{"id":51140,"date":"2020-01-24T12:27:49","date_gmt":"2020-01-24T11:27:49","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/exemplo-de-equacao-de-calor-problema-com-solucao-definicao\/"},"modified":"2020-01-24T12:29:22","modified_gmt":"2020-01-24T11:29:22","slug":"exemplo-de-equacao-de-calor-problema-com-solucao-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/exemplo-de-equacao-de-calor-problema-com-solucao-definicao\/","title":{"rendered":"Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o &#8211; Defini\u00e7\u00e3o"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com a solu\u00e7\u00e3o.\u00a0Neste artigo, existem dois exemplos de solu\u00e7\u00e3o da equa\u00e7\u00e3o do calor.\u00a0Ambos os exemplos est\u00e3o com solu\u00e7\u00e3o.\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>Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o<\/h2>\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>Condu\u00e7\u00e3o de calor em uma grande parede plana<\/h2>\n<p><strong>Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o<\/strong><\/p>\n<p>Considere a parede plana de espessura 2L, na qual h\u00e1\u00a0<strong>gera\u00e7\u00e3o uniforme e constante de calor<\/strong>\u00a0por unidade de volume,\u00a0<strong>q\u00a0<sub>V<\/sub>\u00a0[W \/ m\u00a0<sup>3<\/sup>\u00a0]<\/strong>\u00a0.\u00a0O plano central \u00e9 tomado como a origem de x e a laje se estende a + L \u00e0 direita e &#8211; L \u00e0 esquerda.\u00a0Para\u00a0<a title=\"Condutividade t\u00e9rmica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-condutividade-termica-definicao\/\">condutividade t\u00e9rmica<\/a>\u00a0constante\u00a0k, a forma apropriada da equa\u00e7\u00e3o do calor \u00e9:<\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-heat-generation-equation.png\"><img loading=\"lazy\" class=\"size-medium wp-image-20229 aligncenter lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-heat-generation-equation-300x96.png\" alt=\"equa\u00e7\u00e3o do calor - gera\u00e7\u00e3o de calor - equa\u00e7\u00e3o\" width=\"300\" height=\"96\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-heat-generation-equation-300x96.png\" \/><\/a><\/p>\n<p>A solu\u00e7\u00e3o geral desta equa\u00e7\u00e3o \u00e9:<\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-general-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20230 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-general-solution.png\" alt=\"equa\u00e7\u00e3o de condu\u00e7\u00e3o de calor - solu\u00e7\u00e3o geral\" width=\"290\" height=\"150\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-general-solution.png\" \/><\/a><\/p>\n<p>onde C\u00a0<sub>1<\/sub>\u00a0e C\u00a0<sub>2<\/sub>\u00a0s\u00e3o as constantes de integra\u00e7\u00e3o.<\/p>\n<p>1)<\/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<p><span><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Heat-Conduction-in-a-Large-Plane-Wall-figure.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-20228 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Heat-Conduction-in-a-Large-Plane-Wall-figure-300x276.png\" alt=\"Condu\u00e7\u00e3o de calor em uma grande parede plana\" width=\"300\" height=\"276\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Heat-Conduction-in-a-Large-Plane-Wall-figure-300x276.png\" \/><\/a>Calcule a distribui\u00e7\u00e3o de temperatura, T (x), atrav\u00e9s dessa parede plana e espessa, se:<\/span><\/p>\n<ul>\n<li><span>as temperaturas em ambas as superf\u00edcies s\u00e3o 15,0 \u00b0 C<\/span><\/li>\n<li><span>a espessura desta parede \u00e9 2L = 10 mm.<\/span><\/li>\n<li><span>a condutividade dos materiais \u00e9 k = 2,8 W \/ mK (corresponde ao di\u00f3xido de ur\u00e2nio a 1000 \u00b0 C)<\/span><\/li>\n<li><span>a taxa de calor volum\u00e9trico \u00e9 q\u00a0<\/span><sub><span>V<\/span><\/sub><span>\u00a0= 10\u00a0<\/span><sup><span>6<\/span><\/sup><span>\u00a0W \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><\/li>\n<\/ul>\n<p><span>Nesse caso, as superf\u00edcies s\u00e3o mantidas em determinadas temperaturas T\u00a0<\/span><sub><span>s, 1<\/span><\/sub><span>\u00a0e T\u00a0<\/span><sub><span>s, 2<\/span><\/sub><span>\u00a0.