Physics Phet Lab Report on Pressure within Fluids, Pascals’ Law and the Bernoulli PDF

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Physics Lab report on Pascal's Law and Bernoulli's equation using the Phet Colorado virtual lab....

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Lab #3: Pressure within Fluids, Pascals’ Law and the Bernoulli Principle Your Name Here:

Theory: In class we found that the pressure within a fluid rises based on the depth within the fluid. P = P 0 + ρgy As we probe deeper within a fluid the pressure we find there will depend on any surrounding pressure plus a pressure resulting from a need to support the overlying weight of the fluid. The simulation we will be using will allow for changing the overlying pressure, the density of the fluid and the gravitational constant of the planet, allowing us to probe the relationship. Pascals Law tells us that, for the same depth within a single fluid, the pressure will be the same and that, if the pressure is raised within the fluid, it will be transmitted to all parts of the fluid: P1 = P2 (given same elevation). Continuity of Flow tells us that the volume of fluid passing a point per unit time is the same everywhere, provided the fluid is incompressible. A1v1 = A2v2. Bernoulli’s principle tells us about the behavior of an incompressible, nonviscous fluid. Specifically, P + ρgy + 21 ρv2 = constant Most of the time our focus in the classroom is on what happens within a moving fluid – namely that the pressure drops as the speed increases (given same elevation). In fact, the Bernoulli equation contains the other two (pressure with depth and Pascals law). In this lab we will be using a PhET simulation to probe the relationships we have learned about in class. If you are comfortable downloading and running a .jar file on your computer then you may use the version found here: https://phet.colorado.edu/en/simulation/legacy/fluid-pressure-and-flow If you would prefer not to run an unfamiliar jar file OR your device cannot run a jar file then you may use the Cheerpj online version, just be warned: it runs SLOW. https://phet.colorado.edu/sims/cheerpj/fluid-pressure-and-flow/latest/fluid-pressure-an d-flow.html

Pressure tab: Exercise 1: The first two exercises can be either of the first two tank types. Use the pressure tool to place a pressure meter near the bottom of the tank. Measure the pressure at the

bottom of the square pool as you raise and lower the fluid level inside. Use the ruler to measure the depth. Click and drag the pressure tool to a few locations both inside and outside the liquid and observe the readings. Q1: Is the pressure reported gauge pressure or absolute pressure? What is your evidence? Ans: Absolute pressure at a point inside the liquid is the sum of the atmospheric pressure and the pressure due to the liquid column over that point. Gauge pressure does not consider the atmospheric pressure. Gauge pressure is the difference between the absolute pressure at that point and the atmospheric pressure.

Here as can be observed from the above picture of the pressure measuring instrument, its zero reading is set at atmospheric pressure of 101.300 kPa. Hence this instrument will show the absolute pressure at any point within the liquid.

Q2: If you double or halve the amount of fluid in the container what happens to the pressure within the fluid? Record your result and discuss here.

Ans: The tank is full and the liquid height is 3 m. as shown in the below picture.

Pressure at a point within the tank at a certain depth in the liquid is 129.1 kPa. The tank drain valve is opened and some liquid is drained out until the height of liquid column inside the tank becomes half of its initial height that is 1.5 m as shown in the below figure:

As can be observed, the pressure at this time becomes 114.6 kPa. The ratio of the full height liquid . pressure and the half height liquid pressure is . or 1.13 approximately.

Absolute pressure at a depth of ℎ from the top surface will be represented by the following equation: 𝑃 = 𝑃 + ρ𝑔ℎ If the liquid column height is decreased to half of its previous height, the pressure at the same point becomes: ℎ 𝑃 = 𝑃 + ρ𝑔 2 As can be observed from the above two equations, the pressure shown by the gauge will not be the half of its initial value.

