Introduction
advertisement
Introduction: The flow next to any surface forms a boundary layer, as the flow has zero velocity right at the surface and at some distance out from the surface; it flows at the same velocity as the local outside flow. If this boundary layer flows in parallel layers, with no energy transfer between layers then it is laminar. If there is energy transfer, it is turbulent. All boundary layers start off as laminar. Many influences can act to destabilize a laminar boundary layer, causing it to transition to turbulent. Once the boundary layer transitions, the skin friction goes up as a result of a turbulent boundary layer. Fluid flow is a very interesting phenomenon. The way fluid behaves over certain objects and how it behaves on the surface of an object need to be understood in order to have a detailed knowledge of this effect. Flow of fluids over a body is not made up of linear regions as one would expect, rather the profile of a flow depends on several factors such as adverse pressure gradients, surface roughness, heat and acoustic energy and most importantly the surface in contact. The study of fluid flow allows us to develop more efficient designs in aerodynamics. This experiment will give us a greater insight of fluid motion. http://www.aviation-history.com/theory/lam-flow.htm Calculations (Part A): Observed Boundary Layer thickness at points: A xa = 2 mm B xb = 3 mm C xc = 5 mm D No boundary so we are in transient region 1. Flow Velocity: V1A1 = V2A2 => V1 = V2A2/A1 Where: V2 = √ππ¦ = √(9.81) ∗ (0.004) V2 = 0.198090882 m/s V1 = 0.198090882∗1.776x10^(−3) 1.184x10^(−3) A1 = b1y1 = (0.296)*(0.004) A1 = 1.184x10-3 m2 = 0.29714 V1 = 0.297 m/s A2 = b2y2 = (0.296)*(0.006) A2 = 1.776x10-3 m2 2. Reynolds number at points A,B,C and D: Rex = ππ± πΏ Where: ο· ο· V = V1= 0.297 ms-1 π = 1x10-6 m2s-1 (Kinematic viscosity of water) Rex = ππ± ⁄ πΏ 17828.4 38628.2 63587.96 90330.56 Point A (x =60 mm) B (x =130 mm) C (x =214 mm) D (x =304 mm) 3. Expected boundary layer thickness at points A,B,C and D if the flow was: Using Rex from part 2 and putting in the laminar equation below since Reynolds number is below 500,000, also finding the error using the equation: Error % = ((experimental – theoretical) / (theoretical)) * 100 Point Equation => Laminar Experimental Error (%) A (x =60 mm) B (x =130 mm) C (x =214 mm) D (x =304 mm) 2.09 mm 3.08 mm 3.95 mm 2 mm 3 mm 5 mm No boundary so we are in transient region 4.31 2.6 21 N/A 4.70 mm Pressure Table Pressure Station Number Angle = 0 Angle = 30 Angle = 60 Angle = 90 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 24 198.941 -12.434 0.12954 -12.434 0.000 -74.603 0.000 -99.470 -124.338 -149.205 -161.639 -174.073 -186.507 -12.434 49.735 149.205 198.941 211.374 124.338 49.735 -124.338 0.000 -49.735 0.000 -99.470 -74.603 -74.603 273.543 124.338 0.12954 24.868 24.868 -124.338 24.868 -149.205 -149.205 -149.205 -149.205 -149.205 -149.205 0.000 -149.205 -198.941 -124.338 24.868 223.808 273.543 273.543 24.868 49.735 49.735 -174.073 -124.338 -124.338 348.146 248.676 0.12954 24.868 24.868 -149.205 24.868 -198.941 -223.808 -198.941 -211.374 -198.941 -149.205 24.868 -149.205 -149.205 -149.205 -174.073 -24.868 124.338 348.146 49.735 198.941 49.735 -24.868 -174.073 -149.205 373.013 273.543 0.12954 24.868 24.868 -149.205 24.868 -174.073 -198.941 -198.941 -174.073 -198.941 -174.073 24.868 -149.205 -161.639 -149.205 -149.205 -198.941 -87.036 223.808 49.735 248.676 49.735 223.808 -99.470 -174.073 Normal Force Table Normal Force, Fn Station Number Angle = 0 Angle = 30 Angle = 60 Angle = 90 