Fluids & Bernoulli’s Equation Chapter 10.8. Flow of Fluids There are two types of flow that fluids...
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Transcript of Fluids & Bernoulli’s Equation Chapter 10.8. Flow of Fluids There are two types of flow that fluids...
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
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
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
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: