{"id":47935,"date":"2019-11-07T09:37:34","date_gmt":"2019-11-07T08:37:34","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/o-que-e-o-ciclo-diesel-problema-com-a-solucao-definicao\/"},"modified":"2020-01-26T09:22:01","modified_gmt":"2020-01-26T08:22:01","slug":"o-que-e-o-ciclo-diesel-problema-com-a-solucao-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-o-ciclo-diesel-problema-com-a-solucao-definicao\/","title":{"rendered":"O que \u00e9 o Ciclo Diesel &#8211; Problema com a 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 Diesel &#8211; Problema com a Solu\u00e7\u00e3o.\u00a0Vamos assumir um ciclo Diesel t\u00edpico.\u00a0Calcule as principais caracter\u00edsticas, como temperaturas, press\u00f5es, MEP e efici\u00eancia.\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 Diesel &#8211; Problema com a Solu\u00e7\u00e3o<\/h2>\n<figure id=\"attachment_18421\" class=\"wp-caption alignright\" aria-describedby=\"caption-attachment-18421\"><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-pV-Diagram.png\"><img loading=\"lazy\" class=\"size-medium wp-image-18421 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-pV-Diagram-300x280.png\" alt=\"diagrama pV de um ciclo Diesel ideal\" width=\"300\" height=\"280\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-pV-Diagram-300x280.png\" \/><\/a><figcaption id=\"caption-attachment-18421\" class=\"wp-caption-text\">diagrama pV de um ciclo Diesel ideal<\/figcaption><\/figure>\n<p><strong>Ciclo Diesel &#8211; Problema com a Solu\u00e7\u00e3o<\/strong><\/p>\n<p>Vamos assumir o ciclo Diesel, que \u00e9 um dos\u00a0<strong>ciclos termodin\u00e2micos<\/strong>\u00a0mais comuns\u00a0encontrados em\u00a0<strong>motores de autom\u00f3veis<\/strong>\u00a0.\u00a0Um dos principais par\u00e2metros desses motores \u00e9 a mudan\u00e7a de volumes entre o ponto morto superior (TDC) e o ponto morto inferior (BDC).\u00a0A propor\u00e7\u00e3o desses volumes (\u00a0<em>V\u00a0<\/em><em><sub>1<\/sub><\/em><em>\u00a0\/ V\u00a0<\/em><em><sub>2<\/sub><\/em>\u00a0) \u00e9 conhecida como\u00a0<strong>taxa de compress\u00e3o<\/strong>\u00a0.\u00a0Tamb\u00e9m a rela\u00e7\u00e3o de corte V\u00a0<sub>3<\/sub>\u00a0\/ V\u00a0<sub>2<\/sub>\u00a0, que \u00e9 a rela\u00e7\u00e3o dos volumes no final e no in\u00edcio da fase de combust\u00e3o.<\/p>\n<p>Neste exemplo, vamos assumir o ciclo Diesel com taxa de compress\u00e3o de CR = 20: 1 e taxa de corte \u03b1 = 2. O ar est\u00e1 a 100 kPa = 1 bar, 20 \u00b0 C (293 K) e o volume da c\u00e2mara \u00e9 de 500 cm\u00b3 antes do curso de compress\u00e3o.<\/p>\n<ul>\n<li>Capacidade t\u00e9rmica espec\u00edfica a press\u00e3o constante do ar \u00e0 press\u00e3o atmosf\u00e9rica e \u00e0 temperatura ambiente:\u00a0<strong>c\u00a0<\/strong><strong><sub>p<\/sub><\/strong><strong>\u00a0= 1,01 kJ \/ kgK.<\/strong><\/li>\n<li>Capacidade t\u00e9rmica espec\u00edfica a um volume constante de ar \u00e0 press\u00e3o atmosf\u00e9rica e \u00e0 temperatura ambiente:\u00a0<strong>c\u00a0<\/strong><strong><sub>v<\/sub><\/strong><strong>\u00a0= 0,718 kJ \/ kgK.