{"id":47963,"date":"2019-11-07T13:01:51","date_gmt":"2019-11-07T12:01:51","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-eficiencia-termica-brayton-cycle-definition\/"},"modified":"2020-01-26T13:18:30","modified_gmt":"2020-01-26T12:18:30","slug":"o-que-e-eficiencia-termica-brayton-cycle-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-eficiencia-termica-brayton-cycle-definition\/","title":{"rendered":"O que \u00e9 efici\u00eancia t\u00e9rmica – Brayton Cycle – Definition"},"content":{"rendered":"
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A efici\u00eancia t\u00e9rmica do ciclo de Brayton, para o g\u00e1s ideal, pode ser expressa em termos de temperatura ou raz\u00e3o de press\u00e3o.\u00a0Em geral, a taxa de press\u00e3o \u00e9 o par\u00e2metro principal.\u00a0Engenharia T\u00e9rmica<\/div>\n<\/div>\n
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Efici\u00eancia t\u00e9rmica do ciclo de Brayton<\/h2>\n

Em geral, a\u00a0efici\u00eancia t\u00e9rmica<\/strong><\/a>\u00a0,\u00a0<\/a>\u03b7\u00a0<\/em><\/strong><\/a>th<\/sub><\/em><\/strong><\/a>\u00a0, de qualquer motor de calor \u00e9 definida como a raz\u00e3o entre o\u00a0trabalho<\/a>\u00a0que faz,\u00a0W<\/strong>\u00a0, para o\u00a0calor<\/a>\u00a0de entrada a uma temperatura elevada, Q\u00a0H<\/sub>\u00a0.<\/p>\n

\"f\u00f3rmula<\/a><\/p>\n

A\u00a0efici\u00eancia t\u00e9rmica<\/strong>\u00a0,\u00a0\u03b7\u00a0<\/em><\/strong>th<\/sub><\/em><\/strong>\u00a0, representa a fra\u00e7\u00e3o de\u00a0calor<\/strong>\u00a0,\u00a0Q\u00a0<\/strong>H<\/sub><\/strong>\u00a0, que \u00e9 convertida\u00a0em trabalho<\/strong>\u00a0.\u00a0Como a energia \u00e9 conservada de acordo com a\u00a0primeira lei da termodin\u00e2mica<\/strong><\/a>\u00a0e a energia n\u00e3o pode ser convertida para funcionar completamente, a entrada de calor, Q\u00a0H<\/sub>\u00a0, deve ser igual ao trabalho realizado, W, mais o calor que deve ser dissipado como\u00a0calor residual Q\u00a0<\/strong>C<\/sub><\/strong>\u00a0no meio Ambiente.\u00a0Portanto, podemos reescrever a f\u00f3rmula da efici\u00eancia t\u00e9rmica como:<\/p>\n

\"f\u00f3rmula<\/a><\/p>\n

\"Takaishi,<\/a>
Takaishi, Tatsuo;\u00a0Numata, Akira;\u00a0Nakano, Ryouji;\u00a0Sakaguchi, Katsuhiko (mar\u00e7o de 2008).\u00a0\u201cAbordagem para motores a diesel e g\u00e1s de alta efici\u00eancia\u201d (PDF).\u00a0Revis\u00e3o t\u00e9cnica Mitsubishi Heavy Industries.\u00a045 (1).\u00a0P\u00e1gina visitada em 2011-02-04.<\/figcaption><\/figure>\n

Essa \u00e9 uma f\u00f3rmula muito \u00fatil, mas aqui expressamos a efici\u00eancia t\u00e9rmica usando a primeira lei em termos de\u00a0entalpia<\/a>\u00a0.<\/p>\n

Para calcular a efici\u00eancia t\u00e9rmica do\u00a0ciclo de Brayton<\/strong>\u00a0(compressor \u00fanico e turbina \u00fanica), os engenheiros usam a\u00a0primeira lei da termodin\u00e2mica em termos de entalpia<\/a>\u00a0e n\u00e3o em energia interna.<\/p>\n

