{"id":45057,"date":"2019-10-17T06:15:29","date_gmt":"2019-10-17T05:15:29","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/quest-ce-quotto-cycle-probleme-de-solution-definition\/"},"modified":"2020-02-20T06:18:05","modified_gmt":"2020-02-20T05:18:05","slug":"quest-ce-quotto-cycle-probleme-de-solution-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quotto-cycle-probleme-de-solution-definition\/","title":{"rendered":"Qu&#8217;est-ce qu&#8217;Otto Cycle &#8211; Probl\u00e8me de solution &#8211; D\u00e9finition"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Cycle Otto &#8211; Probl\u00e8me de solution &#8211; Exemple.\u00a0Laissons supposer le cycle Otto.\u00a0Calculez les caract\u00e9ristiques cl\u00e9s telles que la masse d&#8217;air d&#8217;admission, les pressions et le PEM.\u00a0G\u00e9nie thermique<\/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>Cycle Otto &#8211; Probl\u00e8me de solution<\/h2>\n<p>Supposons le\u00a0<a title=\"Cycle Otto - Moteur Otto\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-que-otto-cycle-otto-engine-definition\/\"><strong>cycle d&#8217;Otto<\/strong><\/a>\u00a0, qui est l&#8217;un des\u00a0<strong>cycles thermodynamiques<\/strong>\u00a0les plus courants\u00a0que l&#8217;on puisse trouver dans les\u00a0<strong>moteurs d&#8217;automobiles<\/strong>\u00a0.\u00a0L&#8217;un des param\u00e8tres cl\u00e9s de ces moteurs est la variation des volumes entre le point mort haut (TDC) et le point mort bas (BDC).\u00a0Le rapport de ces volumes (\u00a0<em>V\u00a0<\/em><em><sub>1<\/sub><\/em><em>\u00a0\/ V\u00a0<\/em><em><sub>2<\/sub><\/em>\u00a0) est appel\u00e9\u00a0<strong>taux de compression<\/strong>\u00a0.<\/p>\n<p>Le\u00a0<strong>taux de compression<\/strong>\u00a0d&#8217;un moteur \u00e0 essence ne sera g\u00e9n\u00e9ralement pas beaucoup plus \u00e9lev\u00e9 que 10: 1 en raison du choc potentiel du moteur (auto-allumage) et non inf\u00e9rieur \u00e0 6: 1.\u00a0Par exemple, certains moteurs de voitures de sport peuvent avoir un taux de compression allant jusqu&#8217;\u00e0 12,5: 1 (par exemple, Ferrari 458 Italia).<\/p>\n<figure id=\"attachment_17461\" class=\"wp-caption aligncenter\" 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=\"Cycle Otto - Diagramme 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\">Diagramme pV du cycle d&#8217;Otto.\u00a0La zone d\u00e9limit\u00e9e par la piste cyclable compl\u00e8te repr\u00e9sente le travail total pouvant \u00eatre effectu\u00e9 au cours d&#8217;un cycle.<\/figcaption><\/figure>\n<p>Dans cet exemple, supposons un cycle Otto avec\u00a0<strong>un taux<\/strong>\u00a0de\u00a0<strong>compression<\/strong>\u00a0de\u00a0<strong>CR = 9: 1<\/strong>\u00a0.\u00a0L&#8217;air aspir\u00e9 est \u00e0 100 kPa = 1 bar, 20 \u00b0 C et le volume de la chambre est de 500 cm\u00b3 avant la course de compression.\u00a0La temp\u00e9rature \u00e0 la fin de\u00a0<a title=\"Processus adiabatique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quel-est-le-processus-adiabatique-definition\/\">la dilatation adiabatique<\/a>\u00a0est T\u00a0<sub>4<\/sub>\u00a0= 800 K.<\/p>\n<ul>\n<li>Capacit\u00e9 thermique sp\u00e9cifique \u00e0 pression constante de l\u2019air \u00e0 la pression atmosph\u00e9rique et \u00e0 la temp\u00e9rature ambiante:\u00a0<strong>c\u00a0<\/strong><strong><sub>p<\/sub><\/strong><strong>\u00a0= 1,01 kJ \/ kgK.<\/strong><\/li>\n<li>Capacit\u00e9 thermique sp\u00e9cifique \u00e0 volume constant d\u2019air \u00e0 la pression atmosph\u00e9rique et \u00e0 la temp\u00e9rature ambiante:\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>Calculer:<\/span><\/strong><\/p>\n<ol>\n<li><strong><span>la masse d&#8217;air d&#8217;admission<\/span><\/strong><\/li>\n<li><strong><span>la temp\u00e9rature T\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/li>\n<li><strong><span>la pression p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/li>\n<li><strong><span>la quantit\u00e9 de chaleur ajout\u00e9e par la combustion du m\u00e9lange air-carburant<\/span><\/strong><\/li>\n<li><strong><span>l&#8217;efficacit\u00e9 thermique de ce cycle<\/span><\/strong><\/li>\n<li><strong><span>le d\u00e9put\u00e9 europ\u00e9en<\/span><\/strong><\/li>\n<\/ol>\n<p><strong><span>Solution:<\/span><\/strong><\/p>\n<p><span>1)\u00a0\u00a0<\/span><strong><span>la masse d&#8217;air d&#8217;admission<\/span><\/strong><\/p>\n<p><span>Au d\u00e9but des calculs, nous devons d\u00e9terminer la quantit\u00e9 de gaz dans le cylindre avant la course de compression.