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 -0.0112 0.0000 -0.1876 0.0000 -0.5154 -0.7109 -0.7731 -0.7435 -0.4376 -0.0975 -0.0112 0.0612 0.1345 0.1040 0.5314 0.5719 0.2577 -0.7109 0.0000 -0.2288 0.0000 -0.0520 -0.0672 -0.0919 0.0224 0.0130 -0.3126 0.1144 -0.7731 -0.8531 -0.7731 -0.6863 -0.3751 -0.0780 0.0000 -0.1837 -0.1793 -0.0650 0.0625 1.0294 1.4173 1.5640 0.1288 0.2288 0.1250 -0.0910 -0.1121 -0.1531 0.0224 0.0130 -0.3751 0.1144 -1.0308 -1.2797 -1.0308 -0.9722 -0.5002 -0.0780 0.0224 -0.1837 -0.1345 -0.0780 -0.4376 -0.1144 0.6442 1.9906 0.2577 0.9150 0.1250 -0.0130 -0.1569 -0.1837 0.0224 0.0130 -0.3751 0.1144 -0.9019 -1.1375 -1.0308 -0.8007 -0.5002 -0.0910 0.0224 -0.1837 -0.1457 -0.0780 -0.3751 -0.9150 -0.4510 1.2797 0.2577 1.1438 0.1250 0.1170 -0.0896 -0.2143 Tap Area (m^2) 1 0.0009012 2 0.0005226 3 0.00251415 4 0.0045996 5 0.0051813 6 0.0057177 7 0.0051813 8 0.0045996 9 0.00251415 10 0.0005226 11 0.0009012 12 0.0012312 13 0.0009012 14 0.0005226 15 0.00251415 16 0.0045996 17 0.0051813 18 0.0057177 19 0.0051813 20 0.0045996 21 0.00251415 22 0.0005226 23 0.0009012 24 0.0012312 Normal force at 0 deg 0.8000 0.6000 0.4000 0.2000 0.0000 -0.2000 -0.4000 -0.6000 -0.8000 -1.0000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Normal force Normal Force at 30 deg 1.8000 1.6000 1.4000 1.2000 1.0000 0.8000 0.6000 0.4000 Normal Force at 30 deg 0.2000 0.0000 -0.2000 1 2 3 4 5 6 7 8 9 101112131415161718192021222324 -0.4000 -0.6000 -0.8000 -1.0000 Normal Force at 60deg 2.200 2.000 1.800 1.600 1.400 1.200 1.000 0.800 0.600 0.400 0.200 0.000 -0.200 -0.400 -0.600 -0.800 -1.000 -1.200 -1.400 -1.600 Normal Force at 60deg 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Normal Force at 90 deg 1.600 1.400 1.200 1.000 0.800 0.600 0.400 0.200 0.000 -0.200 -0.400 -0.600 -0.800 -1.000 -1.200 -1.400 Normal Force at 90 deg 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Discussion (Air tunnel) From the pictures in the results, the stagnation pressure, the separation points and the wake region can be identified and they correlate with the theoretical data, because at the separation points the pressure is higher than that of the wake region. This is expected because of the turbulence present in the separation points which results in higher pressure. From theory, it is expected that the reynolds number of the flow will increase as the cylinder changes orientation from 0o to 90o, and this is reflected in the experimental data and calculations. πΆπ· = πΉπ· It is expected that the Cd will increase as the angle of attack is increased from 0o to 90o 0.5ππ£ 2 π΄ since a much higher area is exposed to the flow, due to the eqn: From this eqn, it can be seen that as the area increases, FD will increase, and since FD is related to CD the Drag coefficient will increase as well. However, this is not the case, as seen in the results and calculations since the CD decreases from 0.11 at 0o to 0.013 at 60o. This maybe due to any calculation errors that may have taken place. There are a few sources of error in this laboratory that may contribute to anomalies in the results. Even though these anomalies aren’t huge in this experiment, they are still a source of error. A major source of error that could produce inaccuracies in the calculated results is the absence of manometer readings for atmospheric pressure (i.e. with the air tunnel not turned on). Therefore for the calculations, a sample value for the manometer was used from another group. This is one of the major sources of error in this lab because this reference height of the manometer was not measured in this