{"id":44933,"date":"2019-10-17T00:23:57","date_gmt":"2019-10-16T23:23:57","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/quest-ce-quun-exemple-de-cycle-de-brayton-probleme-de-solution-definition\/"},"modified":"2020-02-19T14:40:05","modified_gmt":"2020-02-19T13:40:05","slug":"quest-ce-quun-exemple-de-cycle-de-brayton-probleme-de-solution-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quun-exemple-de-cycle-de-brayton-probleme-de-solution-definition\/","title":{"rendered":"Qu&#8217;est-ce qu&#8217;un exemple de cycle de Brayton &#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\">Exemple de cycle de Brayton &#8211; Probl\u00e8me de solution.\u00a0Calculez les param\u00e8tres thermodynamiques cl\u00e9s du cycle de Brayton tels que les temp\u00e9ratures, les pressions et les transferts de chaleur.\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 de Brayton &#8211; Probl\u00e8me de solution<\/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=\"Cycle de Brayton - exemple - calcul\" 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>Supposons le\u00a0<strong>cycle de Brayton<\/strong>\u00a0ferm\u00e9\u00a0, qui est l\u2019un des\u00a0<strong>cycles thermodynamiques<\/strong>\u00a0les plus courants\u00a0que l\u2019on puisse trouver dans les moteurs \u00e0 turbine \u00e0 gaz modernes.\u00a0Dans ce cas, supposons une\u00a0<strong>turbine \u00e0 gaz \u00e0 h\u00e9lium<\/strong>\u00a0avec un seul compresseur et une seule turbine.\u00a0L&#8217;un des param\u00e8tres cl\u00e9s de ces moteurs est la temp\u00e9rature maximale \u00e0 l&#8217;entr\u00e9e de la turbine et le taux de compression du compresseur (PR = p\u00a0<sub>2<\/sub>\u00a0\/ p\u00a0<sub>1<\/sub>\u00a0), qui d\u00e9termine l&#8217;efficacit\u00e9 thermique de ce moteur.<\/p>\n<p>Dans cette turbine, l&#8217;\u00e9tage haute pression re\u00e7oit le gaz (point 3 sur la figure) d&#8217;un \u00e9changeur de chaleur:<\/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>le rendement de la turbine isentropique est\u00a0<em>\u03b7\u00a0<\/em><em><sub>T<\/sub><\/em>\u00a0= 0,91 (91%)<\/li>\n<\/ul>\n<p>et \u00e9vacuez-le vers un autre \u00e9changeur de chaleur, o\u00f9 la pression de sortie est (point 4):<\/p>\n<ul>\n<li>p\u00a0<sub>4<\/sub>\u00a0= 2,78 MPa<\/li>\n<li>T\u00a0<sub>4, est<\/sub>\u00a0=?<\/li>\n<\/ul>\n<p>Ainsi, le rapport de pression du compresseur est \u00e9gal \u00e0 PR = 2,41.\u00a0De plus, nous savons que le compresseur re\u00e7oit du gaz (point 1) sur la figure:<\/p>\n<ul>\n<li>p\u00a0<sub>1<\/sub>\u00a0= 2,78 MPa;<\/li>\n<li>T\u00a0<sub>1<\/sub>\u00a0= 299 K (26 \u00b0 C)<\/li>\n<li>le rendement du compresseur isentropique \u03b7\u00a0<sub>K<\/sub>\u00a0= 0,87 (87%).<\/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>Le rapport de capacit\u00e9 thermique,, pour l&#8217;h\u00e9lium est \u00e9gal \u00e0\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=\"Cycle de Brayton - probl\u00e8me avec solution\" 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>Calculer:<\/span><\/strong><\/p>\n<ol>\n<li><span>la chaleur ajout\u00e9e par l&#8217;\u00e9changeur de chaleur (entre 2 \u2192 3)<\/span><\/li>\n<li><span>la temp\u00e9rature de sortie du compresseur du gaz (T\u00a0<\/span><sub><span>2, est<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>le vrai travail effectu\u00e9 sur ce compresseur, lorsque l&#8217;efficacit\u00e9 du compresseur isentropique est \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0,87 (87%)<\/span><\/li>\n<li><span>la temp\u00e9rature de sortie de la turbine du gaz (T\u00a0<\/span><sub><span>4, est<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>le travail r\u00e9el effectu\u00e9 par cette turbine, lorsque le rendement de la turbine isentropique est \u03b7\u00a0<\/span><sub><span>T<\/span><\/sub><span>\u00a0= 0,91 (91%)<\/span><\/li>\n<li><span>l&#8217;efficacit\u00e9 thermique de ce cycle<\/span><\/li>\n<\/ol>\n<p><strong><span>Solution:<\/span><\/strong><\/p>\n<p><span>1) + 2)<\/span><\/p>\n<p><span>D&#8217;apr\u00e8s 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>\u00a0, la chaleur nette ajout\u00e9e est donn\u00e9e par\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>\u00a0mais dans ce cas nous ne savons pas la temp\u00e9rature (T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0) \u00e0 la sortie du compresseur.