{"id":46634,"date":"2019-10-22T21:39:47","date_gmt":"2019-10-22T20:39:47","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/quest-ce-que-la-theorie-des-pompes-equation-de-turbomachine-deuler-definition\/"},"modified":"2020-03-02T16:19:13","modified_gmt":"2020-03-02T15:19:13","slug":"quest-ce-que-la-theorie-des-pompes-equation-de-turbomachine-deuler-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-que-la-theorie-des-pompes-equation-de-turbomachine-deuler-definition\/","title":{"rendered":"Qu&#8217;est-ce que la th\u00e9orie des pompes &#8211; Equation de turbomachine d&#8217;Euler &#8211; D\u00e9finition"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">L&#8217;\u00e9quation de turbomachine d&#8217;Euler, ou parfois appel\u00e9e \u00e9quation de pompe d&#8217;Euler, joue un r\u00f4le central dans les turbomachines, dans la th\u00e9orie des pompes centrifuges \u00e9galement. G\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>Th\u00e9orie de la pompe &#8211; Equations de turbomachine d&#8217;Euler<\/h2>\n<p><strong>L&#8217;\u00e9quation de turbomachine d&#8217;Euler<\/strong>\u00a0, ou parfois appel\u00e9e\u00a0<strong>\u00e9quation de pompe d&#8217;Euler<\/strong>\u00a0, joue un r\u00f4le central dans les\u00a0<strong>turbomachines<\/strong>\u00a0puisqu&#8217;elle relie le\u00a0<strong>travail sp\u00e9cifique Y \u00e0<\/strong>\u00a0la g\u00e9om\u00e9trie et aux vitesses de la roue.\u00a0L&#8217;\u00e9quation est bas\u00e9e sur les concepts de\u00a0<strong>conservation du moment cin\u00e9tique<\/strong>\u00a0et de\u00a0<strong>conservation de l&#8217;\u00e9nergie<\/strong>\u00a0.<\/p>\n<p>Les\u00a0<strong>\u00e9quations de turbomachine d&#8217;Euler<\/strong>\u00a0sont les\u00a0<strong>suivantes<\/strong>\u00a0:<\/p>\n<p><strong>Couple d&#8217;arbre: T\u00a0<sub>arbre<\/sub>\u00a0\u00a0 \u00a0 = \u03c1Q (r\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; r\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0)<\/strong><\/p>\n<p><strong>Puissance de l&#8217;eau: P\u00a0<sub>w<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u03c9.\u00a0T\u00a0<sub>arbre \u00a0 \u00a0 \u00a0\u00a0<\/sub>\u00a0= \u03c1Q (u\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; u\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0)<\/strong><\/p>\n<p><strong>T\u00eate de pompe: H = P\u00a0<sub>w<\/sub>\u00a0\/ \u03c1gQ = (u\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; u\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0) \/ g<\/strong><\/p>\n<p>o\u00f9<\/p>\n<ul>\n<li><strong>R\u00a0<sub>1<\/sub>\u00a0<\/strong>et\u00a0<strong>R\u00a0<sub>2<\/sub><\/strong>\u00a0sont les\u00a0<strong>diam\u00e8tres de la\u00a0<a title=\"Roue - Types de Roues\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/impeller-types-of-impellers\/\">roue<\/a><\/strong>\u00a0\u00e0 l&#8217;entr\u00e9e et \u00e0 la sortie, respectivement.<\/li>\n<li><strong>u\u00a0<sub>1<\/sub><\/strong>\u00a0et\u00a0<strong>u\u00a0<sub>2<\/sub>\u00a0<\/strong>sont les\u00a0<strong>vitesses absolues de la roue<\/strong>\u00a0(u\u00a0<sub>1<\/sub>\u00a0= r\u00a0<sub>1.<\/sub>\u00a0\u03c9) \u00e0 l\u2019entr\u00e9e et \u00e0 la sortie, respectivement.<\/li>\n<li><strong>V\u00a0<sub>t1<\/sub><\/strong>\u00a0et\u00a0<strong>V\u00a0<sub>t2<\/sub><\/strong>\u00a0sont les\u00a0<strong>vitesses tangentielles de l&#8217;\u00e9coulement<\/strong>\u00a0\u00e0 l&#8217;entr\u00e9e et \u00e0 la sortie, respectivement.<\/li>\n<\/ul>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-14957 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\" alt=\"\u00c9quation de turbomachine d'Euler\" width=\"535\" height=\"415\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\" \/><\/a><\/p>\n<p><strong>Les \u00e9quations de turbomachine d&#8217;Euler<\/strong>\u00a0peuvent \u00eatre utilis\u00e9es pour pr\u00e9dire l&#8217;impact du changement de la g\u00e9om\u00e9trie de la roue sur la\u00a0<a title=\"T\u00eate de pompe - Courbe de performance de la pompe centrifuge\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/pump-head-performance-curve\/\">t\u00eate<\/a>\u00a0.\u00a0Peu importe que nous utilisions une pompe ou une turbine.\u00a0Si le couple et la vitesse angulaire ont le\u00a0<strong>m\u00eame signe<\/strong>\u00a0, le fluide (pompe ou compresseur) est en cours de travail.\u00a0Si le couple et la vitesse angulaire sont\u00a0<strong>de signe oppos\u00e9,<\/strong>\u00a0du fluide est extrait (une turbine).