Fluids & Bernoulli’s Equation Chapter 10.8. Flow of Fluids There are two types of flow that fluids can undergo; Laminar flow Turbulent flow.

Fluids & Bernoulli’s Equation

Chapter 10.8

Flow of Flow of FluidsFluids• There are two types of flow that fluids

can undergo;

• Laminar flow• Turbulent flow.

Flow of Flow of FluidsFluids• There are two types of flow that fluids

can undergo;

• Laminar flow• Turbulent flow.

We will be dealing only with laminar flow

Flow of Flow of FluidsFluids• There are two types of flow that fluids

can undergo;

• Laminar flow• Turbulent flow.

We will also be assuming no friction

along the edges

Rate of FlowRate of Flow• Defined as the volume of fluid that

passes a certain section in a given time

Where:• R = the flow rate • A = the cross-sectional area of the pipe • v = the velocity of the fluid.

R = Av

Rate of FlowRate of Flow• Common units for rate of flow

are

• cubic feet per second (ft3/s),

• cubic meters per second (m3/s),

• gallons per second (gps),

• liters per second (L/s).

Rate of FlowRate of FlowFluid in = Fluid out:Flow rate remains

constant even if the radius of the pipe

changes

v2v1

1A

2A

Rate of FlowRate of Flow

Flow rate remains constant even if the

radius of the pipe changes

v2v1

1A

2A

1 1 2 2A v A vR =

ProblemProblem• Water flows through a rubber

hose 2.0 cm in diameter at a velocity of 4.0 m/s. If the hose is coupled into a hose that has a diameter of 3.5 cm, what is the new speed of the fluid?

AnswerAnswerA1v1 = A2 v2 πr2

1v1 = πr22v2

221 1

2 2 22

2.04.0 1.3

3.5

cmv r m mv

s sr cm

Area increases so velocity Area increases so velocity decreasesdecreases

1.0

1.75

Applications

Bernoulli’s Bernoulli’s PrinciplePrinciple• To accelerate a fluid as it goes into the

constriction, the pushing force in the large diameter area must be greater than the pushing force in the constriction.

y

AB

C

At point B, the pushing force in the x direction has increased.

Bernoulli’s Bernoulli’s PrinciplePrinciple• To accelerate a fluid as it goes into the

constriction, the pushing force in the large diameter area must be greater than the pushing force in the constriction.

y

AB

C

PA – PB = ρgy

Bernoulli’s Bernoulli’s PrinciplePrinciple• When the speed of a fluid increases, its

internal pressure (pressure on the sides of the container) decreases.

21.

2P gy v const

Where: const. is some constant, P is the pressure, is the fluid density, y is the height of the fluid, and v is the fluid velocity

Bernoulli’s Bernoulli’s PrinciplePrinciple• We can analyze the flow of a fluid through a system using

this equation.

y1

y2

P1

P2

v1

v2

Bernoulli’s Bernoulli’s PrinciplePrinciple

y1

y2

P1

P2

v1

v2

2 21 1 1 2 2 2

1 1

2 2P gy v P gy v

If no change in If no change in heightheight

y y

P1

P2v1 v2

2 21 1 2 2

1 1

2 2P v P v

ProblemProblem• An oddly shaped

tank is filled with water to a depth of 1.20 m. Calculate the pressure at point B at the bottom of the tank.

A

B

1 .20 m

ProblemProblem• The system is static,

so v is zero. The equation becomes-

A

B

1 .20 m

1 1 2 2P gy P gy

ProblemProblem• Assume the

pressure at point A is zero, and the equation becomes -

A

B

1 .20 m

2 2 2 20 P gy P gy

1 1 2 2P gy P gy

ProblemProblem• Which

coincidentally is the equation for hydrostatic pressure

A

B

1 .20 m

2 2 2 20 P gy P gy

P gh

SolutionSolution

32 2 232

1.0 10 9.8 1.20kg m

P gy x msm

42 1.18 10P x Pa

ProblemProblem• A container of water, diameter 12 cm, has a

small opening near the bottom that can be unplugged so that the water can run out. If the top of the tank is open to the atmosphere, what is the exit speed of the

y2

water leaving through the hole. The water level is 15 cm above the bottom of the container. The center of the 3.0 diameter hole is 4.0 cm from the bottom

SolutionSolution• The water comes out the hole with a speed of

v1. The flow of water in the container, which makes the surface level drop is very slow by comparison. So slow that we can say that it is zero. So v2 = 0.

y

A

1A

2

1v0P

0P = P2

1 5 cm

4 .0 cm

SolutionSolution• The pressure on the top of the surface is the

atmospheric pressure. The surface acting on the water at the opening on the bottom is also the atmospheric pressure (actually it is a tiny bit

y

A

1A

2

1v0P

0P = P2

1 5 cm

4 .0 cm

bigger because it is slightly lower, but the difference is insignificant). So we can let the two pressures equal each other. So P1 = P2

SolutionSolution• Therefore Bernoulli’s equation becomes

y

A

1A

2

1v0P

0P = P2

1 5 cm

4 .0 cm

21 1 2

1

2gy v gy

21 1 2

1

2gy v gy

And simplifies to:

SolutionSolution

1 22 9.8 0.15 0.040 1.5

m mv m m

ss