lab. Measurement errors are also present in the lab. One is the measurement of the ellipse’s major and minor axes as well as the distance between the pressure taps. This is because the only measurement tool available was a rigid 30 cm ruler, which is not a very precise measuring tool, as well as measuring the ellipse through the safety glass could result in parallax errors. Also, reading the manometer height values is another source of error because the human eye is not very accurate when reading off values from analogue devices (i.e. if the manometer display was digital, this error could be eliminated). However, these are minor errors and shouldn’t influence the results by a large margin. In this lab the head loss in the taps and the air tunnel (due to entrance/exit and friction) are not taken into account and this is one source of error that will be present in the labs, however this too is a minor error shouldn’t influence the results by a large amount. Discussion (Hydrogen bubble) From the results it can be observed that the boundary layer thickness increases as the x value increases, i.e. the boundary layer thickness increases further down the flat plate which indicates that the flow will change from laminar to turbulent. This correlates with theory, which states that the boundary layer will increase along the flat plate as the flow transfers from laminar to turbulent. In this lab, only the laminar boundary layer could be measured. From the results, it can be seen that the Reynolds number increases from 17828 to 90330 as x increases from 60 mm to 300 mm, which corellates well with theoretical data since the flow will start to become turbulent as the x value increases. It can also be observed that the %error was about 4.6% at x = 60 mm while it was 21% at x= 214mm, which is expected because the boundary layer is much harder to measure due to the flow becoming turbulent. A major source of error in this experiment is measurement. Once again, the only measurement tool available for use was a 30 cm rigid ruler which was used to measure the small boundary layer thickness. This is a very inaccurate method for measuring the boundary layer thickness and is the major source of error in this lab. Another source of error in this experiment could be that the plate may not have been perfectly parallel to the flow of the water and therefore would result in anomalous values for the boundary layer growth. Conclusion To conclude, this lab successfully observed the pressure variations around an ellipse for different angles of attack, ranging from 0o to 90o, as well as successfully determining the flow speed of the air entering the test section, which from the results, equals to 60.65 m/s. Then, using this velocity the Reynolds number for the flow around the cylinder was successfully determined, which ranges from 1.62 x 105 to 4.04 x 105 from 0o to 90o respectively. The second part of this lab allowed successful observation and measurement of the boundary layer around a plate parallel to the flow, as well as estimating the flow velocity (which is 0.297m/s ) and the boundary grows from 2.09 mm to 4.70 mm from 60 mm to 304 mm respectively. Even though this lab encountered a few difficulties and discrepancies, it was an overall success. MECH3361 Flow lab report Introduction: The boundary layer is the region next to a boundary of an object in which the fluid had had its velocity changed because of the wall shearing resistance. The boundary layer velocity is essentially the same as if a non-viscous fluid were flowing past the object. When fluid passes over