\u00a0Isso corresponde \u00e0\u00a0<\/span><a title=\"Condi\u00e7\u00e3o de limite do Dirichlet - Condi\u00e7\u00e3o de limite do tipo I\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/thermal-conduction\/heat-conduction-equation\/dirichlet-boundary-condition-type-i-boundary-condition\/\"><span>condi\u00e7\u00e3o de limite de Dirichlet<\/span><\/a><span>\u00a0.\u00a0Al\u00e9m disso, esse problema \u00e9 termicamente sim\u00e9trico e, portanto, tamb\u00e9m podemos usar\u00a0<\/span><a title=\"Limite Adiab\u00e1tico - Simetria T\u00e9rmica\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/thermal-conduction\/heat-conduction-equation\/adiabatic-boundary-thermal-symmetry\/\"><span>a condi\u00e7\u00e3o de contorno de simetria t\u00e9rmica<\/span><\/a><span>\u00a0.\u00a0As constantes podem ser avaliadas usando a substitui\u00e7\u00e3o na solu\u00e7\u00e3o geral e t\u00eam a forma:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-boundary-conditions.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20231 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-boundary-conditions.png\" alt=\"equa\u00e7\u00e3o de condu\u00e7\u00e3o de calor - condi\u00e7\u00f5es de contorno\" width=\"428\" height=\"110\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-boundary-conditions.png\" \/><\/a><\/p>\n<p><span>A distribui\u00e7\u00e3o de temperatura resultante e a temperatura da linha central (x = 0) (m\u00e1xima) nesta parede plana nessas condi\u00e7\u00f5es de contorno espec\u00edficas ser\u00e3o:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20232 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-solution.png\" alt=\"equa\u00e7\u00e3o de condu\u00e7\u00e3o de calor - solu\u00e7\u00e3o\" width=\"608\" height=\"210\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-conduction-equation-solution.png\" \/><\/a><\/p>\n<p><span>O\u00a0<\/span><a title=\"Densidade do fluxo de calor - Fluxo t\u00e9rmico\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/introduction-to-heat-transfer\/heat-flux-density-thermal-flux\/\"><span>fluxo de calor<\/span><\/a><span>\u00a0em qualquer ponto, q\u00a0<\/span><sub><span>x<\/span><\/sub><span>\u00a0[Wm\u00a0<\/span><sup><span>-2<\/span><\/sup><span>\u00a0], na parede pode, \u00e9 claro, ser determinado usando a distribui\u00e7\u00e3o de temperatura e com a\u00a0<\/span><a title=\"Lei de Fourier de condu\u00e7\u00e3o t\u00e9rmica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-lei-de-conducao-termica-de-fourier-definicao\/\"><strong><span>lei de Fourier<\/span><\/strong><\/a><span>\u00a0.\u00a0Observe que, com a gera\u00e7\u00e3o de calor, o fluxo de calor n\u00e3o \u00e9 mais independente de x, portanto:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-condution-through-wall.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20233 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-condution-through-wall.png\" alt=\"condu\u00e7\u00e3o de calor atrav\u00e9s da parede\" width=\"315\" height=\"178\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-condution-through-wall.png\" \/><\/a><\/p>\n<p><strong><span>Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o<\/span><\/strong><\/p>\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>Condu\u00e7\u00e3o de calor em uma barra de combust\u00edvel<\/span><\/h2>\n<p><span>A maioria dos PWRs usa o\u00a0<\/span><strong><span>combust\u00edvel de ur\u00e2nio<\/span><\/strong><span>\u00a0, que est\u00e1 na forma de\u00a0<\/span><strong><span>di\u00f3xido<\/span><\/strong><span>\u00a0de\u00a0<strong>ur\u00e2nio<\/strong>\u00a0.\u00a0O di\u00f3xido de ur\u00e2nio \u00e9 um s\u00f3lido semicondutor preto com condutividade t\u00e9rmica muito baixa.\u00a0Por outro lado, o di\u00f3xido de ur\u00e2nio tem um ponto de fus\u00e3o muito alto e um comportamento bem conhecido.