008 Under Pressure Revised 11 10 20 PDF

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Introductory Physics

Hunter College

Under Pressure by: Dr. Islam Hoxha with edits from A.Padilla Revised by R. Marx

Objective: To learn about the pressure in fluids and the factors that influence it, such as density of the fluid, depth and strength of gravity Background The term fluid is used both for liquids and gases. A fluid in equilibrium exerts forces in the walls of the container it is in. Layers of the fluid exert forces on one another. Pressure is defined as the amount of perpendicular force on an area divided by that area, in other words it is force per unit area. P=F/A Pressure is a scalar quantity; the units of pressure in the SI system are pascals: 1 Pa = 1 N/m2 For a fluid at rest, pressure is the same in every direction in a fluid at a given depth;

At a depth of h below the surface of a liquid, the pressure depends on the weight of the liquid above h. Density of a liquid, or gas ( using the letter rho ⍴) or for that matter any object, is defined as the mass over volume , where V=Ah

Since ⍴= mass/volume = m/Ah

m= ⍴Ah. Now force F=weight W= mg = ⍴Ahg.

So substituting terms:

Change of Air Pressure With Altitude In discussion of pressure, both gases and fluids operate by the same mathematical considerations. Let us explore Pressure in our atmosphere. Use this link to obtain the app that provides air pressure at different altitudes. https://www.mide.com/air-pressure-at-altitude-calculator Fill in the middle column with the values that the calculator provides. In the column on the right calculate the change in air pressure for each of this 1000 meter changes. Altitude: meters

Air pressure Pascals

Δ Pressure over 1000 meters

Sea level

101325

XXXXXXXXXXXXXXXX

1000

89874.57

-11450

2000

79495.22

-10379

3000

70108.54

-9386.68

4000

61640.24

-8468.3

5000

54019.91

-7620.33

Explain why the numbers in the right column exhibit the pattern that they do? The numbers indicate as the altitude gets higher; the air pressure drops lower. This is due to decrease in air molecules.

The same logic about pressure in air ( a gas) applied to fluids ( water). It is assumed that density is constant and does not change with depth.

Atmospheric pressure must also be accounted for in dealing with the pressure inside a fluid. So the pressure p, at a depth h.in a fluid is

p=p0+ ρgh This means p-p0= Δp = ρgh P is absolute pressure. The difference ∆p=p-p0 is called gauge pressure It is the pressure of the fluid, independent of the atmospheric pressure , p0. Note that pressure is measured in N/m2 or pascal ( Pa) with 1 Pa=1 N/m2 1 Pa = 1 N/m2 Experiment: Click here for the “Under Pressure “ simulation. https://phet.colorado.edu/sims/html/under-pressure/latest/under-pressure_en.html Get familiar with the icons and buttons. Notice there are 4 different configurations of the fluid confinement in the left

Part 1- Choose the top configuration. There are two valves: The top left fills the container with liquid and the bottom right one discharges fluid from the container. Choose “Atmosphere On” on the top right -Choose: Units-metric and keep them this way for the entire experiment. -Choose fluid density ; 1000 kg/m3 and gravity : Earth ( 9.8 m/s2) -Click the “ruler” icon and also check “grid” In this first step empty the basin by pulling on the bottom valve to release the water. Now place the round pressure gauge at the level of zero Note and record the value of pressure. Next place the gauge at the 1 meter mark. Note and record the pressure. Repeat this for 2 and 3 meters. Level: meters

Air Pressure: Pascals

Change in Pressure: Pascals

0

101.325

XXXXXXXXXX

1

101.337

0.002

2

101.35

0.013

3

101.362

0.012

Examine the values in the column on the right. For the purposes of this lab in which we are looking at pressure in a liquid, how can we treat the changes in air pressure? As the air molecules get increased and denser due to gravitational pull, the pressure increases. This is same for liquid where at the bottom of a container filled with liquid has more pressure.

Now fill the basin with water to the top.

Then, click and drag a pressure gauge to the surface of the liquid. Turn the atmosphere Off. Find the location where the gauge indicates exactly 0 Pa. Then turn the atmosphere back On and read the pressure. This is p0. Record it! 101.325 kPa

Place the pressure gauge at different horizontal locations along the level 1 meter below the surface. ( you can use one gauge per location as you can use up to four of them). Does the pressure change along a horizontal level? Yes

Find the gauge pressure at this 1 meter level. where p = po + pg. Prove this.. Show your calculation. P= 101.325 kPa+

(1000*9.8*1)= 101.325kPa + 9.8kPa= 111.135 kPa

How would you find the gauge pressure using this simulation without making any calculation? Simply adding 9.8 m/s^2 + atmospheric pressure for under 1 meter depth.

Part 2:

Choose configuration 2 Choose Earth gravity,

. Click on “grid”. and fill the container liquid=water

and

atmosphere ON.

Place different gauges along a horizontal line of 2 meters below the surface: two on the left side and two to the right. Repeat along another horizontal line. What do you observe? The pressure on the left-hand and the right-hand side are equal.

Does the pressure depend on the shape of the container? If not, then explained, based on your observations, what determines the pressure in a container. The pressure does not depend on the container shape because liquid can take and mimic the shape of its container. Pressure is directly related to depth. Part 3: Settings: Gravity : 9.8 m/s2, Atmosphere ON, Fluid density =1000 kg/m3 (water),

Top container a) Fill the container to the rim. Use the gauge pressure to measure the pressure at the 6 different depths listed on the table below, starting with depth 0.5m. Since the distance between each meter line is one centimeter, you can use the vertical ruler to locate the 0.5 depths. Finding Density from a Pressure/Depth Graph Trial

Depth Meter Pressure Pas.

