{"id":47975,"date":"2019-11-07T13:53:14","date_gmt":"2019-11-07T12:53:14","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-exemplo-do-ciclo-de-brayton-problema-com-solucao-definicao\/"},"modified":"2020-01-26T13:21:51","modified_gmt":"2020-01-26T12:21:51","slug":"o-que-e-exemplo-do-ciclo-de-brayton-problema-com-solucao-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-exemplo-do-ciclo-de-brayton-problema-com-solucao-definicao\/","title":{"rendered":"O que \u00e9 exemplo do ciclo de Brayton &#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 Ciclo de Brayton &#8211; Problema com a Solu\u00e7\u00e3o.\u00a0Calcule os par\u00e2metros termodin\u00e2micos das chaves de ciclo de Brayton fechados, como temperaturas, press\u00f5es e transfer\u00eancias de calor.\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>Ciclo de Brayton &#8211; Problema com a solu\u00e7\u00e3o<\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-17724 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation-300x196.png\" alt=\"Ciclo de Brayton - exemplo - c\u00e1lculo\" width=\"300\" height=\"196\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation-300x196.png\" \/><\/a>Vamos assumir o\u00a0<strong>ciclo<\/strong>\u00a0fechado de\u00a0<strong>Brayton<\/strong>\u00a0, que \u00e9 um dos\u00a0<strong>ciclos termodin\u00e2micos<\/strong>\u00a0mais comuns\u00a0encontrados nos modernos motores de turbina a g\u00e1s.\u00a0Nesse caso, assuma uma\u00a0<strong>turbina a g\u00e1s h\u00e9lio<\/strong>\u00a0com um \u00fanico compressor e arranjo de turbina \u00fanica.\u00a0Um dos principais par\u00e2metros de tais motores \u00e9 a temperatura m\u00e1xima de entrada da turbina e a raz\u00e3o de press\u00e3o do compressor (PR = p\u00a0<sub>2<\/sub>\u00a0\/ p\u00a0<sub>1<\/sub>\u00a0), que determina a efici\u00eancia t\u00e9rmica desse motor.<\/p>\n<p>Nesta turbina, o est\u00e1gio de alta press\u00e3o recebe g\u00e1s (ponto 3 na figura) de um trocador de calor:<\/p>\n<ul>\n<li>p\u00a0<sub>3<\/sub>\u00a0= 6,7 MPa;<\/li>\n<li>T\u00a0<sub>3<\/sub>\u00a0= 1190 K (917 \u00b0 C))<\/li>\n<li>a efici\u00eancia da turbina isentr\u00f3pica \u00e9\u00a0<em>\u03b7\u00a0<\/em><em><sub>T<\/sub><\/em>\u00a0= 0,91 (91%)<\/li>\n<\/ul>\n<p>e esvazie-o em outro trocador de calor, onde a press\u00e3o de sa\u00edda \u00e9 (ponto 4):<\/p>\n<ul>\n<li>p\u00a0<sub>4<\/sub>\u00a0= 2,78 MPa<\/li>\n<li>T\u00a0<sub>4, \u00e9<\/sub>\u00a0=?<\/li>\n<\/ul>\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>Assim, a raz\u00e3o de press\u00e3o do compressor \u00e9 igual a PR = 2,41.\u00a0Al\u00e9m disso, sabemos que o compressor recebe g\u00e1s (ponto 1) na figura:<\/span><\/p>\n<ul>\n<li><span>p\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 2,78 MPa;<\/span><\/li>\n<li><span>T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 299 K (26 \u00b0 C)<\/span><\/li>\n<li><span>a efici\u00eancia isentr\u00f3pica do compressor \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0,87 (87%).<\/span><\/li>\n<\/ul>\n<p><span>A raz\u00e3o de capacidade t\u00e9rmica,, para o h\u00e9lio \u00e9 igual a\u00a0<\/span><strong><span>= c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1,66<\/span><\/strong><\/p>\n<p><strong><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-17725 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution-300x271.png\" alt=\"Ciclo de Brayton - problema com a solu\u00e7\u00e3o\" width=\"300\" height=\"271\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution-300x271.png\" \/><\/a><span>Calcular:<\/span><\/strong><\/p>\n<ol>\n<li><span>o calor adicionado pelo trocador de calor (entre 2 \u2192 3)<\/span><\/li>\n<li><span>a temperatura de sa\u00edda do compressor do g\u00e1s (T2\u00a0<\/span><sub><span>, \u00e9<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>o trabalho real realizado neste compressor, quando a efici\u00eancia isentr\u00f3pica do compressor \u00e9 \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0,87 (87%)<\/span><\/li>\n<li><span>a temperatura de sa\u00edda da turbina do g\u00e1s (T\u00a0<\/span><sub><span>4, \u00e9<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>o