{"id":4839,"date":"2020-02-11T21:30:03","date_gmt":"2020-02-12T02:30:03","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4839"},"modified":"2026-04-05T20:58:37","modified_gmt":"2026-04-06T01:58:37","slug":"s3eva2019tii_t3-preparacion-de-terreno-en-refineria","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s3eva20\/s3eva2019tii_t3-preparacion-de-terreno-en-refineria\/","title":{"rendered":"s3Eva2019TII_T3 Preparaci\u00f3n de terreno en refiner\u00eda"},"content":{"rendered":"\n<p><em><strong>Ejercicio<\/strong><\/em>: <a href=\"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva20\/3eva2019tii_t3-preparacion-de-terreno-en-refineria\/\" data-type=\"post\" data-id=\"4860\">3Eva2019TII_T3 Preparaci\u00f3n de terreno en refiner\u00eda<\/a><\/p>\n\n\n\n<p>Se requiere usar el <strong>nivel inicial<\/strong> en la matriz, para restar del <strong>nivel requerido<\/strong> que es constante 220.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th>Nivel inicio (m)<\/th><th>0<\/th><th>50<\/th><th>100<\/th><th>150<\/th><th>200<\/th><\/tr><\/thead><tbody><tr><td><strong>0<\/strong><\/td><td>241<\/td><td>239<\/td><td>238<\/td><td>236<\/td><td>234<\/td><\/tr><tr><td><strong>25<\/strong><\/td><td>241<\/td><td>239<\/td><td>237<\/td><td>235<\/td><td>233<\/td><\/tr><tr><td><strong>50<\/strong><\/td><td>241<\/td><td>239<\/td><td>236<\/td><td>234<\/td><td>231<\/td><\/tr><tr><td><strong>75<\/strong><\/td><td>242<\/td><td>239<\/td><td>236<\/td><td>232<\/td><td>229<\/td><\/tr><tr><td><strong>100<\/strong><\/td><td>243<\/td><td>239<\/td><td>235<\/td><td>231<\/td><td>227<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>lo que genera la matriz de diferencias. El valor es positivo indica remoci\u00f3n, el valor negativo indica por rellenar.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table><thead><tr><th><strong>Diferencia<\/strong> (m)<\/th><th>0<\/th><th>50<\/th><th>100<\/th><th>150<\/th><th>200<\/th><\/tr><\/thead><tbody><tr><td><strong>0<\/strong><\/td><td>21<\/td><td>19<\/td><td>18<\/td><td>16<\/td><td>14<\/td><\/tr><tr><td><strong>25<\/strong><\/td><td>21<\/td><td>19<\/td><td>17<\/td><td>15<\/td><td>13<\/td><\/tr><tr><td><strong>50<\/strong><\/td><td>21<\/td><td>19<\/td><td>16<\/td><td>14<\/td><td>11<\/td><\/tr><tr><td><strong>75<\/strong><\/td><td>22<\/td><td>19<\/td><td>16<\/td><td>12<\/td><td>9<\/td><\/tr><tr><td><strong>100<\/strong><\/td><td>23<\/td><td>19<\/td><td>15<\/td><td>11<\/td><td>7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>El volumen se puede calcular por un m\u00e9todo en cada fila, y luego los resultados por columnas por otro m\u00e9todo o el mismo.<br>Por ejemplo Simpson de 1\/3<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I= \\frac{hx}{3}(f(x_0) +4f(x_1)+f(x_2))<\/span>\n\n\n\n<p>con lo que se obtiene:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{fila}(0) = \\frac{50}{3}(21 +4(19)+18) +\\frac{50}{3}(18 +4(16)+14) =24750<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{fila}(25) = = \\frac{50}{3}(21 +4(19)+17) + \\frac{50}{3}(17 +4(15)+13) = 23383,33<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{fila}(50) = \\frac{50}{3}(21 +4(19)+16) + \\frac{50}{3}(16 +4(14)+11) = 22000<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{fila}(75) = \\frac{50}{3}(22 +4(19)+16) + \\frac{50}{3}(16 +4(12)+5) =21850<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> I_{fila}(100) = \\frac{50}{3}(23 +4(19)+15) + \\frac{50}{3}(15 +4(11)+7) = 20483,33<\/span>\n\n\n\n<p>y usando el otro eje, se completa el volumen usando dos veces simpson:<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> Volumen = \\frac{h_y}{3}(f(x_0) +4f(x_1)+f(x_2))<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> Remover = \\frac{25}{3}(24750 +4(23383,33)+22000) +<\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\">+ \\frac{25}{3}(22000 +4(21850)+20483,33)=2251388,89<\/span>\n\n\n\n<p>El signo lo trae desde la diferencia, y muestra el sentido del desnivel.<\/p>\n\n\n\n<p>Se adjunta la gr\u00e1fica de superficie en azul como referencia del signo,&nbsp; respecto al nivel requerido en color verde.<\/p>\n\n\n\n<figure class=\"wp-block-image aligncenter size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"457\" height=\"364\" src=\"http:\/\/blog.espol.edu.ec\/algoritmos101\/files\/2020\/02\/refineriaRemover01.png\" alt=\"refiner\u00eda Remover terreno 01\" class=\"wp-image-18622\" \/><\/figure>\n\n\n\n<p>Error de truncamiento<\/p>\n\n\n\n<p>la cota del error de truncamiento se estima como O(h5)<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> error_{trunca} = -\\frac{h^5}{90} f^{(4)}(z)<\/span>\n\n\n\n<p>para un valor de z entre [a,b]<\/p>\n\n\n\n<p>para cuantificar el valor, se puede usar la diferencia finita \u03944f, pues con la derivada ser\u00eda muy laborioso.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ejercicio: 3Eva2019TII_T3 Preparaci\u00f3n de terreno en refiner\u00eda Se requiere usar el nivel inicial en la matriz, para restar del nivel requerido que es constante 220. Nivel inicio (m) 0 50 100 150 200 0 241 239 238 236 234 25 241 239 237 235 233 50 241 239 236 234 231 75 242 239 236 [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[51],"tags":[58,54],"class_list":["post-4839","post","type-post","status-publish","format-standard","hentry","category-mn-s3eva20","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4839","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=4839"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4839\/revisions"}],"predecessor-version":[{"id":23922,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4839\/revisions\/23922"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}