{"id":45082,"date":"2019-10-17T10:37:26","date_gmt":"2019-10-17T09:37:26","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/quel-est-lexemple-du-processus-isobare-addition-de-chaleur-isobare-definition\/"},"modified":"2020-02-20T10:17:58","modified_gmt":"2020-02-20T09:17:58","slug":"quel-est-lexemple-du-processus-isobare-addition-de-chaleur-isobare-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/quel-est-lexemple-du-processus-isobare-addition-de-chaleur-isobare-definition\/","title":{"rendered":"Quel est l’exemple du processus isobare – Addition de chaleur isobare – D\u00e9finition"},"content":{"rendered":"
\n
Exemple de processus isobare – Addition de chaleur isobare. Supposons un apport de chaleur isobare dans un gaz parfait. Calculez la chaleur ajout\u00e9e par l’\u00e9changeur de chaleur. G\u00e9nie thermique<\/div>\n<\/div>\n
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Exemple de processus isobare – Addition de chaleur isobare<\/span><\/h2>\n
\"premi\u00e8re<\/a>
Le cycle de Brayton id\u00e9al se compose de quatre processus thermodynamiques.\u00a0Deux processus isentropiques et deux processus isobares.<\/span><\/figcaption><\/figure>\n

Supposons le\u00a0\u00a0<\/span>cycle de Brayton id\u00e9al<\/span><\/strong>\u00a0\u00a0qui d\u00e9crit le fonctionnement d’un\u00a0\u00a0moteur thermique \u00e0\u00a0<\/strong><\/span>pression constante<\/span><\/strong>\u00a0\u00a0.\u00a0Les<\/strong>\u00a0\u00a0moteurs \u00e0\u00a0turbine \u00e0 gaz modernes<\/strong>\u00a0et\u00a0\u00a0les moteurs \u00e0 r\u00e9action \u00e0 respiration a\u00e9rienne<\/strong>\u00a0\u00a0suivent \u00e9galement le cycle de Brayton.<\/span><\/p>\n

Le cycle de Brayton id\u00e9al se compose de quatre processus thermodynamiques.\u00a0Deux processus isentropiques et deux processus isobares.<\/span><\/p>\n

    \n
  1. compression isentropique<\/span><\/strong>\u00a0\u00a0– l’air ambiant est aspir\u00e9 dans le compresseur, o\u00f9 il est mis sous pression (1 \u2192 2).\u00a0Le travail requis pour le compresseur est donn\u00e9 par\u00a0\u00a0<\/span>W\u00a0<\/span>C<\/span><\/sub>\u00a0\u00a0= H\u00a0<\/span>2<\/span><\/sub>\u00a0\u00a0– H\u00a0<\/span>1<\/span><\/sub>\u00a0.<\/span><\/strong><\/li>\n
  2. addition de chaleur isobare<\/span><\/strong>\u00a0\u00a0– l’air comprim\u00e9 traverse ensuite une chambre de combustion, o\u00f9 le carburant est br\u00fbl\u00e9 et l’air ou un autre milieu est chauff\u00e9 (2 \u2192 3).\u00a0Il s’agit d’un processus \u00e0 pression constante, car la chambre est ouverte pour entrer et sortir.\u00a0La chaleur nette ajout\u00e9e est donn\u00e9e par\u00a0\u00a0<\/span>Q\u00a0<\/span>add<\/span><\/sub>\u00a0\u00a0= H\u00a0<\/span>3\u00a0<\/span><\/sub>\u00a0– H\u00a0<\/span>2<\/span><\/sub><\/strong><\/li>\n
  3. expansion isentropique<\/span><\/strong>\u00a0\u00a0– l’air chauff\u00e9 sous pression se d\u00e9tend ensuite sur la turbine, c\u00e8de son \u00e9nergie.\u00a0Le travail effectu\u00e9 par turbine est donn\u00e9 par\u00a0\u00a0<\/span>W\u00a0<\/span>T<\/span><\/sub>\u00a0\u00a0= H\u00a0<\/span>4<\/span><\/sub>\u00a0\u00a0– H\u00a0<\/span>3<\/span><\/sub><\/strong><\/li>\n
  4. rejet de chaleur isobare<\/span><\/strong>\u00a0\u00a0– la chaleur r\u00e9siduelle doit \u00eatre rejet\u00e9e afin de fermer le cycle.\u00a0La chaleur nette rejet\u00e9e est donn\u00e9e par\u00a0\u00a0<\/span>Q\u00a0<\/span>re<\/span><\/sub>\u00a0\u00a0= H\u00a0<\/span>4\u00a0<\/span><\/sub>\u00a0– H\u00a0<\/span>1<\/span><\/sub><\/strong><\/li>\n<\/ol>\n

