{"id":48055,"date":"2019-11-08T00:01:41","date_gmt":"2019-11-07T23:01:41","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-exemplo-de-processo-isobarico-adicao-de-calor-isobarico-definicao\/"},"modified":"2021-08-04T09:07:18","modified_gmt":"2021-08-04T08:07:18","slug":"o-que-e-exemplo-de-processo-isobarico-adicao-de-calor-isobarico-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-exemplo-de-processo-isobarico-adicao-de-calor-isobarico-definicao\/","title":{"rendered":"O que \u00e9 exemplo de processo isob\u00e1rico – adi\u00e7\u00e3o de calor isob\u00e1rico – defini\u00e7\u00e3o"},"content":{"rendered":"
Exemplo de processo isob\u00e1rico – adi\u00e7\u00e3o de calor isob\u00e1rico. Vamos supor uma adi\u00e7\u00e3o de calor isob\u00e1rica em um g\u00e1s ideal. Calcule o calor adicionado pelo trocador de calor. Engenharia T\u00e9rmica <\/div><\/div>\n
<\/div>\n
<\/div>\n

Exemplo de processo isob\u00e1rico – adi\u00e7\u00e3o de calor isob\u00e1rico<\/h2>\n
\"primeira<\/a>
O ciclo de Brayton ideal consiste em quatro processos termodin\u00e2micos.\u00a0Dois processos isentr\u00f3picos e dois processos isob\u00e1ricos.<\/figcaption><\/figure>\n

Vamos supor o\u00a0\u00a0ciclo de Brayton ideal<\/strong>\u00a0\u00a0que descreve o funcionamento de uma\u00a0\u00a0m\u00e1quina de calor de\u00a0<\/strong>press\u00e3o constante<\/strong>\u00a0\u00a0.\u00a0Os<\/strong>\u00a0\u00a0motores\u00a0modernos de turbina a g\u00e1s<\/strong>\u00a0e os motores a\u00a0\u00a0jato de respira\u00e7\u00e3o respirat\u00f3ria<\/strong>\u00a0\u00a0tamb\u00e9m seguem o ciclo de Brayton.<\/p>\n

O ciclo de Brayton ideal consiste em quatro processos termodin\u00e2micos.\u00a0Dois processos isentr\u00f3picos e dois processos isob\u00e1ricos.<\/p>\n

    \n
  1. compress\u00e3o isentr\u00f3pica<\/strong>\u00a0\u00a0– 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\u00a0W\u00a0C<\/sub>\u00a0\u00a0= H\u00a02<\/sub>\u00a0\u00a0– H\u00a01<\/sub>\u00a0.<\/strong><\/li>\n
  2. adi\u00e7\u00e3o de calor isob\u00e1rico<\/strong>\u00a0\u00a0– o ar comprimido ent\u00e3o 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, uma vez que a c\u00e2mara est\u00e1 aberta para entrar e sair.\u00a0O calor l\u00edquido adicionado \u00e9 dado por\u00a0\u00a0Q\u00a0add<\/sub>\u00a0\u00a0= H\u00a03\u00a0<\/sub>\u00a0– H\u00a02<\/sub><\/strong><\/li>\n
  3. expans\u00e3o isentr\u00f3pica<\/strong>\u00a0\u00a0– o ar aquecido e pressurizado ent\u00e3o se expande na turbina, desistindo de sua energia.\u00a0O trabalho realizado pela turbina \u00e9 dado por\u00a0\u00a0W\u00a0T<\/sub>\u00a0\u00a0= H\u00a04<\/sub>\u00a0\u00a0– H\u00a03<\/sub><\/strong><\/li>\n
  4. rejei\u00e7\u00e3o de calor isob\u00e1rico<\/strong>\u00a0\u00a0– o calor residual deve ser rejeitado para fechar o ciclo.\u00a0O calor l\u00edquido rejeitado \u00e9 dado por\u00a0\u00a0Q\u00a0re<\/sub>\u00a0\u00a0= H\u00a04\u00a0<\/sub>\u00a0– H\u00a01<\/sub><\/strong><\/li>\n<\/ol>\n

    Suponha uma\u00a0adi\u00e7\u00e3o de calor isob\u00e1rica<\/strong>\u00a0(2 \u2192 3) em um trocador de calor.\u00a0Em turbinas a g\u00e1s t\u00edpicas, o est\u00e1gio de alta press\u00e3o recebe g\u00e1s (ponto 3 na figura; p\u00a03<\/sub>\u00a0=\u00a06,7 MPa<\/strong>\u00a0; T\u00a03<\/sub>\u00a0= 1190 K (917 \u00b0 C)) de um trocador de calor.\u00a0Al\u00e9m disso, sabemos que o compressor recebe g\u00e1s (ponto 1 na figura; p\u00a01<\/sub>\u00a0=\u00a02,78 MPa<\/strong>\u00a0; T\u00a01<\/sub>\u00a0= 299 K (26 \u00b0 C)) e sabemos que a efici\u00eancia isentr\u00f3pica do compressor \u00e9\u00a0\u03b7\u00a0<\/strong>K<\/sub><\/strong>\u00a0= 0,87 (87 %)<\/strong>\u00a0.<\/p>\n

