{"id":46969,"date":"2019-11-02T19:53:20","date_gmt":"2019-11-02T18:53:20","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/exemplo-calculo-dos-paineis-de-isolamento-a-vacuo-definicao\/"},"modified":"2020-01-22T12:45:36","modified_gmt":"2020-01-22T11:45:36","slug":"exemplo-calculo-dos-paineis-de-isolamento-a-vacuo-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/exemplo-calculo-dos-paineis-de-isolamento-a-vacuo-definicao\/","title":{"rendered":"Exemplo – C\u00e1lculo dos pain\u00e9is de isolamento a v\u00e1cuo – Defini\u00e7\u00e3o"},"content":{"rendered":"
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Exemplo – c\u00e1lculo de pain\u00e9is de informa\u00e7\u00f5es a v\u00e1cuo.\u00a0Calcule o fluxo de calor (perda de calor) atrav\u00e9s da parede isolada.\u00a0Use pain\u00e9is de isolamento a v\u00e1cuo com 10 cm de espessura.\u00a0Compare com uma parede nua.\u00a0Engenharia T\u00e9rmica<\/div>\n<\/div>\n
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Exemplo – Pain\u00e9is de isolamento a v\u00e1cuo<\/h2>\n

\"perda<\/a>Uma das principais fontes de\u00a0perda<\/strong>\u00a0de\u00a0calor<\/strong>\u00a0de uma casa \u00e9 atrav\u00e9s das paredes.\u00a0Calcule a taxa de\u00a0fluxo<\/a>\u00a0de\u00a0calor<\/a>\u00a0atrav\u00e9s de uma parede com 3 mx 10 m de \u00e1rea (A = 30 m\u00a02<\/sup>\u00a0).\u00a0A parede tem 15 cm de espessura (L\u00a01<\/sub>\u00a0) e \u00e9 feita de tijolos com\u00a0condutividade t\u00e9rmica<\/a>\u00a0de k\u00a01<\/sub>\u00a0= 1,0 W \/ mK (isolador t\u00e9rmico ruim).\u00a0Suponha que as\u00a0temperaturas<\/a>\u00a0interna e externa\u00a0sejam 22 \u00b0 C e -8 \u00b0 C, e os\u00a0coeficientes de transfer\u00eancia de calor por convec\u00e7\u00e3o<\/a>\u00a0nos lados interno e externo sejam h\u00a01<\/sub>\u00a0= 10 W \/ m\u00a02<\/sup>\u00a0K e h\u00a02<\/sub>\u00a0= 30 W \/ m\u00a02<\/sup>K, respectivamente.\u00a0Observe que esses coeficientes de convec\u00e7\u00e3o dependem muito especialmente das condi\u00e7\u00f5es ambientais e interiores (vento, umidade etc.).<\/p>\n

    \n
  1. Calcule o fluxo de calor (\u00a0perda de calor<\/strong><\/a>\u00a0) atrav\u00e9s desta parede n\u00e3o isolada.<\/li>\n
  2. Agora assuma o\u00a0isolamento t\u00e9rmico<\/strong>\u00a0no lado externo desta parede.\u00a0Use pain\u00e9is de isolamento a v\u00e1cuo de\u00a010 cm de espessura (L\u00a02<\/sub>\u00a0) com a condutividade t\u00e9rmica de k\u00a02<\/sub>\u00a0= 0,013 W \/ mK e calcule o fluxo de\u00a0calor<\/strong><\/a>\u00a0(\u00a0perda de calor<\/strong><\/a>\u00a0) atrav\u00e9s dessa parede composta.<\/li>\n<\/ol>\n

    Solu\u00e7\u00e3o:<\/strong><\/p>\n

    Como foi escrito, muitos dos processos de transfer\u00eancia de calor envolvem sistemas compostos e at\u00e9 envolvem uma combina\u00e7\u00e3o de\u00a0condu\u00e7\u00e3o<\/a>\u00a0e\u00a0convec\u00e7\u00e3o<\/a>\u00a0.\u00a0Com estes sistemas compostos, muitas vezes \u00e9 conveniente trabalhar com um\u00a0coeficiente de transfer\u00eancia total de calor<\/a>\u00a0,<\/strong>\u00a0conhecido como um\u00a0factor-L<\/strong>\u00a0.\u00a0O fator U \u00e9 definido por uma express\u00e3o an\u00e1loga \u00e0\u00a0lei do resfriamento de Newton<\/strong><\/a>\u00a0:<\/p>\n

    \"fator<\/a><\/p>\n

    O\u00a0coeficiente geral de transfer\u00eancia de calor<\/strong>\u00a0est\u00e1 relacionado \u00e0\u00a0resist\u00eancia t\u00e9rmica total<\/a>\u00a0e depende da geometria do problema.<\/p>\n

      \n
    1. parede nua<\/strong><\/li>\n<\/ol>\n

      Assumindo a transfer\u00eancia de calor unidimensional atrav\u00e9s da parede plana e desconsiderando a radia\u00e7\u00e3o, o\u00a0coeficiente geral de transfer\u00eancia de calor<\/strong>\u00a0pode ser calculado como:<\/p>\n