A primeira lei em termos de entalpia \u00e9:<\/p>\n

dH = dQ + Vdp<\/strong><\/em><\/p>\n

Nesta equa\u00e7\u00e3o, o termo\u00a0Vdp<\/em><\/strong>\u00a0\u00e9 um\u00a0trabalho de processo de fluxo.\u00a0<\/strong>Este trabalho, \u00a0\u00a0Vdp<\/em><\/strong>\u00a0, \u00e9 usado para\u00a0sistemas de fluxo aberto<\/strong>\u00a0como uma\u00a0turbina<\/strong>\u00a0ou uma\u00a0bomba<\/strong>\u00a0na qual existe um\u00a0“dp”<\/strong>\u00a0, ou seja, mudan\u00e7a de press\u00e3o.\u00a0N\u00e3o h\u00e1 altera\u00e7\u00f5es no\u00a0volume de controle<\/a>\u00a0.\u00a0Como pode ser visto, essa forma de lei\u00a0simplifica a descri\u00e7\u00e3o da transfer\u00eancia de energia<\/strong>\u00a0.<\/p>\n

Existem express\u00f5es em termos de vari\u00e1veis \u200b\u200bmais familiares, como\u00a0temperatura<\/a>\u00a0e\u00a0press\u00e3o<\/a>\u00a0:<\/p>\n

dH = C\u00a0<\/em><\/strong>p<\/sub><\/em><\/strong>\u00a0dT + V (1-aT) dp<\/em><\/strong><\/p>\n

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Onde\u00a0<\/span>C\u00a0<\/span><\/strong>p<\/span><\/sub><\/strong>\u00a0\u00e9 a\u00a0<\/span>capacidade calor\u00edfica a press\u00e3o constante<\/span><\/strong>\u00a0e\u00a0<\/span>\u03b1<\/span><\/em><\/strong>\u00a0\u00e9 o coeficiente de expans\u00e3o t\u00e9rmica (c\u00fabico).\u00a0Para\u00a0<\/span>o g\u00e1s ideal<\/span><\/strong>\u00a0\u03b1T = 1 e, portanto:<\/span><\/p>\n

dH = C\u00a0<\/span><\/em><\/strong>p<\/span><\/sub><\/em><\/strong>\u00a0dT<\/span><\/em><\/strong><\/p>\n

A press\u00e3o constante<\/span><\/strong>\u00a0, a\u00a0<\/span>mudan\u00e7a de entalpia<\/span><\/strong>\u00a0\u00e9 igual \u00e0\u00a0<\/span>energia<\/span><\/strong>\u00a0transferida do ambiente atrav\u00e9s do aquecimento:<\/span><\/p>\n

Processo isob\u00e1rico (Vdp = 0):<\/span><\/strong><\/p>\n

dH = dQ \u2192 Q = H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong>\u00a0\u00a0\u00a0\u2192 H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong>\u00a0=\u00a0<\/span>C\u00a0<\/span><\/em><\/strong>P<\/span><\/sub><\/em><\/strong>\u00a0(T\u00a0<\/span><\/em><\/strong>3<\/span><\/sub><\/em><\/strong>\u00a0– T\u00a0<\/span><\/em><\/strong>2<\/span><\/sub><\/em><\/strong>\u00a0)<\/span><\/em><\/strong><\/p>\n

Na entropia constante<\/span><\/strong>\u00a0, ou seja, no processo isentr\u00f3pico, a\u00a0<\/span>mudan\u00e7a de entalpia<\/span><\/strong>\u00a0\u00e9 igual ao\u00a0<\/span>trabalho do processo de fluxo<\/span><\/strong>\u00a0realizado no ou pelo sistema:<\/span><\/p>\n

Processo isentr\u00f3pico (dQ = 0):<\/span><\/strong><\/p>\n

dH = Vdp \u2192 W = H\u00a0<\/span><\/strong>4<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u2192 H\u00a0<\/span><\/strong>4<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0=\u00a0<\/span>C\u00a0<\/span><\/em><\/strong>p<\/span><\/sub><\/em><\/strong>\u00a0(T\u00a0<\/span><\/em><\/strong>4<\/span><\/sub><\/em><\/strong>\u00a0– T\u00a0<\/span><\/em><\/strong>3<\/span><\/sub><\/em><\/strong>\u00a0)<\/span><\/em><\/strong><\/p>\n

