Numerical Analysis for Characterization of a Salty Water Meter

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1 Nuerical Aalysis for Characterizatio of a Salty Water Meter José Erique Salias Carrillo Departaeto de Ciecias Básicas Istituto Tecológico de Tehuacá Bolio Arago Perdoo Departaeto de Mecatróica Istituto tecológico de Tehuacá 15/1/014 Key words: Nuerical Aalysis, Nuerical characterizatio, Salty water eter I this paper, It is Calculated ad reported a part of the uerical characterizatio of the proble of a salty water eter. The salty water eter is a priary prototype which it is beig studied i accordace with it s techological applicatios proposed for the salt producers i Zapotitla Salias, a little village i the regio of Tehuaca, i the state of Puebla, Mexico. The searchig of a forula for the area of two seicircle sectios coected with a sei cylidrical sectio with a plate iside of this, It is a easy ad clear proble that has already bee solved by the geoetricias. But fro the poit of view of the uerical ethods, it results iterestig the forula for the area through oly uerical calculatios. Resue E este artículo se calcula y reporta parte de la caracterizació de el problea de u edidor de saliidad del agua. El edidor de saliidad del agua es u prototipo priario el cual es estudiado de acuerdo a sus aplicacioes tecologicas propuestas para los productores de sal de la villa de Sapotitla Salias situado e la regió de Tehuacá, e el estado de Puebla, México. La busqueda de ua forula para el area de dos seccioes seicirculares coectadad por u sei cilidro co ua placa detro de este arreglo, es u problea ya resuelto por los geoetras, pero e este articulo es abordado desde el puto de los étodos uéricos y resulta iteresate el aálisis hecho para esta área, a través de úicaete cálculos uéricos. 35

2 Itroductio The uerical ethods are appropriated for solve itegrals, derivatives, liear ad o liear probles, or dieretial equatios. They have the characteristic of applyig the prograed solutios to obtai the solutio which could be dicult or ipossible for obtai by aother ethod. But ot oly are these cases solved by the prograers, but also the basic probles that had bee already solved by aother ethod. It is the case reported i this paper, the solutio to the proposed proble to d the area of a sei cylidrical closed by the sides ad with a plaar plate iserted i the iddle. It could result trivial fro the poit of view of the classical geoetry. The solutio is for the sei cylidrical surface A c = πrl ad for the two seicircles A ss = πr ad the plaar plate A p p = ab where r = ratio,l = legth of the cilider, a = legth of the plaar plate, b = stregh of the plaar plate Also to solve this proble by the uerical poit of view, it would result useful for future calculatios. We have i id to solve the electroagetic proble to deterie the Electric Field, The Electric Potetial ad to obtai the equatios for the capacity of the device fored by the experietal array ivolved i this priary approxiatio. Aother possibility is to solve the equatio to deterie the resistace for the array workig as a variable resistor. Developet To suppose the geoetry give by the gure 1 where the core of the cylidrical for is closed by the extrees ad a plaar plate is ito the iddle of the sei cylidrical recipiet. I a previous paper (Salias 014) it had obtaied the equatios i ters of sets for the four surfaces supposig the refereces how are showed i the gure, these are the followig sets: S c = CilidricalSurface, f b = F acebefore, f p = F acep osterior, P = M ediuplate S c = (x, y, z) 0 x l, z 0, z + y = r, r y r f b = (x, y, z) x = 0, r y r, (r y ) z 0 f p = (x, y, z) x = l, r y r, (r y ) z 0 P = (x, y, z) (l a) x (l+a), y = 0, b z 0 These sets are obtaied supposig the origi of the coordiated axes as are showed i the gure 3. The goal is lookig for the itegral for for the surface of the device cosidered as the uio of the sets S c fb fp P. Let be a partitio deed for S c give for x = [0, l, l,..., ( 1) l, l ] ad a partitio for z = [ ( r), ( r( 1)),..., ( r), ( r), 0] where x, y, z are easured o the cylidrical surface. These two variables deterie the values for y I the gure we look at the dieretial of area betwee the poits 36

But ot oly are these cases solved by the prograers, but also the basic probles that had bee already solved by aother ethod.