\u00a0A UO\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0\u00e9 prensada em\u00a0<\/span><strong><span>pastilhas cil\u00edndricas<\/span><\/strong><span>\u00a0, essas pastilhas s\u00e3o ent\u00e3o sinterizadas no s\u00f3lido.<\/span><\/p>\n<p><span>Esses\u00a0<\/span><strong><span>pellets cil\u00edndricos<\/span><\/strong><span>\u00a0s\u00e3o ent\u00e3o carregados e encapsulados dentro de uma barra de combust\u00edvel (ou pino de combust\u00edvel), feita de ligas de zirc\u00f4nio devido \u00e0 sua\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/neutron-cross-section\/\"><span>se\u00e7\u00e3o transversal de<\/span><\/a><span>\u00a0absor\u00e7\u00e3o muito baixa\u00a0(ao contr\u00e1rio do a\u00e7o inoxid\u00e1vel).\u00a0A superf\u00edcie do tubo, que cobre os pellets, \u00e9 chamada de\u00a0<\/span><strong><span>revestimento de combust\u00edvel<\/span><\/strong><span>\u00a0.<\/span><\/p>\n<p><span>Veja tamb\u00e9m:\u00a0\u00a0<\/span><a title=\"Condutividade t\u00e9rmica do di\u00f3xido de ur\u00e2nio\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/thermal-conduction\/thermal-conductivity\/thermal-conductivity-of-uranium-dioxide\/\"><span>Condu\u00e7\u00e3o t\u00e9rmica de di\u00f3xido de ur\u00e2nio<\/span><\/a><\/p>\n<p><strong><span>O<\/span><\/strong><span>\u00a0comportamento\u00a0<strong>t\u00e9rmico<\/strong>\u00a0e\u00a0<\/span><strong><span>mec\u00e2nico<\/span><\/strong><span>\u00a0dos\u00a0<\/span><strong><span>pellets<\/span><\/strong><span>\u00a0\u00a0e\u00a0<\/span><strong><span>barras de combust\u00edvel<\/span><\/strong><span>\u00a0constitui uma das tr\u00eas principais disciplinas do projeto.\u00a0<\/span><strong><span>O combust\u00edvel nuclear<\/span><\/strong><span>\u00a0\u00e9 operado sob condi\u00e7\u00f5es muito in\u00f3spitas (t\u00e9rmica, radia\u00e7\u00e3o, mec\u00e2nica) e deve suportar mais do que as condi\u00e7\u00f5es normais de opera\u00e7\u00e3o.\u00a0Por exemplo, as temperaturas no centro dos pellets de combust\u00edvel atingem mais de\u00a0<\/span><strong><span>1000 \u00b0 C<\/span><\/strong><span>\u00a0(1832 \u00b0 F) acompanhadas por libera\u00e7\u00f5es de g\u00e1s de fiss\u00e3o.\u00a0Portanto, o conhecimento detalhado da distribui\u00e7\u00e3o de temperatura em uma \u00fanica barra de combust\u00edvel \u00e9 essencial para a opera\u00e7\u00e3o segura do combust\u00edvel nuclear.\u00a0Nesta se\u00e7\u00e3o, estudaremos\u00a0<\/span><strong><span>a equa\u00e7\u00e3o de condu\u00e7\u00e3o de calor<\/span><\/strong><span>\u00a0em\u00a0<\/span><strong><span>coordenadas cil\u00edndricas<\/span><\/strong><span>usando a condi\u00e7\u00e3o limite de Dirichlet com determinada temperatura da superf\u00edcie (ou seja, usando a condi\u00e7\u00e3o limite de Dirichlet).\u00a0A an\u00e1lise abrangente do perfil de temperatura da barra de combust\u00edvel ser\u00e1 estudada em se\u00e7\u00e3o separada.<\/span><\/p>\n<h2><strong><span>Temperatura na linha central de um pellet de combust\u00edvel<\/span><\/strong><\/h2>\n<p><span>Considere o pellet de combust\u00edvel com raio\u00a0<\/span><strong><span>r\u00a0<\/span><sub><span>U<\/span><\/sub><span>\u00a0= 0,40 cm<\/span><\/strong><span>\u00a0, no qual h\u00e1 gera\u00e7\u00e3o uniforme e constante de calor por unidade de volume,\u00a0<\/span><strong><span>q\u00a0<\/span><sub><span>V<\/span><\/sub><span>\u00a0[W \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0]<\/span><\/strong><span>\u00a0.\u00a0Em vez da taxa de calor volum\u00e9trica q\u00a0<\/span><sub><span>V<\/span><\/sub><span>\u00a0[W \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0], os engenheiros costumam usar a\u00a0<\/span><strong><span>taxa de calor linear, q\u00a0<\/span><sub><span>L<\/span><\/sub><span>\u00a0[W \/ m]<\/span><\/strong><span>\u00a0, que representa a taxa de calor de um metro da barra de combust\u00edvel.\u00a0A\u00a0<\/span><strong><span>taxa linear de calor<\/span><\/strong><span>\u00a0pode ser calculada a partir da taxa volum\u00e9trica de calor por:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/linear-heat-rate-vs-volumetric-heat-rate.