    Supposons un\u00a0<\/span>apport de chaleur isobare<\/span><\/strong>\u00a0(2 \u2192 3) dans un \u00e9changeur de chaleur.\u00a0Dans les turbines \u00e0 gaz typiques, l’\u00e9tage haute pression re\u00e7oit du gaz (point 3 sur la figure; p\u00a0<\/span>3<\/span><\/sub>\u00a0=\u00a0<\/span>6,7 MPa<\/span><\/strong>\u00a0; T\u00a0<\/span>3<\/span><\/sub>\u00a0= 1190 K (917 \u00b0 C)) d’un \u00e9changeur de chaleur.\u00a0De plus, nous savons que le compresseur re\u00e7oit du gaz (point 1 sur la figure; p\u00a0<\/span>1<\/span><\/sub>\u00a0=\u00a0<\/span>2,78 MPa<\/span><\/strong>\u00a0; T\u00a0<\/span>1<\/span><\/sub>\u00a0= 299 K (26 \u00b0 C)) et nous savons que l’efficacit\u00e9 isentropique du compresseur est\u00a0<\/span>\u03b7\u00a0<\/span><\/strong>K<\/span><\/sub><\/strong>\u00a0= 0,87 (87 %)<\/span><\/strong>\u00a0.<\/span><\/p>\n

    Calculez la chaleur ajout\u00e9e par l’\u00e9changeur de chaleur (entre 2 \u2192 3).<\/span><\/strong><\/p>\n

    Solution:<\/span><\/strong><\/p>\n

    D’apr\u00e8s la\u00a0<\/span>premi\u00e8re loi de la thermodynamique<\/span><\/strong>\u00a0, la chaleur nette ajout\u00e9e est donn\u00e9e par\u00a0<\/span>Q\u00a0<\/span><\/strong>add<\/span><\/sub><\/strong>\u00a0= H\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– H\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong>\u00a0ou\u00a0<\/span>Q\u00a0<\/span>add<\/span><\/sub>\u00a0= C\u00a0<\/span>p<\/span><\/sub>\u00a0. (T\u00a0<\/span>3<\/span><\/sub>\u00a0-T\u00a0<\/span>2s<\/span><\/sub>\u00a0),<\/span><\/strong>\u00a0mais dans ce cas, nous ne connaissons pas la temp\u00e9rature (T\u00a0<\/span>2s<\/span><\/sub>\u00a0) \u00e0 la sortie du compresseur.\u00a0Nous allons r\u00e9soudre ce probl\u00e8me en variables intensives.\u00a0Nous devons r\u00e9\u00e9crire l’\u00e9quation pr\u00e9c\u00e9dente (pour inclure\u00a0<\/span>\u03b7\u00a0<\/span><\/strong>K<\/span><\/sub><\/strong>\u00a0) en utilisant le terme (+\u00a0<\/span>h\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong>\u00a0– h\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong>\u00a0) pour:<\/span><\/p>\n

    Q\u00a0<\/span>add<\/span><\/sub>\u00a0=\u00a0<\/span>h\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– h\u00a0<\/span><\/strong>2<\/span><\/sub><\/strong>\u00a0= h\u00a0<\/span><\/strong>3<\/span><\/sub><\/strong>\u00a0– h\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong>\u00a0– (h\u00a0<\/span><\/strong>2s<\/span><\/sub><\/strong>\u00a0– h\u00a0<\/span><\/strong>1<\/span><\/sub><\/strong>\u00a0) \/\u00a0<\/span><\/strong>\u03b7\u00a0<\/span><\/strong>K<\/span><\/sub><\/strong><\/p>\n

    Q\u00a0<\/span>add\u00a0<\/span><\/sub>=<\/span><\/strong>\u00a0c\u00a0<\/span>p<\/span><\/sub>\u00a0(T\u00a0<\/span>3<\/span><\/sub>\u00a0-T\u00a0<\/span>1<\/span><\/sub>\u00a0) – (c\u00a0<\/span>p<\/span><\/sub>\u00a0(T\u00a0<\/span>2s<\/span><\/sub>\u00a0-T\u00a0<\/span>1<\/span><\/sub>\u00a0)\u00a0<\/span>\/\u00a0<\/span><\/strong>\u03b7\u00a0<\/span><\/strong>K<\/span><\/sub><\/strong>\u00a0)<\/span><\/p>\n

    Ensuite, nous calculerons la temp\u00e9rature, T\u00a0<\/span>2s<\/span><\/sub>\u00a0, en utilisant la\u00a0<\/span>relation p, V, T<\/span><\/strong>\u00a0pour le processus adiabatique entre (1 \u2192 2).<\/span><\/p>\n

    \"Relation<\/a><\/p>\n

    Dans cette \u00e9quation, le facteur pour l’h\u00e9lium est \u00e9gal \u00e0\u00a0<\/span>= c\u00a0<\/span>p<\/span><\/sub>\u00a0\/ c\u00a0<\/span>v<\/span><\/sub>\u00a0= 1,66<\/span><\/strong>\u00a0.\u00a0D’apr\u00e8s l’\u00e9quation pr\u00e9c\u00e9dente, la temp\u00e9rature de sortie du compresseur, T\u00a0<\/span>2s<\/span><\/sub>\u00a0, est:<\/span><\/p>\n