\u00a0Ainsi, pour l\u2019aspect conception des turbines et des\u00a0<a title=\"Pompes centrifuges\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-quune-pompe-centrifuge-definition\/\">pompes<\/a>\u00a0, les \u00e9quations d\u2019Euler sont extr\u00eamement utiles.<\/p>\n<div class=\"su-spacer\"><\/div>\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<div class=\"su-spacer\"><\/div>\n<h2><span>Exemple: calcul des performances de la pompe<\/span><\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\"><img loading=\"lazy\" class=\"alignright size-full wp-image-14958 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" alt=\"Calcul de la t\u00eate de pompe\" width=\"264\" height=\"580\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" \/><\/a><span>Dans cet exemple, nous verrons comment pr\u00e9dire<\/span><\/p>\n<ul>\n<li><strong><span>la d\u00e9charge de conception<\/span><\/strong><\/li>\n<li><strong><span>puissance en eau<\/span><\/strong><\/li>\n<li><strong><span>la t\u00eate de pompe<\/span><\/strong><\/li>\n<\/ul>\n<p><span>d&#8217;une pompe centrifuge.\u00a0Ces donn\u00e9es de performance seront d\u00e9riv\u00e9es de l&#8217;\u00a0<\/span><strong><span>\u00e9quation de la turbomachine d&#8217;Euler:<\/span><\/strong><\/p>\n<p><strong><span>Couple d&#8217;arbre:\u00a0<\/span><sub><span>arbre<\/span><\/sub><span>\u00a0T\u00a0\u00a0 \u00a0 = \u03c1Q (r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>Puissance en eau: P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u03c9.\u00a0T\u00a0<\/span><sub><span>arbre \u00a0 \u00a0 \u00a0\u00a0<\/span><\/sub><span>\u00a0= \u03c1Q (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>T\u00eate de pompe: H = P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ \u03c1gQ = (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0) \/ g<\/span><\/strong><\/p>\n<p><span>Les donn\u00e9es suivantes sont fournies pour une pompe \u00e0 eau centrifuge:<\/span><\/p>\n<ul>\n<li><strong><span>diam\u00e8tres de la roue<\/span><\/strong><span>\u00a0\u00e0 l&#8217;entr\u00e9e et \u00e0 la sortie<\/span>\n<ul>\n<li><strong><span>r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 10 cm<\/span><\/strong><\/li>\n<li><strong><span>r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 cm<\/span><\/strong><\/li>\n<\/ul>\n<\/li>\n<li><strong><span>Vitesse = 1500 tr \/ min<\/span><\/strong><span>\u00a0(tours par minute)<\/span><\/li>\n<li><span>l&#8217;angle de lame \u00e0 l&#8217;entr\u00e9e\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 30 \u00b0<\/span><\/strong><\/li>\n<li><span>l&#8217;angle de la lame en sortie\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 \u00b0<\/span><\/strong><\/li>\n<li><span>supposons que les largeurs de lame \u00e0 l&#8217;entr\u00e9e et \u00e0 la sortie soient:\u00a0<\/span><strong><span>b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 4 cm<\/span><\/strong><span>\u00a0.<\/span><\/li>\n<\/ul>\n<p><span>Solution:<\/span><\/p>\n<p><span>Tout d&#8217;abord, nous devons calculer la\u00a0<\/span><strong><span>vitesse radiale de l&#8217;\u00e9coulement<\/span><\/strong><span>\u00a0\u00e0 la sortie.\u00a0D&#8217;apr\u00e8s le diagramme de vitesse, la vitesse radiale est \u00e9gale \u00e0 (nous supposons que le flux entre exactement normal \u00e0 la roue, donc la composante tangentielle de la vitesse est nulle):<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0= u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0bronzage 30 \u00b0 = \u03c9 r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0bronzage 30 \u00b0 = 2\u03c0 x (1500\/60) x 0,1 x bronzage 30 \u00b0 =\u00a0<\/span><strong><span>9,1 m \/ s<\/span><\/strong><\/p>\n<p><span>La composante radiale de la vitesse d&#8217;\u00e9coulement d\u00e9termine la\u00a0<\/span><strong><span>quantit\u00e9 de d\u00e9bit volumique entrant dans la roue<\/span><\/strong><span>\u00a0.\u00a0Ainsi, lorsque nous connaissons\u00a0<\/span><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0\u00e0 l&#8217;entr\u00e9e, nous pouvons d\u00e9terminer\u00a0<\/span><strong><span>le d\u00e9bit<\/span><\/strong><span>\u00a0de cette pompe selon l&#8217;\u00e9quation suivante.\u00a0Ici b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0signifie la largeur de pale de la roue \u00e0 l&#8217;entr\u00e9e.