trabalho real realizado por essa turbina, quando a efici\u00eancia da turbina isentr\u00f3pica \u00e9 \u03b7\u00a0<\/span><sub><span>T<\/span><\/sub><span>\u00a0= 0,91 (91%)<\/span><\/li>\n<li><span>a efici\u00eancia t\u00e9rmica deste ciclo<\/span><\/li>\n<\/ol>\n<p><strong><span>Solu\u00e7\u00e3o:<\/span><\/strong><\/p>\n<p><span>1) + 2)<\/span><\/p>\n<p><span>A partir da\u00a0<\/span><a title=\"Primeira Lei da Termodin\u00e2mica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-a-primeira-lei-da-termodinamica-definicao\/\"><span>primeira lei da termodin\u00e2mica<\/span><\/a><span>\u00a0, o calor l\u00edquido adicionado \u00e9 dado por\u00a0<\/span><strong><span>Q\u00a0<\/span><sub><span>add, ex<\/span><\/sub><span>\u00a0= H\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; H\u00a0<\/span><sub><span>2<\/span><\/sub><\/strong><span>\u00a0\u00a0[kJ] ou Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= C\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0. (T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>,<\/span><\/strong><span>\u00a0mas neste caso n\u00e3o sabemos a temperatura (T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0) na sa\u00edda do compressor.\u00a0Vamos resolver esse problema em\u00a0<\/span><a title=\"Propriedades extensivas e intensivas\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/extensive-and-intensive-properties\/\"><span>vari\u00e1veis \u200b\u200bintensivas<\/span><\/a><span>\u00a0.\u00a0Temos que reescrever a equa\u00e7\u00e3o anterior (para incluir\u00a0<\/span><strong><span>\u03b7\u00a0<\/span><\/strong><strong><sub><span>K<\/span><\/sub><\/strong><span>\u00a0) usando o termo (\u00a0<\/span><strong><span>+ h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><\/strong><span>\u00a0) para:<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>adicione<\/span><\/sub><span>\u00a0= h\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= h\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0&#8211; (h\u00a0<\/span><sub><span>2 s<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) \/ \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub>\u00a0<strong>\u00a0<\/strong><span>\u00a0[kJ \/ kg]<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>add\u00a0<\/span><\/sub><strong><span>=<\/span><\/strong><span>\u00a0c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) &#8211; (c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(t\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>\/<\/span><\/strong><span>\u00a0\u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0)<\/span><\/p>\n<p><span>Em seguida, iremos calcular a temperatura,\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>2s<\/span><\/sub><\/strong><span>\u00a0, utilizando p, V, T Rela\u00e7\u00e3o (a partir de\u00a0<\/span><a title=\"Lei do g\u00e1s ideal\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>lei do g\u00e1s ideal<\/span><\/a><span>\u00a0) para\u00a0<\/span><a title=\"Processo Adiab\u00e1tico\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-processo-adiabatico-definicao\/\"><span>processo adiab\u00e1tico<\/span><\/a><span>\u00a0entre (1 \u2192 2).<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17281 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" alt=\"rela\u00e7\u00e3o p, V, T - processo isentr\u00f3pico\" width=\"209\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" \/><\/a><\/p>\n<p><span>Nesta equa\u00e7\u00e3o, o fator para o h\u00e9lio \u00e9 igual a = c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1,66.