PHYS1146 Lab 6 - Copy

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Report for Experiment #11 Fluid Flow 12/01/23 Abstract The goal of this experiment was to validate Poiseuille's law. This was done by examining the relationship between flow rate and pressure, and the relationship between the flow rate and capillary diameter. In investigation one, a capillary with a diameter of 1.25 mm was used and pressure was adjusted by changing the height of the fluid reservoir. By using the IPL Straight Line Fit calculator, the slope and error in slope was obtained and was 0.0289±0.00129. The y-intercept on the graph was equal to 0.1704 and the y-intercept and error found by using the ILP Straight Line Fit calculator and was equal to 0.161±0.0224. The relationship between flow rate and pressure difference was found to be linear, which validated Poiseuille's law. In investigation two, flow rate through different diameter capillaries was investigated. There was a linear relationship. The slope value was found to be 3.799±0.114 which was close to the true value of four, this also validated Poiseuille’s law.

Introduction A fluid flows due to a difference in pressure and obeys Newton’s laws. Poiseuille’s law of fluid flow states that the flow rate of a fluid is dependent on the length of a tube, the radius, the change in pressure, and the viscosity of the fluid. The difference in pressure and flow rate are directly proportional and flow rate can be defined as the volume of fluid passed per unit of time, in this case volume of water per second. For laminar flow, the relation between resistance and the capillary’s length and diameter is written as: (1) In this equation, resistance is proportional to capillary length. Flow rate is inversely proportional to resistance, so when resistance is doubled flow rate is halved. The flow rate is also proportional to the diameter, so when diameter increases so will the flow rate. The flow rate is inversely proportional to the viscosity of the fluid. These relationships validate Poiseuille’s law because as stated previously the flow rate of a fluid is dependent on the length of a tube, the radius, the change in pressure, and the viscosity of the fluid. In this lab the relationship between flow rate with changes in pressure and the diameter of the capillary were examined to validate Poiseuille’s law. In investigation one the time taken for water to flow through a capillary was measured at different heights. In investigation two the time taken for water to flow through capillaries with different radii at a constant height was measured. The purpose of this was to examine the effects height and radius played on the difference in pressure and to investigate Poiseuille’s law of fluid flow. Investigation 1 The experimental setup consisted of a fluid flow apparatus, a capillary with a 1.25 mm diameter, a ruler, a plastic beaker, a metal beaker, a 100 mL graduated cylinder, a stop watch, a large funnel, a metal rod, and water. The fluid flow apparatus was attached to the metal rod and could be attached to the capillary. The capillary was held by a clamp. The fluid flow apparatus and the position of the capillary could be adjusted. The water reservoir had a scale and a baseline level of 6 was chosen. The height difference of h 2 (height of baseline water level) – h 1 (height of capillary) was equal to 11 cm. The capillary was then attached to the fluid flow apparatus and water was drained out until no air bubbles were left. The height of baseline water level was maintained throughout the experiment by replenishing the water. Once air bubbles were no longer present in the capillary, the 100 mL graduated cylinder with the large funnel was placed under the capillary and water filled the graduated cylinder until the water level reached 50 mL. The amount of time required for the graduated cylinder to reach 50 mL was recorded with an iPhone timer, this was done twice. The average time was calculated using the average function in excel. The flow rate was found by using: (2) The flow rate was found for both trials and the average flow rate was found by using the excel averaging function for the two calculated values. Volume was equal to 50±0.05 mL and the flow rate for t 1 and t 2 were found and then using excel the average flow rate was found. The error in flow rate was calculated by using: (3) R = 128 η L π d 4 Q = ∆ V ∆ t δ Q avg = δ t 2 + δ V 2

Error in time was the standard deviation of both the time values and was calculated in excel using the standard deviation function. Error in volume was equal to half of the smallest value of measurement on the graduated cylinder. The experiment was repeated three more times with a different Δ p each time. The heights of the fluid flow apparatus and the capillary were adjusted to have height differences of 14 cm, 17 cm, and 20 cm. The average flow rate for each Δ p value was calculated by using (3). Table 1: Collected data and flow rate of four different pressures. The baseline height was set to 49.8 cm and the height of the capillary was set to 38.8 cm, 35.8 cm, 32.8 cm, and 29.8 cm for each trial respectively. The average flow rates and error in average flow rates were found to be 0.480±10.232, 0.574±2.317, 0.628±2.262, and 0.743±0.889 mL/s. A graph of average flow rate vs change in pressure was made. Trial 1 2 3 4 Δ p 11 14 17 20 h1 (cm) 38.8 35.8 32.8 29.8 h2 (cm) 49.8 49.8 49.8 49.8 t1 (s) 97.44 85.53 81.15 67.81 t2 (s) 111.91 88.73 78.03 66.77 avg t (s) 104.675 87.13 79.59 67.29 V (mL) 50 50 50 50 Q1 0.513 0.585 0.616 0.737 Q2 0.447 0.564 0.641 0.749 Qavg 0.480 0.574 0.628 0.743 ࠵? Qavg 10.232 2.317 2.262 0.889 ࠵? h1 cm 0.05 0.05 0.05 0.05 ࠵? h2 cm 0.05 0.05 0.05 0.05 ࠵? Δ p 0.071 0.071 0.071 0.071 ࠵? V (mL) 0.05 0.5 0.5 0.5 ࠵? t 10.232 2.263 2.206 0.735

