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Strategic

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Page 1: Strategic
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Proponents : Marc Lester M. PabloClaurence John C. PerezPaul John P. Villamor

Teacher: Alice Y. Caymo

TASK ANALYSIS

Main Objective:To explain Bernoulli’s principle and equation.

Sub-Objectives To define the concept behind Bernoulli’s principle and its application in life. To write Bernoulli’s equation in its general form and describe the equation as it apply to (a.)

Fluid at rest, (b.) Fluid flow at constant pressure and (c.) Flow through horizontal pipe. To apply Bernoulli’s equation to the solution of problem involving absolute pressure (P),

density (ρ), fluid elevation or height (h), and fluid velocity or speed (v).

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Guide Card

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OverviewIn 1738, Swiss Mathematician Physicist Daniel Bernoulli formulated a principle that relates fluid velocity (v) to its internal pressure (P).

The Bernoulli`s principle states that the total energy in a steadily flowing system is constant along the flow path.

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The relationship between the velocity and pressure exerted by a moving liquid is described by the Bernoulli's principle: as the velocity of a fluid increases, the pressure exerted by that fluid decreases.

Have you ever wondered how an airplane can fly? The simple answer is that as air flows around the wing, the plane is pushed up by higher pressure air under the wing, compared to lower pressure over the wing. But to understand this phenomenon more deeply, we must look at a branch of physics known as fluid mechanics, and in particular a principle known as the Bernoulli equation. Not only can this equation predict the air pressure around an airplane's wing, but it can also be used to find

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the force of high winds on a skyscraper, the pressure through a chemical reactor, or even the speed of water coming out of the hose in your backyard.

The Bernoulli equation states that for an ideal fluid (that is, zero viscosity, constant density and steady flow), the sum of its kinetic, potential and thermal energy must not change. This constraint gives rise to a predictable relationship between the velocity (speed) of the fluid, its pressure, and its elevation (relative height). Specifically, given two points along a streamline (an imaginary line tangent to the direction of flow, as shown in figure), the Bernoulli equation states that:

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where v is fluid velocity, ρ is fluid density, h is relative height, and P is pressure. Applying this equation to an example helps to make it clearer. Consider a reservoir located up in the mountains with a pipe leading down to a town at a lower elevation. The pipe delivers water to a hydroelectric plant, and we want to know how fast the water will flow into the plant turbines.

Now let's get back to how Bernoulli's principle applies to the wing of an airplane. When air flow is split around a wing, the air above the wing moves faster than the air below, due to the wing's shape. Since the velocity of the upper air increases, its pressure must decrease to maintain balance — as described in Bernoulli's equation. This results in greater pressure below the wing than above, which forces the wing upwards, enabling flight!

The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one point along its flow. The relationship between these fluids conditions along a streamline always equal the same constant along that streamline in

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an idealized system. An idealized system refers to a fluid that has a constant density (incompressible), and is inviscid.

Assuming that the fluid is inviscid means that it has no viscosity. Therefore, certain effects of viscosity, such as water sticking to the walls of its pipe or container, do not apply.

Because the Bernoulli equation is equal to a constant at all points along a streamline, we can equate two points on a streamline. Using information on the system at one point, we can solve for information at another.

According to the Venturi effect, a fluid's pressure decreases as its velocity increases. The Bernoulli equation puts this relationship into mathematical terms and includes a term for fluid height. To illustrate this relationship, consider water moving down a water slide. At the top where you load, the water is slow moving, pushed only by the water behind it. When the slide drops, the water rushes down quickly, increasing speed as it falls. Thus, the velocity is also affected by gravity through height. When all these terms are related and scaled for density and gravity, we have the Bernoulli equation, where v is fluid velocity, ρ is fluid density, h is relative height, and P is pressure. Notice the constant has units of pressure as well.

