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Fluid Mechanics Tuesday Week 6 MECH3261- Flow Laboratory Introduction Fluid mechanics involves the study of fluids and the forces that act upon them. It is a field that relies heavily on mathematical models and experimental data to describe fluid flow phenomena. It is for this reason that the following experiment is conducted; fluid behaviour is modeled in the provided apparatus in order to provide a greater insight into concepts that define fluid mechanics, including but not limited to the growth of boundary layers, pressure changes with varying orientations of an object as well as changes in velocity at different points in the flow. Indeed, the study of fluid mechanics has a great significance in the physical world, allowing for the development of more effective and efficient real-life applications. Experiment objectives The experiment is divided into two parts. The first part involves the use of a wind tunnel, while the second part employs hydrogen bubble apparatus. Each experiment respectively aims to: 1. Observe air flow and behaviour of boundary layers around an object a wind tunnel, and to analyse the effects of force and pressure due to the changing angles of attack. 2. Calculate flow velocity and boundary layer thickness for an open channel flow, thus observing the development of the boundary layer. Apparatus - Wind Tunnel (as supplied by the lab) - The test subject - Elliptical Prism - Compass - Manometers (attached to the test subject) - Hydrogen bubble apparatus consisting of an electrode - Flow channel with weir - Ruler Method Part A- Wind tunnel experiment 1. Allow the wind tunnel to run at a desirable pressure 2. Secure the test subject at the open end of the wind tunnel at an attack angle of 0°, i.e. with the plane of the minor axis normal to the air flow. 3. Record the manometer readings (in our case, take a photo for each angle) 4. Rotate the test subject at increments of 30° (Up to 90°), whilst taking a photo of the manometer and making necessary observations. 5. Calculate the relevant data. Fluid Mechanics Tuesday Week 6 Figure 1a. The ellipse and tap set up Flow y y x Characteristic length Ο΄ Note that: - Characteristic length is calculated based on simple geometry when the elliptical block’s orientation is changed. - We take the height of the ellipse to be 30cm or 0.3m Figure 1b. Wind tunnel apparatus A1 A2 Fluid Mechanics Tuesday Week 6 Method Part B- Hydrogen bubble apparatus 1. Set up the hydrogen bubble flow bench and measured all distances as shown in Figure 2a. 2. Inserted electrode into the first slot (marked A). 3. Switched the machine on to allow for the formation of hydrogen bubbles by hydrolysis. The hydrogen bubbles mapped the location and size of the boundary layer. 4. Measured and recorded the estimated boundary layer thickness δm (see Table 2.1). 5. Moved the electrode to the next slot and repeated steps 3 to 5 for the remaining slots. 6. Measured the depth of water upstream and above the crest of the weir as shown in Figure 2b. A B D C 57mm 127mm 212mm 302mm Figure 2a. Apparatus set-up (top view) 6mm Figure 2b. Depth of water in apparatus set up (side view) 3mm 320mm Direction of flow Fluid Mechanics Tuesday Week 6 Results Part A- Wind tunnel experiment Table 1.1 Manometer readings Imperial (Inches) Tap Number TAP 1 TAP 2 Pref 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 θ=0 1.6 2.5 2.5 2.6 2.6 2.9 2.6 3 3.1 3.2 3.4 3.4 3.45 2.6 2.4 1.9 1.6 1.6 2 2.4 3.1 3.3 3.25 3.2 2.9 2.9 2.85 (Let θ be the attack angle) θ = 30 1.25 1.95 2.5 2.4 2.4 2.4 3.2 2.4 3.2 3.3 2.9 2.4 1.9 1.45 2.4 1.55 2.2 3.2 3.7 4 4.3 3.2 2.85 2.5 3.2 3.1 3.15 θ = 60 1.1 1.9 2.5 2.5 2.5 3.3 2.5 3.4 2.6 1.7 1.3 1.1 1.15 2.5 2.7 3.4 3.45 3.4 3.3 3.35 3.3 3.5 3.5 3.5 3.5 3.4 3.3 θ = 90 1.05 1.5 2.5 2.5 2.5 3.4 2.5 3.2 1.7 1.15 1.1 1.65 2.