1

0.5

106.556 kPa

2

1.0

111.337kPa

3

1.5

116.445kPa

4

2.0

120.96kPa

5

2.5

126.277kPa

6

3.0

130.634kPa

●

Plot P ( y axis) vs h ( x axis)

● What is the physical meaning of your slope? It is liquid density times gravitational pull. As we go downer the pressure goes greater. ●

What is the y-intercept for your data? It should be atmospheric pressure

1.013 × 105 N/m2 It is the pressure of atmosphere on liquid. The push from above that is enforced on the liquid below. ● Calculate the density of the fluid, from your slope and the value of g. (show calculation) pg=delta y/deltax => (111.337kPa-106.556 kPa)/ (1-0.5)= 9.564 m/s^2 dencity= pressure/gh=> density= (106.556 kPa-101.325 kPa )/ 9.81m/s^2 *0.5m= 1066.5 kg/m^3 ● Compare your calculated density with the known value. Or, calculate the percent error of your calculated density with the known value for water.

(1000/1066)*100%= 6.2

Part 4: Repeat step 3, for a Mystery Planet. Use the bottom container. This time, water is the fluid. Its density would be 1000 kg/m3 . In this case, determine the gravity, g, at the surface of the mystery planet. Trial

Depth Meter Pressure Pas.

1

0.5

217.241kPa

2

1.0

227.332kPa

3

1.5

237.424kPa

4

2.0

247.515kPa

5

2.5

257.387kPa

6

3.0

266.6kPa

● Plot P vs h.

● What is the physical meaning of your slope? It is liquid density times gravitational pull. As we go downer the pressure goes greater ● What is the y-intercept for your data? You should have atmospheric pressure, about1.013 × 105 N/m2 It is the pressure of atmosphere on liquid. The push from above that is enforced on the liquid below.

● Calculate the gravity of this mystery planet, from your slope (show calculation) M=deltay/deltax= (227.332kPa - 217.241kPk)/ (1-0.5)= 20.182 m/s^2 ● What planet could this be? Jupiter

Part 5 Pascal’s Principle

Choose the third configuration/container Set gravity=9.8 m/s2, atmosphere ON, fluid density=1000 kg/m3, and choose grid. Place a pressure gauge along the 2 meter mark on each side and record the pressure reading. 114.108kPa Then click and drag 250 kg weight in the well on the left. As you drag it, a dashed receiving box will appear above the fluid on the left side. Drop it in. Record the new pressure on both sides. Record the pressures 250, 500, 750 and 1000 kg? Remove the pressure gauges and return the masses to the top.Consider the next measurements approximations. Place the vertical ruler against the fluid on the left so that the 3 centimeter mark is set at the top of the fluid.(narrow tube). Each centimeter is equivalent to one meter of depth, with 5 small markings at 0.2 centimeters. Again place the masses sequentially in the dashed boxes and record the change in depth on the left side. Return the masses to the top. Place the vertical ruler on the right side ( wide tube) with the 3 centimeter mark at the top of the fluid. Again, place the masses in the dashed boxes sequentially and record the increase in the height of the fluid.

Create a Data Table to Show Your Results:

Mass [kg]

Pressure In (left tank)Pin

Pressure Out (right tank) Pout

Depth Decrease(m)

Height Increase(m)

Depth /Height

0

114.108kPa

114.108kPa

0

0

0

250

114.52kPa

114.52kPa

0.3m

0.05m

6

500

114.945kPa

114.945kPa

0.4m

0.1m

4

750

115.365kPa

115.365kPa

0.5m

0.2m

2.5

1000

115.755kPa

115.755kPa

0.6m

0.3m

2

Compare the pressure changes on the left and right sides as the masses are added to the fluid. How does the addition of masses affect the changes in pressure on both sides? The pressure increased equally in both side by adding the masses.

How does this relate to Pascal's Law?

Now force F=weight W= mg = ⍴Ahg. As the mass increases the pressure increases. Calculate the ratio of depth change on the left to height change on the right. After each mass increase. Record it. What pattern do you see? The ratio decreases as the mases are added. Assume that each of these sides is a tube and the cross-section line you see is the diameter. The approximate measurement of the diameters on the left and right are 1.75 cm and 8 cm respectively. 0.04m*pi*0.05m=0.0063m^3 0.00875m*pi*0.3= 0.008m^3 By approximately how much of a factor does the volume increase on the right compare to the volume decrease on the left side? 0.008/0.0063= 1.3

Now calculate the cross-sectional area of the left and right tubes in centimeters squared using the approximate diameter values provided Right (Larger)___________50.3__cm^2_________ Left (smaller)_______2.4cm^2__________ Recall that ΔP = F x A

Based on this, determine how much greater the force upward on the right tube must be compared to the force downward on the left side. FA (left)=FA (right) F(right)=F(left)A/A= 50.3cm^2/ 2.4cm^2= 20.958 N Why is this information so important to mechanics at auto repair shops?

To use in hydraulic jacks or automobile shocks, it is used to adjust one side respectful to other.

Post Lab Questions 1. What are the “bends”? What does this have to do with pressure? Can whales get the “bends”? some gases become more soluble under pressure as a result Nitrogen can dissolve in blood stream and cause sickness or form bobbles in muscles and cause horrible spasm pain. Bends also can damage the lungs so all mammals are prone to change in pressure including marine mammals.

2. Briefly summarize the debate over how planes can fly as it relates to pressure. Planes use positive pressure system where the pressure inside the plane is equal or greater than the pressure outside to keep the integrity of the plane intact and prevent the doors to claps inward. That’s why jet fighters during parachute can push the cockpit cover miles away from the plane before jumping out....