\u00a0A partir da equa\u00e7\u00e3o anterior segue-se que a temperatura de sa\u00edda do compressor, t\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0, \u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17429 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\" alt=\"processo isob\u00e1rico - exemplo\" width=\"492\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\" \/><\/a><\/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-divider su-divider-style-dotted\"><\/div>\n<p><span>Usando esta temperatura e a\u00a0<\/span><a title=\"Efici\u00eancia isentr\u00f3pica - Turbina \/ Compressor \/ Bocal\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-processes\/isentropic-process\/isentropic-efficiency-turbinecompressornozzle\/\"><span>efici\u00eancia isentr\u00f3pica do compressor<\/span><\/a><span>\u00a0, podemos calcular o calor adicionado pelo trocador de calor:<\/span><strong><span>Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0=<\/span><\/strong><span>\u00a0c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) &#8211; (c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(t\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>\/<\/span><\/strong><span>\u00a0\u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0) = 5200. (1190 &#8211; 299) &#8211; 5200. (424-299) \/0.87 = 4,633 MJ \/ kg &#8211; 0,747 MJ \/ kg =\u00a0<\/span><strong><span>3,886 MJ \/ kg<\/span><\/strong><\/p>\n<p><span>3)<\/span><\/p>\n<p><span>O trabalho realizado no g\u00e1s pelo compressor no processo de compress\u00e3o isentr\u00f3pica \u00e9:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C, S<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0= C\u00a0<\/span><\/em><\/strong><strong><em><sub><span>p<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0(t\u00a0<\/span><\/em><\/strong><strong><em><sub><span>2s<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>1<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0) = 5200 x (424 &#8211; 299) = 0,650 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>O trabalho real feito no g\u00e1s pelo compressor na compress\u00e3o adiab\u00e1tica \u00e9 ent\u00e3o:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C, verdadeiro<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0= c\u00a0<\/span><\/em><\/strong><strong><em><sub><span>p<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0(t\u00a0<\/span><\/em><\/strong><strong><em><sub><span>2s<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>1<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0).\u00a0\u03b7\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C<\/span><\/sub><\/em><\/strong>\u00a0<strong><em><span>= 5200 x (424 &#8211; 299) \/ 0,87 = 0,747 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>4)<\/span><\/p>\n<p><span>A temperatura de sa\u00edda da turbina do g\u00e1s,\u00a0<\/span><strong><span>T4\u00a0<\/span><sub><span>, \u00e9<\/span><\/sub><\/strong><span>\u00a0, pode ser calculada usando a mesma rela\u00e7\u00e3o\u00a0\u00a0<\/span><strong><span>p, V, T<\/span><\/strong><span>\u00a0\u00a0que em 2), mas entre os estados 3 e 4:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17281 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" alt=\"rela\u00e7\u00e3o p, V, T - processo isentr\u00f3pico\" width=\"209\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" \/><\/a><\/p>\n<p><span>A partir da equa\u00e7\u00e3o anterior, segue que a temperatura de sa\u00edda do g\u00e1s, T\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0, \u00e9:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17283 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\" alt=\"processo isentr\u00f3pico - exemplo\" width=\"503\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\" \/><\/a><\/p>\n<p><span>5)<\/span><\/p>\n<p><span>O trabalho realizado pela turbina a g\u00e1s na expans\u00e3o isentr\u00f3pica \u00e9 ent\u00e3o:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>T, s<\/span><\/sub><\/em><\/strong><em><span>\u00a0= c\u00a0<\/span><\/em><em><sub><span>p<\/span><\/sub><\/em><em><span>\u00a0(\u00a0<\/span><\/em><em><sub><span>T3<\/span><\/sub><\/em><em><span>\u00a0&#8211;\u00a0<\/span><\/em><em><sub><span>T4s<\/span><\/sub><\/em><em><span>\u00a0) = 5200 x (1190 &#8211; 839) =\u00a0<\/span><\/em><strong><em><span>1,825 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>O verdadeiro trabalho realizado pela turbina a g\u00e1s na expans\u00e3o adiab\u00e1tica \u00e9 ent\u00e3o:<\/span><\/p>\n<p><strong><em><span>WT\u00a0<\/span><\/em><\/strong><strong><em><sub><span>, real<\/span><\/sub><\/em><\/strong><em><span>\u00a0=\u00a0<\/span><\/em><em><sub><span>cp<\/span><\/sub><\/em><em><span>\u00a0(\u00a0<\/span><\/em><em><sub><span>T3<\/span><\/sub><\/em><em><span>\u00a0&#8211;\u00a0<\/span><\/em><em><sub><span>T4s<\/span><\/sub><\/em><em><span>\u00a0).