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20236 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/linear-heat-rate-vs-volumetric-heat-rate.png\" alt=\"taxa de calor linear vs taxa de calor volum\u00e9trico\" width=\"125\" height=\"49\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/linear-heat-rate-vs-volumetric-heat-rate.png\" \/><\/a><\/p>\n<p><span>A linha central \u00e9 tomada como a origem da coordenada r.\u00a0Devido \u00e0 simetria na dire\u00e7\u00e3o z e na dire\u00e7\u00e3o azimutal, podemos separar as vari\u00e1veis \u200b\u200be simplificar esse problema para\u00a0<\/span><strong><span>um problema unidimensional<\/span><\/strong><span>\u00a0.\u00a0Assim, resolveremos apenas a temperatura em fun\u00e7\u00e3o do raio\u00a0<\/span><strong><span>T (r)<\/span><\/strong><span>\u00a0.\u00a0Para\u00a0<\/span><a title=\"Condutividade t\u00e9rmica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-condutividade-termica-definicao\/\"><span>condutividade t\u00e9rmica<\/span><\/a><span>\u00a0constante\u00a0, k, a forma apropriada da\u00a0<\/span><strong><span>equa\u00e7\u00e3o de calor cil\u00edndrica<\/span><\/strong><span>\u00a0\u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20237 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-2.png\" alt=\"equa\u00e7\u00e3o do calor - cil\u00edndrica - 2\" width=\"503\" height=\"178\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-2.png\" \/><\/a><\/p>\n<p><span>A solu\u00e7\u00e3o geral desta equa\u00e7\u00e3o \u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-general-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20238 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-general-solution.png\" alt=\"equa\u00e7\u00e3o do calor - cil\u00edndrica - solu\u00e7\u00e3o geral\" width=\"323\" height=\"149\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-general-solution.png\" \/><\/a><\/p>\n<p><span>onde C\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0e C\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0s\u00e3o as constantes de integra\u00e7\u00e3o.<\/span><\/p>\n<p><strong><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-20223 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet-244x300.png\" alt=\"Condu\u00e7\u00e3o t\u00e9rmica - pellet de combust\u00edvel\" width=\"244\" height=\"300\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet-244x300.png\" \/><\/a><span>Calcule a distribui\u00e7\u00e3o de temperatura, T (r)<\/span><\/strong><span>\u00a0, neste sedimento de combust\u00edvel, se:<\/span><\/p>\n<ul>\n<li><span>as temperaturas na superf\u00edcie do sedimento de combust\u00edvel s\u00e3o\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>U<\/span><\/sub><span>\u00a0= 420 \u00b0 C<\/span><\/strong><\/li>\n<li><span>raio da pastilha de combust\u00edvel\u00a0<\/span><strong><span>r\u00a0<\/span><sub><span>U<\/span><\/sub><span>\u00a0= 4 mm<\/span><\/strong><span>\u00a0.<\/span><\/li>\n<li><span>a condutividade m\u00e9dia do material \u00e9\u00a0<\/span><strong><span>k = 2,8 W \/ mK<\/span><\/strong><span>\u00a0(corresponde ao di\u00f3xido de ur\u00e2nio a 1000 \u00b0 C)<\/span><\/li>\n<li><span>a taxa de calor linear \u00e9\u00a0<\/span><strong><span>q\u00a0<\/span><sub><span>L<\/span><\/sub><span>\u00a0= 300 W \/ cm<\/span><\/strong><span>\u00a0e, portanto, a taxa de calor volum\u00e9trica \u00e9 q\u00a0<\/span><sub><span>V<\/span><\/sub><span>\u00a0= 597 x 10\u00a0<\/span><sup><span>6<\/span><\/sup><span>\u00a0W \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><\/li>\n<\/ul>\n<p><span>Neste caso, a superf\u00edcie \u00e9 mantida a temperaturas dadas T\u00a0<\/span><sub><span>U<\/span><\/sub><span>\u00a0.\u00a0Isso corresponde \u00e0\u00a0<\/span><a title=\"Condi\u00e7\u00e3o de limite do Dirichlet - Condi\u00e7\u00e3o de limite do tipo I\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/thermal-conduction\/heat-conduction-equation\/dirichlet-boundary-condition-type-i-boundary-condition\/\"><span>condi\u00e7\u00e3o de limite de Dirichlet<\/span><\/a><span>\u00a0.