{"id":49169,"date":"2019-11-14T20:21:40","date_gmt":"2019-11-14T19:21:40","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-a-formula-do-momento-equacao-do-momento-definicao\/"},"modified":"2020-02-02T09:22:27","modified_gmt":"2020-02-02T08:22:27","slug":"o-que-e-a-formula-do-momento-equacao-do-momento-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-formula-do-momento-equacao-do-momento-definicao\/","title":{"rendered":"O que \u00e9 a f\u00f3rmula do momento &#8211; Equa\u00e7\u00e3o do momento &#8211; Defini\u00e7\u00e3o"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">F\u00f3rmula do Momento &#8211; Equa\u00e7\u00e3o do Momento.\u00a0A segunda equa\u00e7\u00e3o de conserva\u00e7\u00e3o que devemos considerar no volume de controle \u00e9 a f\u00f3rmula do momento.\u00a0Engenharia T\u00e9rmica<\/div>\n<\/div>\n<div class=\"su-divider su-divider-style-dotted\"><\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2>F\u00f3rmula do Momento &#8211; Equa\u00e7\u00e3o do Momento<\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Conservation-of-Momentum-Fluids-min-2.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14358 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Conservation-of-Momentum-Fluids-min-2-300x210.png\" alt=\"Conserva\u00e7\u00e3o do Momento - Fluidos\" width=\"300\" height=\"210\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Conservation-of-Momentum-Fluids-min-2-300x210.png\" \/><\/a>Assumimos que o fluido seja\u00a0<strong>est\u00e1vel<\/strong>\u00a0e\u00a0<strong>incompress\u00edvel<\/strong>\u00a0.\u00a0Para determinar a taxa de mudan\u00e7a de momento de um fluido, consideraremos um tubo de\u00a0<strong>fluxo<\/strong>\u00a0(\u00a0<strong>volume de controle<\/strong>\u00a0), como fizemos na\u00a0<a title=\"Equa\u00e7\u00e3o de Bernoulli - Princ\u00edpio de Bernoulli\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-equacao-de-bernoulli-principio-de-bernoulli-definicao\/\">equa\u00e7\u00e3o de Bernoulli<\/a>\u00a0.\u00a0Nesse\u00a0<strong>volume de controle,<\/strong>\u00a0qualquer altera\u00e7\u00e3o no momento do fluido dentro de um volume de controle \u00e9 devida \u00e0 a\u00e7\u00e3o de for\u00e7as externas no fluido dentro do volume.<\/p>\n<p>Como pode ser visto na figura, o\u00a0<strong>m\u00e9todo do volume de controle<\/strong>\u00a0pode ser usado para analisar a lei de conserva\u00e7\u00e3o do momento no fluido.\u00a0O volume de controle \u00e9 uma\u00a0<strong>superf\u00edcie imagin\u00e1ria que<\/strong>\u00a0encerra um volume de interesse.\u00a0O volume de controle pode ser fixo ou m\u00f3vel, e pode ser r\u00edgido ou deform\u00e1vel.\u00a0Para determinar todas as for\u00e7as que atuam nas superf\u00edcies do volume de controle, temos que resolver as leis de conserva\u00e7\u00e3o nesse volume de controle.<\/p>\n<p>A primeira equa\u00e7\u00e3o de conserva\u00e7\u00e3o que devemos considerar no volume de controle \u00e9 a\u00a0<a title=\"Equa\u00e7\u00e3o de continuidade\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-equacao-de-continuidade-definicao\/\">equa\u00e7\u00e3o de continuidade<\/a>\u00a0(\u00a0<a title=\"Lei de conserva\u00e7\u00e3o da massa\" href=\"https:\/\/www.nuclear-power.com\/laws-of-conservation\/law-of-conservation-of-matter\/\">a lei da conserva\u00e7\u00e3o da massa<\/a>\u00a0).