\u00a0\u03b7\u00a0<\/span><\/em><em><sub><span>T<\/span><\/sub><\/em>\u00a0<em><span>= 5200 x (1190 &#8211; 839) x 0,91 =\u00a0<\/span><\/em><strong><em><span>1,661 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>6)<\/span><\/p>\n<p><span>Como foi derivado na se\u00e7\u00e3o anterior, a\u00a0<\/span><strong><span>efici\u00eancia t\u00e9rmica<\/span><\/strong><span>\u00a0de um ciclo de Brayton ideal \u00e9 uma fun\u00e7\u00e3o da\u00a0<\/span><strong><span>raz\u00e3o<\/span><\/strong><span>\u00a0de\u00a0<strong>press\u00e3o<\/strong>\u00a0e\u00a0<\/span><strong><span>\u03ba<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17696 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\" alt=\"efici\u00eancia t\u00e9rmica - ciclo de brayton - raz\u00e3o de press\u00e3o - equa\u00e7\u00e3o\" width=\"187\" height=\"66\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\" \/><\/a><\/p>\n<p><span>Portanto<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><em><sub><span>th<\/span><\/sub><\/em><\/strong><em><span>\u00a0=<\/span><\/em><span>\u00a00,295 =<\/span><strong><span>\u00a029,5%<\/span><\/strong><\/p>\n<p><span>A efici\u00eancia t\u00e9rmica tamb\u00e9m pode ser calculada usando o trabalho e o calor (sem \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0):<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><\/strong><em><sub><strong><span>th, s<\/span><\/strong>\u00a0<\/sub><\/em><em><span>= (\u00a0<\/span><\/em><em><span>WT\u00a0<\/span><\/em><em><sub><span>, s<\/span><\/sub><\/em><em><span>\u00a0&#8211; WC\u00a0<\/span><\/em><em><sub><span>, s<\/span><\/sub><\/em><em><span>\u00a0)\u00a0<\/span><\/em><strong><em><span>\/<\/span><\/em><\/strong><sub><span>\u00a0adi\u00e7\u00e3o de<\/span><\/sub><span>\u00a0Q<sub>\u00a0, s<\/sub>\u00a0= (1,825 &#8211; 0,650) \/ 3,983<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a0\u00a00,295 =<\/span><strong><span>\u00a029,5%<\/span><\/strong><\/p>\n<p><span>Finalmente, a efici\u00eancia t\u00e9rmica, incluindo a efici\u00eancia isentr\u00f3pica de turbina \/ compressor, \u00e9:<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><em><sub><span>th, real\u00a0<\/span><\/sub><\/em><\/strong><em><span>\u00a0= (\u00a0<\/span><\/em><em><span>WT\u00a0<\/span><\/em><em><sub><span>, real<\/span><\/sub><\/em><em><span>\u00a0\u00a0&#8211;<\/span><\/em><span>\u00a0\u00a0WC<\/span><em><sub><span>\u00a0, real<\/span><\/sub><\/em><em><span>\u00a0)\u00a0<\/span><\/em><strong><em><span>\/<\/span><\/em><\/strong><span>\u00a0Q<\/span><sub><span>\u00a0add<\/span><\/sub><span>\u00a0= (1,661 &#8211; 0,747) \/ 3,886\u00a0<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a0\u00a00,235 =<\/span><strong><span>\u00a023,5%<\/span><\/strong><\/p>\n<\/div>\n<\/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>Exemplo de Ciclo de Brayton &#8211; Problema com a Solu\u00e7\u00e3o.\u00a0Calcule os par\u00e2metros termodin\u00e2micos das chaves de ciclo de Brayton fechados, como temperaturas, press\u00f5es e transfer\u00eancias de calor.\u00a0Engenharia T\u00e9rmica Ciclo de Brayton &#8211; Problema com a solu\u00e7\u00e3o Vamos assumir o\u00a0ciclo\u00a0fechado de\u00a0Brayton\u00a0, que \u00e9 um dos\u00a0ciclos termodin\u00e2micos\u00a0mais comuns\u00a0encontrados nos modernos motores de turbina a g\u00e1s.\u00a0Nesse caso, assuma &#8230; <a title=\"O que \u00e9 exemplo do ciclo de Brayton &#8211; problema com solu\u00e7\u00e3o &#8211; defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-exemplo-do-ciclo-de-brayton-problema-com-solucao-definicao\/\" aria-label=\"More on O que \u00e9 exemplo do ciclo de Brayton &#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>O que \u00e9 exemplo do ciclo de Brayton - problema com solu\u00e7\u00e3o - defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Exemplo de Ciclo de Brayton - Problema com a Solu\u00e7\u00e3o. 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