    \"processus<\/a><\/p>\n

    D’apr\u00e8s la\u00a0<\/span>loi<\/span><\/a>\u00a0des\u00a0<\/span>gaz parfaits,<\/span><\/a>\u00a0nous savons que la chaleur sp\u00e9cifique molaire d’un\u00a0gaz parfait monoatomique<\/a>\u00a0est:<\/span><\/p>\n

    C\u00a0<\/span><\/em><\/strong>v<\/span><\/sub><\/em><\/strong>\u00a0= 3 \/ 2R = 12,5 J \/ mol K<\/span><\/em><\/strong>\u00a0et<\/span>\u00a0C\u00a0<\/span><\/em><\/strong>p<\/span><\/sub><\/em><\/strong>\u00a0= C\u00a0<\/span><\/em><\/strong>v<\/span><\/sub><\/em><\/strong>\u00a0+ R = 5 \/ 2R = 20,8 J \/ mol K<\/span><\/em><\/strong><\/p>\n

    Nous transf\u00e9rons les capacit\u00e9s calorifiques sp\u00e9cifiques en unit\u00e9s de\u00a0<\/span>J \/ kg K via:<\/span><\/strong><\/p>\n

    c\u00a0<\/span><\/em><\/strong>p<\/span><\/sub><\/em><\/strong>\u00a0= C\u00a0<\/span><\/em><\/strong>p<\/span><\/sub><\/em><\/strong>\u00a0.\u00a01 \/ M (poids molaire de l’h\u00e9lium) = 20,8 x 4,10\u00a0<\/span><\/em><\/strong>-3<\/span><\/sup><\/em><\/strong>\u00a0= 5200 J \/ kg K<\/span><\/em><\/strong><\/p>\n

    En utilisant cette temp\u00e9rature et l’\u00a0<\/span>efficacit\u00e9 du compresseur isentropique,<\/span><\/a>\u00a0nous pouvons calculer la chaleur ajout\u00e9e par l’\u00e9changeur de chaleur:<\/span><\/p>\n

    Q\u00a0<\/span>add<\/span><\/sub>\u00a0=<\/span><\/strong>\u00a0c\u00a0<\/span>p<\/span><\/sub>\u00a0(T\u00a0<\/span>3<\/span><\/sub>\u00a0-T\u00a0<\/span>1<\/span><\/sub>\u00a0) – (c\u00a0<\/span>p<\/span><\/sub>\u00a0(T\u00a0<\/span>2s<\/span><\/sub>\u00a0-T\u00a0<\/span>1<\/span><\/sub>\u00a0)\u00a0<\/span>\/\u00a0<\/span><\/strong>\u03b7\u00a0<\/span><\/strong>K<\/span><\/sub><\/strong>\u00a0) = 5200. (1190 – 299) – 5200. (424-299) \/0,87 = 4,633 MJ \/ kg – 0,747 MJ \/ kg =\u00a0<\/span>3,886 MJ \/ kg<\/span><\/strong><\/p>\n<\/div>\n

     <\/p>\n

    \"Processus<\/a>
    Processus isobare – principales caract\u00e9ristiques<\/span><\/figcaption><\/figure>\n
    \"La<\/a>
    Pour une masse fixe de gaz \u00e0 pression constante, le volume est directement proportionnel \u00e0 la temp\u00e9rature Kelvin.\u00a0Source: grc.nasa.gov La politique de la NASA sur le droit d’auteur stipule que \u00able mat\u00e9riel de la NASA n’est pas prot\u00e9g\u00e9 par le droit d’auteur sauf indication contraire\u00bb<\/p>\n

    <\/span><\/figcaption><\/figure>\n

    <\/div>\n
    <\/div>\n
    \n

    ……………………………………………………………………………………………………………………………….<\/p>\n

    Cet article est bas\u00e9 sur la traduction automatique de l’article original en anglais. Pour plus d’informations, voir l’article en anglais. Pouvez vous nous aider Si vous souhaitez corriger la traduction, envoyez-la \u00e0 l’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":"

    Exemple de processus isobare – Addition de chaleur isobare. Supposons un apport de chaleur isobare dans un gaz parfait. Calculez la chaleur ajout\u00e9e par l’\u00e9changeur de chaleur. G\u00e9nie thermique Exemple de processus isobare – Addition de chaleur isobare Le cycle de Brayton id\u00e9al se compose de quatre processus thermodynamiques.\u00a0Deux processus isentropiques et deux processus isobares. … 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":"\nQuel est l'exemple du processus isobare - Addition de chaleur isobare - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"Exemple de processus isobare - Addition de chaleur isobare. Supposons un apport de chaleur isobare dans un gaz id\u00e9al. Calculez la chaleur ajout\u00e9e par l'\u00e9changeur de chaleur. 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