{"id":49407,"date":"2019-11-18T20:21:41","date_gmt":"2019-11-18T19:21:41","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/exemple-de-loi-sur-les-gaz-parfaits-compression-des-gaz-definition\/"},"modified":"2019-11-18T20:21:41","modified_gmt":"2019-11-18T19:21:41","slug":"exemple-de-loi-sur-les-gaz-parfaits-compression-des-gaz-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/exemple-de-loi-sur-les-gaz-parfaits-compression-des-gaz-definition\/","title":{"rendered":"Exemple de loi sur les gaz parfaits – Compression des gaz – D\u00e9finition"},"content":{"rendered":"
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Exemple: Loi des gaz parfaits – Compression des gaz \u00e0 l’int\u00e9rieur d’une pression.\u00a0Quelle est la pression finale du gaz \u00e0 l’int\u00e9rieur du pressuriseur?\u00a0Supposons que le gaz est id\u00e9al.\u00a0G\u00e9nie thermique<\/span><\/div>\n<\/div>\n
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Exemple: Loi des gaz parfaits – Compression des gaz dans un pressuriseur<\/span><\/h2>\n
\"pressuriseur\"<\/a>
Un pressuriseur est un composant cl\u00e9 des REP.<\/span><\/figcaption><\/figure>\n

La pression dans le circuit primaire<\/span><\/strong>\u00a0des\u00a0<\/span>REP<\/span><\/a>\u00a0est maintenue par un\u00a0<\/span>pressuriseur<\/span><\/strong><\/a>\u00a0, une cuve s\u00e9par\u00e9e qui est connect\u00e9e au circuit primaire (jambe chaude) et partiellement remplie d\u2019eau chauff\u00e9e \u00e0 la\u00a0<\/span>temp\u00e9rature de saturation<\/span><\/strong><\/a>\u00a0(point d\u2019\u00e9bullition) de la pression souhait\u00e9e par un syst\u00e8me\u00a0<\/span>\u00e9lectrique<\/span><\/strong>\u00a0immerg\u00e9.\u00a0chauffe<\/strong>\u00a0.\u00a0Pendant le chauffage de l’installation, le pressuriseur peut \u00eatre rempli d’azote au lieu de\u00a0<\/span>vapeur satur\u00e9e<\/span><\/a>\u00a0.<\/span><\/p>\n

Supposons qu’un pressuriseur contient\u00a0<\/span>12 m\u00a0<\/span><\/strong>3<\/span><\/sup><\/strong>\u00a0d’azote \u00e0\u00a0<\/span>20 \u00b0 C<\/span><\/strong>\u00a0et\u00a0<\/span>15 bars<\/span><\/strong>\u00a0.\u00a0La temp\u00e9rature est \u00e9lev\u00e9e \u00e0\u00a0<\/span>35 \u00b0 C<\/span><\/strong>\u00a0et le volume est r\u00e9duit \u00e0\u00a0<\/span>8,5 m\u00a0<\/span><\/strong>3<\/span><\/sup><\/strong>\u00a0.\u00a0Quelle est la pression finale du gaz \u00e0 l’int\u00e9rieur du pressuriseur?\u00a0Supposons que le gaz est id\u00e9al.<\/span><\/p>\n

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

Puisque le gaz est id\u00e9al, nous pouvons utiliser la loi des gaz parfaits pour relier ses param\u00e8tres, \u00e0 la fois dans l’\u00a0<\/span>\u00e9tat initial i<\/span><\/strong>\u00a0et dans l’\u00a0<\/span>\u00e9tat final f<\/span><\/strong>\u00a0.\u00a0Donc:<\/span><\/p>\n

p\u00a0<\/span>init<\/span><\/sub>\u00a0V\u00a0<\/span>init<\/span><\/sub>\u00a0= nRT\u00a0<\/span>init<\/span><\/sub><\/em><\/strong><\/p>\n

et<\/span><\/p>\n

p\u00a0<\/span>final<\/span><\/sub>\u00a0V\u00a0<\/span>final<\/span><\/sub>\u00a0= nRT\u00a0<\/span>final<\/span><\/sub><\/em><\/strong><\/p>\n

En divisant la deuxi\u00e8me \u00e9quation par la premi\u00e8re \u00e9quation et en r\u00e9solvant pour\u00a0<\/span>p\u00a0<\/span>f<\/span><\/sub><\/em><\/strong>\u00a0on obtient:<\/span><\/p>\n

p\u00a0<\/span>final<\/span><\/sub>\u00a0= p\u00a0<\/span>init<\/span><\/sub>\u00a0T\u00a0<\/span>final<\/span><\/sub>\u00a0V\u00a0<\/span>init<\/span><\/sub>\u00a0\/ T\u00a0<\/span>init<\/span><\/sub>\u00a0V\u00a0<\/span>final<\/span><\/sub><\/em><\/strong><\/p>\n

Notez que nous ne pouvons pas convertir les unit\u00e9s de volume et de pression en unit\u00e9s SI de base, car elles s\u2019annulent.\u00a0D’autre part, nous devons utiliser Kelvins au lieu de degr\u00e9s Celsius.\u00a0Donc T\u00a0<\/span>init<\/span><\/sub>\u00a0= 293 K et T\u00a0<\/span>final<\/span><\/sub>\u00a0= 308 K.<\/span><\/p>\n

Il s’ensuit que la pression r\u00e9sultante \u00e0 l’\u00e9tat final sera:<\/span><\/p>\n

p\u00a0<\/span>final<\/span><\/sub><\/em><\/strong>= (15 bars) x (308 K) x (12 m\u00a0<\/span><\/em>3<\/span><\/sup>\u00a0) \/ (293 K) x (8,5 m\u00a0<\/span>3<\/span><\/sup><\/em>\u00a0) =\u00a0<\/span><\/em>22 bars<\/span><\/em><\/strong><\/p>\n<\/div>\n<\/div>\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","protected":false},"excerpt":{"rendered":"

Exemple: Loi des gaz parfaits – Compression des gaz \u00e0 l’int\u00e9rieur d’une pression.\u00a0Quelle est la pression finale du gaz \u00e0 l’int\u00e9rieur du pressuriseur?\u00a0Supposons que le gaz est id\u00e9al.\u00a0G\u00e9nie thermique Exemple: Loi des gaz parfaits – Compression des gaz dans un pressuriseur Un pressuriseur est un composant cl\u00e9 des REP. La pression dans le circuit primaire\u00a0des\u00a0REP\u00a0est … 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":"\nExemple de loi sur les gaz parfaits - Compression des gaz - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"Exemple: Loi des gaz parfaits - Compression des gaz \u00e0 l'int\u00e9rieur d'une pression. Quelle est la pression finale du gaz \u00e0 l'int\u00e9rieur du pressuriseur? Supposons que le gaz est id\u00e9al. 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