<\/strong><\/li>\n<li><strong>\u03ba = c\u00a0<\/strong><strong><sub>p<\/sub><\/strong><strong>\u00a0\/ c\u00a0<\/strong><strong><sub>v<\/sub><\/strong><strong>\u00a0= 1,4<\/strong><\/li>\n<\/ul>\n<p><strong><span>Calcular:<\/span><\/strong><\/p>\n<ol>\n<li><strong><span>a massa de entrada de ar<\/span><\/strong><\/li>\n<li><strong><span>a temperatura T\u00a0<\/span><\/strong><strong><sub><span>2<\/span><\/sub><\/strong><\/li>\n<li><strong><span>a press\u00e3o p\u00a0<\/span><\/strong><strong><sub><span>2<\/span><\/sub><\/strong><\/li>\n<li><strong><span>a temperatura T\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/li>\n<li><strong><span>a quantidade de calor adicionada pela queima da mistura combust\u00edvel-ar<\/span><\/strong><\/li>\n<li><strong><span>a efici\u00eancia t\u00e9rmica deste ciclo<\/span><\/strong><\/li>\n<li><strong><span>o deputado<\/span><\/strong><\/li>\n<\/ol>\n<p><strong><span>Solu\u00e7\u00e3o:<\/span><\/strong><\/p>\n<p><span>1)<\/span><\/p>\n<p><span>No in\u00edcio dos c\u00e1lculos, temos que determinar a quantidade de g\u00e1s no cilindro antes do curso de compress\u00e3o.\u00a0Usando a lei do g\u00e1s ideal, podemos encontrar a massa:<\/span><\/p>\n<p><em><span>pV = mR\u00a0<\/span><\/em><em><sub><span>espec\u00edfico<\/span><\/sub><\/em><em><span>\u00a0T<\/span><\/em><\/p>\n<p><span>Onde:<\/span><\/p>\n<ul>\n<li><em><span>p<\/span><\/em><span>\u00a0\u00e9 a press\u00e3o absoluta do g\u00e1s<\/span><\/li>\n<li><em><span>m<\/span><\/em><span>\u00a0\u00e9 a massa da subst\u00e2ncia<\/span><\/li>\n<li><em><span>T<\/span><\/em><span>\u00a0\u00e9 a temperatura absoluta<\/span><\/li>\n<li><em><span>V<\/span><\/em><span>\u00a0\u00e9 o volume<\/span><\/li>\n<li><em><span>R\u00a0<\/span><\/em><em><sub><span>espec\u00edfico<\/span><\/sub><\/em><span>\u00a0\u00e9 a constante de g\u00e1s espec\u00edfica, igual \u00e0 constante de g\u00e1s universal dividida pela massa molar (M) do g\u00e1s ou mistura.\u00a0Para ar seco R\u00a0<\/span><sub><span>espec\u00edfico<\/span><\/sub><span>\u00a0= 287,1 J.kg\u00a0<\/span><sup><span>-1<\/span><\/sup><span>\u00a0.K\u00a0<\/span><sup><span>-1<\/span><\/sup><span>\u00a0.<\/span><\/li>\n<\/ul>\n<p><span>Portanto<\/span><\/p>\n<p><strong><span>m<\/span><\/strong><span>\u00a0= p\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0\/ R\u00a0<\/span><sub><span>espec\u00edfico<\/span><\/sub><span>\u00a0T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= (100000 \u00d7 500 \u00d7 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0) \/ (287,1 \u00d7 293) =\u00a0<\/span><strong><span>5,95 \u00d7 10\u00a0<\/span><\/strong><strong><sup><span>-4<\/span><\/sup><\/strong><strong><span>\u00a0kg<\/span><\/strong><\/p>\n<p><strong><span>2)<\/span><\/strong><\/p>\n<p><strong><span>Nesse problema, todos os volumes s\u00e3o conhecidos:<\/span><\/strong><\/p>\n<ul>\n<li><span>V\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0= 500 \u00d7 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0(0,5l)<\/span><\/li>\n<li><span>V\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>m\u00e1x<\/span><\/sub><span>\u00a0\/ CR = 25 \u00d7 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0m\u00a0<\/span><sup><span>3<\/span><\/sup><\/li>\n<\/ul>\n<p><span>Observe que (V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0) x n\u00famero de cilindros = cilindrada total do motor<\/span><\/p>\n<p><span>Como o processo \u00e9 adiab\u00e1tico, podemos usar a seguinte rela\u00e7\u00e3o p, V, T para processos