\u00a0Na forma mais simples, \u00e9 representado pela seguinte equa\u00e7\u00e3o:<\/p>\n<p>\u2211\u1e41\u00a0<sub>dentro<\/sub>\u00a0= \u2211\u1e41\u00a0<sub>fora<\/sub><\/p>\n<p>Soma das taxas de fluxo de massa entrando por unidade de tempo = Soma das taxas de fluxo de massa saindo por unidade de tempo<\/p>\n<p>A segunda equa\u00e7\u00e3o de conserva\u00e7\u00e3o que devemos considerar no volume de controle \u00e9 a<strong>\u00a0f\u00f3rmula\u00a0<\/strong>do<strong>\u00a0momento\u00a0.\u00a0<\/strong>Na forma mais simples, a<strong>\u00a0f\u00f3rmula<\/strong>\u00a0do<strong>\u00a0momento<\/strong>\u00a0pode ser representada pela seguinte equa\u00e7\u00e3o:<\/p>\n<p>&#8211;<br \/>\n<a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Formula-fluids.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-14362 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Formula-fluids.png\" alt=\"F\u00f3rmula Momentum - fluidos\" width=\"559\" height=\"185\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Formula-fluids.png\" \/><\/a><\/p>\n<h2><\/h2>\n<h2>Escolhendo um volume de controle<\/h2>\n<p>Um volume de controle pode ser selecionado como qualquer volume arbitr\u00e1rio atrav\u00e9s do qual o fluido flui.\u00a0Esse volume pode ser est\u00e1tico, em movimento e at\u00e9 deformar durante o fluxo.\u00a0Para resolver qualquer problema, precisamos resolver as leis b\u00e1sicas de conserva\u00e7\u00e3o neste volume.\u00a0\u00c9 muito importante conhecer todas as velocidades de fluxo relativas \u00e0 superf\u00edcie de controle e, portanto, \u00e9 muito importante definir exatamente os limites do volume de controle durante uma an\u00e1lise.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<h2><span>Exemplo: a for\u00e7a que atua sobre um cotovelo defletor<\/span><\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-control-volume.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14368 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-control-volume-300x255.png\" alt=\"Equa\u00e7\u00e3o do momento - volume de controle\" width=\"300\" height=\"255\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-control-volume-300x255.png\" \/><\/a><strong><span>Um cotovelo<\/span><\/strong><span>\u00a0(digamos da tubula\u00e7\u00e3o prim\u00e1ria) \u00e9 usado para desviar o fluxo de \u00e1gua a uma velocidade de\u00a0<\/span><strong><span>17 m \/ s<\/span><\/strong><span>\u00a0.\u00a0O di\u00e2metro da tubula\u00e7\u00e3o \u00e9 igual a\u00a0<\/span><strong><span>700 mm<\/span><\/strong><span>\u00a0.\u00a0A press\u00e3o manom\u00e9trica no interior do tubo \u00e9 de cerca de\u00a0<\/span><strong><span>16 MPa<\/span><\/strong><span>\u00a0\u00e0 temperatura de 290 \u00b0 C.\u00a0Fluido \u00e9 de densidade constante\u00a0<\/span><strong><span>\u2374 ~ 720 kg \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><\/strong><span>\u00a0(a 290 \u00b0 C).\u00a0O \u00e2ngulo do cotovelo \u00e9 de\u00a0<\/span><strong><span>45 \u00b0<\/span><\/strong><span>\u00a0.<\/span><\/p>\n<p><span>Calcule a\u00a0<\/span><strong><span>for\u00e7a na parede<\/span><\/strong><span>\u00a0de um cotovelo defletor (ou seja, calcule o vetor F\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0).<\/span><\/p>\n<p><strong><span>Premissas:<\/span><\/strong><\/p>\n<ul>\n<li><span>O fluxo \u00e9 constante.<\/span><\/li>\n<li><span>As perdas por atrito s\u00e3o insignificantes.<\/span><\/li>\n<li><span>O peso do cotovelo \u00e9 insignificante.<\/span><\/li>\n<li><span>O peso da \u00e1gua no cotovelo \u00e9 insignificante.<\/span><\/li>\n<\/ul>\n<p><span>Tomamos o cotovelo como o\u00a0<\/span><strong><span>volume de controle<\/span><\/strong><span>\u00a0.