\u00a0Nous allons r\u00e9soudre ce probl\u00e8me en\u00a0<\/span><a title=\"Propri\u00e9t\u00e9s \u00e9tendues et intensives\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/extensive-and-intensive-properties\/\"><span>variables intensives<\/span><\/a><span>\u00a0.\u00a0Nous devons r\u00e9\u00e9crire l&#8217;\u00e9quation pr\u00e9c\u00e9dente (pour inclure\u00a0<\/span><strong><span>\u03b7\u00a0<\/span><\/strong><strong><sub><span>K<\/span><\/sub><\/strong><span>\u00a0) en utilisant le terme (\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) pour:<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>add<\/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>2s<\/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>Ensuite, nous calculerons la temp\u00e9rature,\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>2s<\/span><\/sub><\/strong><span>\u00a0, en utilisant la relation p, V, T (de la\u00a0<\/span><a title=\"Loi du gaz parfait\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>loi des gaz parfaits<\/span><\/a><span>\u00a0) pour le\u00a0<\/span><a title=\"Processus adiabatique\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quel-est-le-processus-adiabatique-definition\/\"><span>processus adiabatique<\/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=\"Relation p, V, T - processus isentropique\" 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>Dans cette \u00e9quation, le facteur pour l&#8217;h\u00e9lium est \u00e9gal \u00e0 = c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1,66.\u00a0D&#8217;apr\u00e8s l&#8217;\u00e9quation pr\u00e9c\u00e9dente, la temp\u00e9rature de sortie du compresseur, T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0, est:<\/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=\"processus isobare - exemple\" 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<span>En utilisant cette temp\u00e9rature et l&#8217;\u00a0<\/span><a title=\"Efficacit\u00e9 isentropique - Turbine \/ compresseur \/ buse\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-processes\/isentropic-process\/isentropic-efficiency-turbinecompressornozzle\/\"><span>efficacit\u00e9 du compresseur isentropique,<\/span><\/a><span>\u00a0nous pouvons calculer la chaleur ajout\u00e9e par l&#8217;\u00e9changeur de chaleur:<\/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>Le travail effectu\u00e9 sur le gaz par le compresseur dans le processus de compression isentropique est:<\/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>Le vrai travail effectu\u00e9 sur le gaz par le compresseur dans la compression adiabatique est alors:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C, r\u00e9el<\/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>La temp\u00e9rature de sortie de turbine du gaz,\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>4, est<\/span><\/sub><\/strong><span>\u00a0, peut \u00eatre calcul\u00e9e en utilisant la m\u00eame\u00a0\u00a0<\/span><strong><span>relation p, V, T<\/span><\/strong><span>\u00a0\u00a0qu&#8217;en 2) mais entre les \u00e9tats 3 et 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=\"Relation p, V, T - processus isentropique\" 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>De l&#8217;\u00e9quation pr\u00e9c\u00e9dente, la temp\u00e9rature de sortie du gaz, T\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0, est:<\/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=\"processus isentropique - exemple\" 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>Le travail effectu\u00e9 par la turbine \u00e0 gaz dans l&#8217;expansion isentropique est alors:<\/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(T\u00a0<\/span><\/em><em><sub><span>3<\/span><\/sub><\/em><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><em><sub><span>4s<\/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>Le v\u00e9ritable travail effectu\u00e9 par la turbine \u00e0 gaz dans l&#8217;expansion adiabatique est alors:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>T, r\u00e9el<\/span><\/sub><\/em><\/strong><em><span>\u00a0= c\u00a0<\/span><\/em><em><sub><span>p<\/span><\/sub><\/em><em><span>\u00a0(T\u00a0<\/span><\/em><em><sub><span>3<\/span><\/sub><\/em><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><em><sub><span>4s<\/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>Comme il a \u00e9t\u00e9 d\u00e9riv\u00e9 dans la section pr\u00e9c\u00e9dente, l&#8217;\u00a0<\/span><strong><span>efficacit\u00e9 thermique<\/span><\/strong><span>\u00a0d&#8217;un cycle de Brayton id\u00e9al est fonction du\u00a0<\/span><strong><span>rapport<\/span><\/strong><span>\u00a0de\u00a0<strong>pression<\/strong>\u00a0et de\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=\"efficacit\u00e9 thermique - cycle de Brayton - rapport de pression - \u00e9quation\" 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>par cons\u00e9quent<\/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>L&#8217;efficacit\u00e9 thermique peut \u00e9galement \u00eatre calcul\u00e9e en utilisant le travail et la chaleur (sans \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>W\u00a0<\/span><\/em><em><sub><span>T, s<\/span><\/sub><\/em><em><span>\u00a0&#8211; W\u00a0<\/span><\/em><em><sub><span>C, s<\/span><\/sub><\/em><em><span>\u00a0)\u00a0<\/span><\/em><strong><em><span>\/<\/span><\/em><\/strong><span>\u00a0Q<\/span><sub><span>\u00a0add, s<\/span><\/sub><span>\u00a0= (1,825 &#8211; 0,650) \/ 3,998<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a0\u00a00,295 =<\/span><strong><span>\u00a029,5%<\/span><\/strong><\/p>\n<p><span>Enfin, l&#8217;efficacit\u00e9 thermique, y compris l&#8217;efficacit\u00e9 turbine \/ compresseur isentropique, est:<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><em><sub><span>e, r\u00e9el\u00a0<\/span><\/sub><\/em><\/strong><em><span>\u00a0= (\u00a0<\/span><\/em><em><span>W\u00a0<\/span><\/em><em><sub><span>T, r\u00e9el<\/span><\/sub><\/em><em><span>\u00a0\u00a0&#8211; W\u00a0<\/span><\/em><em><sub><span>C, r\u00e9el<\/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\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<\/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","protected":false},"excerpt":{"rendered":"<p>Exemple de cycle de Brayton &#8211; Probl\u00e8me de solution.\u00a0Calculez les param\u00e8tres thermodynamiques cl\u00e9s du cycle de Brayton tels que les temp\u00e9ratures, les pressions et les transferts de chaleur.\u00a0G\u00e9nie thermique Cycle de Brayton &#8211; Probl\u00e8me de solution Supposons le\u00a0cycle de Brayton\u00a0ferm\u00e9\u00a0, qui est l\u2019un des\u00a0cycles thermodynamiques\u00a0les plus courants\u00a0que l\u2019on puisse trouver dans les moteurs \u00e0 turbine &#8230; <a title=\"Qu&#8217;est-ce qu&#8217;un exemple de cycle de Brayton &#8211; Probl\u00e8me de solution &#8211; D\u00e9finition\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quun-exemple-de-cycle-de-brayton-probleme-de-solution-definition\/\" aria-label=\"En savoir plus sur Qu&#8217;est-ce qu&#8217;un exemple de cycle de Brayton &#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;un exemple de cycle de Brayton - Probl\u00e8me de solution - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"Exemple de cycle de Brayton - Probl\u00e8me de solution. 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