a thin plate, a velocity gradient is generated between the fluid in the free stream and the fluid next to the plate. The fluid in the free stream has a uniform velocity, whereas the flow touching the plate has zero velocity due to the no-slip condition. Aim: To qualitatively analyse and measure the pressure profile around an ellipse at various angles to the flow (Wind Tunnel Experiment 1) To visualise the flow pattern associated with the boundary layer and analyse the interaction between a fluid and the surface of a thin, flat plate as the fluid passes by. (Hydrogen bubble apparatus Experiment 2) Methods: Wind Tunnel Experiment 1. 2. 3. 4. 5. 6. Record the size of the major and minor axes and the locations of the pressure tappings. Change the orientation of the ellipse for different angles of attack, firstly to 0o Switch on the wind tunnel and allow a steady flow to develop in the tunnel. Observe and record the manometer readings for different pressure tappings. Repeat Step 2-4, but change the angle of attack of the ellipse for 300, 600, 900. Determine the flow speed of the air in the test section using both Bernoulli’s equation method and the stagnation pressure methods. 7. Calculate the pressure profile around an ellipse at different angles of attack. Hydrogen bubble boundary layer thickness Experiment 1. Measure the length of 3 different downstream locations from the leading edge. 2. Measure the height of the downstream weir. 3. Switch on the Hydrogen bubble apparatus; place the hydrogen generating wire in the first location just behind the leading edge. 4. Measure and record the height of the boundary layer thickness at this first location. 5. Repeat the Step 3-4, but change the position of the wire in the other 2 places. Results Wind Tunnel The manometer readings (in inch) are tabulated in a table below: The manometer reading Station The angle of The angle of The angle of The angle of Number Attack = 0 Attack = 30 Attack = 60 Attack = 90 Reference reading (inch): 13.15 TAP1 12.20 12.00 11.80 11.70 TAP2 13.00 12.70 12.30 12.10 1 13.20 13.20 13.20 13.20 2 13.20 13.20 13.20 13.20 3 13.50 13.90 14.00 14.00 4 13.20 13.30 13.30 13.30 5 13.50 13.90 14.10 13.70 6 13.60 14.10 13.30 12.40 7 13.70 13.60 12.50 11.70 8 13.80 13.10 12.00 11.60 9 13.80 12.70 11.80 11.70 10 13.80 12.20 11.80 12.30 11 13.20 13.05 13.10 13.20 12 12.70 12.20 13.30 14.20 13 12.40 12.90 14.00 14.10 14 12.10 13.80 14.10 14.00 15 12.30 14.30 14.00 14.00 16 12.80 14.70 14.00 14.00 17 13.15 15.00 14.00 14.10 18 13.90 14.00 14.00 14.10 19 14.00 13.80 14.10 14.20 20 13.60 13.70 14.05 14.20 21 13.80 13.90 14.20 14.20 22 13.40 13.90 14.20 14.10 23 13.50 13.90 14.10 14.00 24 13.50 13.90 14.10 14.00 25 13.40 13.80 13.90 13.90 *According to lab instructor, readings for station 17 are to be disregarded. There is a major damage at the tube labeled 17 and hence may affect the entire results if it is put into account. We will need to know information about velocity at tap 2 (smaller rectangular area). This can be found by apply Bernoulli equation[1] and continuity equation[2] (for both equations few terms cancel out): π1 π + π1 2 2 = π2 π + π2 2 2 …….(1) A1V1 = A2V2 ……..