Published on: **Mar 4, 2016**

Source: www.slideshare.net

- 1. 1 UNIVERSITY OF GAZIANTEP ENGINEERING FACULTY MECHANICAL ENGINEERING DEPARTMENT ME 416 ENGINEERING LABORATORY PRESSURE LOSS EXPERIMENTATION SUBMITTED TO: Res.Assist ALPEREN TOZLU SUBMITTED BY: (GROUP 7) HALİL İBRAHİM GÜLMEZ MESUT GÜNGÖR FIRAT GÜNYEL YAŞAR GÜRBÜZ EMRE KARACA ERDİ KARAÇAL DECEMBER 2013
- 2. 2 INTRODUCTION In this experiment, static pressure loss and flow rate in a circular pipe in which air flows, are measured by using pressure tappings and pitot tube. In order to determine the flow rate, local air velocities are calculated by using dynamic pressure of air flow equations with the theory of pitot tube which allow to transverse across the cross-section of the pipe. With the rotation of fan, an air flow in a variable speed exists through the circular pipe. To calculate the variable local air velocities, we start to transverse the pitot tube across the pipe cross section from the pipe wall to its centerline axis. The inclined well type alcohol manometer readings of pitot tube and static pressure tappings are recorded for later calculations. The dynamic pressures of air are calculated by converting the alcohol head to the dynamic pressure. The dynamic pressure of air helps us to determine local air velocities in different pitot tube radius position changes. We have to find a velocity equation to find flow rate and a local velocity graph in different radiuses can help to express an equation for flow rate in the pipe. We can easily calculate mean velocity, reynold number and mass flow rate by using some formulas below: = ∗ ∗ = 92 = 20° = 15° = 820 / = ∗ → , = 1,094 @ 293 → = 15,78 ∗ 10 / Diameter of the pipe (d) is 104 mm, = ∗ ∗ ℎ ∗ sin equal to = ∗ ∗ Therefore; = ∗ ∗ ∗ ∗ = 2 ∗ ∗ ∫ ( ) ∗ ∗ ̇ = ∗ = = ∗
- 3. 3 CALCULATION Firstly, we calculated specific gravity of air and then dynamic pressures at distances which we took from experiment. Then, we started with selection of r distances. In our experiment r=52mm is our pipe wall and the center of the pipe is r=0. We have calculated all required data by using upward formulas at Excel and Wolfram Mathematica. After calculation in Excel , we got a table-1 to show our calculation results . At the pipe center; r= 0 and h=18 mm alcohol so; = ∗ ∗ ℎ ∗ sin → = 820 ∗ 9,81 ∗ 0,018 ∗ sin 15 = 33,47 = 2 ∗ 820 ∗ 9,81 ∗ 18 ∗ 10 ∗ sin 15 1,094 = 8,28 / Then others calculation results shown at Table 1: r (mm) h (mm) h (m) Pdyn (Pa) U (m/s) 0 18 0,018 37,47 8,28 10 17 0,017 35,39 8,04 27 16 0,016 33,31 7,80 35 13 0,013 27,06 7,03 42 11 0,011 22,90 6,47 44 10 0,01 20,82 6,17 47 9 0,009 18,74 5,85 49 7 0,007 14,57 5,16 51,5 3,5 0,0035 7,29 3,65 52 0 0 0 0 Table -1 Pipe is symmetric so ; we know that these results are the same for other side of pipe cross section.
- 4. 4 The next step is drawing the U( r ).This means that Velocity is a function of pipe radius of r. Plot -1 We have drawn the U( r ).As we said before we assume that center of pipe is r=0 and pipe wall is r=52 and r= -52. After this step we must find the equation of our plot. To do this, one side of the graph is enough and the plotting is shown at Plot-2. Plot-2 *******Red line at the Plot-2 is the nearly polynomial equivalence of our experimental result. R² = 0,737 a value which closest to 1 and this R² = 0,737 shows us how our polynomial lines fit the experimental data. When R² closest to 1 this means that the equation of plot is good fit the data. We tried to make bigger R² but finally Umean value could not found at the intervals of data. 0 1 2 3 4 5 6 7 8 9 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 U( r=mm ) m/s U( r ) y = -0,004x2 + 0,159x + 7,659 R² = 0,737 0 1 2 3 4 5 6 7 8 9 10 0 10 20 30 40 50 60 U(r) U(r) Polinom. (U(r))
- 5. 5 For example, as it seen on the Plot-3, this red line fits the data better than the plot-2.Also Plot-3 has higher R² value. This R² is closer 1.But if we use this equation at the end of our calculations, we can not find the mean velocity at the intervals of data .So that, we choose Plot-2 and its equation to calculate other necessary data flow rate, mean velocity etc. Plot-3 We can find the equation of velocity of air at any distance (r), with respect to fitting line of plot “u” versus” r” ( look to plot -2) y = -0,004x2 + 0,159x + 7,659 ℎ Therefore; Flow rate = 2 ∗ ∗ ∫ ( ) ∗ ∗ Pdyn = ρgh = 1 2 ρU ( ) = −0,004 ∗ + 0,159 ∗ + 7,659 = 2 ( ) ∗ ∗ = 2 (−0,004 ∗ + 0,159 ∗ + 7,659) ∗ ∗ , = , / y = -1E-05x4 + 0,001x3 - 0,027x2 + 0,186x + 8,212 R² = 0,871 0 1 2 3 4 5 6 7 8 9 10 0 10 20 30 40 50 60 U(r) U(r) Polinom. (U(r))
- 6. 6 Mass Flow Rate ̇ = ∗ ̇ =0,0651*1,094 -> ̇ = 0,07 kg/s Mean Velocity = = , ∗ . = , / Reynolds Number =15,68∗ 10 = ∗ = 7,664 ∗ ∗ , ∗ = 50832,65 Re > 2300 flow is turbulent. The next step is finding the position of Mean Velocity of 7,66 m/s. From Wolfram Mathematica : Input : Solve[7.664 == −0.004 ∗ + 0.159 ∗ + 7.659, ] Output : {{ → 0.0314}, { → 39.71}} The mean velocity is at { → . } mm. ; x refer to radius of pipe r.Mean velocity is 39.71 mm above and below the center of pipe. Uncertainty ± 0,001 = g*ρ*sin(15) =2081.83 = ∗ = 2,08 Pa At r=0 (center of pipe) Max Dynamic pressure Pdyn = g*ρ*hmax*sin(15) = 37,47 Pa X= ∗ , , = %5,55 of Full Scale
- 7. 7 CONCLUSION In this experiment, has been used two data which are the height of manometer and the radial distance of pipe cross section. For this measurement, is utilized by static pressure tapping and pitot tube. We measured the pressure values at any radial distance from wall to center of pipe by using the well known incline manometer. So, during the experiment we used only pressure and radial distance of its center. After that, we calculated other unknown values by using experimental data such as mean velocity, volume flow rate, Reynolds number and also characteristic of flow. Consequently, we indicated how to change velocity of flow from wall to center at any point of cross section of pipe. We have sketched a velocity-radius graph and derived flow rate equation. It would help to estimate mean velocity and Reynolds number too by their specific formulas. However, after we had calculated these values, we realized that the position of mean velocity was not correct with regard to our calculation results. Normally, we know that it is center of cross section. And so, we thought that some human and fixed errors could be introduced in the experiment and maybe some measurement values were incorrect.