      \"coeficiente<\/a><\/p>\n

      O\u00a0<\/span>coeficiente geral de transfer\u00eancia de calor<\/span><\/strong>\u00a0\u00e9 ent\u00e3o:<\/span><\/p>\n

      U = 1 \/ (1\/10 + 0,15 \/ 1 + 1\/30) = 3,53 W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0K<\/span><\/p>\n

      O fluxo de calor pode ser calculado simplesmente como:<\/span><\/p>\n

      q = 3,53 [W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0K] x 30 [K] = 105,9 W \/ m\u00a0<\/span>2<\/span><\/sup><\/p>\n

      A perda total de calor atrav\u00e9s desta parede ser\u00e1:<\/span><\/p>\n

      q\u00a0<\/span>perda<\/span><\/sub>\u00a0= q.\u00a0A = 105,9 [W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0] x 30 [m\u00a0<\/span>2<\/span><\/sup>\u00a0] = 3177W<\/span><\/p>\n

        \n
      1. parede comp\u00f3sita com isolamento t\u00e9rmico<\/span><\/strong><\/li>\n<\/ol>\n

        Assumindo a transfer\u00eancia de calor unidimensional atrav\u00e9s da parede composta plana, sem resist\u00eancia ao contato t\u00e9rmico e sem considerar a radia\u00e7\u00e3o, o\u00a0<\/span>coeficiente geral de transfer\u00eancia de calor<\/span><\/strong>\u00a0pode ser calculado como:<\/span><\/p>\n

        \"coeficiente<\/a><\/p>\n

        \"pain\u00e9is<\/a>O\u00a0<\/span>coeficiente geral de transfer\u00eancia de calor<\/span><\/strong>\u00a0\u00e9 ent\u00e3o:<\/span><\/p>\n

        U = 1 \/ (1\/10 + 0,15 \/ 1 + 0,1 \/ 0,013 + 1\/30) = 0,125 W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0K<\/span><\/p>\n

        O fluxo de calor pode ser calculado simplesmente como:<\/span><\/p>\n

        q = 0,125 [W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0K] x 30 [K] = 3,76 W \/ m\u00a0<\/span>2<\/span><\/sup><\/p>\n

        A perda total de calor atrav\u00e9s desta parede ser\u00e1:<\/span><\/p>\n

        q\u00a0<\/span>perda<\/span><\/sub>\u00a0= q.\u00a0A = 3,76 [W \/ m\u00a0<\/span>2<\/span><\/sup>\u00a0] x 30 [m\u00a0<\/span>2<\/span><\/sup>\u00a0] = 113 W<\/span><\/p>\n

        Como pode ser visto, uma adi\u00e7\u00e3o de isolador t\u00e9rmico causa uma diminui\u00e7\u00e3o significativa nas perdas de calor.\u00a0Deve ser adicionado, uma adi\u00e7\u00e3o da pr\u00f3xima camada de isolador t\u00e9rmico n\u00e3o causa economias t\u00e3o altas.\u00a0Isso pode ser visto melhor no m\u00e9todo de resist\u00eancia t\u00e9rmica, que pode ser usado para calcular a transfer\u00eancia de calor atrav\u00e9s de\u00a0<\/span>paredes compostas<\/span><\/strong>\u00a0.\u00a0A taxa de transfer\u00eancia constante de calor entre duas superf\u00edcies \u00e9 igual \u00e0 diferen\u00e7a de temperatura dividida pela\u00a0<\/span>resist\u00eancia t\u00e9rmica<\/span><\/a>\u00a0total\u00a0entre essas duas superf\u00edcies.<\/span><\/p>\n

        \"resist\u00eancia<\/a><\/p>\n<\/div>\n<\/div>\n

        ……………………………………………………………………………………………………………………………….<\/p>\n

        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","protected":false},"excerpt":{"rendered":"

        Exemplo – c\u00e1lculo de pain\u00e9is de informa\u00e7\u00f5es a v\u00e1cuo.\u00a0Calcule o fluxo de calor (perda de calor) atrav\u00e9s da parede isolada.\u00a0Use pain\u00e9is de isolamento a v\u00e1cuo com 10 cm de espessura.\u00a0Compare com uma parede nua.\u00a0Engenharia T\u00e9rmica Exemplo – Pain\u00e9is de isolamento a v\u00e1cuo Uma das principais fontes de\u00a0perda\u00a0de\u00a0calor\u00a0de uma casa \u00e9 atrav\u00e9s das paredes.\u00a0Calcule a taxa … 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":"\nExemplo - C\u00e1lculo dos pain\u00e9is de isolamento a v\u00e1cuo - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Exemplo - c\u00e1lculo de pain\u00e9is de isolamento a v\u00e1cuo. Calcule o fluxo de calor (perda de calor) atrav\u00e9s da parede isolada. 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