\u00a0En utilisant la loi du gaz parfait, nous pouvons trouver la masse:<\/span><\/p>\n<p><strong><em><span>pV = mR\u00a0<\/span><\/em><em><sub><span>sp\u00e9cifique<\/span><\/sub><\/em><em><span>\u00a0T<\/span><\/em><\/strong><\/p>\n<p><span>o\u00f9:<\/span><\/p>\n<ul>\n<li><em><span>p<\/span><\/em><span>\u00a0est la\u00a0<\/span><a title=\"Pression absolue\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-pressure-physics\/absolute-pressure\/\"><span>pression absolue<\/span><\/a><span>\u00a0du gaz<\/span><\/li>\n<li><em><span>m<\/span><\/em><span>\u00a0est la masse de substance<\/span><\/li>\n<li><em><span>T<\/span><\/em><span>\u00a0est la\u00a0<\/span><a title=\"\u00c9chelle Kelvin - Temp\u00e9rature absolue\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-temperature-physics\/kelvin-scale-absolute-temperature\/\"><span>temp\u00e9rature absolue<\/span><\/a><\/li>\n<li><em><span>V<\/span><\/em><span>\u00a0est le volume<\/span><\/li>\n<li><em><span>R\u00a0<\/span><\/em><em><sub><span>sp\u00e9cifique<\/span><\/sub><\/em><span>\u00a0\u00a0est la constante de gaz sp\u00e9cifique, \u00e9gale \u00e0 la constante de gaz universelle divis\u00e9e par la masse molaire (M) du gaz ou du m\u00e9lange.\u00a0Pour l&#8217;air sec R\u00a0<\/span><sub><span>sp\u00e9cifique<\/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>par cons\u00e9quent<\/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>sp\u00e9cifique<\/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>Dans ce probl\u00e8me, tous les volumes sont connus:<\/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\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(0,5 l)<\/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>Notez que (V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0) x nombre de cylindres = cylindr\u00e9e totale du moteur.<\/span><\/p>\n<p><span>2)\u00a0\u00a0<\/span><strong><span>la temp\u00e9rature T\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/p>\n<p><span>Puisque le processus est adiabatique, nous pouvons utiliser la relation p, V, T suivante pour les\u00a0<\/span><a title=\"Processus adiabatique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quel-est-le-processus-adiabatique-definition\/\"><span>processus adiabatiques<\/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=\"formule adiabatique - exemple\" 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>Donc<\/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 pression p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><\/p>\n<p><span>Encore une fois, nous pouvons utiliser la\u00a0<\/span><a title=\"Loi du gaz parfait\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>loi du gaz parfait<\/span><\/a><span>\u00a0pour trouver la pression au d\u00e9but de la course de puissance comme:<\/span><\/p>\n<p><strong><span>p\u00a0<\/span><\/strong><strong><sub><span>3<\/span><\/sub><\/strong><span>\u00a0= mR\u00a0<\/span><sub><span>sp\u00e9cifique<\/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 quantit\u00e9 de chaleur ajout\u00e9e<\/span><\/strong><\/p>\n<p><span>Pour calculer la quantit\u00e9 de chaleur ajout\u00e9e par la combustion du m\u00e9lange air-carburant, Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0, nous devons utiliser la\u00a0<\/span><a title=\"Premi\u00e8re loi de la thermodynamique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quelle-est-la-premiere-loi-de-la-thermodynamique-definition\/\"><span>premi\u00e8re loi de la thermodynamique<\/span><\/a><span>\u00a0pour le\u00a0<\/span><a title=\"Processus isochorique - Processus