Chapter 2

Example One: Standing Wavein a Duct

The first example (see Figure 1.1.a) is a simple model of a 1D standingwave in a rigid walled duct. A 1D standing wave is created when a sourceof constant velocity is operated at one end of a closed tube. The soundemitted from the source experiences multiple reflections from each end ofthe tube. The resulting forward and back propagating waves combine toform a “standing wave” of high amplitude.

This problem introduces the very simple geometry of a long rectangle.One of the dimensions is much larger than the other two, enabling theassumption of 1D plane propagating waves to be valid, which simplifiesthe theoretical analysis at low frequencies (below the cut-on frequency ofhigher order modes). Velocity boundary conditions, the required directionof normals and meshing are introduced. How the accuracy of results canbe affected by mesh resolution is also demonstrated. Results obtained fromthe numerical model are then compared to the analytical solution.

Open up a new project (Files > New) and save it (Files > Save) inyour working directory as “stand.gid”. This will create a folder in which allof the files generated using GiD will automatically be saved.

The first step is to construct the rectangular tube depicted in Figure 2.1.Open up an auxiliary window from which coordinates can be easily

entered (Utilities > Tools > Coordinates Window).Chose the option of creating a point (Geometry > Create > Point).Enter the eight points in Table 2.1 by typing their coordinates in the

coordinate window.

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2. Example One: Standing Wave in a Duct

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43

8

76

5

Figure 2.1: Rectangular tube.

Table 2.1: Coordinates of the duct vertices.

point coordinates1 (0,0,0)2 (1,0,0)3 (1,1,0)4 (0,1,0)5 (1,0,10)6 (1,1,10)7 (0,1,10)8 (0,0,10)

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Click “apply” after entering the coordinates of each point (see Fig-ure 2.2). Click “close” once all points have been entered.

Figure 2.2: Entering the point coordinates.

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2. Example One: Standing Wave in a Duct

Using the trackball (View > Rotate > Trackball)∗, rotate the coor-dinate system until all eight points can clearly be seen in a 3D view.

Centre the image and ensure that the zoom is reasonable for the windowsize. First use (View > Zoom) then click and hold the left mouse buttonwhile dragging the mouse over the desired zoom area.

Construct a rectangular prism by joining the nodes as depicted in Fig-ure 2.3. To do this, select the option to create a line (Geometry > Create> Straight Line). Right click the mouse and select Contextual > JoinC-a (or hit “ctrl-a”). Click on the points to join the lines. To start aline from a point different to the finish point of the last line, right clickthe mouse and select Contextual > Escape (or hit the “esc” key) beforeclicking the first point of the new line.

Figure 2.3: Joining the points to form a prism.

∗The trackball can also be found as an icon to the left hand side of the screen

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Now that the basic geometry has been defined, the surfaces of the bound-ary problem need to be defined. To do this select Geometry > Create> NURBS surface > By contour. Click on the four lines defining theedges of one of the prism NURBS† surfaces. Upon selection they will behighlighted in red. If an incorrect line is accidently selected it can be un-selected by clicking on it once more. Once the four lines are highlightedhit “esc”. a purple rectangle will appear inside of the original boundary,indicating the existence of a surface. Use the same procedure to define eachof the other five prism surfaces as depicted in Figure 2.4.

Figure 2.4: Constructing the prism surfaces.

†NURBS are non-uniform rational B-splines. They are a type of curves that caninterpolate a set of points. NURBS can also be defined by their control polygon, anotherset of points that the curve approximates smoothly.

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2. Example One: Standing Wave in a Duct

The next step is to check the surface normals. Select Utilities > SwapNormals > Surfaces > Select. Arrows show the direction of the surfacenormals. Some of the surfaces may point into the prism whilst others maybe pointing out (see Figure 2.5).

Figure 2.5: “Draw Normals” environment.

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For an internal boundary element problem all surfaces must face out.Click on all of the surfaces (the pink lines) that have an inward pointingnormal until all surfaces are oriented in the correct direction (see Figure 2.6).Hit “esc” to leave the “Draw Normals” mode.

Figure 2.6: All normals are oriented correctly for an internal BEM problem.

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2. Example One: Standing Wave in a Duct

The next step is to define the problem as a Helm3D BEM problem. Todo this select (Data > Problem type > Helm3d). Once the problemtype has been chosen, the boundary conditions need to be defined. The onlyboundary conditions required for the standing wave problem are velocityboundary conditions.

A velocity of 1+0i (ie. magnitude 1, 0 phase shift) is needed to representthe piston at the entrance of the tube.

To set this select (Data > Conditions). Select “Velocity” from thedrop down menu. Set real velocity component to 1 and the imaginaryvelocity components to zero. Click on “assign” and select the surface atz=0 (as depicted in Figure 2.7). Select “finish”.

All other surfaces have been automatically set to zero velocity, which isthe default boundary condition.

Further information about already applied boundary conditions can beobtained by selecting the “Draw” menu on the Conditions window. Forexample, selecting “colors” will show applied boundary conditions by colour.

Figure 2.7: Setting the z=0 velocity boundary condition.

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Select (Data > Problem data). Select the “General” tab, and givethe project a title of <stand>. Ensure the “Analysis type” is selected to be“Internal”, “Loading type” is “Radiation” and that “Symmetry” is “None”(see Figure 2.8).

Figure 2.8: Defining the general problem data.

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2. Example One: Standing Wave in a Duct

Next select the “Fluid properties” tab and set the density and speed ofsound are to 1.21 and 343 respectively (see Figure 2.9).

Figure 2.9: Defining the properties problem data.