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 Metric (Meters) Tap Number TAP 1 TAP 2 Pref 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 θ=0 0.04064 0.0635 0.0635 0.06604 0.06604 0.07366 0.06604 0.0762 0.07874 0.08128 0.08636 0.08636 0.08763 0.06604 0.06096 0.04826 0.04064 0.04064 0.0508 0.06096 0.07874 0.08382 0.08255 0.08128 0.07366 0.07366 0.07239 θ = 30 0.03175 0.04953 0.0635 0.06096 0.06096 0.06096 0.08128 0.06096 0.08128 0.08382 0.07366 0.06096 0.04826 0.03683 0.06096 0.03937 0.05588 0.08128 0.09398 0.1016 0.10922 0.08128 0.07239 0.0635 0.08128 0.07874 0.08001 θ = 60 0.02794 0.04826 0.0635 0.0635 0.0635 0.08382 0.0635 0.08636 0.06604 0.04318 0.03302 0.02794 0.02921 0.0635 0.06858 0.08636 0.08763 0.08636 0.08382 0.08509 0.08382 0.0889 0.0889 0.0889 0.0889 0.08636 0.08382 θ = 90 0.02667 0.0381 0.0635 0.0635 0.0635 0.08636 0.0635 0.08128 0.04318 0.02921 0.02794 0.04191 0.0635 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 0.0889 Fluid Mechanics Tuesday Week 6 Table 1.2 values used in calculation process Wind Tunnel Dimensions: Section Length (m) Area (m^2) Given values Air density (ρ air) Air dynamic viscosity (μ air) Water density (ρ water) Water kinematic Viscosity (ν water) Tap1 0.6 0.36 Tap2 0.3 0.09 1.2kg / m^3 1.8 x 10^-5 Pa.s 998kg/m^3 1x10^-6 m^2/s Calculating the velocity of the flow in the wind tunnel: Assuming a steady, incompressible and frictionless flow, the velocity can be found by applying Bernoulli’s equations between section 1 and 2 of the wind tunnel: π1 π π1 π + + π1² 2 π1² 2 + ππ§1 = = π2 π + π2 π π2² + π2² => 2 2 + ππ§2 and z1 = z2 (no elevation difference) π2² − π1² = 2(π1−π2) πρair V1A1= V2A2 Since flow is continuous, (A1 corresponds to Tap1, A2 corresponds to Tap2) => V1 = (V2A2)/A1 => π22 (1 − Now π΄22 2(π1−π2) π΄1 πρair 2) = P1-P2 = ρwaterg(h2-h1) therefore, π22 = 2(π1−π2) (π΄12 ) πρair (π΄12 −π΄22 ) Since flow is continuous, (A1 corresponds to Tap1) Now letting h1= tap 1 value and h2 = tap 2 value, we can calculate the pressure difference and hence the velocity at each angle of attack. Angle of Attack (degrees) Pressure different (P1-P2) (Pa) 0 223.8080868 30 174.0729564 60 90 198.9405216 111.9040434 Velocity 2 (m/s) 19.94695574 17.59156143 Characteristic length (m) 0.04 0.05 Reynolds Number 53191.88197 58638.53811 Table 2.2 Pressure difference and velocity of each angle of attack 18.80617022 14.10462767 0.0866 0.1 108574.2894 94030.85111 Fluid Mechanics Tuesday Week 6 Reynolds number for each angle of attack was calculated using π π = ρVD μ Where V is the velocity, D is characteristic length, ρ is the air density and μ is the air viscosity. Calculating the pressures for each angle of attack: In order to find the forces we must find the pressures for each tap in each angle case. This is done by using the pressure difference equation on the previous equation and subbing in the water level for the reading for the height difference. The coordinates of the taps were also calculated using Solidworks. We also need to calculate the area normal to each tap to find the force. The area was assumed to be the arc length from the two points on either side of the tap multiplied by the height of the elliptical body. For example, the area assumed for tap 6 would be the arc length from point 7 to point 5 times by 30cm. The following results are calculated using excel Table 2.3 TAP 1 2 3 4 Pressure for different angles (Pa) 0° 30° 60° 646.5567 596.821565 621.68913 646.5567 596.821565 621.68913 721.1594 596.821565 820.6296516 646.5567 795.762086 621.68913 90° 621.68913 621.68913 845.497217 621.68913 Coordinates x y 49.5 2.82 49 3.98 44 9.5 34 14.66 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 746.027 770.8945 795.7621 845.4972 845.4972 857.931 646.5567 596.8216 472.4837 397.881 397.881 497.3513 596.8216 770.8945 820.6297 808.1959 795.7621 721.1594 721.1594 708.7256 795.762086 422.748608 285.977 273.543217 410.314826 621.68913 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 870.364782 18 0 -18 -34 -44 -49 -49.5 -50 -49.5 -49 -44 -34 -18 0 18 34 44 49 49.5 50 596.821565 795.762086 820.629652 721.159391 596.821565 472.483739 360.579695 596.821565 385.447261 547.086434 795.762086 920.099912 994.702608 1069.3053 795.762086 708.725608 621.68913 795.762086 770.894521 783.328304 845.4972168 646.5566952 422.7486084 323.2783476 273.5432172 285.9769998 621.68913 671.4242604 845.4972168 857.9309994 845.4972168 820.6296516 833.0634342 820.6296516 870.364782 870.364782 870.364782 870.364782 845.4972168 820.6296516 18.66 20 18.66 14.66 9.5 3.98 2.82 0 -2.82 -3.98 -9.5 -14.66 -18.66 -20 -18.66 -14.66 -9.5 -3.98 -2.82 0 Area (m^2) 0.0001231 0.0002596 0.0005524 0.000827 0.0010325 0.0001692 0.0010325 0.000827 0.0005524 0.0002596 0.0001231 0.0001692 0.0001231 0.0002596 0.0005524 0.000827 0.0010325 0.00108 0.0010325 0.000827 0.0005524 0.0002596 0.0001231 0.0001692 Fluid Mechanics Tuesday Week 6 Calculating the drag forces due to pressure for each angle of attack: In knowing the pressures and normal areas, we can find the normal forces using F=P*A Table 2.4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 TOTAL CD Normal Forces for different