\u00a0Al\u00e9m disso, esse problema \u00e9 termicamente sim\u00e9trico e, portanto, tamb\u00e9m podemos usar\u00a0<\/span><a title=\"Limite Adiab\u00e1tico - Simetria T\u00e9rmica\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/heat-transfer\/thermal-conduction\/heat-conduction-equation\/adiabatic-boundary-thermal-symmetry\/\"><span>a condi\u00e7\u00e3o de contorno de simetria t\u00e9rmica<\/span><\/a><span>\u00a0.\u00a0As constantes podem ser avaliadas usando a substitui\u00e7\u00e3o na solu\u00e7\u00e3o geral e t\u00eam a forma:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-boundary-conditions.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20239 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-boundary-conditions.png\" alt=\"equa\u00e7\u00e3o do calor - cil\u00edndrica - condi\u00e7\u00f5es de contorno\" width=\"337\" height=\"140\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-boundary-conditions.png\" \/><\/a><\/p>\n<p><span>A distribui\u00e7\u00e3o de temperatura resultante e a temperatura da linha central (r = 0) (m\u00e1xima) neste sedimento cil\u00edndrico de combust\u00edvel nessas condi\u00e7\u00f5es de contorno espec\u00edficas ser\u00e3o:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20240 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-solution.png\" alt=\"equa\u00e7\u00e3o do calor - cil\u00edndrica - solu\u00e7\u00e3o\" width=\"322\" height=\"162\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cylindrical-solution.png\" \/><\/a><\/p>\n<p><span>O\u00a0<\/span><strong><span>fluxo de calor radial<\/span><\/strong><span>\u00a0em qualquer raio, q\u00a0<\/span><sub><span>r<\/span><\/sub><span>\u00a0[Wm\u00a0<\/span><sup><span>-1<\/span><\/sup><span>\u00a0], no cilindro pode, \u00e9 claro, ser determinado usando a distribui\u00e7\u00e3o de temperatura e com a\u00a0<\/span><a title=\"Lei de Fourier de condu\u00e7\u00e3o t\u00e9rmica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-lei-de-conducao-termica-de-fourier-definicao\/\"><span>lei de Fourier<\/span><\/a><span>\u00a0.\u00a0Observe que, com a gera\u00e7\u00e3o de calor, o fluxo de calor n\u00e3o \u00e9 mais independente de r.<\/span><\/p>\n<p><span>A figura a seguir mostra a distribui\u00e7\u00e3o de temperatura no pellet de combust\u00edvel em v\u00e1rios n\u00edveis de pot\u00eancia.<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Temperature-distribution-nuclear-fuel.png\"><img loading=\"lazy\" class=\"aligncenter size-large wp-image-20224 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Temperature-distribution-nuclear-fuel-1024x465.png\" alt=\"Distribui\u00e7\u00e3o de temperatura - combust\u00edvel nuclear\" width=\"669\" height=\"304\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Temperature-distribution-nuclear-fuel-1024x465.png\" \/><\/a><\/p>\n<p><span>______<\/span><\/p>\n<p><span>A temperatura em um reator operacional varia de ponto a ponto dentro do sistema.\u00a0Como consequ\u00eancia, h\u00e1 sempre\u00a0<\/span><strong><span>uma vara de combust\u00edvel<\/span><\/strong><span>\u00a0e\u00a0<\/span><strong><span>um volume local<\/span><\/strong><span>\u00a0, que est\u00e1\u00a0<\/span><strong><span>mais quente\u00a0<\/span><\/strong><span>\u00a0do que todo o resto.\u00a0Para limitar esses\u00a0<\/span><strong><span>locais quentes,<\/span><\/strong><span>\u00a0os\u00a0<\/span><a title=\"Fatores de canal quente - fatores de pico\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power\/reactor-physics\/reactor-operation\/normal-operation-reactor-control\/hot-channel-factors-peaking-factors\/\"><strong><span>limites de pot\u00eancia m\u00e1xima<\/span><\/strong><\/a><span>\u00a0devem ser introduzidos.\u00a0Os limites de pot\u00eancia m\u00e1xima est\u00e3o associados a uma\u00a0<\/span><strong><span>crise de ebuli\u00e7\u00e3o<\/span><\/strong><span>\u00a0e \u00e0s condi\u00e7\u00f5es que podem causar o derretimento do pellet de combust\u00edvel.