isom\u00e9trique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quun-processus-isochorique-processus-isometrique-definition\/\"><span>processus isochore<\/span><\/a><span>\u00a0, qui indique que Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= \u2206U, donc:<\/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 temp\u00e9rature \u00e0 la fin de la course de compression peut \u00eatre d\u00e9termin\u00e9e en utilisant la relation p, V, T pour les processus adiabatiques entre les points\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=\"processus adiabatique - formule\" 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>ensuite<\/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>l&#8217;efficacit\u00e9 thermique<\/span><\/strong><\/p>\n<p><a title=\"Efficacit\u00e9 thermique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-que-lefficacite-thermique-definition\/\"><strong><span>Efficacit\u00e9 thermique<\/span><\/strong><\/a><span>\u00a0pour un cycle Otto:<\/span><\/p>\n<p><span>Comme d\u00e9riv\u00e9 dans la section pr\u00e9c\u00e9dente, l&#8217;efficacit\u00e9 thermique d&#8217;un cycle d&#8217;Otto est fonction du taux de compression et de \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=\"Cycle d'Otto - efficacit\u00e9 - exemple\" 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 pression moyenne effective<\/span><\/strong><\/p>\n<p><span>Le\u00a0<\/span><strong><span>d\u00e9put\u00e9 europ\u00e9en a<\/span><\/strong><span>\u00a0\u00e9t\u00e9 d\u00e9fini comme:<\/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=\"pression moyenne effective - d\u00e9finition\" 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>Si cette \u00e9quation, le volume de d\u00e9placement est \u00e9gal \u00e0 V\u00a0<\/span><sub><span>max<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>min<\/span><\/sub><span>\u00a0.\u00a0Le travail net pour un cycle peut \u00eatre calcul\u00e9 en utilisant la chaleur ajout\u00e9e et l&#8217;efficacit\u00e9 thermique:<\/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 =\u00a0<\/span><strong><span>304,7 J<\/span><\/strong><\/em><br \/>\n<strong><em><span>MEP<\/span><\/em><\/strong><em><span>\u00a0= 304,7 \/ (<\/span><\/em><span>\u00a0500 \u00d7 10<\/span><sup><span>\u00a0-6<\/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>Cet article est bas\u00e9 sur la traduction automatique de l&#8217;article original en anglais. Pour plus d&#8217;informations, voir l&#8217;article en anglais. Pouvez vous nous aider Si vous souhaitez corriger la traduction, envoyez-la \u00e0 l&#8217;adresse: translations@nuclear-power.com ou remplissez le formulaire de traduction en ligne. Nous appr\u00e9cions votre aide, nous mettrons \u00e0 jour la traduction le plus rapidement possible. Merci<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Cycle Otto &#8211; Probl\u00e8me de solution &#8211; Exemple.\u00a0Laissons supposer le cycle Otto.\u00a0Calculez les caract\u00e9ristiques cl\u00e9s telles que la masse d&#8217;air d&#8217;admission, les pressions et le PEM.\u00a0G\u00e9nie thermique Cycle Otto &#8211; Probl\u00e8me de solution Supposons le\u00a0cycle d&#8217;Otto\u00a0, qui est l&#8217;un des\u00a0cycles thermodynamiques\u00a0les plus courants\u00a0que l&#8217;on puisse trouver dans les\u00a0moteurs d&#8217;automobiles\u00a0.\u00a0L&#8217;un des param\u00e8tres cl\u00e9s de ces moteurs &#8230; <a title=\"Qu&#8217;est-ce qu&#8217;Otto Cycle &#8211; Probl\u00e8me de solution &#8211; D\u00e9finition\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quotto-cycle-probleme-de-solution-definition\/\" aria-label=\"En savoir plus sur Qu&#8217;est-ce qu&#8217;Otto Cycle &#8211; Probl\u00e8me de solution &#8211; D\u00e9finition\">Lire la suite<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[8],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Qu&#039;est-ce qu&#039;Otto Cycle - Probl\u00e8me de solution - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"Cycle Otto - Probl\u00e8me de solution - Exemple. 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