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**The Continuity Equation Dynamic Fluids**

Why would you put your thumb over the end of a garden hose?

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Mass Flow Rate Since an ideal fluid is incompressible, a fluid entering one end of a pipe at a certain rate (kg/s) must leave the other at the same rate. As long as the pipe has no leaks. That rate is called the mass flow rate and is expressed in kg/s

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**Mass Flow Rate Constant**

From Density Formula Mass Flow Rate Constant

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Continuity Equation Same, incompressable, fluid so roe drops out!

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Water enters the tube below from the left side at 4 m/s with an opening of radius 5 cm. The tube narrows to half the radius. With what speed will water leave the right side? What would happen if the water entered the right side at 4 m/s?

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**The Bernoulli family : Swiss mathematicians in the eighteenth century**

Daniel Bernoulli (1700–1782), developer of Bernoulli's principle Jakob Bernoulli (1654–1705), also known as Jean or Jacques, after whom Bernoulli numbers are named Johann Bernoulli (1667–1748) Nicolaus I Bernoulli (1687–1759) Nicolaus II Bernoulli (1695–1726) The mathematical ideas developed by the family members include: Bernoulli differential equation Bernoulli distribution Bernoulli inequality Bernoulli number Bernoulli polynomials Bernoulli process Bernoulli trial Bernoulli's principle

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**Fluid flow is best described by Bernoulli’s Principle**

2 assumptions Ideal fluid (incompressible) Non-viscous fluid (laminar flow). No friction. This is viscous

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**Two observations about flowing fluids in a pipe**

When encountering a region of reduced cross-sectional area, the pressure always drops! This obeys ∑F=ma. The fluid in A1 can only speed up (accelerate) due to an unbalanced force pushing it. P2 must be way greater than P1.

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2nd 2. If a fluid moves to a higher elevation the pressure at the lower level is greater than that at the higher level. We learned that in the study of static fluids. P=ρgh

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**Consider these 2 things happening at once**

Wouldn’t this create a dramatic drop in pressure?!

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**Based on Work/Energy Theorem**

Pressure in any fluid is caused by collision forces which are non-conservative. Non-conservative forces produce work that is dependant on the path. Net work ≠ 0

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**= W + + K U const . 1 × + + + F d mv mgh const . 2 1 + r + r = PAd Vv**

Vgh const . 2 1 PV + r Vv 2 + r Vgh = const . 2 1 + r 2 + r = P v gh const . 2

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**The Bernoulli Equation**

Shows the relationship between: Pressure p Height h Speed v for an ideal fluid through any tube of flow P1 + ½ v12 + gh1 = P2 + ½ v2 2 + gh 2

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P1 + ½ v12 + gh1 = P2 + ½ v2 2 + gh 2 P1 + ½ v12 + gh = a constant1

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**Prairie dogs do not suffocate in their burrows**

Prairie dogs do not suffocate in their burrows. The effect of air speed on pressures creates ample circulation. The animal maintains different shapes to the 2 entrances of it’s burrows and because of this the air, ρ=1.29kg/m3, blows past the different openings at different speeds. Assuming the openings are at the same vertical level, find the difference in air pressure between the openings and indicate which way the air circulates.

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2 ways to pump water! If the well is shallow they both work but if the well is deep only one does, which one?

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**Streamlines: show speed pictorially**

Streamlines: show speed pictorially. The closer together, the faster the fluid is moving.

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The Venturi Meter uses the height a tube of mercury is raised to find speed and pressure in a pipe

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**Commercial Venturi Meters**

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**P + ½ ρv2 + ρgh = P + ½ ρv2 + ρgh P = 150,000 Pa P = 200,000 Pa**

200,000 + ½ x1000 x 4^ x10x0 = 150,000 + ½ 1000 v^ x10x5 50, = V^ ,000 58,000 – 50, = 500 V^2 8000/ = V^2 = V^2 v = 4 m/s, so the speed doesn’t change in this case. The drop in pressure P is balanced by an increase in ρgh only. Av must = Av: since the pipe doesn’t get thinner or thicker, v must be the same. P = 150,000 Pa What would happen to v in and v out if the height were raised above 5 m? P = 200,000 Pa Flow would decrease, top pressure would drop. H= 5 meters V = 4 m/s How high would you have to raise it to stop flow? 20 m where pgh = 200, 000 Pa.

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**P + ½ ρv2 + ρgh = P + ½ ρv2 + ρgh P = 150,000 Pa P = 200,000 Pa**

200,000 + ½ x1000 x 4^ x10x0 = 150,000 + ½ 1000 v^ x10x5 50, = V^ ,000 58,000 – 50, = 500 V^2 8000/ = V^2 = V^2 v = 4 m/s, so the speed doesn’t change in this case. The drop in pressure P is balanced by an increase in ρgh only. Av must = Av: since the pipe doesn’t get thinner or thicker, v must be the same. P = 150,000 Pa What would happen to v in and v out if the height we raised above 5 m? Both would decrease by the same amount How high would you have to raise it to stop flow? 20 m where pgh = 200, 000 Pa. P = 200,000 Pa H= 5 meters V = 4 m/s

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NNPC FSTP ENGINEERS Physics Course Code: Lesson 7.

NNPC FSTP ENGINEERS Physics Course Code: Lesson 7.

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