adiab\u00e1ticos:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-process-formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17558 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-process-formula.png\" alt=\"\" width=\"154\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-process-formula.png\" \/><\/a><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-formula-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17563 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-formula-example.png\" alt=\"\" width=\"155\" height=\"78\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/adiabatic-formula-example.png\" \/><\/a><\/p>\n<p><span>portanto<\/span><\/p>\n<p><span>T\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.\u00a0CR\u00a0<\/span><sup><span>k &#8211; 1<\/span><\/sup><span>\u00a0= 293.\u00a020\u00a0<\/span><sup><span>0,4<\/span><\/sup><span>\u00a0= 971 K<\/span><\/p>\n<p><span>3)<\/span><\/p>\n<p><span>Novamente, podemos usar a lei do g\u00e1s ideal para encontrar a press\u00e3o no final do curso de compress\u00e3o como:<\/span><\/p>\n<p><strong><span>P\u00a0<\/span><\/strong><strong><sub><span>2<\/span><\/sub><\/strong><span>\u00a0= mR\u00a0<\/span><sub><span>espec\u00edfico<\/span><\/sub><span>\u00a0T\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0\/ V\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 5,95 x 10\u00a0<\/span><sup><span>-4<\/span><\/sup><span>\u00a0x 287,1 x 971\/25 x 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0= 6.635.000 Pa =\u00a0<\/span><strong><span>66,35 barra<\/span><\/strong><\/p>\n<p><span>4)<\/span><\/p>\n<p><span>Como o processo 2 \u2192 3 ocorre a press\u00e3o constante, a equa\u00e7\u00e3o ideal de estado do g\u00e1s fornece<\/span><\/p>\n<p><span>T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0= (V\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0\/ V\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0) x T\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 1942 K<\/span><\/p>\n<p><span>Para calcular a quantidade de calor adicionada pela queima da mistura combust\u00edvel-ar,\u00a0<\/span><sub><span>adicione<\/span><\/sub><span>\u00a0Q\u00a0, precisamos usar a primeira lei da termodin\u00e2mica para o processo isob\u00e1rico, que afirma:<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= mc\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; T\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0) = 5,95 x 10\u00a0<\/span><sup><span>-4<\/span><\/sup><span>\u00a0x 1010 x 971 =\u00a0<\/span><strong><span>583,5 J<\/span><\/strong><\/p>\n<p><span>5)<\/span><\/p>\n<p><span>Efici\u00eancia t\u00e9rmica para este ciclo Diesel:<\/span><\/p>\n<p><span>Como foi derivado na se\u00e7\u00e3o anterior, a efici\u00eancia t\u00e9rmica do ciclo Diesel \u00e9 uma fun\u00e7\u00e3o da taxa de compress\u00e3o, da taxa de corte e \u03ba:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-Thermal-Efficiency2.