\u00a0No entanto, considera\u00e7\u00f5es metal\u00fargicas\u00a0<\/span><strong><span>imp\u00f5em limites superiores<\/span><\/strong><span>\u00a0\u00e0 temperatura do revestimento do combust\u00edvel e do sedimento de combust\u00edvel.\u00a0Acima dessas\u00a0<\/span><a title=\"O que \u00e9 temperatura - F\u00edsica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-temperatura-fisica-definicao\/\"><span>temperaturas<\/span><\/a><span>existe o perigo de o combust\u00edvel ser danificado.\u00a0Um dos principais objetivos no projeto de reatores nucleares \u00e9 fornecer a remo\u00e7\u00e3o do calor produzido no n\u00edvel de pot\u00eancia desejado, garantindo que a temperatura m\u00e1xima do combust\u00edvel e a temperatura m\u00e1xima do revestimento estejam sempre abaixo desses valores predeterminados.<\/span><\/p>\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>Distribui\u00e7\u00e3o de temperatura no revestimento de combust\u00edvel<\/span><\/h2>\n<p><strong><span>O<\/span><\/strong><span>\u00a0\u00a0revestimento \u00e9 a camada externa das barras de combust\u00edvel, situada entre o\u00a0<\/span><strong><span>l\u00edquido de arrefecimento<\/span><\/strong><span>\u00a0do\u00a0\u00a0<strong>reator<\/strong>\u00a0\u00a0e o\u00a0\u00a0<\/span><strong><span>combust\u00edvel nuclear<\/span><\/strong><span>\u00a0\u00a0(isto \u00e9,\u00a0\u00a0<\/span><strong><span>granulados de combust\u00edvel<\/span><\/strong><span>\u00a0).\u00a0\u00c9 feito de um material resistente \u00e0 corros\u00e3o com se\u00e7\u00e3o transversal de baixa absor\u00e7\u00e3o para n\u00eautrons t\u00e9rmicos, geralmente\u00a0\u00a0<\/span><strong><span>liga de zirc\u00f4nio<\/span><\/strong><span>\u00a0.\u00a0<\/span><strong><span>O revestimento<\/span><\/strong><span>\u00a0\u00a0evita que produtos de fiss\u00e3o radioativa escapem da matriz de combust\u00edvel para o l\u00edquido de arrefecimento do reator e os contaminem.\u00a0O revestimento constitui uma das barreiras na\u00a0abordagem de\u00a0&#8216;\u00a0<\/span><strong><span>defesa em profundidade<\/span><\/strong><span>\u00a0&#8216;.<\/span><\/p>\n<p><span>Considere o revestimento de combust\u00edvel do raio interno\u00a0\u00a0<\/span><strong><span>r\u00a0<\/span><sub><span>Zr, 2<\/span><\/sub><span>\u00a0\u00a0= 0,408 cm<\/span><\/strong><span>\u00a0\u00a0e raio externo\u00a0\u00a0<\/span><strong><span>r\u00a0<\/span><sub><span>Zr, 1<\/span><\/sub><span>\u00a0\u00a0= 0,465 cm<\/span><\/strong><span>\u00a0.\u00a0Em compara\u00e7\u00e3o com o pellet de combust\u00edvel, quase n\u00e3o h\u00e1 gera\u00e7\u00e3o de calor no revestimento do combust\u00edvel (o revestimento \u00e9\u00a0\u00a0<\/span><a title=\"Libera\u00e7\u00e3o de energia da fiss\u00e3o\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power\/fission\/energy-release-from-fission\/\"><span>levemente aquecido pela radia\u00e7\u00e3o<\/span><\/a><span>\u00a0).\u00a0Todo o calor gerado no combust\u00edvel deve ser transferido por\u00a0\u00a0<\/span><strong><span>condu\u00e7\u00e3o<\/span><\/strong><span>\u00a0\u00a0atrav\u00e9s do revestimento e, portanto, a superf\u00edcie interna \u00e9 mais quente que a superf\u00edcie externa.<\/span><\/p>\n<p><span>Para encontrar a distribui\u00e7\u00e3o de temperatura atrav\u00e9s do revestimento, precisamos resolver a\u00a0\u00a0<\/span><strong><span>equa\u00e7\u00e3o de condu\u00e7\u00e3o de calor<\/span><\/strong><span>\u00a0.\u00a0Devido \u00e0 simetria na dire\u00e7\u00e3o z e na dire\u00e7\u00e3o azimutal, podemos separar as vari\u00e1veis \u200b\u200be simplificar esse problema para um problema unidimensional.\u00a0Assim, resolveremos apenas a temperatura em fun\u00e7\u00e3o do raio T (r).\u00a0Neste exemplo, assumiremos que n\u00e3o h\u00e1 estritamente nenhuma gera\u00e7\u00e3o de calor dentro do revestimento.