angles (N) 0° 30° 60° 0.079598 0.0734754 0.076536876 0.167837 0.15492683 0.161382116 0.398375 0.32968985 0.453323544 0.534697 0.65808853 0.514131663 0.770276 0.61622052 0.872979066 0.130435 0.13464295 0.109397393 0.821627 0.84730321 0.436489533 0.699219 0.59639273 0.267348465 0.467061 0.32968985 0.151107848 0.222707 0.12265041 0.074235773 0.079598 0.04439139 0.076536876 0.100982 0.10098221 0.113604985 0.058168 0.04745286 0.104090151 0.103285 0.14201626 0.22270732 0.219793 0.43958647 0.467060622 0.411305 0.76091486 0.678653796 0.616221 1.0270342 0.860141138 0.832566 1.15484973 0.886280024 0.847303 0.82162736 0.898654921 0.668371 0.5861101 0.719784329 0.439586 0.34342693 0.480797699 0.187203 0.20656911 0.225934963 0.088783 0.09490573 0.104090151 0.119916 0.13253915 0.138850537 9.064915 9.76548661 9.094119788 3.164302 3.50624776 1.649501944 90° 0.07653688 0.16138212 0.46706062 0.51413166 0.82162736 0.07152906 0.29527233 0.22621793 0.22666177 0.16138212 0.10715163 0.14726572 0.10715163 0.22593496 0.4807977 0.71978433 0.89865492 0.93999396 0.89865492 0.71978433 0.4807977 0.22593496 0.10715163 0.14726572 9.22812595 2.5770198 The total forces are also calculated and by using the following equation as well as the values from previous calculations, the drag coefficient CD was calculated for each angle case. ∑Fx = FD and πΆπ· = *Note that A is frontal area for each orientation of the ellipse. πΉπ· 1 πv2 π΄ 2 Fluid Mechanics Tuesday Week 6 1200 1000 Pressure (Pa) 800 Angle of attack = 0° Angle of attack = 30° 600 Angle of attack = 60° 400 Angle of attack = 90° 200 0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324 Tap Figure 1c. Pressure at each tap for varying angles of attack 1.4 1.2 Force (N) 1 0.8 Angle of attack = 0° Angle of attack = 30° 0.6 Angle of attack = 60° Angle of attack = 90° 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Tap Figure 1d. Force distribution at each tap for different angles of attack Fluid Mechanics Tuesday Week 6 Results Part B- Hydrogen bubble apparatus At the crest of the weir, the flow is defined as critical with Froude number = 1, and thus using the equation: π πΉπ = √ππ¦ We can calculate the free stream velocity V2 = √(9.81 x 0.003) = 0.1716m/s Using the continuity equation V1A1 = V2A2, we can find V1 (the upstream velocity). However, since the width of channel remains constant at 0.32m, the equation can be reduced to: V1 = V2 ( Y2 0.003 ) = 0.1716 ( ) = 0.0858m/s Y1 0.006 The Reynolds number Rex can then be calculated for each section of the flow, using: π ππ₯ = ππ₯ π Where x is the distance from the beginning of the boundary layer, V is the upstream velocity = 0.0858m/s and νs is the kinematic viscosity of water = 1 x 10-06 m2/s. Based on the calculated Rex, we can go on to calculate the boundary layer thickness for both the laminar and turbulent flow regimes using the relations: 1 πΏπππππππ = 4.65π₯π ππ₯ −2 πΏπ‘π’πππ’ππππ‘ = 0.38π₯π ππ₯ −1/5 x (m) 0.057 0.127 0.212 0.302 Rex 4890.6 10896.6 18189.6 25911.6 δm (m) 0.004 0.008 0.011 N/A δl (m) 0.00379 0.005657 0.007309 0.008724 δt (m) 0.003961 0.007518 0.011328 0.015035 Table 2.1 A comparison of measured and theoretical boundary layers Note that no boundary layer was observed at point D. This is later addressed in the discussion. Fluid Mechanics Tuesday Week 6 We can also calculate the rate of boundary layer growth at each section of the flow using the following relation: πΏπ ππ = tan π₯π xi (m) 57 70 85 90 mi 0.070291 0.114786 0.130139 N/A δi (m) 4 8 11 N/A Table 2.2 Boundary layer growth rate 0.016 Boundary layer thickness (m) 0.014 0.012 0.01 0.008 δ laminar δ turbulent 0.006 δ measured 0.004 0.002 0 0.057 0.127 0.212 0.302 Distance from leading edge (m) Figure 2c. Theoretical and measured boundary layer thickness at different points in the flow Note that the graph discontinues for the measured boundary layer thickness, as no boundary was observed at point D. Theoretical values for laminar and turbulent thicknesses at that point however were calculated, and so the graph continues for that distance. Fluid Mechanics Tuesday Week 6 Discussion Part A Most force vector plots roughly showed a consistent or obvious pattern however there were some adjacent taps which showed large variations in pressure. When observing the force distribution over the ellipse, the profile at θ = 60° was most erratic out of the four angles of attack tested, showing small irregularities in magnitude of normal forces and direction of forces toward the front of the ellipse (relative to the wind tunnel). There were also some discrepancies with theoretical expectations. For example, the plots should have been symmetrical but they weren’t due to experimental errors which could have included: the body was not a perfect ellipse, tapping locations were not exact, nor symmetrical (in our calculations they were assumed symmetrical), and the leading edge tap may not have been exactly centered. Theoretically, increasing the angle of attack should increase the total drag force on the elliptical cylinder. In the experimental results, the total drag force increased from α = 0° to α = 60°, however it then decrease when α = 90° where it should have been at a maximum. Thus there were experimental errors which could have been due to turbulence, or the manometer not functioning correctly. Further, the increasing drag force is due to the increased projected area, which was observed in our experiment. Several improvements could have been made to the experiment. Taking photos of the manometer levels and simply reading off images creates large room for error. Perhaps an electronic manometer counter would have significantly increased the accuracy of readings. Also, the arc lengths that were calculated were straight lines from point to point and not a curve. This then affects the calculated area and hence the forces etc. Overall the experiment gives a good understanding on the behaviour of fluid flow, and shows the effects on the orientation of a certain shape within in a flow. This understanding leads to better knowledge on design and hence better engineering practice. Part B Based on the calculation of the Reynolds number and assuming that for an Rex of greater than 5x10^5 the flow is turbulent, we find that the flow throughout the open-channel in the experiment remains laminar. One would therefore be inclined to use the relation for δlaminar to calculate the theoretical boundary layer thickness. In the results however, both relations (laminar and turbulent) have been used to calculate boundary layer thickness as tabulated in Table 2.1. It is interesting to note that although the flow is determined to be laminar, the measured boundary layer thickness actually holds almost the exact values as the calculated δturbulent. This leads to the idea that there has been some form of calculation error, which is almost expected with every experiment. The calculations relied primarily on the measured depths of the flow channel, which in itself is open to all sorts of human error, especially considering the fact that the depths dealt with were in millimeters and therefore difficult to read off a regular ruler. In calculating the upstream velocity, the cross-sectional area of the channel was assumed constant when in reality, it was slightly less due to the placement of the plate in the middle of the channel. As no measurements were taken of the width of the plate, its area was eliminated from the calculations thus affecting calculated velocity, which in turn was used to calculate the Reynolds number at each point in the flow. Fluid Mechanics Tuesday Week 6 The measurements of the boundary layer were also inaccurate, as the thickness was estimated based on the point where we determined where the hydrogen bubbles paralleled out to the plate. Not only was this difficult to observe because of the poor visibility of the hydrogen bubbles, but the constant movement of the bubbles and their eventual diffusion into the stream made that parallel point difficult to determine. At the fourth point (point D), no boundary layer was observed. This was accounted for by the abrupt ending of the plate, which did not allow for enough length over which a boundary layer could be determined. The rate of boundary layer growth was found to increase as the distance across the plate increased, which is expected at the boundary layer behaves in such a manner that it eventually broadens as the flow becomes turbulent and therefore mixes with the free stream. Conclusion The experiments conducted allowed for an observation of flow and the behaviour of boundary layers around an object in a wind tunnel, and the effects of angle of attack on the force and pressure on the object at different points, as well as helping to model boundary layer growth over a plate in an open channel flow. Although both experiments were open to many sources of error, they provide a valuable insight into the behaviour of fluids and demonstrate the importance of mathematical modeling in fluid mechanics.