
6421 views  Technical Discussions   Jeff Phillips
 Jeff Phillips
 01:19 May281998 Near Zone Calculation I would like to find out the formula for calcualtion of the near zone of a rectangular crystal. I am aware of the calculation for a circular crystal: NZ=(D^2f)/4v But I am not aware of the formula for a rectangular crystal. Any help on this matter would be appreciated. Thank you, Jeff Phillips jeffreyp@deakin.edu.au
 
   Michael Trinidad Consultant, LMATS Pty Ltd , Australia, Joined Jan 2003, ^{138}   02:10 May291998 Re: Near Zone Calculation Most of the reference books I have refer to the diameter of the crystal for the calculations however I have a copy of the AAp 7002.0081 Non Destructive Inspection Methods General Data a RAAF document based on the American Tech order 33B11.The formula for square or rectangular crystals D = the maximum diagonal of the transducer element in inches instead of diameter If you need the page you can either get it from your local RAAF base page 411, para 137 or email me and I'll fax it down. : I would like to find out the formula for calcualtion of the near zone of a rectangular crystal. : I am aware of the calculation for a circular crystal: : NZ=(D^2f)/4v : But I am not aware of the formula for a rectangular crystal. : Any help on this matter would be appreciated. : Thank you, : Jeff Phillips : jeffreyp@deakin.edu.au
 
   Manuel Haces
 Manuel Haces
 02:31 May301998 Re: Near Zone Calculation Jeff Phillips, jeffreyp@deakin.edu.au I found a good reference of the near zone calculations, in the book Ultrasonic Testign of Materials, 3rd Rev., J.Krautkämer and H.Krautkrämer, Chapter 4: Wave Physics of Sound Field. I think the material will help you quite a lot.Regards: Manuel Haces.
 
   Rolf Diederichs Director, NDT.net, Germany, Joined Nov 1998, ^{608}   04:58 Jun011998 Re: Near Zone Calculation : I would like to find out the formula for calcualtion of the near zone of a rectangular crystal. : I am aware of the calculation for a circular crystal: : NZ=(D^2f)/4v : But I am not aware of the formula for a rectangular crystal. : Any help on this matter would be appreciated. : Thank you, : Jeff Phillips : jeffreyp@deakin.edu.au I post this message on behalf of Ed Ginzel. During my stay at the 7th ECNDT in Copenhagen a network error occurred and Ed Ginzel's message could not reach the forum. His message includes also diagrams which mailinglist readers can only view through the forum webpage (Figure Comparing Sound Pressure for Circular and Square Elements). It was a very busy time at the ECNDT conference so that my participation on the forum was impossible. Also the Compuserve Network in Denmark was not very good. So much the more I am happy seeing a lot of interesting discussions have been happened during this time. Rolf Diederichs  Ed Ginzel's message: Inresponse to the question about Near Zone of a rectangular element I found it useful to refer to the Krautkramer 'Ultrasonic Testing of Materials' (I have the third English edition). On page 72 a discussion of the differences between circular and rectangular oscillators is given. For the simple case of a square the equation for the near zone (N) is N=1.35D^2/4lambda The writer then goes on to explain that as the side ratio decreases the end of the near field is characterised less clearly by a maximum of the sound pressure. I assume the D in the equation for the square element is the length of a side of the square but it is not defined in the text. This would follow in that the near zone is extended beyond the distance it would be for a circular element. The constant 1.35 closely approximates the difference between the diagonal of a square and the diameter of a circle having its diameter equal to the length of a side of the square (ideally 1.4). In about 1985 Dr. Udo Schlengerman produced a booklet (The Krautkramer Branson Booklet), that provided a guide to the basics of ultrasonic testing. In it he provided a pair of graphs of the sound pressure vs. distance for a circular and a rectangular element. See attached drawing. This illustrates how quickly the sound pressure fluctuations flatten out. Clearly other factors will enter into the situation too. To assume that all parts of the element will vibrate with the same displacement is ideal. The reality is that the end points on a high aspect ratio rectangle are not damped in the same way the mid point are. Strong lobes and larger transverse components in solids can occur and add to the fact that the Near Zone peak is poorly defined.
 
   Udo Schlengermann Consultant,  Standards Consulting, Germany, Joined Nov 1998, ^{177}   05:09 Jun011998 Re: Near Zone Calculation Dear Jeff,the existence of a near zone with ultrasonic transducers has always the same physical reason: diffraction by the boundaries. Therefore the extension and the shape of this zone are determined by the shape of the transducer and the wavelength in the material, resp. frequency and sound velocity. To normalize the equations for sound pressure this influence of shape, frequency and velocity is eliminated by using the relative distance Z = (z v)/(aa f), where z is the distance, v sound velocity, a half the size of the transducer and f is frequency. The condition to be fulfilled when using this normalization is: size a must be much greater than the wavelength, which is always met in nondestructive testing in the megahertz range. The nearfield distance is defined as the distance to the last pressure maximum on the acoustical axis. Expressed in relative distances this means N = k (aa f)/ v, with a dimensionless factor k. For a circular, plane transducer this factor is k = 1.0, which together with the radius a of the disk transducer  gives the equation you know: Ncirc = (aa f)/ v = (DD f)/ (4v). For a rectangular transducer having a large side 2a and a small side 2b the same notation can be used: Nrect = k (aa f)/ v, where the factor k is a function of the side ratio (b/a). This ratio is always between 0 (line source) and 1 (square source). There is no simple equation to calculate the factor k depending on (b/a), but I can offer you a diagram which I made long time ago, showing this factor k over the ratio (b/a). The values of k were determined by me on the following way: 1. by numerical calculation the sound pressure on the acoustical axis was calculated for rectangular transducers with varying ratio (b/a). 2. from these values the location of the last pressure maximum was determined to a precison of 5 decimals expressed in relative distances (aa f)/ v. 3. these values for N depending on (b/a) gives the curve shown in the attached diagram. To give some examples: a) The nearfield distance of a line source of length 2a is N( 0) = 0.988 (aa f)/ v. b) the nearfield length of a transducer with (b/a) = 0.5 is N(0.5) = 1.014 (aa f/ v). c) the nearfield length of a transducer with (b/a) = 1 is N(1.0) = 1.368 (aa f) v). Diagram by separate mail kind regards Udo Schlengermann
 
   Udo Schlengermann Consultant,  Standards Consulting, Germany, Joined Nov 1998, ^{177}   03:03 Jun021998 Nearfield distance of rectangular transducers 3. The near zone distance is defined as the distance of the last pressure maximum on the acoustical axis. But for side ratios (b/a) < 0.6 this last maximum is not the absolute peak. Fig. 4.26 in the Krautkramer book (4th revision) shows that for smaller (b/a) the absolute peak is at much shorter distances. This makes it very difficult to identify the near field distance of these rectangles with (b/a) < 0.6. The Krautkramer booklet shows only the behaviour of a square transducer (b/a) = 1. Udo Schlengermann (uschlengermann@krautkramer.de)
 
  