Exercise 2: Change the fluid to have the density of gasoline (700 kg/m3). Measure the pressure at the bottom of the tank and record in table 1. Now change the fluid to have a density of 1400 kg/m3 . Measure again the pressure at the bottom of the tank and record in table 1. Finally, change the gravitational acceleration to be 19.6 m/s2. Record the pressure in table 1. Table 1: Run

Fluid density (kg/m3) Gravitational acceleration (m/s2)

Pressure (kPa)

1

700

9.8

121.9

2

1400

9.8

142.5

3

1400

19.6

284.9

Q3: Did doubling the density at the same gravitational acceleration double the pressure? Would you expect it to have doubled the pressure? Why or why not?

Ans: No, doubling the density does not double the pressure. As can be observed, the initial pressure is 121.9 kPa. After doubling the density, ratio of the new liquid pressure to that of the . earlier one is . or 1.17 approximately. Absolute pressure at a depth of ℎ from the top surface will be represented by the following equation: 𝑃 = 𝑃 + ρ𝑔ℎ Where ρ is the liquid density. If we substitute the same with 2ρ in the above equation, the equation becomes as follows: 𝑃 = 𝑃 + 2ρ𝑔ℎ Since the atmospheric pressure remains the same and only the pressure due to the liquid density gets doubled, the total pressure will not get doubled.

Q4: When you doubled the gravitational acceleration at the same density, did it double the pressure? Would you have expected it to double the pressure? Why or why not? Ans: No, doubling the gravitational acceleration does not double the pressure. As can be observed, the pressure at the original gravitational acceleration value was 142.5 kPa. After the gravitational acceleration is doubled, the new pressure value becomes 284.9 kPa. Absolute pressure at a depth of ℎ from the top surface will be represented by the following equation: 𝑃 = 𝑃 + ρ𝑔ℎ Doubling the gravitational acceleration changes the above equation to the following form: 𝑃 = 𝑃 + ρ2𝑔ℎ Again, as the atmospheric pressure remains the same and only the pressure due to the

gravitational acceleration gets doubled; the total pressure will not get doubled. However, if the pressure measuring tool measured the difference between the absolute and atmospheric pressure, that is the gauge pressure, doubling any of the three quantities, that is height, density or gravitational acceleration would have doubled the measured pressure reading. Repeat the above measurements (for experiment 2) with the atmosphere turned off. Table 2: Run

Fluid density (kg/m3) Gravitational acceleration(m/s2)

Pressur e(kPa)

1

700

9.8

10.3

2

1400

9.8

20.5

3

1400

19.6

41.1

Q5: Did doubling the density at the same gravitational acceleration double the pressure? Would you expect it to? Why or why not? Ans: Yes, as expected and explained in the answer to the previous question, doubling the density at the same gravitational acceleration doubled the measured pressure. With gravitational acceleration set back to 9.8 m/s2 and the atmosphere turned back on, take readings at the ground level, 1 m above the ground and 2 m above the ground. Use this information to calculate the density of the air. You will need to move the ruler and pressure measuring tool out of the fluid. Q6: What do you find for the density of air? Is this value reasonable? Show your work. Ans: Pressure at the ground level is 101.325 kPa. Pressure at 1 m above the ground is 101.313 kPa and pressure at 2 m above the ground is 101.301 kPa. Assume height of air column measured from the ground level is ℎ. Hence according to the formula of pressure at a point, we can write: 𝑃 = ℎρ𝑔 Where ρ and 𝑔 are the air density and gravitational acceleration respectively. At the ground level, pressure measured was 101.325 kPa or 101.325 × 10 Pa. Hence the above equation becomes: 101.325 × 10 = ℎρ𝑔 101.325 × 10 𝑜𝑟, ℎ = ρ

At 1 m above the ground level, the equation will become: 101.313 × 10 = (ℎ − 1)ρ𝑔 101.325 − 1 ρ𝑔 𝑜𝑟, 101.313 × 10 =  ρ𝑔 𝑜𝑟, 101.313 × 10 = 101.325 × 10 − ρ𝑔 𝑜𝑟, ρ𝑔 = (101.325 − 101.313) × 10 𝑜𝑟, ρ𝑔 = 0.012 × 10 0.012 × 10 𝑜𝑟, ρ = 𝑔 ρ=

0.012 × 10 9.8

ρ = 0.001225 × 10

Hence the density of air is 1.225 kg/m3.