3 (a) 1 Las piezas del trasductor (b) Los ejes de referecia e el trasductor (c) 3 The Nuerical Partitios 37

4 (x i, y j, z j ), (x i, y (j+1), z (j+1) ), (x (i+1), y j, z j ), (x (i+1), y (j+1), z (j+1) ), ad a approxiatio for A is A = x i ( yj ) + ( z j ) The the itegral give by A = da = i=1 j=1 x i ( yj ) + ( z j ) S Where y j = r zj, y j+1 = r zj+1 Ad x i = x i+1 x i, z j = z j+1 z j, y j = y j+1 y j = r z j+1 r zj Therefore a approxiatio for the itegral is A = i=1 j=1 x i ( yj ) + ( z j ) or A = ( r ) i=1 j=1 (x i+1 x i ) (z j+1 z j ) + z j+1 r zj Or i a siplied for expressio (z j+1 z j ) + r z j+1 zj A = i=1 j=1 (x i+1 x i ) (r zj+1 )(r zj ) By other had, the two areas of seicircular for ca be partitioed by the followig for: For x = 0, either it is the seicircular surface ear the origi or for x = l, if it is the surface far frot the origi, ad for the z i, y j for both surfaces x i = 0,or x i = l; z i = [ r y j = [ r, ( 1)r,..., ( 1 r +r usig the approxiatio A = 1 i=0, ( 1)r,..., ( i)r ), 0 r +r 1 j=0 z i y j χ( z i + y j r)..., 1r, 0r ], ] ad the area could have bee calculated { 0 if z Whereχ( zi + y j r) = i + yj r 1 if zi + y j r Now, oly is ecessary to coplete the surface with the plate part. The plate iside the body of the device has the ext partitios: For x [ l a, l+a ] with z [ b, 0] Where the dieretial of area it is give by A = x i z j = (x i+1 x i )(z j+1 z j ) Ad the area ca be obtaied addig the icreets with A = 1 1 i=0 j=0 (x i+1 x i )(z j+1 z j ) But x i = l a + ia co i [0, ] Ad z j = b + jb co j [0, ] The area ca be rewritte as follow A = 1 (( ) 1 l a i=0 j=0 + (i+1)a ( ) ) (( ) ( l a + ia b + (j+1)(b) ) ( b ) 1 i=0 1 j=0 ( a Nuerical Results: b + jb We otai for a partitio of = 10,y = 10 the values for A 1, A, A 3.reported ito the table 1 for the true values we used l = 1.c, r = 4c, a = 4.5, b = 3.7c ad we applied the expresios A t1 = πrl, A t = πr, A t3 = ab )) = 38

) i=1 j=1 (x i+1 x i ) (z j+1 z j ) + z j+1 r zj Or i a siplied for expressio (z j+1 z j ) + r z j+1 zj A = i=1 j=1 (x i+1 x i ) (r zj+1 )(r zj ) By other had, the two areas of seicircular for ca be

5 Table 1: Nuerical Results of the aproxiatios with their errors Set Area c True Area c Error c A * A A E-14 ad the values of the Errors are calculated by the forulas E 1 = A 1 A t1, E = A A t, E 3 = A 3 A t3 Coclusios: We have three expressios for the four surfaces A 1 is for the cylidrical surface A 1 = i=1 j=1 (x i+1 x i ) (y j+1 y j ) + r + y j+1 + yj (r yj+1 )(r yj ) A for the seicircles at x = 0 ad x = l A = 1 1 i=0 j=0 z i y j χ( zi + y j r) Ad A 3 for the plate iside the body of the eter A 3 = 1 1 i=0 j=0 (x i+1 x i ) (z j+1 z j ) = 1 i=0 1 j=0 ( a ) ( b ) The uerical calculatios for = 10, = 10 give A 1 = c, A =.08c, A 3 = 16.65c ad the true values are A t1 = c, A t = c, A t3 = 16.65c The errors are of ErrorA 1 = c, ErrorA = c aderrora 3 = E 14c. The true expresios are for the seicilidera c = πrl, for the two seicircles A ss = πr, ad for the plaar plate A pp = ab. The greatest error is obtaied i the seicircular sectios, it could be because i the aproxiatio half of the eliied poits add the half of area o the Total Area. Ackowledgets I Ackowledge to the Istituto Tecologico de Tehuaca, the istitutio which otivates the writig of this article ad which give e the resourses for cotiue with this research. Refereces 39

surface A 1 = i=1 j=1 (x i+1 x i ) (y j+1 y j ) + r + y j+1 + yj (r yj+1 )(r yj ) A for the seicircles at x = 0 ad x = l A = 1 1 i=0 j=0 z i y j χ( zi + y j r) Ad A 3 for the plate iside the body of

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The Binomial Multi- Section Transformer

The Binomial Multi- Section Transformer 4/15/21 The Bioial Multisectio Matchig Trasforer.doc 1/17 The Bioial Multi- Sectio Trasforer Recall that a ulti-sectio atchig etwork ca be described usig the theory of sall reflectios as: where: Γ ( ω