{"id":40743,"date":"2019-09-25T03:11:38","date_gmt":"2019-09-25T02:11:38","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/que-es-el-ciclo-de-otto-problema-con-la-solucion-definicion\/"},"modified":"2020-01-11T10:26:02","modified_gmt":"2020-01-11T09:26:02","slug":"que-es-el-ciclo-de-otto-problema-con-la-solucion-definicion","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/es\/que-es-el-ciclo-de-otto-problema-con-la-solucion-definicion\/","title":{"rendered":"Qu\u00e9 es el ciclo de Otto &#8211; Problema con la soluci\u00f3n &#8211; Definici\u00f3n"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Ciclo Otto &#8211; Problema con la soluci\u00f3n &#8211; Ejemplo.\u00a0Supongamos el ciclo de Otto.\u00a0Calcule caracter\u00edsticas clave como la masa de aire de admisi\u00f3n, presiones y MEP.\u00a0Ingenieria termal<\/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 Otto &#8211; Problema con la soluci\u00f3n<\/h2>\n<p>Supongamos el\u00a0<a title=\"Ciclo Otto - Motor Otto\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-ciclo-de-otto-motor-de-otto-definicion\/\"><strong>ciclo Otto<\/strong><\/a>\u00a0, que es uno de los\u00a0<strong>ciclos termodin\u00e1micos<\/strong>\u00a0m\u00e1s comunes\u00a0que se pueden encontrar en\u00a0<strong>los motores de autom\u00f3viles<\/strong>\u00a0.\u00a0Uno de los par\u00e1metros clave de tales motores es el cambio en los vol\u00famenes entre el punto muerto superior (TDC) y el punto muerto inferior (BDC).\u00a0La relaci\u00f3n de estos vol\u00famenes (\u00a0<em>V\u00a0<\/em><em><sub>1<\/sub><\/em><em>\u00a0\/ V\u00a0<\/em><em><sub>2<\/sub><\/em>\u00a0) se conoce como la\u00a0<strong>relaci\u00f3n de compresi\u00f3n<\/strong>\u00a0.<\/p>\n<p>La\u00a0<strong>relaci\u00f3n de compresi\u00f3n<\/strong>\u00a0en un motor de gasolina generalmente no ser\u00e1 mucho mayor que 10: 1 debido a un potencial golpe del motor (autoencendido) y no menor a 6: 1.\u00a0Por ejemplo, algunos motores de autos deportivos pueden tener una relaci\u00f3n de compresi\u00f3n de hasta 12.5: 1 (por ejemplo, Ferrari 458 Italia).<\/p>\n<figure id=\"attachment_17461\" class=\"wp-caption alignright\" aria-describedby=\"caption-attachment-17461\"><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-Cycle-PV-Diagram.png\"><img loading=\"lazy\" class=\"size-medium wp-image-17461 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-Cycle-PV-Diagram-300x245.png\" alt=\"Otto Cycle - Diagrama PV\" width=\"300\" height=\"245\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-Cycle-PV-Diagram-300x245.png\" \/><\/a><figcaption id=\"caption-attachment-17461\" class=\"wp-caption-text\">Diagrama pV del ciclo Otto.\u00a0El \u00e1rea delimitada por la ruta completa del ciclo representa el trabajo total que se puede hacer durante un ciclo.<\/figcaption><\/figure>\n<p>En este ejemplo, supongamos un ciclo Otto con una\u00a0<strong>relaci\u00f3n<\/strong>\u00a0de\u00a0<strong>compresi\u00f3n<\/strong>\u00a0de\u00a0<strong>CR = 9: 1<\/strong>\u00a0.\u00a0El aire de admisi\u00f3n est\u00e1 a 100 kPa = 1 bar, 20 \u00b0 C, y el volumen de la c\u00e1mara es de 500 cm\u00b3 antes de la carrera de compresi\u00f3n.\u00a0La temperatura al final de\u00a0<a title=\"Proceso adiab\u00e1tico\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-proceso-adiabatico-definicion\/\">la expansi\u00f3n adiab\u00e1tica<\/a>\u00a0es T\u00a0<sub>4<\/sub>\u00a0= 800 K.<\/p>\n<ul>\n<li>Capacidad calor\u00edfica espec\u00edfica a presi\u00f3n constante del aire a presi\u00f3n atmosf\u00e9rica y temperatura ambiente:\u00a0<strong>c\u00a0<\/strong><strong><sub>p<\/sub><\/strong><strong>\u00a0= 1,01 kJ \/ kgK.