\u00a0O volume de controle \u00e9 mostrado na imagem.\u00a0A equa\u00e7\u00e3o do momento \u00e9 uma equa\u00e7\u00e3o vetorial, portanto possui tr\u00eas componentes.\u00a0Tomamos as coordenadas x e z como mostrado e resolveremos o problema separadamente, de acordo com essas coordenadas.<\/span><\/p>\n<p><span>Primeiro, vamos considerar o componente na\u00a0<\/span><strong><span>coordenada x<\/span><\/strong><span>\u00a0.\u00a0A conserva\u00e7\u00e3o da equa\u00e7\u00e3o do momento linear se torna:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-x-coordinate.png\"><img loading=\"lazy\" class=\"size-full wp-image-14367 alignleft lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-x-coordinate.png\" alt=\"Equa\u00e7\u00e3o do momento - coordenada x\" width=\"603\" height=\"371\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-x-coordinate.png\" \/><\/a><\/p>\n<p><span>Segundo, vamos considerar o componente na\u00a0<\/span><strong><span>coordenada y<\/span><\/strong><span>\u00a0.\u00a0A conserva\u00e7\u00e3o da equa\u00e7\u00e3o do momento linear se torna:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-y-coordinate.png\"><img loading=\"lazy\" class=\"size-full wp-image-14365 alignleft lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-y-coordinate.png\" alt=\"Equa\u00e7\u00e3o do momento - coordenada y\" width=\"566\" height=\"383\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-y-coordinate.png\" \/><\/a><\/p>\n<p><span>A for\u00e7a final que atua na parede de um cotovelo defletor ser\u00e1:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-resulting-force.png\"><img loading=\"lazy\" class=\"size-full wp-image-14366 alignleft lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-resulting-force.png\" alt=\"Equa\u00e7\u00e3o do momento - for\u00e7a resultante\" width=\"294\" height=\"51\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-equation-resulting-force.png\" \/><\/a><\/p>\n<div class=\"su-spacer\"><\/div>\n<\/div>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2><span>Exemplo: jato de \u00e1gua atingindo uma placa estacion\u00e1ria<\/span><\/h2>\n<p><strong><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Equation-Water-Jet.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14373 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Equation-Water-Jet-300x218.png\" alt=\"Equa\u00e7\u00e3o de Momentum - Jato de \u00c1gua\" width=\"300\" height=\"218\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Momentum-Equation-Water-Jet-300x218.png\" \/><\/a><span>Uma placa estacion\u00e1ria<\/span><\/strong><span>\u00a0(por exemplo, l\u00e2mina de um moinho de \u00e1gua) \u00e9 usada para desviar o fluxo de \u00e1gua a uma velocidade de\u00a0<\/span><strong><span>1 m \/ se<\/span><\/strong><span>\u00a0a um \u00e2ngulo de\u00a0<\/span><strong><span>90 \u00b0<\/span><\/strong><span>\u00a0.\u00a0Ocorre \u00e0 press\u00e3o atmosf\u00e9rica e a vaz\u00e3o m\u00e1ssica \u00e9 igual a\u00a0<\/span><strong><span>Q = 1 m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s<\/span><\/strong><span>\u00a0.<\/span><\/p>\n<ol>\n<li><strong><span>Calcule a for\u00e7a da press\u00e3o.<\/span><\/strong><\/li>\n<li><strong><span>Calcule a for\u00e7a do corpo.<\/span><\/strong><\/li>\n<li><strong><span>Calcule a for\u00e7a total.<\/span><\/strong><\/li>\n<li><strong><span>Calcule a for\u00e7a resultante.