ρ = 1.225

Air being a homogeneous mixture of different gases and existing on a huge volume, this low density is quite justifiable. Exercise 3: Change to the third tank type. You will no longer have the ability to add or remove fluid, but you will gain 3 weights which fit perfectly into the opening on the left hand side. Place a pressure gauge at the bottom of each side of the tank. Write the pressure on each side in table 3. Q7: What does Pascal’s law suggest should happen when you add a mass such that it adds pressure to the left-hand side? Ans: According to the Pascal’s law, the pressure of an incompressible static enclosed fluid within any point is same at the same depth measured from the same reference level and if the pressure is increased at a certain point, the additional pressure gets transmitted to all the points uniformly resulting an equal increment in pressure at every points within the liquid. Thus in this condition also, the pressure of all the points at the same depth remains equal. Now, add a 250 g mass to the hole on the left-hand side. Record the pressures in table 3. Q8: What happened when you added a 250 g mass to the left-hand side? Did it match your expectations? Explain. Ans: The fluid used here is assumed to be incompressible. The fluid is in static condition and is enclosed within the tank. Hence abiding by the Pascal’s law, the increment in pressure by adding the external mass 250 g on the liquid gets transmitted uniformly within the liquid. Since the two pressure measuring tools are placed at the bottom of the liquid at the same depth, both the tools show same increased pressure. Pressure before adding the mass was 132.2 kPa which changed to 132.6 kPa after adding the mass.

Q9: Now predict before you try: what will happen to the pressures when you add another 250 g mass? How about the 500 g mass? Show calculations here. Ans: Assume the area of the left-hand side tank open surface is A. Pressure at a depth h from this surface is expressed by the following equation: 𝑃 = 𝑃 + ℎρ𝑔 𝑜𝑟, 𝑃. 𝐴 = 𝐴. (𝑃 + ℎρ𝑔) 𝑜𝑟, 𝑃. 𝐴 = 𝐴. 𝑃 + 𝐴. ℎρ𝑔 Since pressure multiplied by the area is the force acting on that area, the above equation can be written as: 𝐹 = 𝐹 + 𝐴ℎρ𝑔 Where 𝐹 is the force due to atmospheric pressure. Adding a 250-gm mass will exert a force of 0.250 x 9.8 N on the same liquid column. Hence the force on the liquid column will become: 𝐹 = 𝐹 + 𝐴ℎρ𝑔 + 0.250 × 9.8 Divide both sides by area A:

The equation becomes:

𝐹 𝐹 𝐴ℎρ𝑔 0.250 × 9.8 + = + 𝐴 𝐴 𝐴 𝐴

2.45 𝐴 As found from the simulation 𝑃 = 132.6 kPa or 132.6 × 10 Pa and 𝑃 = 132.2 kPa or 132.2 × 10 Pa. Substituting these values into the above equation, we get: 2.45 132.6 × 10 = 132.2 × 10 + 𝐴 2.45 = (132.6 − 132.2) × 10 𝑜𝑟, 𝐴 𝐴 = 0.006125 𝑚 .×. Pa or 400 Adding another 250 gm should increase the pressure at the same point by .   Pa. So, the pressure should become (132.2 × 10 + 400) Pa or 132.6 × 10 Pa. The simulation shows exactly the same pressure. 𝑃 = 𝑃 + ℎρ𝑔 +

Adding another 250 gm mass will increase the pressure by 400 Pa and the new pressure reading will become (132.6 × 10 + 400) Pa or 133.0 × 10 Pa. Adding another 500 gm mass will increase the pressure by 800 pa and the new pressure reading will be (133.0 × 10 + 800) Pa or 133.8 Pa. Now actually add the second 250 g mass and then the 500 g mass and record the result in Table 3.