    Calcule o calor adicionado pelo trocador de calor (entre 2 \u2192 3).<\/strong><\/p>\n

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

    Pela\u00a0primeira lei da termodin\u00e2mica<\/strong>\u00a0, o calor l\u00edquido adicionado \u00e9 dado por\u00a0Q\u00a0<\/strong>add<\/sub><\/strong>\u00a0= H\u00a0<\/strong>3<\/sub><\/strong>\u00a0– H\u00a0<\/strong>2<\/sub><\/strong>\u00a0ou\u00a0Q\u00a0add<\/sub>\u00a0= C\u00a0p<\/sub>\u00a0. (T\u00a03<\/sub>\u00a0-T\u00a02s<\/sub>\u00a0),<\/strong>\u00a0mas neste caso n\u00e3o sabemos a temperatura (T\u00a02s<\/sub>\u00a0) na sa\u00edda do compressor.\u00a0Vamos resolver esse problema em vari\u00e1veis \u200b\u200bintensivas.\u00a0Temos que reescrever a equa\u00e7\u00e3o anterior (para incluir\u00a0\u03b7\u00a0<\/strong>K<\/sub><\/strong>\u00a0) usando o termo (+\u00a0h\u00a0<\/strong>1<\/sub><\/strong>\u00a0– h\u00a0<\/strong>1<\/sub><\/strong>\u00a0) para:<\/p>\n

    Q\u00a0add<\/sub>\u00a0=\u00a0h\u00a0<\/strong>3<\/sub><\/strong>\u00a0– h\u00a0<\/strong>2<\/sub><\/strong>\u00a0= h\u00a0<\/strong>3<\/sub><\/strong>\u00a0– h\u00a0<\/strong>1<\/sub><\/strong>\u00a0– (h\u00a0<\/strong>2s<\/sub><\/strong>\u00a0– h\u00a0<\/strong>1<\/sub><\/strong>\u00a0) \/\u00a0<\/strong>\u03b7\u00a0<\/strong>K<\/sub><\/strong>\u00a0\u00a0<\/strong><\/p>\n

    Q\u00a0add\u00a0<\/sub>=<\/strong>\u00a0c\u00a0p<\/sub>\u00a0(T\u00a03<\/sub>\u00a0-T\u00a01<\/sub>\u00a0) – (c\u00a0p<\/sub>\u00a0(T\u00a02s<\/sub>\u00a0-T\u00a01<\/sub>\u00a0)\u00a0\/\u00a0<\/strong>\u03b7\u00a0<\/strong>K<\/sub><\/strong>\u00a0)<\/p>\n

    Em seguida, calcularemos a temperatura, T\u00a02s<\/sub>\u00a0, usando\u00a0p, V, Rela\u00e7\u00e3o T<\/strong>\u00a0para o processo adiab\u00e1tico entre (1 \u2192 2).<\/p>\n

    \"rela\u00e7\u00e3o<\/a><\/p>\n

    Nesta equa\u00e7\u00e3o, o fator para h\u00e9lio \u00e9 igual a\u00a0= c\u00a0p<\/sub>\u00a0\/ c\u00a0v<\/sub>\u00a0= 1,66<\/strong>\u00a0.\u00a0Da equa\u00e7\u00e3o anterior segue que a temperatura de sa\u00edda do compressor,\u00a0T2s<\/sub>\u00a0, \u00e9:<\/p>\n

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

    Da\u00a0Lei do G\u00e1s Ideal<\/a>\u00a0, sabemos que o calor espec\u00edfico molar de um\u00a0g\u00e1s monoat\u00f4mico ideal<\/a>\u00a0\u00e9:<\/p>\n

    C\u00a0<\/em><\/strong>v<\/sub><\/em><\/strong>\u00a0= 3 \/ 2R = 12,5 J \/ mol K<\/em><\/strong>\u00a0e\u00a0C\u00a0<\/em><\/strong>p<\/sub><\/em><\/strong>\u00a0= C\u00a0<\/em><\/strong>v<\/sub><\/em><\/strong>\u00a0+ R = 5 \/ 2R = 20,8 J \/ mol K<\/em><\/strong><\/p>\n

    Transferimos as capacidades de calor espec\u00edficas em unidades de\u00a0J \/ kg K via:<\/strong><\/p>\n

    c\u00a0<\/em><\/strong>p<\/sub><\/em><\/strong>\u00a0= C\u00a0<\/em><\/strong>p<\/sub><\/em><\/strong>\u00a0.\u00a01 \/ M (peso molar de h\u00e9lio) = 20,8 x 4,10\u00a0<\/em><\/strong>-3<\/sup><\/em><\/strong>\u00a0= 5200 J \/ kg K<\/em><\/strong><\/p>\n