A\u00a0<\/span>entalpia<\/span><\/strong>\u00a0pode ser transformada em uma\u00a0vari\u00e1vel\u00a0<\/span>intensiva<\/span><\/strong>\u00a0ou\u00a0<\/span>espec\u00edfica<\/span><\/strong>\u00a0dividindo-se pela\u00a0<\/span>massa<\/span><\/a>\u00a0.\u00a0<\/span>Os engenheiros usam mais a\u00a0<\/span><\/strong>entalpia espec\u00edfica<\/span><\/strong><\/a>\u00a0na an\u00e1lise termodin\u00e2mica do que a pr\u00f3pria entalpia.<\/span><\/p>\n

Agora, vamos assumir o\u00a0<\/span>ciclo de Brayton ideal<\/span><\/strong>\u00a0que descreve o funcionamento de um motor de calor com press\u00e3o constante.\u00a0Os modernos motores de turbina a g\u00e1s e os motores a jato de respira\u00e7\u00e3o tamb\u00e9m seguem o ciclo de Brayton.\u00a0Esse ciclo consiste em quatro processos termodin\u00e2micos:<\/span><\/p>\n

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  1. \n
    \"Ciclo<\/a>
    Ciclo de Brayton – diagrama de Ts<\/span><\/figcaption><\/figure>\n

    compress\u00e3o isentr\u00f3pica – o ar ambiente \u00e9 aspirado para o compressor, onde \u00e9 pressurizado (1 \u2192 2).\u00a0O trabalho necess\u00e1rio para o compressor \u00e9 dado por\u00a0<\/span>W\u00a0<\/span><\/strong>C<\/span><\/sub><\/strong>\u00a0= H\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong>\u00a0.<\/span><\/strong><\/li>\n

  2. adi\u00e7\u00e3o de calor isob\u00e1rico – o ar comprimido passa por uma c\u00e2mara de combust\u00e3o, onde o combust\u00edvel \u00e9 queimado e o ar ou outro meio \u00e9 aquecido (2 \u2192 3).\u00a0\u00c9 um processo de press\u00e3o constante, j\u00e1 que a c\u00e2mara est\u00e1 aberta para entrar e sair.\u00a0O calor l\u00edquido adicionado \u00e9 dado por\u00a0<\/span>Q\u00a0<\/span><\/strong>add<\/span><\/sub><\/strong>\u00a0= H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong><\/li>\n
  3. expans\u00e3o isentr\u00f3pica – o ar aquecido e pressurizado se expande na turbina, gasta sua energia.\u00a0O trabalho realizado pela turbina \u00e9 dado por\u00a0<\/span>W\u00a0<\/span><\/strong>T<\/span><\/sub><\/strong>\u00a0= H\u00a0<\/span><\/strong>4<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong><\/li>\n
  4. rejei\u00e7\u00e3o de calor isob\u00e1rica – o calor residual deve ser rejeitado para fechar o ciclo.\u00a0O calor l\u00edquido rejeitado \u00e9 dado por\u00a0<\/span>Q\u00a0<\/span><\/strong>re<\/span><\/sub><\/strong>\u00a0= H\u00a0<\/span><\/strong>4<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong><\/li>\n<\/ol>\n

    Como pode ser visto, podemos descrever e calcular completamente esses ciclos (da mesma forma para o ciclo de Rankine) usando entalpias.<\/span><\/p>\n

    Efici\u00eancia t\u00e9rmica – Ciclo de Brayton<\/span><\/strong><\/p>\n

    A efici\u00eancia t\u00e9rmica desse ciclo simples de Brayton, para o g\u00e1s ideal, agora pode ser expressa em termos de temperatura:<\/span><\/p>\n

    \"efici\u00eancia<\/a><\/p>\n

    Onde<\/span><\/p>\n