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Fig 1. - Average flow rate at different Δ p The slope on the graph was found to be 0.0281. By using the IPL Straight Line Fit calculator, the slope and error in slope was obtained and was 0.0289±0.00129. The y-intercept on the graph was equal to 0.1704 and the y-intercept and error found by using the ILP Straight Line Fit calculator and was equal to 0.161±0.0224. It was expected that the y-intercept would be at zero because with no pressure difference there should be no flow. The y-intercept calculated, and its uncertainty did not overlap with zero which would indicate there was some inaccuracy in the measurements of flow. The data does show a strong linear relationship between pressure and flow, confirming that Poiseuille’s law is valid. There may have been random errors in the recording of the time and the estimation of the volume at which the time was taken. Fig 2. - Slope and error in slope values obtained by the IPL Straight Line Fit calculator.

Investigation 2 The experimental setup consisted of a fluid flow apparatus, capillaries with a 1.5, 1.25, 1.0, and 0.5 mm diameters, a ruler, a plastic beaker, a metal beaker, a 100 mL graduated cylinder, a 50 mL graduated cylinder a stop watch, a large funnel, a metal rod, and water. The fluid flow apparatus was attached to the metal rod and could be attached to the capillary. The capillary was held by a clamp. The fluid flow apparatus and the position of the capillary could be adjusted. The water reservoir had a scale and a baseline level of 6 was chosen. The height difference of h 2 (height of baseline water level) – h 1 (height of capillary) was equal to 20 cm besides the last trial where height difference was equal to 50 cm. Data for the 1.25 mm diameter trial was taken from the values calculated in investigation one. After copying and pasting the data from investigation one, the capillary was changed to the 1.5 mm diameter and the capillary was then attached to the fluid flow apparatus and water was drained out until no air bubbles were left. The height of baseline water level was maintained throughout the experiment by replenishing the water. Once air bubbles were no longer present the 100 mL graduated cylinder with the large funnel was placed under the capillary and water filled the graduated cylinder until the water level reached 50 mL. The amount of time required for the graduated cylinder to reach 50 mL was recorded with an iPhone timer, this was done three times. The average time was calculated using the average function in excel. The flow rate was found by using (2) and average flow rate was found by using the averaging function on excel for the three flow rate values. The volume was equal to 50±0.05 mL and the error in flow rate was calculated by using (3). This process was repeated with a capillary of 1.0 mm diameter. After values for the 1.0 mm diameter were found, the capillary was changed to the 0.5 mm diameter. However, in this trial the volume was changed to 10 mL rather than 50 mL and the height difference was set to 50 cm. This was done in order to obtain data in a reasonable time. After these changes the process for collecting data was repeated. Table 2: Collected data and flow rate of four different diameters. Trial 1 2 3 4 d(mm) 0.5 1 1.25 1.5 ࠵? d (cm) 0.02 0.02 0.02 0.02 Δ p (cm) 50 20 20 20 h1 (cm) 19 29.8 29.8 29.8 h2 (cm) 69 49.8 49.8 49.8 t1 (s) 167.16 75.06 67.81 33.12 t2 (s) 172.7 74.31 66.77 33.43 t3 (s) 171.57 75.41 x 33.43 avg t (s) 170.477 74.927 67.29 33.327 V (mL) 10 50 50 50 Q1 0.060 0.666 0.737 1.510 Q2 0.058 0.673 0.749 1.496 Q3 0.058 0.663 x 1.496

Due to the pressure difference in the 0.5 mm trial being different than the other trials, the flow rate was corrected by multiplying the calculated value by . After this correction the Δ p was the same in all trials and Poiseuille’s equation could be rewritten as: (4) In this equation C is equal to . By taking the logarithm on both sides the equation could be rewritten as: (5) The natural log for average flow rate and diameter was calculated by using the excel function. Error in natural log of average flow rate and natural log of the diameter was calculated using: (6) This equation was provided by the lab procedure and A represented a measured quantity. The diameters for each trial were 0.5± 0.02 mm, 1.0± 0.02 mm, 1.25± 0.02 mm, and 1.5± 0.02 mm. The average time for the trials were found to be 170.477 s, 74.927 s, 67.290 s, and 33.327 s. The average flow rates were found to be 0.023±2.928 mL/s, 0.667±0.752 mL/s, 0.743±0.889 mL/s, and 1.500±0.531 mL/s. The natural log of the flow rates were found to be –3.752±124.76, -0.404±1.127, -0.297±1.197, and 0.406±0.354. The natural log of the diameters were found to be –0.693±0.04, 0.000±0.02, 0.223±0.016, and 0.405±0.013. A graph of ln Q vs ln d was then made in excel. Qavg 0.023 0.667 0.743 1.500 ࠵? Qavg 2.928 0.752 0.889 0.531 lnQavg -3.752 -0.404 -0.297 0.406 ࠵? lnQavg 124.76 1.127 1.197 0.354 lnd -0.693 0.000 0.223 0.405 ࠵? lnd 0.04 0.02 0.016 0.013 ࠵? h1 0.05 0.05 0.05 0.05 ࠵? h2 0.05 0.05 0.05 0.05 ࠵? Δ p 0.071 0.071 0.071 0.071 ࠵? V (mL) 0.05 0.5 0.5 0.5 ࠵? t 2.927 0.562 0.735 0.179 2 5 Q = Cd n π ∆ p 128 η L ln Q = ln ( Cd n ) = n ln d + ln C δ ln A = δ A A