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Finally select the “Frequencies” tab and set the “Frequency start” and“Frequency stop” to both be 25.725, corresponding to the second resonancefrequency of the tube, leaving the “Frequency interval” to be zero (seeFigure 2.10). Ignore the other tabs giving default values for output andCHIEF points, click “Accept” and then “Close” to finish.

Figure 2.10: Defining the frequency problem data.

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2. Example One: Standing Wave in a Duct

The next step is to mesh the boundary of the tube. Select (Mesh >Generate mesh). A dialog box will appear asking you to “Enter the size ofelements to be generated”. Type in <2>. A dialog box will appear whichstates that 52 triangle elements have been created. Press “OK” and themesh will appear (Figure 2.11).

Figure 2.11: View of the meshed tube.

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A solution to the problem can now be generated by selecting (Calculate> Calculate). A dialog box will appear telling you once the solution isdone (Figure 2.12). Click “OK”.

Figure 2.12: Process information dialog box.

To review the results, once must enter the postprocessor. To do thiseither click “Postprocess” on the dialog box, or select (Files > Postpro-cess). To display the pressure over the boundary of the duct select (Viewresults > Contour fill > Press amp) (Figure 2.13).

Figure 2.13: Pressure amplitude over boundary of duct.

As expected, a standing wave corresponding to the second resonancefrequency of the tube is observed. To see how the real and imaginary

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2. Example One: Standing Wave in a Duct

pressure components vary along the tube or to see the total sound pressurelevel in dB, select (View results > Contour fill) and chose the desiredparameter.

Although the BEM problem was solved only at one frequency, the so-lution can be swept over a frequency range in order to see the frequencydependance of the solution. To do this, you first need to return to the pre-processor (Files > Preprocess), or click on the icon in the toolbar. Thefrequency range can then be changed within the problem data environment(Data > Problem data). Select the “Frequencies” tab and change “fre-quency start” to 1, “frequency stop” to “110” and “frequency interval” to asuitable interval such as 1 (the smaller the increment, the higher the reso-lution of the result, but the necessary computational effort also increases).

If the problem data or conditions are changed the model must alwaysbe remeshed before the new solution is obtained. To do this select (Mesh> Generate mesh) or type control-g. A warning will appear alerting youthat the old mesh will be erased and asking you whether to continue withthe mesh. Click “OK” for this and the following two dialog boxes.

It is possible to extract frequency dependent data from the model forfurther analysis. We wish to find the pressure at a location on the surfaceof the piston (and hence calculate the impedance). To do this, we mustfirst find the number of a node located on the surface of the piston. Use(Utilities > List > Nodes) and select a node on the surface of the piston(Figure 2.14).

Once the node of interest is selected, hit “esc” and a dialog box willappear showing the node number (Figure 2.15), node 3 in this case.

Select (Data > Problem data), and select the “Output points” tab.Enter the node number in the “Output node” box and hit “Accept”.

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Figure 2.14: .

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2. Example One: Standing Wave in a Duct

Figure 2.15: .

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Figure 2.16: .

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2. Example One: Standing Wave in a Duct

Next generate a new solution to the problem (Calculate > Calculate).A greater period of time will elapse before the dialog box telling you thatthe solution is done appears. This is due to a separate solution having tobe generated at each frequency increment.

Once the solution is complete you can once again enter the postprocessor(Files > Postprocess). To set the to frequency at which to review yourresults, (View results > Default Analysis / Step > frequency) andthen the desired frequency.

The pressure at the output node, ie. the point of excitation at eachfrequency, can be viewed by opening the “output.fdat” file (using any simpletext editor) which will have appeared in your working folder (the folder inwhich you have saved “stand”). The first line contains the headings of eachcolumn of data. Appendix B gives a description of the file format used. Ofgreatest interest in this case are the first column which is the frequency andthe fifth column which is the pressure amplitude (in Pascals) of the outputpoint (in our case the point of excitation). As the boundary condition atthe point of excitation necessitates unit velocity amplitude at this location,the specific acoustic impedance, which is the ratio between the acousticpressure and the particle velocity is simply the magnitude of the pressureat this point.

The theoretical resonance frequencies of the system are simply the res-onances of an open-closed duct and are given by:

fn = nc

2l (2.1)

where n is the mode number, c is the speed of sound and l is the lengthof the duct. The analytical specific acoustic impedance at the excitationlocation is:

Zs = 0− iρc cot(kl) = p

v(2.2)

where i =√−1, ρ is the density of the medium, k is the wavenumber, p

acoustic pressure and v is the particle velocity. By solving the BEM problemover a range of frequencies as you have done, the theoretical specific acousticimpedance and the BEM specific acoustic impedance (the ratio between theacoustic pressure and the particle velocity) at the point of excitation canbe obtained. An example comparing the theory and BEM solutions wasconstructed using Matlab (see Figure 2.17).

Try comparing your BEM results with the theoretical solution usingyour preferred graphing package.

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Extension

0 10 20 30 40 50 60 70 80 90 10010

0

101

102

103

104

105

106

Frequency [Hz]

Spe

cific

aco

ustic

impe

danc

e [k

g.s−

1 .m−

2 ]

TheoryBEM

Figure 2.17: Harmonic response of an open-closed acoustic duct at the pointof excitation.

2.1 ExtensionAn important point to consider in BEM problems is mesh size. In orderfor a BEM solution to be accurate, there must be a sufficient number ofelements per wavelength. Hence at higher frequency, the mesh densitymust be greater. Try experimenting with your mesh density and analysisfrequency to see how these affect your solution. Compare the BEM resultswith the theoretical result for each case. It has been proposed that foraccurate BEM results to be obtained there should be a minimum of sixelements per wavelength. Does this hypothesis hold true for the case of anopen-closed duct?

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