Table 3 Mass (g)

Pressure Left(kPa)

Pressure Right (kPa)

250 g

132.6

132.6

250 g + 250 g

133.0

133.0

All 3

133.8

133.8

Q10: Did the result match your expectation? If not, why? Ans: The results exactly matched the expected outcomes.

Flow Tab: Change over to the flow tab. You do not need to have the particles shown if they contribute to the simulation running slowly. In this tab you can create a situation like the standard one portrayed when deriving the Bernoulli formula (each pipe end has a handle which can be used to change its elevation). The pipe, at various points, has handles which can be used to adjust the size of the pipe at that point. Set the flow rate close to 6000 L/s. Turn on the flux meter option. We aren’t really interested in this tool for the flux value, but instead because it shows the area. If you drag the tool around you will be able to measure the area in various places. The pressure tool will show you the pressure in a particular location and the velocity tool will tell you the velocity of the fluid at that location. Exercise #4: First move the pressure sensor around the interior of the tube. Q11: What do you notice? Is the pressure the same everywhere? Why or why not? Ans: The pressure at all the points at the same heights within the fluid is same. Since the cross-sectional area of the pipe is uniform, the velocity of the fluid is also uniform. Thus from Bernoulli’s equation we can conclude that the pressure at the same heights within the fluid will be the same.

If you have resized/reshaped the tube reset the simulation. Now, measure the pressure and the velocity inside the center of the tube (centered both vertically and horizontally).

Q12: What was the pressure at the center of the tube? What was the velocity?

Ans: Pressure at the center of the tube is 120.840 kPa and the speed is 1.9 m/s.

Change the shape at the middle of the pipe to be larger. Use the Flux Meter to find the area of this region compared to the original size. Q13: Record the original area and expanded area here. Use Bernoulli’s Equation in combination with continuity of flow to calculate the expected pressure in this region. Then verify and record the actual value found with the pressure meter. Show your work. Ans: After resizing, the new cross-sectional area becomes 6.9 m/s2 . Assume the velocity at this time becomes V2. The continuity equation can be written as: 3.1 × 1.9 = 6.9 × 𝑉 3.1 × 1.9 𝑜𝑟, 𝑉 = 6.9 𝑜𝑟, 𝑉 = 0.85 𝑚/𝑠

The Bernoulli’s equation for this flow can be written as: 1 1 𝑃 + ρ𝑔𝑦 + ρ𝑉 = 𝑃 + ρ𝑔𝑦 + ρ𝑉 2 2 Since 𝑦 = 𝑦 , the potential energy term will get cancelled from both sides of the equation. 1 1 120.840 × 10 + × 1000 × (1.9)  = 𝑃 + × 1000 × (0.9) 2 2 1 1   𝑃 = 120.840 × 10 + × 1000 × (1.9) − × 1000 × (0.9) 2 2 𝑃 = 122.24 × 10 Hence the estimated pressure at this point is 122.24 × 10  Pa or 122.24 kPa. The simulation shows the pressure value as 122.281 kPa and the velocity as 0.9 m/s which is very close to our calculated value. The small deviation may be attributed to the error in accurately positioning the pressure tool at the middle of the flow. Now narrow the pipe instead. Again, use the Flux Meter to find the area of this region. Q14: Record the narrowed area here. Use Bernoulli’s Equation to calculate the expected pressure in the region. Then verify and record the actual value found with the pressure meter. Show your work.