<\/span><\/strong><\/li>\n<\/ol>\n<p><strong><span>Solu\u00e7\u00e3o<\/span><\/strong><\/p>\n<ol>\n<li><span>A\u00a0<\/span><strong><span>for\u00e7a da press\u00e3o<\/span><\/strong><span>\u00a0\u00e9 zero, pois a press\u00e3o na entrada e nas sa\u00eddas do volume de controle \u00e9 atmosf\u00e9rica.<\/span><\/li>\n<li><span>Como o volume de controle \u00e9 pequeno, podemos ignorar a\u00a0<\/span><strong><span>for\u00e7a<\/span><\/strong><span>\u00a0do\u00a0<strong>corpo<\/strong>\u00a0devido ao peso da gravidade.<\/span><\/li>\n<li><strong><span>F\u00a0<\/span><sub><span>x<\/span><\/sub><\/strong><span>\u00a0= \u03c1.Q. (w\u00a0<\/span><sub><span>1x<\/span><\/sub><span>\u00a0&#8211; w\u00a0<\/span><sub><span>2x<\/span><\/sub><span>\u00a0) = 1000.\u00a01\u00a0(1 &#8211; 0) =\u00a0<\/span><strong><span>1000 N<\/span><\/strong><br \/>\n<strong><span>F\u00a0<\/span><sub><span>y<\/span><\/sub><\/strong><span>\u00a0=\u00a0<\/span><strong><span>0<\/span><\/strong><br \/>\n<strong><span>F<\/span><\/strong><span>\u00a0=\u00a0<\/span><strong><span>(1000, 0)<\/span><\/strong><\/li>\n<li><span>A\u00a0<\/span><strong><span>for\u00e7a resultante<\/span><\/strong><span>\u00a0no plano \u00e9 da mesma magnitude, mas na dire\u00e7\u00e3o oposta \u00e0 for\u00e7a total\u00a0<\/span><strong><span>F<\/span><\/strong><span>\u00a0(atrito e peso s\u00e3o negligenciados).<\/span><\/li>\n<\/ol>\n<p><span>O jato de \u00e1gua exerce sobre a placa a for\u00e7a de 1000 N na dire\u00e7\u00e3o x.<\/span><\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div>\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.<\/p>\n<p>Este artigo \u00e9 baseado na tradu\u00e7\u00e3o autom\u00e1tica do artigo original em ingl\u00eas. Para mais informa\u00e7\u00f5es, consulte o artigo em ingl\u00eas. Voc\u00ea pode nos ajudar. Se voc\u00ea deseja corrigir a tradu\u00e7\u00e3o, envie-a para: translations@nuclear-power.com ou preencha o formul\u00e1rio de tradu\u00e7\u00e3o on-line. Agradecemos sua ajuda, atualizaremos a tradu\u00e7\u00e3o o mais r\u00e1pido poss\u00edvel. Obrigado.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>F\u00f3rmula do Momento &#8211; Equa\u00e7\u00e3o do Momento.\u00a0A segunda equa\u00e7\u00e3o de conserva\u00e7\u00e3o que devemos considerar no volume de controle \u00e9 a f\u00f3rmula do momento.\u00a0Engenharia T\u00e9rmica F\u00f3rmula do Momento &#8211; Equa\u00e7\u00e3o do Momento Assumimos que o fluido seja\u00a0est\u00e1vel\u00a0e\u00a0incompress\u00edvel\u00a0.\u00a0Para determinar a taxa de mudan\u00e7a de momento de um fluido, consideraremos um tubo de\u00a0fluxo\u00a0(\u00a0volume de controle\u00a0), como fizemos na\u00a0equa\u00e7\u00e3o &#8230; <a title=\"O que \u00e9 a f\u00f3rmula do momento &#8211; Equa\u00e7\u00e3o do momento &#8211; Defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-formula-do-momento-equacao-do-momento-definicao\/\" aria-label=\"More on O que \u00e9 a f\u00f3rmula do momento &#8211; Equa\u00e7\u00e3o do momento &#8211; Defini\u00e7\u00e3o\">Ler mais<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[14],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>O que \u00e9 a f\u00f3rmula do momento - Equa\u00e7\u00e3o do momento - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"F\u00f3rmula do Momento - Equa\u00e7\u00e3o do Momento. 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A segunda equa\\u00e7\\u00e3o de conserva\\u00e7\\u00e3o que devemos considerar no volume de controle \\u00e9 a f\\u00f3rmula do momento. 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