    Usando esta temperatura e a\u00a0efici\u00eancia do compressor isentr\u00f3pico<\/a>\u00a0, podemos calcular o calor adicionado pelo trocador de calor:<\/p>\n

    Q\u00a0add<\/sub>\u00a0=<\/strong>\u00a0c\u00a0p<\/sub>\u00a0(T\u00a03<\/sub>\u00a0-T\u00a01<\/sub>\u00a0) – (c\u00a0p<\/sub>\u00a0(T\u00a02s<\/sub>\u00a0-T\u00a01<\/sub>\u00a0)\u00a0\/\u00a0<\/strong>\u03b7\u00a0<\/strong>K<\/sub><\/strong>\u00a0) = 5200. (1190 – 299) – 5200. (424-299) \/ 0,87 = 4,633 MJ \/ kg – 0,747 MJ \/ kg =\u00a03,886 MJ \/ kg<\/strong> <\/p><\/div><\/div>

    \n
    \"Processo<\/a>
    Processo isob\u00e1rico – caracter\u00edsticas principais<\/figcaption><\/figure>\n
    \"A<\/a>
    Para uma massa fixa de g\u00e1s a press\u00e3o constante, o volume \u00e9 diretamente proporcional \u00e0 temperatura Kelvin.\u00a0Fonte: grc.nasa.gov A pol\u00edtica de direitos autorais da NASA afirma que \u201co material da NASA n\u00e3o \u00e9 protegido por direitos autorais, a menos que indicado\u201d.<\/figcaption><\/figure>\n<\/div><\/div>
    <\/span>Exemplo: Pist\u00e3o sem fric\u00e7\u00e3o - Calor - Entalpia<\/div>
    \n
    \"Entalpia<\/a>
    Calcule a temperatura final, se 3000 kJ de calor forem adicionados.<\/figcaption><\/figure>\n

    Um pist\u00e3o sem atrito \u00e9 usado para proporcionar uma press\u00e3o constante de\u00a0500 kPa<\/strong>\u00a0num cilindro contendo vapor (\u00a0vapor sobreaquecido<\/a>\u00a0) de um volume de\u00a02 m\u00a03<\/sup><\/strong>\u00a0\u00a0a\u00a0500 K<\/strong>\u00a0.\u00a0Calcule a temperatura final, se\u00a03000 kJ<\/strong>\u00a0de\u00a0calor<\/strong>\u00a0forem adicionados.<\/p>\n

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

    Usando\u00a0tabelas de vapor<\/a>\u00a0, sabemos que a\u00a0entalpia espec\u00edfica<\/strong>\u00a0desse vapor (500 kPa; 500 K) \u00e9 de cerca de\u00a02912 kJ \/ kg<\/strong>\u00a0.\u00a0Como nessa condi\u00e7\u00e3o o vapor tem densidade de 2,2 kg \/ m\u00a03<\/sup>\u00a0, sabemos que h\u00e1 cerca de\u00a04,4 kg de vapor<\/strong>\u00a0no pist\u00e3o com entalpia de 2.912 kJ \/ kg x 4,4 kg =\u00a012812 kJ<\/strong>\u00a0.<\/p>\n

    Quando usamos simplesmente\u00a0Q = H\u00a02<\/sub>\u00a0– H\u00a01<\/sub><\/strong>\u00a0, a entalpia de vapor resultante ser\u00e1:<\/p>\n

    H\u00a02<\/sub>\u00a0= H\u00a01<\/sub>\u00a0+ Q =\u00a015812 kJ<\/strong><\/p>\n

    A partir das\u00a0tabelas de vapor<\/strong>\u00a0, esse vapor superaquecido (15812 \/ 4,4 = 3593 kJ \/ kg) ter\u00e1 uma temperatura de\u00a0828 K (555 \u00b0 C)<\/strong>\u00a0.\u00a0Visto que nesta entalpia o vapor tem densidade de 1,31 kg \/ m\u00a03<\/sup>\u00a0, \u00e9 \u00f3bvio que ele se expandiu em cerca de 2,2 \/ 1,31 = 1,67 (+ 67%).\u00a0Portanto, o volume resultante \u00e9 2 m\u00a03<\/sup>\u00a0x 1,67 = 3,34 m\u00a03<\/sup>\u00a0e \u2206V = 3,34 m\u00a03<\/sup>\u00a0– 2 m\u00a03<\/sup>\u00a0= 1,34 m\u00a03<\/sup>\u00a0.<\/p>\n

    A\u00a0parte\u00a0p\u2206V<\/strong>\u00a0da entalpia, ou seja, o trabalho realizado \u00e9:<\/p>\n

    W = p\u2206V = 500 000 Pa x 1,34 m\u00a03<\/sup>\u00a0= 670 kJ<\/strong> <\/p><\/div><\/div> <\/div>

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