Lab4ME211FXC

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1 ME 211 Fluid Mechanics Experiment #4 Fluid Dynamic Drag Fany Xelhua Cabrera 1 November 2022 Lab section AB
2 Abstract The importance of this experiment, fluid dynamic drag, is to examine the fundamental concepts involved when a solid sphere falls through a liquid. The main purpose of this lab is to calculate the drag force exerted on a solid sphere, smooth and rough, falling through a liquid. It is also to see how the drag coefficient and Reynolds number are related to each other. Introduction In this experiment, our focus was to see how a solid sphere behaves when it falls through a liquid medium. It is also to measure the drag force, and to see how drag coefficient and Reynolds number are related to each other. Different sized and weighted steel spheres are dropped in glycerin. The time is measured for how long it takes to travel the specified distance. The drag force is calculated along with drag coefficient, and Reynolds number. Drag coefficient and Reynolds number are graphed to show how the two are related. The measured and calculated drag forces are also compared with each other to see how accurate the experiment is. Theory In order to find Reynolds number and drag force coefficient, we considered buoyant force, force of gravity, and drag force. We took the time measurement of the ball falling from the top to the bottom of the tube, and with all these values we were able to solve for Reynolds number and theoretical drag force using Stokes law. Figure 1: Schematic of the forces exerted on a smooth solid sphere falling within a surrounding liquid in the direction of gravity. D is the drag force, W is the weight, g is gravity, ? 𝑏 is the buoyant force.
3 The drag force on a sphere of diameter d , which moves with instantaneous velocity U , is given by, ? = ? 𝐷 𝐴? 𝑓𝑙𝑢𝑖𝑑 ? 2 2 (1) A = 𝜋𝑑 2 4 , and ? 𝐷 is drag coefficient. Reynolds number is defined by, 𝑅? = ? 𝑓𝑙𝑢𝑖𝑑 ?? 𝜇 𝑓𝑙𝑢𝑖𝑑 (2) When Re << 1, the drag force is given by Stokes law ? = 3?𝜇 𝑓𝑙𝑢𝑖𝑑 ? ? (3) ? 𝑓𝑙𝑢𝑖𝑑 is the density of the fluid and 𝜇 𝑓𝑙𝑢𝑖𝑑 is the viscosity of the fluid. If Re is greater than 1, equation (3) cannot be used to determine the drag force D. Combining equation (1) (3), we deduce the classic formula for creeping (low Re ) flow of a sphere moving with terminal velocity in a fluid ? 𝐷 = 24 ∕ 𝑅? ⇒ log(24) − log(𝑅?) (3𝑎) Which means that a graph of log( ? 𝐷 ) vs. log( Re ) will be a straight line. This theoretical trend could be different than the one found from the experiments, though. According to Archimedes principle, the buoyant force on the sphere is given by ? 𝑏 = ? 𝑓𝑙𝑢𝑖𝑑 ??? 3 6 (4) If a is the instantaneous acceleration of the sphere (generally a 0), then the force balance on the sphere can be written as ? − ? − ? 𝑏 = ? 𝑎 (5) Using equations (1) and (4), we can rewrite equation (5) as ?? − ? 𝐷 𝐴? 𝑓 ? 2 2 − ? 𝑓𝑙𝑢𝑖𝑑 ??? 3 /6 = ?𝑎 (6) When a sphere is moving with terminal velocity ( a = 0), equation (5) may be used to determine the drag force D = W - ? 𝑏 . When Re << 1 and terminal velocity is known, equation (3) can be used to evaluate the viscosity 𝜇 𝑓𝑙𝑢𝑖𝑑 . The value of 𝜇 𝑓𝑙𝑢𝑖𝑑 is needed in order to check whether Re << 1. Apparatus/Procedure In this experiment, we used a large vertical plastic clear cylinder filled with glycerin, stainless steel spheres of different diameters, ruler to measure fall distance, and a timer. There were two lines marked on the cylinder to distinguish the start and end of the