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P + ρgh + ½ ρv2 = constant

The Bernoulli equation can also be interpreted as an expression of conservation of energy, which can be formulated as

where KE is kinetic energy, PE is potential energy, and W is the work done on the system. Imagine a block of ice sliding down the water slide at some velocity. The block has a kinetic energy equal to one-half its mass times its velocity squared, or

where KE is kinetic energy and m is mass. The block also has a certain potential energy described by

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where PE is the potential energy and h is the height. Finally, the work is being done on the block by the force of the water pressure behind it (with a force F=PA, where P is water pressure, and A is the area of the face of the ice that is getting pushed along). When the block is pushed a distance equivalent to its own width, Δx, then the work done on the block is

where V is volume. The equation for conservation of energy becomes

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Dividing this equation by the volume, and recalling that density, ρΔ equals mass divided by volume, it reduces to

which is the Bernoulli equation.

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There are also other formulas such as:

For a stationary liquid, such both v1 and v2 are zero:P2 – P1 = ρg(h1 – h2)

If the pressure is constant, (P1 = P2):v = √2gh

Rate of liquid flowR= vA= A√2gh

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For a horizontal pipe, (h1 = h2):P1 + ½ ρv1

2 = P2 + ½ ρv22

Activity Card

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Activity One: Ping-Pong Ball

Purpose: The purpose of this activity is to apply Bernoulli's Principle to understand why birds, kites, and planes can fly.

Materials

• 1 clean funnel with a narrow opening of 1 cm or less

• 1 ping- pong ball

Procedure

1. Bend your head back so that you will be able to blow the ping-pong ball toward the ceiling.

2. Put the ball in the top of the funnel and blow hard and fast into the stem of the funnel.

3. Record in your science journal what happened to the ball.

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4. Bend your head down so that you will be able to blow through the funnel straight down toward the floor.

5. Hold the ball inside the funnel close to the hole (temporarily) and take a deep breath.

6. Let go of the ball as you blow hard through the stem of the funnel until you use all air in your lungs.

7. Record what happened to the table tennis ball.

What Happened?

Activity Two: The Spool

Purpose: The purpose of this activity is to learn how the air pressure phenomenon works.

Materials

• thread spool

• cardboard, 7 cm by 7 cm, lightweight but firm

• pin

Procedure

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1. Cut a piece of cardboard (from the back of a notebook) so that it measures 7 cm by 7 cm.

2. Stick a pin through the centre of the cardboard.

3. Place the spool over the pin so that the pin goes into the hole in the spool.

4. Hold the cardboard against the spool vertically. Blow firmly through the hole in the top of the spool and observe what happens to the cardboard.

5. Release your hand from underneath the cardboard.

6. Record your observations in the science journal.

Conclusion

• Explain why the cardboard did not fall once you removed your hand.

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Activity Three: The Cups

Find two disposable cups and some string. Cut the string into two -1 foot long pieces. Using tape, attach the end of one piece of string to the bottom of one of the

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cups, and attach the other end to the edge of a table. Repeat this with the other cup. Position the cups so they hang off the table two inches apart and at the same height. You might need to adjust the spacing between the cups. Blow between the two cups. What happened? Why are the cups drawn together?.

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Questions

1) A crack in a water tank has a cross-sectional area of 1 cm3. At what rate is the water lost from the tank if the water level in the tank is 4 cm above the opening?

2) The water flowing in a constriction of a Venturi tube has a velocity of v1 = 4 m/s. If h= 8 cm, what will be the exit velocity v2 when it flows into the larger tube?

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3) Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000 Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), there is no change in height. Use the Bernoulli equation to calculate the velocity of the water exiting the nozzle. (Hint: The density of water is 1000 kg/m3 and gravity g is 9.8 m/s2. Pay attention to units!)]

4) Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 Pa. The refinery needs the ethanol to be at a pressure of 2 atm (202600 Pa) on a lower level. How far must the pipe drop in height in order to achieve this pressure? Assume the velocity does not change. (Hint: Use the Bernoulli equation. The density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. Pay attention to units!)

5) Water at a gauge pressure of 3.8 atm at street level flows in to an office building at a speed of 0.06 m/s through a pipe 5.0 cm in diameter. The pipes taper down to 2.6cm in diameter by the top floor, 20 m above. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipe and ignore viscosity.