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Fig. 3 – Natural log of flow rates at different natural log diameters. The slope of this graph was representative of n and was found to be 3.797. The slope and error in slope was found using the IPL Straight Line Fit calculator was found to be 3.799±0.114. The expected result for perfect laminar flow described by Poiseuille’s Law is a value of n equals four. The result obtained was close to the expected result indicating most of the flow was in perfect laminar flow and it was unlikely there was much turbulent flow. Possible random errors would include inaccuracy of the measurement of the height, the time, and the volume of fluid at which the time was taken. Fig 4. - Slope and error in slope values obtained by the IPL Straight Line Fit calculator. Conclusion These experiments were conducted to explore the validity of Poiseuille’s law, by measuring the effect of changes in pressure and capillary diameter on the flow rate of a fluid. In investigation one a fluid

flow apparatus was set to a height of 49.8 cm and a capillary with a diameter of 1.25 mm was set to four different heights of 38.8, 35.8, 32.8, and 29.8 cm. The rate at which fluid flowed out of the capillary was measured for each of these height differences. A graph of average flow rate vs Δ p was made to validate Poiseuille’s law, the slope and error in slope was obtained and was 0.0289±0.00129. The y-intercept on the graph was equal to 0.1704 and the y-intercept and error found by using the ILP Straight Line Fit calculator and was equal to 0.161±0.0224. The y-intercept and uncertainty did not overlap with zero, which was expected. This could have been due to inaccuracy in measurements of flow. The graphical data did show a linear relationship between flow and pressure, which validated Poiseuille’s law. In investigation two the fluid flow apparatus was set to a constant height of 20 cm, except for the final trial where the height was 50 cm, and each trial was done with a capillary of different diameters. The diameters used were 1.5, 1.25, 1.0, and 0.5 mm. A graph of the average flow rates vs diameters was made, and the slope was found using the IPL Straight Line Fit calculator was found to be 3.799±0.114. The expected result for perfect laminar was four. The result was close to four and this also validated Poiseuille’s law. Possible improvements for this lab would be to ensure that the capillary was held straight so gravity would not affect the flow rate, increase the number of trials, and improving the accuracy of the volume and time measurements. Questions 1. You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and ⅝ inch inner diameter. Compare the rate at which water flows through the two types of hoses (i.e., the ratio of rates). a. Q=Av b. Q 1/2 =Q 5/8 c. d. 0.64 2. How does the flow rate Q through a capillary change if the viscosity of the fluid in question doubles? How will it change if the fluid’s density doubles instead? a. When the viscosity of a fluid doubles the flow rate decreases by half. When a fluid's density doubles the flow rate will not change. 3. Calculate the ratio Rtube/Rcap for the setup in Investigation 1, where Rtube is the resistance of the tube connecting the reservoir and capillary and Rcap is the resistance of the capillary. What conclusion can you draw from this? a. A 1 2 v 1 2 = A 5 8 v 5 8 v 5 8 v 1 2 = A 1 2 A 5 8 = π ( 0.5 2 ) 2 π ( 0.625 2 ) 2 = R tube = 67 cm (0.7 cm ) 4 = 279.05

b. c. d. The capillary has a higher resistance than the tube therefore … 4. Now compute the ratio Δ ptube/ Δ pcap, where Δ ptube is the pressure drop over the tube and Δ pcapis the pressure drop over the capillary. a. = 6566 b. = 686 c. 9.57:1 5. Why must the capillary in this experiment be positioned horizontally? a. The capillary was positioned vertically so no external forces would act on the capillary and affect the flow rate of the water. Acknowledgments References [1] Hyde, Batishchev, and Altunkaynak, Introductory Physics Laboratory, pp 80-86, Macmillan Higher Education, 2022. [2] IPL Lab Report Guide, https://web.northeastern.edu/ipl/wp-content/uploads/2017/09/IPL-Lab-Report- Guide.pdf [3] Giancoli, Physics Principles with Applications, Pearson, 2021. R cap = 23.9 cm ( 0.125 cm ) 4 = 97894.4 R tube R cap = 279.05 97894.4 = 0.00285 ∆ p tube = (1000)(9.8) ( 0.67 ) ∆ p cap = (1000)(9.8)(0.07) 6566 686 =

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