Ans: In this case, the new cross-sectional A 2 is 0.9 m 2 . Assume the new velocity as V 2. Then the continuity equation becomes: 3.1 × 1.9 = 0.9 × 𝑉 3.1 × 1.9 𝑜𝑟, 𝑉 = 0.9 𝑜𝑟, 𝑉 = 6.54 Hence the approximate value of the velocity is 6.54 m/s. Use the Bernoulli’s equation to find the pressure at this place: 1 1 120.840 × 10 + × 1000 × (1.9) = 𝑃 + × 1000 × (6.54) 2 2 1 1 𝑃 = 120.840 × 10 + × 1000 × (1.9) − × 1000 × (6.54) 2 2 𝑃 = 101.259 × 10  Hence the calculated approximate value of pressure is 101.259 kPa. Value of pressure obtained from the simulator is 103.427 kPa and velocity is 6.6 m/s. Deviation in pressure may be attributed to the placement error of the pressure tool.

Q15: Can you narrow the tube enough that the pressure at the center becomes lower than the pressure outside the tube? What if you also change the volume flow rate? What would happen if you drilled a hole at this location? Ans: Yes, by narrowing the pipe enough, the pressure within the fluid can be made lower than the outside pressure of the pipe which is the atmospheric pressure of 101.325 kPa approximately. The below screenshot shows the same:

Decreasing the volume flow rate will increase the pressure and vice versa. If a hole is drilled at this position, the water will come out gushingly through this hole, pressure of this stream will become atmospheric and the velocity will reduced by the amount equivalent to the pressure drop. Water Tower Tab: In this tab we are presented with a tower containing fluid. We will not be making use of the hose option (it appears to be broken in the online version of the lab). You may change the height of the tower and the level of fluid within the tower. Experiment 5: We will be investigating how far the fluid travels from the tower. First, fill the tower and set the fill line to “manual” (so the tower will empty). Set up the tape measure directly under the hole so you can measure the distance the water travels in the air. Open the hole and watch how the water behaves over time.

Q16: How does the distance from the tower change as the tank empties? Is this expected? Why or why not?

Ans: The distance travelled by the water stream reduces as the tank empties. This result is expected. Initially at rest, the whole energy of stored water was potential. As the tank continues to leak water, the center of mass of the stored water reduces. Since potential energy varies directly with the distance measured from the ground here, the potential energy of the stored water also decreases. While water comes out from the tank, this potential energy continues to convert in kinetic energy until it becomes zero at the ground level. So a decreasing potential energy consequently the decreasing kinetic energy indicates lowering the speed of water jet from the tank. As speed decreases, length covered by the water stream also decreases. Next, fill the tower again and set the fill line to “match leaking”. Record the height of your tower, the height of the center of the hole, and the height of the water in the table below.

Run 1

Height of Tower (m) 15

Height of Hole (m) 15.5

Height of Water (m) 25

Q17: Predict how far the water should travel from the tower using Bernoulli’s Equation (to find the initial horizontal velocity) and Projectile Motion. Does your prediction match the measured value? Show your work. Ans: We assume bottom of the tank as the reference line for measuring the potential energy. Height of the tanks is (Height of Water- Height of the tower). Hence the height is (25-15) m or 10 m. Since height of the tower is 15 m and that of the center of hole is 15.5 m, height of the center of the hole from our chosen reference level becomes 0.5 m. We apply the Bernoulli’s equation along a horizontal line passing through the center of the hole. At this point, for the water within the tank, 𝑦 = 0.5, 𝑃 = ℎρ𝑔 where h is the height of the top most water surface from the midpoint of the hole. Hence ℎ = (10 − 0.5)𝑚 = 9.5 𝑚 and 𝑣=0 Assume the velocity of the water jet 𝑣 . Gauge pressure and y value for the water jet stream is zero. We use gauge pressure in the Bernoulli’s equation here: 𝑃 + ρ𝑔𝑦 =

1  ρ𝑣 2 

1 𝑜𝑟, 9.5 × 1000 × 9.8 + 1000 × 9.8 × 0.5 = × 1000 × 𝑣 2 𝑜𝑟, 𝑣 = 14 So, the velocity is 14 m/s.

Now explore with the pressure gauge. Q18: What do...