<\/strong><\/li>\n<li>Capacidad calor\u00edfica espec\u00edfica a volumen constante de aire a presi\u00f3n atmosf\u00e9rica y 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>la masa de aire de admisi\u00f3n<\/span><\/strong><\/li>\n<li><strong><span>la temperatura T\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/li>\n<li><strong><span>la presi\u00f3n p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/li>\n<li><strong><span>la cantidad de calor agregado al quemar la mezcla de combustible y aire<\/span><\/strong><\/li>\n<li><strong><span>la eficiencia t\u00e9rmica de este ciclo<\/span><\/strong><\/li>\n<li><strong><span>el eurodiputado<\/span><\/strong><\/li>\n<\/ol>\n<p><strong><span>Soluci\u00f3n:<\/span><\/strong><\/p>\n<p><span>1)\u00a0\u00a0<\/span><strong><span>la masa del aire de admisi\u00f3n<\/span><\/strong><\/p>\n<p><span>Al comienzo de los c\u00e1lculos tenemos que determinar la cantidad de gas en el cilindro antes de la carrera de compresi\u00f3n.\u00a0Usando la ley de los gases ideales, podemos encontrar la masa:<\/span><\/p>\n<p><strong><em><span>pV = mR\u00a0<\/span><\/em><em><sub><span>espec\u00edfico<\/span><\/sub><\/em><em><span>\u00a0T<\/span><\/em><\/strong><\/p>\n<p><span>d\u00f3nde:<\/span><\/p>\n<ul>\n<li><em><span>p<\/span><\/em><span>\u00a0es la\u00a0<\/span><a title=\"Presi\u00f3n absoluta\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-pressure-physics\/absolute-pressure\/\"><span>presi\u00f3n absoluta<\/span><\/a><span>\u00a0del gas<\/span><\/li>\n<li><em><span>m<\/span><\/em><span>\u00a0es la masa de sustancia<\/span><\/li>\n<li><em><span>T<\/span><\/em><span>\u00a0es la\u00a0<\/span><a title=\"Escala Kelvin - Temperatura absoluta\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-temperature-physics\/kelvin-scale-absolute-temperature\/\"><span>temperatura absoluta<\/span><\/a><\/li>\n<li><em><span>V<\/span><\/em><span>\u00a0es el volumen<\/span><\/li>\n<li><em><span>R\u00a0<\/span><\/em><em><sub><span>espec\u00edfica<\/span><\/sub><\/em><span>\u00a0\u00a0es la constante de gas espec\u00edfica, igual a la constante de gas universal dividida por la masa molar (M) del gas o mezcla.\u00a0Para aire 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>por lo tanto<\/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\u00a0<\/span><sup><span>10-6<\/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>En este problema se conocen todos los vol\u00famenes:<\/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>m\u00e1x\u00a0<\/span><\/sub><strong><span>=<\/span><\/strong><span>\u00a0500 \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>3<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0= V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0\/ CR = 55,56 \u00d7 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0m\u00a0<\/span><sup><span>3<\/span><\/sup><\/li>\n<\/ul>\n<p><span>Tenga en cuenta 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 = desplazamiento total del motor.<\/span><\/p>\n<p><span>2)\u00a0\u00a0<\/span><strong><span>la temperatura T\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/p>\n<p><span>Dado que el proceso es adiab\u00e1tico, podemos usar la siguiente relaci\u00f3n p, V, T para\u00a0<\/span><a title=\"Proceso adiab\u00e1tico\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-proceso-adiabatico-definicion\/\"><span>procesos adiab\u00e1ticos<\/span><\/a><span>\u00a0:<\/span><\/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=\"f\u00f3rmula adiab\u00e1tica - ejemplo\" 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>as\u00ed<\/span><\/p>\n<p><strong><span>T\u00a0<\/span><sub><span>3<\/span><\/sub><\/strong><span>\u00a0= T\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0.