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Additional Questions1) Water circulates throughout a house in a hot-water heating system. If water is

pumped out at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm diameter pipe on the second floor 5.0 m above? Assume pipes do not divide into branches.

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2) What is the lift due to Bernoulli's Principle on a wing of 70.0 m2 area if the air passes over the top and bottom surfaces at speeds of 340 m/s and 290 m/s respectively?

3) A can has an inside diameter of 12.2 cm.  (a)  How far below the water level do you want make a 0.0500 cm diameter hole to have a velocity of 2.25 m/s?  (b)  How far must the water drop for the speed to drop by 1/2?

4) If the wind blows at 200 km/h over your house during Hurricane Bonnie, what is the net force on the roof if the area is 288 m2?

5) Wind at 20.0 km/h blows over the 12.0 cm diameter opening of a chimney.  The box where the fuel burns is 1.00 m x 1.50 m.  What is the pressure difference in the chimney? If the wind blows at 200 km/h over your house during Hurricane Bonnie, what is the net force on the roof if the area is 288 m2?

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Vocabulary/Definitions

Bernoulli’s Principle:

In fluid dynamics, Bernoulli's principle states that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Named after Dutch-Swiss mathematician Daniel Bernoulli who published his principle in his book Hydrodynamica in 1738. Also called the Bernoulli effect.

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inviscid flow: Flow in which one can ignore the effects of fluid viscosity.

streamline: A line tangent to the flow of a fluid at any given instant.

Venturi effect: The reduction in fluid pressure that results when a fluid flows through a constricted section of pipe. As a fluid's velocity increases, its pressure decreases, and vice versa. Named after Italian physicist Giovanni Battista Venturi (1746-1822).

References http://www.teachengineering.org/view_lesson.php?url=http://

www.teachengineering.org/collection/cub_/lessons/cub_bernoulli/cub_bernoulli_lesson01.xml#assessment

http://www.teachengineering.org/collection/cub_/lessons/ cub_bernoulli/cub_bernoulli_lesson01_bepworksheetas_draft4_tedl_dwc.pdf

http://www.mysciencesite.com/ Activities_Showing_Bernoulli_s_Principle.pdf

http://www.csp.science.ubc.ca/life/StudentSamples/Website2/examples.html http://www.angelfire.com/nc3/pweb/lessons/bernoull.htm

Answer KeyActivity CardActivity 1: Bernoulli's Principle states that when the speed of a moving fluid (air) increases, the pressure on its edges decreases. The ball clings to the funnel when it is pointed toward the ceiling when the air is blown hard and fast through the stem of the funnel. Still air exerts more pressure around the ball than that around a stream of moving air. The ball clings to the funnel when it is pointed toward the floor because the air moves away from it faster, creating a low-pressure area in the center.

Activity 2:

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Activity CardActivity 1: Bernoulli's Principle states that when the speed of a moving fluid (air) increases, the pressure on its edges decreases. The ball clings to the funnel when it is pointed toward the ceiling when the air is blown hard and fast through the stem of the funnel. Still air exerts more pressure around the ball than that around a stream of moving air. The ball clings to the funnel when it is pointed toward the floor because the air moves away from it faster, creating a low-pressure area in the center.

Activity 2:

Assessment Card:1. The area A= 1cm2 = 10-4 m2 and the height h= 4m. Direct substitution into Equation R = A√2gh

gives us:R = A√2gh = (10-4 m2) √ (2)(9.8m/s2)(4 m)

= (10-4 m2) (8.85 m/s) = 8.85 x 10-3 m3/s

2. The difference in pressure is:P2 – P1 = ρghUsing the equation where height is zero, we have:P2 – P1 = ½ρv1

2 - ½ ρv22

Combining these two eqns, we obtain:ρgh = ½ρv1

2 - ½ ρv22

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Assessment Card:1. The area A= 1cm2 = 10-4 m2 and the height h= 4m. Direct substitution into Equation R = A√2gh

gives us:R = A√2gh = (10-4 m2) √ (2)(9.8m/s2)(4 m)