\u00a0Para condutividade t\u00e9rmica constante, k, a forma apropriada da equa\u00e7\u00e3o de calor cil\u00edndrica \u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20241 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding.png\" alt=\"equa\u00e7\u00e3o do calor - revestimento\" width=\"190\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding.png\" \/><\/a><\/p>\n<p><span>A solu\u00e7\u00e3o geral desta equa\u00e7\u00e3o \u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-general-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20242 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-general-solution.png\" alt=\"equa\u00e7\u00e3o do calor - revestimento - solu\u00e7\u00e3o geral\" width=\"245\" height=\"133\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-general-solution.png\" \/><\/a><\/p>\n<p><span>onde C\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0\u00a0e C\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0\u00a0s\u00e3o as constantes de integra\u00e7\u00e3o.<\/span><\/p>\n<p><span>1)<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-20223 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet-244x300.png\" alt=\"Condu\u00e7\u00e3o t\u00e9rmica - pellet de combust\u00edvel\" width=\"244\" height=\"300\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Thermal-conduction-fuel-pellet-244x300.png\" \/><\/a><span>Calcule a distribui\u00e7\u00e3o de temperatura, T (r), neste revestimento de combust\u00edvel, se:<\/span><\/p>\n<ul>\n<li><span>a temperatura na superf\u00edcie interna do revestimento \u00e9 T\u00a0<\/span><sub><span>Zr, 2<\/span><\/sub><span>\u00a0\u00a0= 360 \u00b0 C<\/span><\/li>\n<li><span>a temperatura do l\u00edquido de arrefecimento do reator nesta coordenada axial \u00e9 T a\u00a0<\/span><sub><span>granel<\/span><\/sub><span>\u00a0\u00a0= 300 \u00b0 C<\/span><\/li>\n<li><span>o coeficiente de transfer\u00eancia de calor (convec\u00e7\u00e3o; fluxo turbulento) \u00e9 h = 41 kW \/ m\u00a0<\/span><sup><span>2<\/span><\/sup><span>\u00a0.K.<\/span><\/li>\n<li><span>a condutividade m\u00e9dia do material \u00e9 k = 18 W \/ mK<\/span><\/li>\n<li><span>a taxa linear de calor do combust\u00edvel \u00e9 q\u00a0<\/span><sub><span>L<\/span><\/sub><span>\u00a0\u00a0= 300 W \/ cm e, portanto, a taxa volum\u00e9trica de calor \u00e9 q\u00a0<\/span><sub><span>V<\/span><\/sub><span>\u00a0\u00a0= 597 x 10\u00a0<\/span><sup><span>6<\/span><\/sup><span>\u00a0\u00a0W \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><\/li>\n<\/ul>\n<p><span>A partir da rela\u00e7\u00e3o b\u00e1sica para transfer\u00eancia de calor por convec\u00e7\u00e3o, podemos calcular a superf\u00edcie externa do revestimento como:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/newtons-law-cladding.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20269 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/newtons-law-cladding.png\" alt=\"newtons law - revestimento\" width=\"423\" height=\"299\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/newtons-law-cladding.png\" \/><\/a><\/p>\n<p><span>Como pode ser visto, tamb\u00e9m neste caso, fornecemos temperaturas de superf\u00edcie T\u00a0<\/span><sub><span>Zr, 1<\/span><\/sub><span>\u00a0\u00a0e T\u00a0<\/span><sub><span>Zr, 2<\/span><\/sub><span>\u00a0.\u00a0Isso corresponde \u00e0 condi\u00e7\u00e3o de limite de Dirichlet.\u00a0As constantes podem ser avaliadas usando a substitui\u00e7\u00e3o na solu\u00e7\u00e3o geral e t\u00eam a forma:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-general-solution-cladding.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20264 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-general-solution-cladding.png\" alt=\"equa\u00e7\u00e3o do calor - solu\u00e7\u00e3o geral - revestimento\" width=\"352\" height=\"53\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-general-solution-cladding.png\" \/><\/a><\/p>\n<p><span>Resolvendo C\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0\u00a0e C\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0\u00a0e substituindo na solu\u00e7\u00e3o geral, obtemos ent\u00e3o:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-solution.