<\/span><\/p>\n<p><strong><span>Q<\/span><\/strong><span>\u00a0=\u00a0<\/span><strong><span>2\u03c0.r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r1<\/span><\/sub>\u00a0<\/strong><span>= 2\u00b6 x 0,1 x 0,04 x 9,1 =\u00a0<\/span><strong><span>0.229 m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s<\/span><\/strong><\/p>\n<p><span>Afin de calculer la\u00a0<\/span><strong><span>puissance en eau (P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0)<\/span><\/strong><span>\u00a0requise, nous devons d\u00e9terminer la\u00a0<\/span><strong><span>vitesse d&#8217;\u00e9coulement tangentiel de sortie V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0, car on a suppos\u00e9 que la vitesse tangentielle d&#8217;entr\u00e9e V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0est \u00e9gale \u00e0 z\u00e9ro.<\/span><\/p>\n<p><span>La vitesse d&#8217;\u00e9coulement radial de sortie r\u00e9sulte de la\u00a0<\/span><strong><span>conservation de Q<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><strong><span>Q = 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0\u00a0\u21d2\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0= Q \/ 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 0,229 \/ (2\u03c0 x 0,2 x 0,04) =\u00a0<\/span><strong><span>\u00a04,56 m \/ s<\/span><\/strong><\/p>\n<p><span>\u00c0 partir de la figure (\u00a0<\/span><strong><span>triangle de vitesse<\/span><\/strong><span>\u00a0), l&#8217;angle de la lame de sortie, \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0, peut \u00eatre facilement repr\u00e9sent\u00e9 comme suit.<\/span><\/p>\n<p><strong><span>cot \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0) \/ V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><\/p>\n<p><span>et donc la vitesse d&#8217;\u00e9coulement tangentiel de sortie V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0est:<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0=\u00a0<\/span><strong><span>u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>r2<\/span><\/sub><span>\u00a0.\u00a0lit b\u00e9b\u00e9 20 \u00b0<\/span><\/strong><span>\u00a0= \u03c9 r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>r2<\/span><\/sub><span>\u00a0.\u00a0lit b\u00e9b\u00e9 20 \u00b0 = 2\u03c0 x 1500\/60 x 0,2 &#8211; 4,56 x 2,75 = 31,4 &#8211; 12,5 =\u00a0<\/span><strong><span>18,9 m \/ s.<\/span><\/strong><\/p>\n<p><span>La puissance en eau requise est alors:<\/span><\/p>\n<p><strong><span>P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0= \u03c1 Q u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub>\u00a0<\/strong><span>= 1000 [kg \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0] x 0,229 [m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s] x 31,4 [m \/ s] x 18,9 [m \/ s] = 135900 W =\u00a0<\/span><strong><span>135,6 kW<\/span><\/strong><\/p>\n<p><span>et la t\u00eate de pompe est:<\/span><\/p>\n<p><strong><span>H \u2248 P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ (\u03c1 g Q)<\/span><\/strong><span>\u00a0= 135900 \/ (1000 x 9,81 x 0,229) =\u00a0<\/span><strong><span>60,5 m<\/span><\/strong><\/p>\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>L&#8217;\u00e9quation de turbomachine d&#8217;Euler, ou parfois appel\u00e9e \u00e9quation de pompe d&#8217;Euler, joue un r\u00f4le central dans les turbomachines, dans la th\u00e9orie des pompes centrifuges \u00e9galement. G\u00e9nie thermique Th\u00e9orie de la pompe &#8211; Equations de turbomachine d&#8217;Euler L&#8217;\u00e9quation de turbomachine d&#8217;Euler\u00a0, ou parfois appel\u00e9e\u00a0\u00e9quation de pompe d&#8217;Euler\u00a0, joue un r\u00f4le central dans les\u00a0turbomachines\u00a0puisqu&#8217;elle relie le\u00a0travail sp\u00e9cifique &#8230; <a title=\"Qu&#8217;est-ce que la th\u00e9orie des pompes &#8211; Equation de turbomachine d&#8217;Euler &#8211; D\u00e9finition\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/fr\/quest-ce-que-la-theorie-des-pompes-equation-de-turbomachine-deuler-definition\/\" aria-label=\"En savoir plus sur Qu&#8217;est-ce que la th\u00e9orie des pompes &#8211; Equation de turbomachine d&#8217;Euler &#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 que la th\u00e9orie des pompes - Equation de turbomachine d&#039;Euler - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"L\u2019\u00e9quation de turbomachine d\u2019Euler, ou parfois appel\u00e9e \u00e9quation de pompe d\u2019Euler, joue un r\u00f4le central dans les turbomachines, dans la th\u00e9orie des pompes centrifuges \u00e9galement. 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