= (10-4 m2) (8.85 m/s) = 8.85 x 10-3 m3/s

2. The difference in pressure is:P2 – P1 = ρghUsing the equation where height is zero, we have:P2 – P1 = ½ρv1

2 - ½ ρv22

Combining these two eqns, we obtain:ρgh = ½ρv1

2 - ½ ρv22

Assessment Card:4. ½ 𝜌𝑣12 + 𝜌𝑔ℎ1+ 𝑃1 = ½ 𝜌𝑣2

2 + 𝜌𝑔ℎ2 + 𝑃2 Since the velocity does not change (v1=v2), the velocity term can be subtracted from both sides 𝜌𝑔ℎ1 + 𝑃1 = 𝜌𝑔ℎ2 + 𝑃2 Rearrange algebraically to solve for change in height 𝑃1 − 𝑃 2= ℎ2 − ℎ1 = Δ = −13.1 ℎ 𝑚𝑒𝑡𝑒𝑟𝑠

ρg

5. By continuity equation:

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Assessment Card:4. ½ 𝜌𝑣12 + 𝜌𝑔ℎ1+ 𝑃1 = ½ 𝜌𝑣2

2 + 𝜌𝑔ℎ2 + 𝑃2 Since the velocity does not change (v1=v2), the velocity term can be subtracted from both sides 𝜌𝑔ℎ1 + 𝑃1 = 𝜌𝑔ℎ2 + 𝑃2 Rearrange algebraically to solve for change in height 𝑃1 − 𝑃 2= ℎ2 − ℎ1 = Δ = −13.1 ℎ 𝑚𝑒𝑡𝑒𝑟𝑠

ρg

5. By continuity equation:

Enrichment Card:

1. First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1.

v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 / (0.013 m)2 = 1.2 m/sTo find pressure, we use Bernoulli’s equation:P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)

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Enrichment Card:

1. First calculate the flow speed on the second floor, calling it v2, using the equation of continuity. We can call the basement point 1.

v2 = (v1A1) / (A2) = (v1π(r1)2) / (π(r2)2) = (0.50 m/s)(0.020 m)2 / (0.013 m)2 = 1.2 m/sTo find pressure, we use Bernoulli’s equation:P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)

Enrichment Card:3. a)  v1  =  (  2g ( y2  -  y1 ) ) 1/2                 Solve for y2  -  y1.    

0.0500 cm is much smaller than 12.2 cm.                         v1

2         ( 2. 25 m /s )2 y2  -  y1  =  -------  =  ---------------------  =  0.258 m                     2g          2 x 9.80 m /s2     (b)                     v1

2         ( 1.13 m /s)2 y2  -  y1  =  -------  =  ---------------------  =  0.0651 m                     2g          2 x 9.80 m/s 2    The water must drop from a difference of 0.258 m to a difference of 0.0651m.  

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Enrichment Card:3. a)  v1  =  (  2g ( y2  -  y1 ) ) 1/2                 Solve for y2  -  y1.    

0.0500 cm is much smaller than 12.2 cm.                         v1

2         ( 2. 25 m /s )2 y2  -  y1  =  -------  =  ---------------------  =  0.258 m                     2g          2 x 9.80 m /s2     (b)                     v1

2         ( 1.13 m /s)2 y2  -  y1  =  -------  =  ---------------------  =  0.0651 m                     2g          2 x 9.80 m/s 2    The water must drop from a difference of 0.258 m to a difference of 0.0651m.  

Enrichment Card:5. 20.0 km         1 h             1000 m

-----------  x  ----------  x  ------------  =  5.56 m/s      1 h            3600 s          1 km

Pb  +  ½ ρvb2   = Pt  +  ½ ρ vt

2                 Solve for Pb  -  Pt.    vb  =  0 

Pb   -  Pt   =   ½ ρ vt2   -  ½ ρ vb

2  


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