\u00a0CR\u00a0<\/span><sup><span>\u03ba &#8211; 1<\/span><\/sup><span>\u00a0= 800.\u00a09\u00a0<\/span><sup><span>0.4<\/span><\/sup><span>\u00a0=\u00a0<\/span><strong><span>1926 K<\/span><\/strong><\/p>\n<p><span>3)\u00a0\u00a0<\/span><strong><span>la presi\u00f3n p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/p>\n<p><span>Nuevamente, podemos usar la\u00a0<\/span><a title=\"Ley del gas ideal\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>ley de gas ideal<\/span><\/a><span>\u00a0para encontrar la presi\u00f3n al comienzo de la carrera de potencia como:<\/span><\/p>\n<p><strong><span>p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><span>\u00a0= mR\u00a0<\/span><sub><span>espec\u00edfico<\/span><\/sub><span>\u00a0T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0\/ V\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0= 5.95 \u00d7 10\u00a0<\/span><sup><span>-4<\/span><\/sup><span>\u00a0x 287.1 x 1926 \/ 55.56 \u00d7 10\u00a0<\/span><sup><span>-6<\/span><\/sup><span>\u00a0= 5920000 Pa =\u00a0<\/span><strong><span>59.2 bar<\/span><\/strong><\/p>\n<p><span>4)\u00a0\u00a0<\/span><strong><span>la cantidad de calor agregado<\/span><\/strong><\/p>\n<p><span>Para calcular la cantidad de calor agregado al quemar la mezcla de combustible y aire, Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0, tenemos que usar la\u00a0<\/span><a title=\"Primera ley de la termodin\u00e1mica\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-la-primera-ley-de-la-termodinamica-definicion\/\"><span>primera ley de la termodin\u00e1mica<\/span><\/a><span>\u00a0para el\u00a0<\/span><a title=\"Proceso isoc\u00f3rico - Proceso isom\u00e9trico\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-proceso-isocorico-proceso-isometrico-definicion\/\"><span>proceso isocr\u00f3rico<\/span><\/a><span>\u00a0, que establece que Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= \u2206U, por lo tanto:<\/span><\/p>\n<p><strong><span>Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= mc\u00a0<\/span><sub><span>v<\/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)<\/span><\/strong><\/p>\n<p><span>La temperatura al final de la carrera de compresi\u00f3n se puede determinar utilizando la relaci\u00f3n p, V, T para procesos adiab\u00e1ticos entre los puntos\u00a0<\/span><strong><span>1 \u2192 2.<\/span><\/strong><\/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=\"proceso adiab\u00e1tico - f\u00f3rmula\" 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><strong><span>T\u00a0<\/span><sub><span>2<\/span><\/sub><\/strong><span>\u00a0= T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.\u00a0CR\u00a0<\/span><sup><span>\u03ba &#8211; 1<\/span><\/sup><span>\u00a0= 293.