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Fluids & Bernoulli’s Equation Chapter 10.8

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Flow of Fluids There are two types of flow that fluids can undergo; Laminar flow Turbulent flow.

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Flow of Fluids There are two types of flow that fluids can undergo; Laminar flow Turbulent flow. We will be dealing only with laminar flow

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Flow of Fluids There are two types of flow that fluids can undergo; Laminar flow Turbulent flow. We will also be assuming no friction along the edges

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Rate 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

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Rate of Flow Common units for rate of flow are cubic feet per second (ft 3 /s), cubic meters per second (m 3 /s), gallons per second (gps), liters per second (L/s).

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Rate of Flow Fluid in = Fluid out: Flow rate remains constant even if the radius of the pipe changes

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Rate of Flow Flow rate remains constant even if the radius of the pipe changes R =

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Problem 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?

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Answer A 1 v 1 = A 2 v 2 πr 2 1 v 1 = πr 2 2 v 2 Area increases so velocity decreases 1.0 1.75

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Applications

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Bernoulli’s Principle 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. At point B, the pushing force in the x direction has increased.

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Bernoulli’s Principle 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. P A – P B = ρgy

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Bernoulli’s Principle When the speed of a fluid increases, its internal pressure (pressure on the sides of the container) decreases. 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

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Bernoulli’s Principle We can analyze the flow of a fluid through a system using this equation.

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Bernoulli’s Principle

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If no change in height

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Problem 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.

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Problem The system is static, so v is zero. The equation becomes-

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

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Problem Which coincidentally is the equation for hydrostatic pressure

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Solution

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Problem 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 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

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Solution The water comes out the hole with a speed of v 1. 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 v 2 = 0.

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Solution 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 bigger because it is slightly lower, but the difference is insignificant). So we can let the two pressures equal each other. So P 1 = P 2

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Solution Therefore Bernoulli’s equation becomes And simplifies to:

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Solution

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