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20265 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-solution.png\" alt=\"equa\u00e7\u00e3o do calor - revestimento - solu\u00e7\u00e3o\" width=\"381\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/heat-equation-cladding-solution.png\" \/><\/a><\/p>\n<p><strong><span>\u2206T &#8211; superf\u00edcie de revestimento &#8211; refrigerante<\/span><\/strong><\/p>\n<p><span>O conhecimento detalhado da geometria, raio externo do revestimento, taxa de calor linear, coeficiente de transfer\u00eancia de calor por convec\u00e7\u00e3o e temperatura do l\u00edquido de refrigera\u00e7\u00e3o determina\u00a0\u00a0<\/span><strong><span>\u2206T\u00a0<\/span><\/strong><span>\u00a0entre o l\u00edquido de arrefecimento (T a\u00a0<\/span><sub><span>granel<\/span><\/sub><span>\u00a0) e a superf\u00edcie do revestimento (T\u00a0<\/span><sub><span>Zr, 1<\/span><\/sub><span>\u00a0).\u00a0Portanto, podemos calcular a temperatura da superf\u00edcie do revestimento (T\u00a0<\/span><sub><span>Zr, 1<\/span><\/sub><span>\u00a0) simplesmente usando a\u00a0\u00a0<\/span><strong><span>Lei de Newton<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/%E2%88%86T-cladding-surface-coolant-newtons-law.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20266 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/%E2%88%86T-cladding-surface-coolant-newtons-law.png\" alt=\"\u2206T - superf\u00edcie de revestimento - refrigerante - newtons law\" width=\"425\" height=\"306\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/\u2206T-cladding-surface-coolant-newtons-law.png\" \/><\/a><\/p>\n<p><strong><span>\u2206T no revestimento do combust\u00edvel<\/span><\/strong><\/p>\n<p><span>O conhecimento detalhado da geometria, raio externo e interno do revestimento, taxa de calor linear e temperatura da superf\u00edcie do revestimento (T\u00a0<\/span><sub><span>Zr, 1<\/span><\/sub><span>\u00a0) determina\u00a0\u00a0<\/span><strong><span>\u2206T\u00a0<\/span><\/strong><span>\u00a0entre as superf\u00edcies externa e interna do revestimento.\u00a0Portanto, podemos calcular a temperatura da superf\u00edcie interna do revestimento (T\u00a0<\/span><sub><span>Zr, 2<\/span><\/sub><span>\u00a0) simplesmente usando a\u00a0\u00a0<\/span><strong><span>Lei de Fourier<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/%E2%88%86T-in-fuel-cladding-fouriers-law.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20267 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/%E2%88%86T-in-fuel-cladding-fouriers-law.png\" alt=\"\u2206T no revestimento de combust\u00edvel - lei de fouriers\" width=\"370\" height=\"460\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/\u2206T-in-fuel-cladding-fouriers-law.png\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\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>Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com a solu\u00e7\u00e3o.\u00a0Neste artigo, existem dois exemplos de solu\u00e7\u00e3o da equa\u00e7\u00e3o do calor.\u00a0Ambos os exemplos est\u00e3o com solu\u00e7\u00e3o.\u00a0Engenharia T\u00e9rmica Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o Condu\u00e7\u00e3o de calor em uma grande parede plana Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o Considere a parede &#8230; <a title=\"Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o &#8211; Defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/exemplo-de-equacao-de-calor-problema-com-solucao-definicao\/\" aria-label=\"More on Exemplo de equa\u00e7\u00e3o de calor &#8211; Problema com solu\u00e7\u00e3o &#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>Exemplo de equa\u00e7\u00e3o de calor - Problema com solu\u00e7\u00e3o - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Exemplo de equa\u00e7\u00e3o de calor - Problema com a solu\u00e7\u00e3o. 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