(2) We have A1 = 0.36m2 and A2 = 0.09m2. By rearranging and substituting V1, A1 and A2 from continuity into Bernoulli Equation, the equation reduces to: V2 = 1.4605 ( π1−π2 ππππ Angle of Attack (degrees) Pressure different (P1-P2) (Pa) Velocity 2 (m/s) Characteristic length (m) Reynolds Number ) 1/2 where P1-P2 = ρH2O * g * (H1-H2) 0 198.9405 18.805 0.04 50146.58 30 174.073 17.590 0.05 58634.78 60 124.3378 14.867 0.0866 85830.02 90 99.4703 13.397 0.1 88647.46 *characteristic length is calculated based on simple geometry when the elliptical block’s orientation is changed. *Reynolds Number is calculated based on the equation of Re = ππππ∗ π2 πΏπβ π (refer to Lab handout for the value of ππππ and ). We are needed to calculated coefficient of drag and lift, thus cord area for each segment on the ellipse has to be calculated (done in solidworks apps). Areas obtained are then multiplied by the pressures (see Appendix 1 for Table of Pressure) to get the Force in X and Y direction (see Appendix 2 for Table of Force). Note that X-component is parallel to the flow. Resolving forces to X and Y components could be done by knowing the angles of the tangential streamline. These could be done by using Solidwoks apps. After we get the list of Fx and Fy for one particular Angle of Attack, the summation of them (∑Fx and ∑Fy) are to be used to find CD and CL by applying the following equations: ∑Fx = FD and πΆπ· = πΉπ· 1 πv2 π΄ 2 , ∑Fy = FL and πΆπΏ = πΉπΏ 1 πv2 π΄ 2 *Note that A is frontal area for each orientation of the ellipse. Drag and Lift Coefficients and Forces are evident in a table below: AOA (degree) characteristic length (m) height of ellipse (m) frontal area (m^2) FD (N) FL (N) CD CL 0 0.04 0.3 0.012 -1.7840 -5.0872 -0.70067 -1.99802 30 0.05 0.3 0.015 1.9766 -10.6311 0.709814 -3.81772 60 0.0866 0.3 0.02598 2.0542 -5.6596 0.596219 -1.64266 90 0.1 0.3 0.03 1.9554 -2.9839 0.605268 -0.92363 Boundary layer experiment The result from the experiment is shown below, plate length, Xi (m) Point 1 Point 2 Point 3 0.058 0.128 0.213 experimental b.layer, δ (m) 0.001 0.003 0.005 The calculation methods to find the boundary layer thickness are as follow, For laminar, πΏπ = 4.65π₯π ππ₯ −1/2 (1) For turbulence, πΏπ = 0.38π₯π ππ₯ −1/5 (2) To apply these equations, we need to find Reynolds number by the equation π ππ₯ = ππ₯ π , where V is the free stream velocity. This can be calculated by the fact that when the flow is over the weir, the flow is in critical condition and the Froude number for this flow is 1. With that fact we can find the velocity of the flow over the weir since the equation of Froude number is given by, πΉπ = π √ππ¦ With Fr = 1, g = 9.81 and y = 0.001m (flow depth over the weir), we can solve the equation to find the velocity which is V = 0.099 m/s. By using continuity equation, we can find the free stream velocity. π1 = π2 = π1 π΄1 = π2 π΄2 And since the width is constant along the flow, we can reduce the eq to π1 π1 = π2 π2 where Y1 = 0.001m and Y2 = 0.007m and V1 = 0.099m/s By solving the equation, V2 = 0.01415m/s which V2 is free stream velocity. By using the equation (1), (2) and Reynolds number equation, we can solve the boundary layer thickness for three different positions at the plate, Xi (m) Point 1 Point 2 Point 3 0.058 0.128 0.213 Reynolds no 820.70 1811.20 3013.95 δexperiment (m) 0.001 0.003 0.005 δlaminar(m) 0.009414315 0.013985555 0.018041163 δ turbulence(m) 0.005759366 0.010849264 0.01630557 Graph of boudary layer thickness vs distance from leading edge 0.02 0.018 boundary layer thickness (m) 0.016 0.014 experiment boundary layer thickness 0.012 laminar boundary layer thickness 0.01 0.008 turbulance boundary layer thickness 0.006 0.004 0.002 0 0 0.05 0.1 0.15 0.2 distance from leading edge (m) 0.25 Discussion For the Wind Tunnel experiment, the relationship between Angle of Attack and Drag Coefficient could be seen clearly. Generally speaking, by increasing the angle of attack, we are actually increasing the exposed area of the ellipse. If