curve ball curves?

• Why a sailboat can move against the wind?

• Why an airplane wing has lift?

Bernoulli’s Principle – Where the velocity of the fluid is high, the pressure is low and where the velocity is low, the pressure is high

Why is that the case?

Consider the amount of along it can replenish the air every 15 minutes in a room of volume 300 mwork necessary to move blue-colored fluid from position in (a) to position in (b)

At point 1(left end), work is done on blue fluid to move it:

W1 = F1Δl1

= P1A1Δl1

At point 2 (right end), negative work is done by white fluid against motion of blue fluid:

W2 = -P2A2Δl2

Also, negative work is done to lift the fluid against gravity.

W3 = -mg(y2 - y1)

Net work = W1 + W2 + W3

= P along it can replenish the air every 15 minutes in a room of volume 300 m1A1Δl1 - P2A2Δl2 - mgy2 + mgy1

Since W = ∆KE

½m2v22 – ½m1v12= P1A1Δl1 - P2A2Δl2 – m2gy2 + m1gy1

Since m = ρV = ρAΔl

½ρA2Δl2v22 – ½ρA1Δl1v12 = P1A1Δl1 - P2A2Δl2 – ρA2Δl2gy2 + ρA1Δl1gy1

Since A1Δl1 = A2Δl2

½ρv22 – ½ρv12 = P1 - P2 - ρgy2 + ρgy1

P1+ ½ρv12 + ρgy1 = P2 + ½ρv22 + ρgy2

P + ½ along it can replenish the air every 15 minutes in a room of volume 300 mρv2 + ρgy = constant

P = Pressure (Pa)

ρ = density (kg/m3)

v = fluid velocity (m/s)

g = gravitational acceleration (m/s2)

y = vertical height above reference point (m)

Water circulates throughout a house in a hot-water heating system. If the water is pumped 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 2nd floor 5.0 m above?

Returning to the original questions… system. If the water is pumped 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 2

When pitch is thrown with a spin on it, spin on side A slows down air molecules causing higher pressure. Spin on side B speeds up air molecules causing lower pressure. Net force in left direction

When air moves over the top of the wing, the streamlines get forced together causing them to speed up (Av = Av). The streamlines underneath are moving slower. Slower molecules underneath have higher pressure than molecules above wing causing net for up.

When sails are set at angle, the wind hits the front of the sail lowering the pressure, and still air behind the sail has higher pressure. Net force pushes sail to left. Water pushes against keel to the right giving a net force into the wind.

What is the speed at which the water leaves the spigot? sail lowering the pressure, and still air behind the sail has higher pressure. Net force pushes sail to left. Water pushes against keel to the right giving a net force into the wind.

• Assumptions

• the size of the spigot is small compared to the diameter of the jar, so v2 will be insignificant compared to v1

• P1 = P2 because both are open to atmosphere

P1+ ½ρv12 + ρgy1 = P2 + ½ρv22 + ρgy2