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4 timing period during the fall of each object. We dropped the balls one at a time at the top opening of the tube and used a phone timer to record how long it took for the different size balls to reach the bottom of the tube. Our results were recorded on a Table 1 and Table 2 shown below. Results In this experiment, we calculated the W (weight), U (velocity), 𝑭 𝒃 (buoyant force), D (drag force), 𝝁 𝒇𝒍𝒖𝒊𝒅 (viscosity), ? ? (drag coefficient), and Re (Reynolds number). First, we measured the time of the fall distance for each drop. Once we had our values for weight and buoyant force, we measured D. We calculated and recorded value D based on equation (3). For most of the drops of the spheres, the measured D and calculated D (eq. (3)) were very close in value. Table 1: Smooth Spheres Drop d (in) d (m) W (N) 𝝆 𝒇𝒍𝒖𝒊𝒅 (kg/ ? 3 ) t (s) L (m) U (m/s) 𝑭 𝒃 (N) D = W- 𝑭 𝒃 (N) 𝝁 𝒇𝒍𝒖𝒊𝒅 (Pa.s) ? ? Re D (N) Eq. (3) 1 0.0625 0.0015875 0.0002 1260 28.31 0.6128 0.02165 2.589E-5 1.741E-4 0.538 296.30 0.0805 1.743E-4 2 0.0625 0.0015875 0.0002 1260 28.33 0.6128 0.02163 2.589E-5 1.741E-4 0.538 298.51 0.0804 1.741E-4 3 0.0625 0.0015875 0.0002 1260 28.35 0.6128 0.02162 2.589E-5 1.741E-4 0.538 298.51 0.0804 1.740E-4 4 0.125 0.003175 0.0013 1260 7.58 0.6128 0.08084 2.071E-4 0.00109 0.451 33.468 0.7171 0.00109 5 0.125 0.003175 0.0013 1260 7.66 0.6128 0.08 2.071E-4 0.00109 0.455 34.120 0.7034 0.00109 6 0.125 0.003175 0.0013 1260 7.57 0.6128 0.08085 2.071E-4 0.00109 0.451 33.463 0.7172 0.00109 7 0.1875 0.0047625 0.0043 1260 3.89 0.6128 0.15753 6.991E-4 0.00360 0.509 12.923 1.8572 0.00360 8 0.1875 0.0047625 0.0043 1260 3.89 0.6128 0.15753 6.991E-4 0.00360 0.509 12.923 1.8572 0.00360 9 0.1875 0.0047625 0.0043 1260 3.70 0.6128 0.16562 6.991E-4 0.00360 0.484 11.688 2.0534 0.00360 10 0.25 0.00635 0.0103 1260 2.75 0.6128 0.22284 0.001657 0.008643 0.648 8.7225 2.7515 0.008642 11 0.25 0.00635 0.0103 1260 2.65 0.6128 0.23125 0.001657 0.008643 0.625 8.1070 2.9604 0.008650 12 0.25 0.00635 0.0103 1260 2.55 0.6128 0.24031 0.001657 0.008643 0.601 7.5019 3.1992 0.008644 13 0.3125 0.0079375 0.0200 1260 1.92 0.6128 0.31917 0.003237 0.016763 0.702 5.2781 4.5471 0.016762 14 0.3125 0.0079375 0.0200 1260 1.83 0.6128 0.33486 0.003237 0.016763 0.669 4.7942 5.0060 0.016759 15 0.3125 0.0079375 0.0200 1260 2.04 0.6128 0.30039 0.003237 0.016763 0.746 5.9595 4.0272 0.016764 16 0.375 0.009525 0.0346 1260 1.71 0.6128 0.35836 0.005593 0.029007 0.902 5.0335 4.7681 0.029018 17 0.375 0.009525 0.0346 1260 1.54 0.6128 0.39792 0.005593 0.029007 0.812 4.0807 5.8813 0.029006 18 0.375 0.009525 0.0346 1260 1.60 0.6128 0.383 0.005593 0.029007 0.844 4.4067 5.4462 0.029019
5 Figure 2: ? 𝐷 vs. Re log-log graph for smooth spheres Table 2: Rough Spheres Drop d (in) d (m) W (N) 𝝆 𝒇𝒍𝒖𝒊𝒅 (kg/ ? 3 ) t (s) L (m) U (m/s) 𝑭 𝒃 (N) D = W- 𝑭 𝒃 (N) 𝝁 𝒇𝒍𝒖𝒊𝒅 (Pa.s) ? ? Re D (N) Eq. (3) 1 0.1875 0.0047625 0.0043 1260 4.04 0.6128 0.15168 6.991E-4 0.00360 0.529 13.905 1.7260 0.00360 2 0.1875 0.0047625 0.0043 1260 3.91 0.6128 0.15673 6.991E-4 0.00360 0.512 13.066 1.8369 0.00360 3 0.1875 0.0047625 0.0043 1260 3.93 0.6128 0.15593 6.991E-4 0.00360 0.514 13.184 1.8204 0.00360 4 0.25 0.00635 0.0103 1260 2.5 0.6128 0.24512 0.001657 0.008643 0.589 7.2079 3.3297 0.008641 5 0.25 0.00635 0.0103 1260 2.57 0.6128 0.23844 0.001657 0.008643 0.606 7.6236 3.1481 0.008648 6 0.25 0.00635 0.0103 1260 2.67 0.6128 0.22951 0.001657 0.008643 0.629 8.2209 2.9194 0.008640 Figure 3: ? 𝐷 vs. Re log-log graph for rough spheres 1 10 100 1000 0.01 0.1 1 10 Drag coefficient, CD Reynolds number, Re 𝐶 D vs. Re log-log (smooth spheres) 1 10 100 1 10 Drag coefficient, CD Reynolds number, Re 𝐶 D vs. Re log-log (rough spheres)