we go back to the equation of CD= 1 2 πΉπ· πv2 π΄ , it is apparent that Area (denoted as A) is inversely proportional to CD. Thus CD increases as angle of attack changes from 0-90 degree. To further visualize this, one can also imagine when we have a larger exposed area, there will be more resistances that give rise to a higher value of Drag Force. As FD is directly proportional to CD, therefore CD increases too for a large exposed area. The experimental data shows a graph as follows: Cd vs angle of attack 1 Cd 0.5 0 0 20 40 60 -0.5 -1 Angle of Attack (degree) 80 Cd vs angle of 100attack Based on the graph, there is an overshoot particularly at AOA=30, which violates the relationship that we have discussed above. This is due to the errors that had occurred during the experiment. One of the errors that had occurred in this experiment is that students just only estimate the readings of manometers. These approximations may be less or more than the exact one and thus may result the pressure and drag force to be different than the actual one. Moreover, the manometer used is old and the scale is hard to read. These would also stand as one of the errors. Another source of error is that we had used a density of air of 1.2 kg/m3. This might not be the case in our experiment. A slight different of ρair will affect the value of Reynolds Number, FD, FL and hence the values of CD and CL. For the Hydrogen Bubble experiment, there is an apparent finding that is the boundary layers are developed when the flow strike the flat plate (note that the flat plate has no elevation, there is no pressure change-zero pressure gradient). The boundary layers are developed due to frictions as the flow moves across the plate. It can be seen clearly during the experiment, where the flow velocity profile is unchanged before it strikes the plate surface at the leading edge. After that, the flow velocity profile closest to the plate begins to slow. At a certain distance away from the plate, the velocity profile is back to normal. This is how the boundary layers form. Based on the graph of measured thickness vs those turbulent and laminar assumptions, our findings are likely to behave more to boundary layers of a turbulent flow. As there are differences in values between measured and turbulent flow assumptions, we could explain this by observing the errors that might affect our results. One of the errors that had occurred in this experiment is that students are not able to measure the boundary layer thickness correctly. All those data were taken by an approximation. Our group was using a millimeter ruler and we were not able to get the boundary layer thickness in exact except as a whole number (ie: 1mm). Besides, there is also error in measuring the depth of the upstream flow and just after the weir. Ruler might not be put perfectly perpendicular to the base of the basin. This might give lesser value of the depth of both flows (upstream and downstream) and hence producing less accurate upstream velocity. Conclusion . Appendix 1 (TABLE OF PRESSURE) Table of Pressure Tap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 AOA = 0 -12.4338 -12.4338 -87.0365 -12.4338 -87.0365 -111.9040 -136.7716 -161.6392 -161.6392 -161.6392 -12.4338 111.9040 186.5067 261.1094 211.3743 87.0365 0.0000 -186.5067 -211.3743 -111.9040 -161.6392 -62.1689 -87.0365 -87.0365 AOA = 30 -12.4338 -12.4338 -186.5067 -37.3013 -186.5067 -236.2419 -111.9040 12.4338 111.9040 236.2419 24.8676 236.2419 62.1689 -161.6392 -285.9770 -385.4473 0 -211.3743 -161.6392 -136.7716 -186.5067 -186.5067 -186.5067 -186.5067 AOA = 60 -12.4338 -12.4338 -211.3743 -37.3013 -236.2419 -37.3013 161.6392 285.9770 335.7121 335.7121 12.4338 -37.3013 -211.3743 -236.2419 -211.3743 -211.3743 0 -211.3743 -236.2419 -223.8081 -261.1094 -261.1094 -236.2419 -236.2419 AOA = 90 -12.4338 -12.4338 -211.3743 -37.3013 -136.7716 186.5067 360.5797 385.4473 360.5797 211.3743 -12.4338 -261.1094 -236.2419 -211.3743 -211.3743 -211.3743 0 -236.2419 -261.1094 -261.1094 -261.1094 -236.2419 -211.3743 -211.3743 Appendix 2 TABLE OF FORCES Area (m^2) 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 AOA = 0 Fx (N) 0.0100 0.0053 0.1621 0.0199 0.0731 0.0000 -0.1149 -0.2585 -0.3010 -0.0683 -0.0100 0.1378 -0.1499 -0.1103 -0.3936 -0.1392 0.0000 0.0000 -0.1776 -0.1790 -0.3010 -0.0263 -0.0699 0.1072 Total: -1.7840 N Tap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Fy (N) -0.0050714 -0.0038285 -0.1470443 -0.0536209 -0.4449963 -0.6398337 -0.6992799 -0.6970721 -0.2730823 -0.049771 -0.0050714 8.4398E-18 0.07607133 0.08039925 0.35710766 0.37534653 0 -1.0663896 -1.0807053 -0.4825884 -0.2730823 -0.0191427 -0.0355 -6.564E-18 -5.0872 N *note that we do not consider anything at tap 17. All information regarding tap 17 are disregarded due to some damages on the tube. Appendix 2 TABLE OF FORCES Tap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Area (m^2) 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 Total: AOA = 30 Fx (N) 0.0100 0.0053 0.3473 0.0597 0.1567 0.0000 -0.0940 0.0199 0.2084 0.0998 0.0200 0.2909 -0.0500 0.0683 0.5325 0.6165 0 0.0000 -0.1358 -0.2188 -0.3473 -0.0788 -0.1499 0.2296 Fy (N) -0.0050714 -0.0038285 -0.315095 -0.1608628 -0.9535635 -1.3507601 -0.5721381 0.05362093 0.189057 0.07274218 0.01014284 1.7817E-17 0.02535711 -0.049771 -0.4831457 -1.6622489 0 -1.2085749 -0.8264217 -0.5898303 -0.315095 -0.057428 -0.0760713 -1.407E-17 1.9766 N -10.6311 N *note that we do not consider anything at tap 17. All information regarding tap 17 are disregarded due to some damages on the tube. Appendix 2 TABLE OF FORCES Tap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Area (m^2) 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 Total: AOA = 60 Fx (N) 0.0100 0.0053 0.3936 0.0597 0.1984 0.0000 0.1358 0.4574 0.6251 0.1418 0.0100 -0.0459 0.1699 0.0998 0.3936 0.3381 0 0.0000 -0.1984 -0.3580 -0.4862 -0.1103 -0.0699 0.1072 2.0542 N Fy (N) -0.0050714 -0.0038285 -0.3571077 -0.1608628 -1.2078471 -0.2132779 0.82642172 1.23328145 0.56717099 0.10337047 0.00507142 -2.813E-18 -0.0862142 -0.0727422 -0.3571077 -0.9115559 0 -1.2085749 -1.2078471 -0.9651768 -0.441133 -0.0803993 -0.0355 -6.564E-18 -5.6596 N *note that we do not consider anything at tap 17. All information regarding tap 17 are disregarded due to some damages on the tube. Appendix 2 TABLE OF FORCES Tap 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Area (m^2) 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 0.0009012 0.0005226 0.00251415 0.0045996 0.0051813 0.0057177 0.0051813 0.0045996 0.00251415 0.0005226 0.0009012 0.0012312 Total: AOA = 90 Fx (N) 0.0100 0.0053 0.3936 0.0597 0.1149 0.0000 0.3029 0.6165 0.6714 0.0893 -0.0100 -0.3215 0.1898 0.0893 0.3936 0.3381 0 0.0000 -0.2193 -0.4177 -0.4862 -0.0998 -0.0699 0.1072 Fy (N) -0.0050714 -0.0038285 -0.3571077 -0.1608628 -0.6992799 1.06638958 1.84355615 1.66224892 0.60918366 0.06508511 -0.0050714 -1.969E-17 -0.096357 -0.0650851 -0.3571077 -0.9115559 0 -1.3507601 -1.3349889 -1.1260396 -0.441133 -0.0727422 -0.0355 -6.564E-18 1.9554 N -2.9839 N *note that we do not consider anything at tap 17. All information regarding tap 17 are disregarded due to some damages on the tube.