\u00a09\u00a0<\/span><sup><span>0.4<\/span><\/sup><span>\u00a0=\u00a0<\/span><strong><span>706 K<\/span><\/strong><\/p>\n<p><span>luego<\/span><\/p>\n<p><strong><span>Q\u00a0<\/span><\/strong><strong><sub><span>add<\/span><\/sub><\/strong><span>\u00a0= mc\u00a0<\/span><sub><span>v<\/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 \u00d7 10\u00a0<\/span><sup><span>-4<\/span><\/sup><span>\u00a0x 718 x 1220 =\u00a0<\/span><strong><span>521.2 J<\/span><\/strong><\/p>\n<p><span>5)\u00a0\u00a0<\/span><strong><span>la eficiencia t\u00e9rmica<\/span><\/strong><\/p>\n<p><a title=\"Eficiencia t\u00e9rmica\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-la-eficiencia-termica-definicion\/\"><strong><span>Eficiencia t\u00e9rmica<\/span><\/strong><\/a><span>\u00a0para un ciclo Otto:<\/span><\/p>\n<p><span>Como se dedujo en la secci\u00f3n anterior, la eficiencia t\u00e9rmica de un ciclo Otto es una funci\u00f3n de la relaci\u00f3n de compresi\u00f3n y \u03ba:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-cycle-efficiency-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17564 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-cycle-efficiency-example.png\" alt=\"Ciclo Otto - eficiencia - ejemplo\" width=\"309\" height=\"172\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Otto-cycle-efficiency-example.png\" \/><\/a><\/p>\n<p><span>6)\u00a0\u00a0<\/span><strong><span>la presi\u00f3n efectiva media<\/span><\/strong><\/p>\n<p><span>El\u00a0<\/span><strong><span>MEP<\/span><\/strong><span>\u00a0se defini\u00f3 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=\"presi\u00f3n efectiva media - definici\u00f3n\" 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>En esta ecuaci\u00f3n, el volumen de desplazamiento es igual a V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0.\u00a0El trabajo neto para un ciclo se puede calcular utilizando el calor agregado y la eficiencia 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>521.2 x 0.5847 = 304.7\u00a0<\/span><strong><span>J<\/span><\/strong><\/em><br \/>\n<strong><em><span>MEP<\/span><\/em><\/strong><em><span>\u00a0= 304.7 \/ (<\/span><\/em><span>\u00a0500 \u00d7<\/span><sup><span>\u00a010-6<\/span><\/sup><span>\u00a0&#8211; 55.56 \u00d7 10<\/span><sup><span>\u00a0-6\u00a0<\/span><\/sup><em><span>) = 685.6 kPa =\u00a0<\/span><strong><span>6.856 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 art\u00edculo se basa en la traducci\u00f3n autom\u00e1tica del art\u00edculo original en ingl\u00e9s. Para m\u00e1s informaci\u00f3n vea el art\u00edculo en ingl\u00e9s. Puedes ayudarnos. Si desea corregir la traducci\u00f3n, env\u00edela a: translations@nuclear-power.com o complete el formulario de traducci\u00f3n en l\u00ednea. Agradecemos su ayuda, actualizaremos la traducci\u00f3n lo antes posible. Gracias.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Ciclo Otto &#8211; Problema con la soluci\u00f3n &#8211; Ejemplo.\u00a0Supongamos el ciclo de Otto.\u00a0Calcule caracter\u00edsticas clave como la masa de aire de admisi\u00f3n, presiones y MEP.\u00a0Ingenieria termal Ciclo Otto &#8211; Problema con la soluci\u00f3n Supongamos el\u00a0ciclo Otto\u00a0, que es uno de los\u00a0ciclos termodin\u00e1micos\u00a0m\u00e1s comunes\u00a0que se pueden encontrar en\u00a0los motores de autom\u00f3viles\u00a0.\u00a0Uno de los par\u00e1metros clave de &#8230; <a title=\"Qu\u00e9 es el ciclo de Otto &#8211; Problema con la soluci\u00f3n &#8211; Definici\u00f3n\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-ciclo-de-otto-problema-con-la-solucion-definicion\/\" aria-label=\"M\u00e1s en Qu\u00e9 es el ciclo de Otto &#8211; Problema con la soluci\u00f3n &#8211; Definici\u00f3n\">Leer m\u00e1s<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[16],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Qu\u00e9 es el ciclo de Otto - Problema con la soluci\u00f3n - Definici\u00f3n<\/title>\n<meta name=\"description\" content=\"Ciclo Otto - Problema con la soluci\u00f3n - Ejemplo. 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