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-18420 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-Thermal-Efficiency2.png\" alt=\"\" width=\"300\" height=\"65\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Diesel-cycle-Thermal-Efficiency2.png\" \/><\/a><span>Onde<\/span><\/p>\n<ul>\n<li><span>\u03b7\u00a0<\/span><sub><span>Diesel<\/span><\/sub><span>\u00a0\u00e9 a efici\u00eancia t\u00e9rmica m\u00e1xima de um ciclo Diesel<\/span><\/li>\n<li><span>\u03b1 \u00e9 a\u00a0raz\u00e3o de\u00a0corte V\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0\/ V\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0(ou seja, a raz\u00e3o de volumes no final e no in\u00edcio da fase de combust\u00e3o)<\/span><\/li>\n<li><span>CR \u00e9 a taxa de compress\u00e3o<\/span><\/li>\n<li><strong><span>\u03ba = c\u00a0<\/span><\/strong><strong><sub><span>p<\/span><\/sub><\/strong><strong><span>\u00a0\/ c\u00a0<\/span><\/strong><strong><sub><span>v<\/span><\/sub><\/strong><strong><span>\u00a0= 1,4<\/span><\/strong><\/li>\n<\/ul>\n<p><span>Para este exemplo:<\/span><\/p>\n<p><span>\u03b7\u00a0<\/span><sub><span>Diesel<\/span><\/sub><span>\u00a0= 0,6467 = 64,7%<\/span><\/p>\n<p><span>6)<\/span><\/p>\n<p><span>O MEP foi definido como:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/mean-effective-pressure-definition.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17565 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/mean-effective-pressure-definition.png\" alt=\"\" width=\"300\" height=\"63\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/mean-effective-pressure-definition.png\" \/><\/a><\/p>\n<p><span>Nesta equa\u00e7\u00e3o, o volume de deslocamento \u00e9 igual a V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0.\u00a0O trabalho l\u00edquido para um ciclo pode ser calculado usando o calor adicionado e a efici\u00eancia t\u00e9rmica:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>net<\/span><\/sub><\/em><\/strong><em><span>\u00a0=\u00a0<\/span><\/em><em><span>Q\u00a0<\/span><\/em><em><sub><span>add<\/span><\/sub><\/em><em><span>\u00a0.\u00a0\u03b7\u00a0<\/span><\/em><em><sub><span>Otto<\/span><\/sub><\/em><em><span>\u00a0=\u00a0<\/span><\/em><em><span>583,5 x 0,6467 =\u00a0<\/span><strong><span>377,3 J<\/span><\/strong><\/em><\/p>\n<p><strong><em><span>MEP<\/span><\/em><\/strong><em><span>\u00a0= 377,3 \/ (<\/span><\/em><span>\u00a0500 \u00d7 10<\/span><sup><span>\u00a0-6<\/span><\/sup><span>\u00a0&#8211; 25 \u00d7 10<\/span><sup><span>\u00a0-6\u00a0<\/span><\/sup><em><span>) = 794,3 kPa =\u00a0<\/span><strong><span>7,943 bar<\/span><\/strong><\/em><\/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 Ciclo Diesel &#8211; Problema com a Solu\u00e7\u00e3o.\u00a0Vamos assumir um ciclo Diesel t\u00edpico.\u00a0Calcule as principais caracter\u00edsticas, como temperaturas, press\u00f5es, MEP e efici\u00eancia.\u00a0Engenharia T\u00e9rmica Ciclo Diesel &#8211; Problema com a Solu\u00e7\u00e3o diagrama pV de um ciclo Diesel ideal Ciclo Diesel &#8211; Problema com a Solu\u00e7\u00e3o Vamos assumir o ciclo Diesel, que \u00e9 um dos\u00a0ciclos termodin\u00e2micos\u00a0mais &#8230; <a title=\"O que \u00e9 o Ciclo Diesel &#8211; Problema com a Solu\u00e7\u00e3o &#8211; Defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-o-ciclo-diesel-problema-com-a-solucao-definicao\/\" aria-label=\"More on O que \u00e9 o Ciclo Diesel &#8211; Problema com a 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 o Ciclo Diesel - Problema com a Solu\u00e7\u00e3o - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Exemplo de Ciclo Diesel - Problema com a Solu\u00e7\u00e3o. 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