Fluids

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# Fluids - PowerPoint PPT Presentation

Fluids. Buoyant Force. The pressure force on the sides of the volume balances the weight of fluid in the volume. The force remains even without the original fluid. This is the buoyant force. Equals the fluid weight Directed upward Acts on the volume. F b = mg. Sinking and Rising.

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## PowerPoint Slideshow about 'Fluids' - wolfe

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Presentation Transcript

### Fluids

Buoyant Force
• The pressure force on the sides of the volume balances the weight of fluid in the volume.
• The force remains even without the original fluid.
• This is the buoyant force.
• Equals the fluid weight
• Directed upward
• Acts on the volume

Fb = mg

Sinking and Rising
• An object in a fluid displaces a volume that had some mass.
• If the object is heavier than the fluid it sinks.
• If the object is lighter it rises.

Fb = rVg

Fb = rVg

W= mg

W= mg

Archimedes’ Principle
• An object suspended in a fluid has less apparent weight due to buoyancy.

FT = mg - rVg

Fb = rVg

W= mg

An iceberg has an average density of 86% of seawater.

What fraction of the iceberg is underwater?

The buoyant force is the weight of water displaced by the iceberg: Fb = rwaterVsubg.

The weight is the total weight of the ice: Wi = riceViceg.

Find the ratio of Vsub/Vice

We know the ratio rice/rwater

rwaterVsubg = riceViceg

Vsub/Vice = rice/rwater = 0.86

Iceberg
Flow Rate
• Streamlines in a fluid represent the path of a particle in the fluid.
• Groups for fluid flow
• Cross sectional area
• The flow rate measures fluid movement.
• mass per time
• density times area times velocity
Conservation of Mass
• The mass into a tube must flow out at the same rate.
• This is called the continuity equation.
• For constant density it only requires the area and velocity.
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

Canyon
Fluid Energy
• The kinetic energy in a fluid is the same as for any other mass: K = ½ mv2.
• The change in potential energy is: U = mgh.
• The work done on a fluid is due to pressure.
• Pressure acting on a volume: W = PADx = PV.
• From the work energy principle:
Bernoulli’s Equation
• The volume element is somewhat arbitrary in a moving fluid.
• Mass divided by volume is density
• Divide by volume and separate states on each end
• Bernoulli’s equation is equivalent to conservation of energy for fluids.
Lift
• If the height doesn’t change much, Bernoulli becomes:
• Where speed is higher, pressure is lower.
• Speed is higher on the long surface of the wing – creating a net force of lift.

FL