6 Discussion/Questions 1. Why do you think we use data for the smallest sphere to obtain the viscosity through Stokes’ law? The smallest spheres are easiest to time how long it takes them to fall, making the velocity calculations the most accurate. The viscosity calculation is based on velocity, which would in turn also be more accurate with the smaller balls. 2. If you released the spheres close to the cylinder’s wall, would you expect the same results? No, I would expect slightly different results. An object moving through a tube causes the fluid to move in a conical shape, pushing the fluid around it away from it. The velocity, this means, would be smaller closer to the walls than the middle. This would throw off the drag coefficient calculation and change our values. It was best to try to release the balls as close to the middle as possible. 3. If some small air bubbles attach themselves to your sphere in a given test, will the resulting data lie above or below the expected curve? Give reasons for your answer. I would expect the data to lie above the curve because the air bubbles would slow the ball down slightly. 4. What do you conclude from the shape of the ? 𝐷 ( Re ) curve? Overall, it seems that as the Reynolds number increases, the drag coefficient decreases more and more. 5. Why is the ? 𝐷 versus Re plot drawn on log-log scale? If it wasn’t drawn on a log -log scale, the graph would look very meaningless and askew. Because the drag coefficient on the smaller spheres are so much higher than the bigger ones, we have to account for the difference. 6. Does roughness have an effect on drag force? Compare with values of smooth spheres and explain any differences. Yes, roughness clearly made a difference in the calculated drag coefficient, as can be seen in tables 1 and 2 of the same size balls. The drag force on the rough surface spheres was always higher than on the smooth ones, as there is most likely more surface ar ea for the fluid to “push back” on the sphere. Conclusion I think the results of the measured D and Calculated D from equation (3) were almost the same values for each drop including smooth and rough spheres. The experiment was overall successful. Figure 2 and figure 3 showed the relation between Reynolds number and drag coefficient to be linear in a log-log graph.
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7 Some errors that could have affected the results would be human error with the timer. I think that is why there was fractional differences in both values of D. The time could have also affected U, velocity but everything on Table 1 and Table 2 showed precise and accurate values. References Lab Manual, University of Illinois at Chicago, Illinois. September 2019. White, Frank M. Fluid Mechanics Eight Edition, New York. 2016. Sample Calculation Calculations are in regard to Trial 1: drop 1. Determining the weight ? = ?? = 7800 ?? ? 3 (2.0948? − 9 ? 3 ) = 1.6339? − 5 ?? ∗ 9.8𝑁 1?? = 0.00016 = 0.0002 𝑁 Buoyant Force ? 𝑏 ? 𝑏 = ? 𝑓𝑙𝑢𝑖𝑑 ??? 3 6 = (1260?? ? 3 ) (9.81 ? 𝑠 2 ) ?(0.0015875 ?) 3 6 = 2.589? − 5 𝑁 Viscosity 𝜇 𝑓𝑙𝑢𝑖𝑑 ? = 3?𝜇 𝑓𝑙𝑢𝑖𝑑 ? ? 1.741? − 4 𝑁 = 3?𝜇 𝑓𝑙𝑢𝑖𝑑 (0.02165 ? 𝑠 )(0.0015875 ?) 1.741? − 4 = 3.2392? − 4𝜇 𝑓𝑙𝑢𝑖𝑑 𝜇 𝑓𝑙𝑢𝑖𝑑 = 0.5375 Pa. s Reynolds number Re 𝑅? = ? ??𝑢𝑖? ?? 𝜇 ??𝑢𝑖? = (1260?? ? 3 ) (0.02165 ? 𝑠 )(0.0015875 ?) 0.538 𝑃𝑎.𝑠 = 0.0805 Drag coefficient ? 𝐷 ? 𝐷 = 24 𝑅? = 24 0.0805 = 296.30 Drag Force D ? = 3?𝜇 𝑓𝑙𝑢𝑖𝑑 ? ? = 3?(0.5375 𝑃𝑎. 𝑠) (0.02165 ? 𝑠 ) (0.0015875 ?) = 1.743? − 4